Energy transfers physics is the accounting of joules as they move: energy sits in stores and travels between them along named pathways, which are mechanically, electrically, by heating and by radiation. A transfer names the store the joules leave, the pathway they take and the store they arrive in. Add the branches up and the account must balance.

A 2200 W kettle left to run for a hundred seconds takes in 220 000 joules. Not one of them vanishes. They end up in a short list of places, and the whole of this topic is learning to write that list down.

The catch is that one place on the list is never measured. You name the destinations you can count, add them up, deduct them from the supply, and whatever is missing has gone to the surroundings — which is how an energy account finds a loss that nobody put a meter on.

Energy Transfers in Physics: What Is a Store and What Is a Pathway?

A store is somewhere joules are held and can be counted; a pathway is the named route they travel while moving from one store to another.

The distinction sounds like bookkeeping, and that is exactly what it is. Stores are nouns: the thermal store of a litre of water, the gravitational potential store of a load held ten metres up. Pathways are adverbs — the joules moved mechanically, or by heating. Ask “which store?” and “by which pathway?” as two separate questions and most of the confusion in this topic disappears.

Older teaching listed kinds of energy and gave each a name, which quietly invited the question of what happens to a kind of energy when it stops being that kind. The prior question of what energy actually is is worth settling first. Stores and pathways answer the follow-up: not what energy is, but where it is held and how it got there.

Two consequences land immediately. Several items on the old list of energy kinds turn out to be pathways rather than stores, and electrical working is the clearest case, because no joules are ever held “as electricity”. And a store or a pathway can be named while carrying nothing at all.

The Four Pathways Energy Can Travel By

There are four named pathways: mechanically, electrically, by heating and by radiation. Each one names how the joules moved, not what they became.

• Mechanically — a force acts through a distance. A winch cable lifting a load moves joules mechanically into the load’s gravitational potential store.
• Electrically — a current does work in a circuit. This is the supply route for both machines in the simulator below, and it is the pathway most often mistaken for a store.
• By heating — a temperature difference drives joules across a boundary by conduction or convection. Conduction, convection and radiation are the three mechanisms; the first two sit inside this pathway and the third gets a name of its own, immediately below.
• By radiation — joules leave as electromagnetic waves, which need no material to travel through. Keeping this branch separate is what lets the radiated joules be computed and read on their own.

Naming a pathway does not oblige it to carry anything. On the simulator’s kettle the mechanical pathway is named and carries exactly 0 J; on its winch the radiated branch is named and carries exactly 0 J. At most three of the four names ever do the work at once — exactly three on each of the simulator’s five presets, and fewer still if you push a slider until another branch empties — so a device that uses all four at once is not something this model promises.

The Energy Account Formula

One identity carries the whole topic: the joules supplied equal the joules into the store you are keeping plus the joules that end up in the surroundings.

• E_in — the joules supplied, in joules (J). On both machines in the simulator they arrive electrically.
• P — the power supplied, in watts (W), which is joules per second. Power is a rate, so it only becomes an amount of energy once a time is attached to it.
• t — the run time, in seconds (s).
• E_store — the joules that reached the store you are keeping, in joules (J).
• E_surroundings — the joules that ended up in the thermal store of everything else, in joules (J).

Three branch formulae do the arithmetic underneath that identity: one for a thermal store, one for a gravitational potential store, and the net Stefan-Boltzmann form for the radiated branch.

• m — a mass in kilograms (kg): of water heated, or of load lifted.
• c — specific heat capacity, in J/(kg·K). The simulator states 4181 J/(kg·K) for liquid water near room temperature as a model value, not as a measurement of a particular sample.
• dT — the temperature rise, in kelvin (K). A rise, not a temperature.
• g — standard gravity, 9.80665 m/s², exact by definition.
• h — the height risen, in metres (m).
• sigma — the Stefan-Boltzmann constant, 5.670374419e-8 W/(m²·K⁴), fixed by the SI defining constants.
• A and eps — the radiating area in m² and the emissivity, which has no unit. The simulator states 0.045 m² and 0.30 for its kettle body as assumptions about its own model, not as specifications of any product.
• T_surf and T_room — absolute temperatures in kelvin (K). The simulator states surroundings at 293.15 K, which is 20.00 °C, and assumes the body reaches the water temperature.

Work the thermal branch by hand a few times, then check it against the Water Heating Energy Calculator, which takes a mass, a specific heat capacity and a temperature rise and returns the joules that reach the store. How one branch compares with the supply as a ratio is a different question from how many joules each branch carries, and the efficiency calculator is the tool that answers that one.

How the Account Is Built, Branch by Branch

Build it in four moves: total the supply, compute every branch you can name, deduct those from the supply, and hand the remainder to the surroundings.

1. Total the supply. For the simulator’s kettle preset, 2200 W for 100 s gives 220 000 J, which the four-significant-figure display rounds to 220.0 kJ.
2. Compute the store branch. One kilogram of water rising 50 K at the stated 4181 J/(kg·K) takes 209 050 J, printed as 209.1 kJ, and it travels by heating.
3. Compute every other named branch. With the stated 0.045 m² at emissivity 0.30 and a body at 343.15 K against surroundings at 293.15 K, the radiated branch over 100 s is 496.1 J.
4. Deduct, and name the remainder. 220 000 minus 209 050 minus 496.07 leaves 10 453.93 J, which goes by heating into the thermal store of the surroundings.

Five settings make the pattern visible. Three of them heat water and two lift a load; the last two rows have no radiated branch at all, because nothing in the model radiates on that machine.

Rows one and three are the pair worth staring at. Both put an identical 209 050 J into the water’s thermal store, and the slower, gentler one sends nearly three times as much to the surroundings — 30.95 kJ against 10.95 kJ. Radiation is a rate multiplied by a time, so holding a surface hot for 480 s instead of 100 s radiates 4.8 times the joules, exactly the ratio of the two run times.

Draw that account with branch widths proportional to the joules and the shape of it becomes obvious at a glance.

The simulator below is that account with the settings under your hand. Choose the machine, set the power and the run time, set the mass and either the temperature rise or the lift height, and the diagram redraws with every branch width recomputed and every figure printed beside it.

Why the Last Branch Is a Deduction, Not a Measurement

The surroundings branch is never measured. It is defined as whatever is left when every branch you can name is taken off the supply.

So the balance line reads zero because it was built to read zero. Add the store branch, the radiated branch and the remainder back together and the supply comes back exactly, at every setting, because the remainder was constructed as that difference in the first place. Presenting a tautology as a result that came out right would be dishonest. It is the method, not a coincidence.

That sounds like a weakness, and it is the point. Nobody can meter the warmed body of a kettle, the worktop underneath it, the warm air above it or the steam leaving the spout. What you can do is insist that the joules add up, and let the arithmetic tell you how large the unmeasured branch has to be. Conservation of energy is what licenses the deduction: nothing is permitted to simply disappear, so a shortfall has to be somewhere.

Being honest about what sits inside that branch matters more than the figure. For the kettle it holds the warmed metal, the warmed worktop, the warm air and every joule the steam carries away, all lumped together, because none of those is modelled separately. The kettle body’s own thermal store is not tracked at all, and the model simply assumes the outer surface reaches the water temperature. That choice is deliberately generous: the final temperature is applied across the whole run, when a real surface starts cool and warms, so the radiated branch is best read as an upper bound.

Four numbers in that account are the simulator’s stated assumptions rather than measurements: 4181 J/(kg·K) for water, surroundings at 293.15 K, a radiating area of 0.045 m² and an emissivity of 0.30. Change the area or the emissivity and the radiated branch moves with them, while the remainder quietly absorbs the difference. So the honest form of the result is “with the simulator’s stated 0.045 m² at emissivity 0.30, the radiated branch is 496.1 J” — dropping the assumptions would turn it into a claim about kettles, which this page does not make.

One state in the model is not a tautology, and it is the one worth pushing the sliders into. Ask for a temperature rise that needs more joules than the supply can provide and the account becomes impossible: the named branches demand more than arrived, and the remainder would have to be negative. An honest model reports the shortfall and clamps the branch at zero, because you cannot take more out of a store than went into it.

Common Mistakes About Energy Transfers

Every mistake below is cured the same way: write the account down, name each store and each pathway separately, and read what the arithmetic says rather than what the words suggest. The headings quote the sentences as people usually say them.

“The energy was used up”

Nothing was used up; it moved. The 220 000 J the simulator’s kettle takes in are all still there afterwards — with its stated 1.00 kg, 50 K rise and 4181 J/(kg·K), 209 050 of them are in the water and the rest is spread through the kitchen. What has been spent is not the energy but how concentrated it was, because joules thinly spread through a room already at room temperature are very hard to do anything with.

“Electricity is a kind of energy you can store”

Electrically names a route, not a place. A charged battery holds a chemical store and a charged capacitor holds an electrostatic one, but no device anywhere holds joules “as electricity”, and that is why an energy account lists electrically as a supply pathway and never as a destination. Asking which store, separately from which pathway, kills this one outright.

“The missing joules are lost”

They are not missing and they are not lost — they are in the thermal store of the surroundings, and the account says exactly how many. “Lost” is a statement about what you can still do with them, not about whether they exist. Call the branch what it is and the bookkeeping stays honest.

“A slower, gentler run must be kinder”

Not in this model, and not in a real kitchen either. The store depends on mass, specific heat capacity and temperature rise; the losses depend on how long the surface stays hot. At 500 W for 480 s the water gains the same 209 050 J as it does at 2200 W for 100 s, and 20 000 J more go to the surroundings — every joule of the extra supply.

Where Energy Accounting Earns Its Keep

Any machine with one supply and more than one destination is an energy account waiting to be written down. The method does not care what the machine is: name the stores, name the pathways, deduct, and read the remainder.

The two machines in the simulator are deliberately ordinary. A kettle sends most of its joules into a thermal store and the rest into the room; a winch sends most of its joules into a gravitational potential store and the rest into the room. Same three lines, different store, different pathway into it. Swap the kettle for a motor and only the names on the lines change: decide which store the machine exists to fill, name the pathways out, and the branch nobody meters is still the one you reach by deduction.

The picture that goes with this accounting is older than any simulator. A Sankey diagram is a flow diagram in which the width of the arrows is proportional to the flow rate of the depicted extensive property, so a branch’s width is its quantity and a small branch cannot pretend to be a large one. The form is named after the Irish engineer Captain Matthew Henry Phineas Riall Sankey, who used one in 1898 in a figure showing the energy efficiency of a steam engine.

In practice the remainder is the branch that pays for itself. It is the number that tells you whether to insulate the tank, shorten the run or buy a different machine — and on the low-power row above, the remainder nearly trebles from 10.45 kJ to 28.57 kJ while the store does not budge. A magnitude check is worth running too: 10.95 kJ into a kitchen is about what a 100 W lamp puts out in under two minutes, which is why nobody notices it.

How Energy Transfers Relate to Conversion, Conservation and Stores

A transfer says how the joules moved; a transformation says which kind of store they ended up in. They describe the same event along different axes, and neither one replaces the other.

Take the winch. “Electrically in, mechanically into a gravitational potential store” is the transfer account, and it names routes. The conversion account asks instead what became what, example by example, and that is the sibling question to this one rather than a rival answer. If you can name the pathway, you can usually name the transformation; the reverse is not reliable, which is why the pathway is the better place to start.

Conservation sits underneath both. It is what forces the account to close and therefore what makes the deduction legitimate; without it the remainder would be a guess rather than an arithmetic consequence. Everything else on this page is bookkeeping built on that one guarantee.

Worked Problems

Every figure below is recomputed from the givens stated inside its own problem, and the four model values are named as the simulator’s assumptions wherever they are used.

Frequently Asked Questions

Significant figures physics is the practice of keeping only the digits that a measurement’s own resolution supports, the last of them being the first digit you are unsure of. Multiplying or dividing keeps the fewer significant figures; adding or subtracting keeps the coarser decimal place. A figure count is a coarse stand-in for precision, never the precision itself.

Write 5.00 and 4.99 on a page and subtract one from the other. Three digits each went in, and 0.01 came out carrying one. Nothing failed there — the subtraction simply spent them.

That is the half a counting rule never mentions. Learning which digits to count takes five minutes; knowing what they are worth, and spotting the moment the rule hands you a digit your readings cannot support, is the part that changes an answer. This page does both, and then works eight problems.

Significant Figures in Physics: What Is a Significant Figure?

A significant figure is one of the digits of a measurement that its own resolution supports, and the last of them is the first digit you are not sure of.

Read a scale and you write down every mark you can see, plus one digit you estimate. Write 9.99 and you are claiming that the first two digits are solid, that the last 9 is your best reading of the gap between marks, and that nothing after it is worth saying. Everything in this topic follows from that one habit.

So a figure count is a claim about a written number rather than about the quantity it measures. The unit does the other job, and how a reading gets its unit in the first place is a separate step that this page assumes you already have.

Two consequences arrive straight away. A digit you did not measure may not be written down, and a digit you did measure may not be thrown away — which is why a trailing zero is sometimes a real measurement and sometimes only spacing.

How to Count the Figures in a Written Number

Count from the written string, left to right, starting at the first digit that is not a zero and stopping at the last digit written down. The only difficulty is the zeros, and there are exactly four cases.

• Leading zeros never count. In 0.053 they only position the decimal point, so that reading has two significant figures.
• Captive zeros always count. A zero with digits on both sides was measured like any other digit, so 10.053 has five.
• Trailing zeros after a decimal point count. Nobody writes one by accident, so 0.04070 has four and 5.0000 has five.
• Trailing zeros in a bare integer are ambiguous. Written as it stands, 2500 could be two, three or four figures, and the writing gives you no way to tell.

Once you have the count, the place of that last digit is fixed, and so is the claim it makes. Take the decade of the value, step back by the figure count, and allow half a unit of whatever place you land in.

• x — the value as written, in whatever unit the measurement carries (m, A, Pa, and so on). Significant figures are unit-free: the count never changes when the unit does.
• n — the significant figure count. A whole number, 1 or more, with no unit.
• E — the decade of the value, the whole-number power of ten just below it. No unit.
• m — the leading mantissa, abs(x) divided by 10^E, so it always sits between 1 and 10. No unit.
• p — the place of the last significant digit, as a power of ten. No unit, and legitimately negative.
• u — the half-unit of that place: the resolution the quote claims, in the same unit as x.
• dp — the decimal places the quote carries, which is just -p, and which can also be negative.

Work the half-unit and the percentage by hand the first few times, until the two steps become automatic. Then hand a written value and its figure count to the Significant Figures Calculator, which prices a quote both ways round and will also tell you how many figures a target precision needs. The table below prices eight written forms, and it is the one block here that answers the counting question on its own.

The last row is the one worth staring at. One written form, three readings, and a hundredfold spread in what it claims — so a bare integer ending in zeros is not a measurement until somebody says which zeros count. Scientific notation fixes it permanently, because it writes the significant digits and the decade in separate places.

One asymmetry in the rule is worth naming rather than hiding. A written zero has no significant figures at all, by the counting rule above, yet writing 0.0 plainly does assert a resolution. The honest move is to give the bound separately instead of pricing the digits: zero, to within whatever the readings allow.

OpenStax’s University Physics states the counting rule the same way — start at the first measured value on the left and count through the last digit written on the right — and settles the zeros in one line: zeros are significant except when they serve only as placeholders. It also keeps the two propagation rules of the next section firmly apart, which is the most useful thing a textbook can do here.

The simulator below is this arithmetic with the readings under your hand. Dial two measured values, give each its own figure count, choose one of the four operations, and watch two things at once: which digits of each reading are significant, and how far the answer’s last kept digit is from the digit the two readings actually support.

The Two Rules Are Different Rules

Multiplication and division keep the smaller of the two figure counts. Addition and subtraction keep the coarser decimal place instead, which is a different question with a different answer.

Treating the two as one rule is the commonest mistake in this topic. Why does the multiplication rule work at all? Because relative half-widths add: a product’s bound, as a percentage, is exactly the sum of the two inputs’ percentages, and keeping the worse input’s figure count is a rough way of honouring the larger percentage while ignoring the smaller.

Addition behaves differently because the places line up rather than the percentages. Add 1.0 and 9.999: the exact sum is 10.999, the coarser place is the first decimal, so the answer is 11.0 — three significant figures out of a reading that only carried two.

That gained figure is real, not an artefact. The bound on the sum is 0.05 plus 0.0005, which is 0.0505, and a quote to the first decimal claims 0.05 — so two of the three digits are supported outright and the third is marginal. A product could never do this: the multiplication rule can only ever take figures away.

What a Figure Count Is Worth: The Precision Formula

A figure count sets the decade of the precision, and the leading digits decide where inside that decade you land — which means one count can be worth anything across a factor of ten.

• u_rel — the relative precision the quote claims, as a percentage. It depends on the figure count and the mantissa only, never on the decade.
• U — the half-width of the bound the inputs actually support, in the unit of the answer. A worst case, not a standard deviation.
• u_rule — the half-unit of the place the rule’s answer stops at, in the same unit.
• ratio — U divided by u_rule. No unit. One is the break-even point, and two is where the last digit has nothing left behind it.

Put numbers on it. Three figures on a value whose leading digits are 1.00 claims 0.5000 %. The same three figures on 9.99 claim 0.05005 %. One count, a factor of 9.99 in what it buys.

Now the scene that defeats three wrong intuitions at once. Multiply 1.00 by 9.99, both quoted to three figures, and the rule keeps three and hands you 9.99.

Follow the bound instead. 0.5000 % plus 0.05005 % is 0.55005 % — carried to five figures here so the next step reproduces — and 0.55005 % of 9.99 is a half-width of 0.05495, as a worst case. The rule’s third digit claims a half-unit of 0.005, so the honesty ratio is 0.05495 divided by 0.005, which is 10.99.

Read that as two whole figures. A half-width of 0.05495 is larger even than the 0.05 a one-decimal quote would claim, so the only honest answer is 1e+1 — one figure where the rule gave three. Multiplying by something written 1.00 is not multiplying by 1.

Swap the leading 1 for a leading 9 and the same count behaves. 9.99 times 9.99 is 99.8001, three figures gives 99.8, and the bound is 0.05005 % plus 0.05005 %, or 0.1001 %, which on 99.8001 is a half-width of 0.0999. Against the 0.05 that three figures claims there, the ratio is 1.998 — five and a half times better behaved, at the identical figure count.

That second identity is exact for a product and only a floor for a quotient. Division widens the bound, because the smallest allowed divisor sits under the largest allowed numerator: the corners (x + u_x) / (y – u_y) and (x – u_x) / (y + u_y) give that same sum of percentages divided by 1 – u_y^2 / y^2. Take the corners for a division, then, and keep the sum for a product.

That ratio is the one check worth running on an answer the rule has given you. At or below 1 every digit the rule kept is supported; between 1 and 2 the last digit is marginal; at 2 or above it is not supported at all. Eight cases, two readings each, show how widely one rule can land.

Two of the eight are conservative and three are badly over 1.5, so neither verdict is “the rule”. The ratio column is rounded to two decimal places while the verdict is read off the unrounded value, which is why row six shows 1.00 and still reads marginal: its true ratio is 1.0001, a hair over the line.

Published constants are the standing illustration of why a digit count is not the whole story. They differ from one another in precision by many orders of magnitude and are quoted with an explicit uncertainty rather than left to a digit count — the guide to physics constants carries that material, and no constant’s value or uncertainty appears on this page.

One housekeeping note, since a page about digits should not exempt itself. Every relative precision and half-width printed here is given to four significant figures, which is this site’s convention, and the honesty ratio to two decimal places. Where one of those figures carries an extra digit, it is flagged where it happens, for the reason flagged there.

Where the Rules Go Wrong

The rules fail in three recognisable places: a subtraction of two near-equal readings, a reading too small to reach the answer, and an apparent tie that your calculator and your marking scheme round in opposite directions.

Subtracting two near-equal readings

Take 5.00 minus 4.99. Each reading is around 0.1 % precise, and the answer, 0.01, is 100 % precise — one subtraction has cost a factor of exactly 1000. The rule gives one figure; the bound supports none at all, because the lower end of it is exactly zero and the two readings may be identical.

No version of the rule can rescue that, and this is where the arithmetic has to be believed over the recipe. A difference can leave the bound large compared with the result: the half-widths add while the values cancel, so the percentage can grow without limit.

The exact-cancellation case is instructive rather than broken. Subtract 5.00 from 5.00 and the answer is 0 with no significant figures, and no relative precision at all — but the absolute bound is still real at plus or minus 0.01, so the honest statement is “zero, to within 0.01”.

A reading too small to matter

Add 1.0 and 0.00990 and the exact sum is 1.0099, which the coarser place rounds straight back to 1.0. The second reading’s three carefully written figures bought exactly none of the answer.

The relative precisions say the same thing twice over: the sum comes out 4.951 % precise against 5.000 % for the coarser reading on its own. Measuring the small quantity better would not move either figure. In practice this is the moment to improve the coarse reading or to stop measuring the fine one.

An apparent tie, rounded two ways

Round 0.045 to one figure. The taught rule looks at the written digits, sees a 5, and rounds away from zero to 0.05. Ask software to do it from the stored binary value and you can get 0.04, because the double nearest 0.045 sits just below it.

This is not a rare curiosity. The same split appears at 0.615, where the digits give 0.62 and the double gives 0.61, and at 1.005, where the digits give 1.01 and the double gives 1.00. On a value that is not a tie, such as 10.999 or 999.9, both routes agree — so the tie is the cause.

The tie also reaches into the answer itself. Multiply 5 by 5, one figure each, and the exact product is 25; one figure of it is a tie, so the taught rule gives 30, written 3e+1 to show the single figure. The bound runs from 20.25 to 30.25, the honesty ratio is exactly 1, and that one figure is 20 % precise: a one-in-five answer, which is worth saying out loud.

How Many Figures Should You Keep?

Decide the precision you need, then keep the smallest figure count that delivers it, rounding the continuous answer up.

The count that comes back is not a whole number, and rounding up is exact advice rather than caution: the ceiling is always enough, and one figure fewer is never enough. There is nothing to be gained by rounding down and nothing lost by rounding up except digits.

The edge can be sharp. A target of 0.5 % on a value whose leading digits are 9.99 computes as 2.00043 figures, which ceils to three. That count is printed to six figures on purpose: at four it would read 2.000 and hide the very thing that decides the answer.

Two figures on 9.99 deliver 0.5005 %, and a miss by five ten-thousandths of a percentage point is still a miss. Relax the target to 0.51 % and the count falls to 1.992, so two figures will do.

Then write the answer so the count can be read back off it. A plain decimal is fine when it round-trips: 20 to three figures is 20.0, and nobody can misread that. Two figures has to be written 2.0e+1, because the plain 20 counts as one or two and hides the very thing you decided.

One habit carries all of this: keep full precision through the working and round once, at the end — the unit conversions guide sets that rule out for a chain of conversions, which is where it costs the most.

And one boundary, because the words are similar. None of this is the quantum limit on what can be measured at all: a figure count is a statement about how a number was written, while the Heisenberg uncertainty principle is a statement about nature, and the two are not the same bound.

Worked Problems

Every reading below is a written string taken as given, and nothing here measures or specifies any instrument. Checking what an answer’s last digit is worth is one step inside a wider method, and the guide to solving physics problems is where the rest of that method lives.

Frequently Asked Questions

Efficiency formula physics reduces to one division: efficiency equals the useful output divided by the total input, multiplied by 100 for a percentage. The ratio carries no unit and never exceeds 1, and when stages run in series their efficiencies multiply rather than add or average.

Hand four stages in series the figures 38 %, 96 %, 92 % and 88 %, and only 29.53 % of the input survives to the far end. Nothing there is faulty. Three of those four pass on almost everything they are given, and the total still lands below thirty per cent.

That gap is where exam marks go, and where money goes. The division itself takes one keystroke; the judgement around it — which figure counts as the useful output, whether to add or to multiply, when to round — is the part worth learning. Everything below is built out of one division and the arithmetic that follows from it.

What Is Efficiency in Physics?

Efficiency is the fraction of what went into a device that came back out in the form you actually wanted, written as a decimal between 0 and 1 or multiplied by 100 and quoted as a percentage.

Two numbers go in. One is everything the device was supplied with; the other is the part of that supply which did the job you wanted done. Divide the second by the first and you are finished — and the answer comes out bare, with no joules and no watts clinging to it.

The published definition of energy conversion efficiency says the same thing: the ratio between a machine’s useful output and its input, in energy terms, with the result sitting somewhere between 0 and 1. The input and the output can be chemical, electrical, mechanical, light or heat, and the one requirement is that both are the same kind of quantity.

Why does a single fraction earn a name of its own? Because it is the only figure that lets you hold a kettle and a crane up against each other. Both sides have to be counted in the same unit before you can, and what a joule actually counts is where that counting starts.

Three claims carry the rest of this page, and each section below is the evidence for one of them.

• Claim 1. Efficiency is one division, and the arithmetic is never the difficult part. Choosing which figure is “useful” is, because that depends entirely on what the device is for.
• Claim 2. Stage efficiencies multiply. A chain of them is therefore never better than its worst single link, and is usually worse than it.
• Claim 3. A percentage handed to you has probably already been rounded, and multiplying a rounded percentage moves the answer in the second decimal place.

The Efficiency Formula

The efficiency formula is one division followed by one multiplication: divide the useful output by the total input, then multiply by 100 to read the result as a percentage.

• eta — the efficiency. A pure number with no unit at all, running from 0 to 1 inclusive; multiply by 100 to quote it as a percentage.
• E_useful — the useful energy output, in joules (J). The part of the supply that arrived in the form the device exists to deliver.
• E_input — the total energy input, in joules (J). Everything that went in, useful and wasted together.
• E_waste — the energy that did not arrive as the useful output, in joules (J).
• P_useful and P_input — the same two quantities measured as powers, in watts (W), over one shared interval.
• eta1, eta2, eta3, eta4 — the efficiency of each separate stage of a chain. Pure numbers.
• eta_total — the overall efficiency of the whole chain. Also a pure number.

The waste follows from the efficiency and needs no second measurement, which is worth knowing because the waste is usually the figure that changes a decision.

Work it by hand on paper the first few times, so the substitution becomes automatic, and then hand the two figures to the Efficiency Calculator, which runs the same relation in whichever of the three directions you are missing a number and prints the waste alongside the percentage. The table below applies it to four scenarios, each one a set of stated givens rather than a measurement of any real product.

Read the last column rather than the fourth. A percentage is the kind of figure a reader half expects; the 22800 J going somewhere other than the water on the kettle row is not, and it is the number that would show up on a bill. Every row is worked line by line further down the page.

The simulator below pins the input at 1000 units and gives you the useful output as a slider, so the first stage’s efficiency is the definition itself sitting under your hand. Three more sliders add stages behind it, and the lab page that hosts this simulator adds eight preset buttons for the scenes this page works through, the four-stage chain among them.

Energy Form and Power Form Give the Same Number

Measure both quantities as energies in joules, or both as powers in watts, and the efficiency comes out identical — because a shared time interval sits above and below the line and cancels out of the ratio exactly.

Take a lamp given 12.00 W and delivering 9.000 W as light. The division gives 0.7500, which is 75.00 %. Now run the same lamp for 60.0 s: the light carries 540.0 J and the supply delivered 720.0 J, and 540.0 / 720.0 is 0.7500 again.

Nothing was approximated there. Multiplying both figures by the same 60.0 s multiplies the top and the bottom of one fraction by the same thing, so the quotient cannot budge. Double both sides, halve both sides, switch from seconds to hours — the answer is untouched.

In practice the power form is the one you meet first, because supplies are rated in watts. Getting a watt out of a measurement is a separate step that this page assumes you already have, and where a power comes from in the first place is the guide that does it.

One rule survives all of this: the two figures must be the same kind of quantity in the same unit. Joules over joules, watts over watts. Kilojoules over joules is the single commonest slip and it reads a thousand times out, with no warning in the answer to tell you.

Deciding Which Number Counts as Useful

The formula never decides which figure is the useful output — what the device exists to do decides it, and the very same joules can be the useful output in one problem and the waste in the next.

A lamp is for light, so the heat coming off it is waste. An electric heater is for heat, so that heat is the entire point, and the joules a lamp would be wasting are the ones you are paying for. Swap the device and the numerator moves, while the arithmetic sits perfectly still.

That is why a worked answer should name its numerator out loud. “Useful output = the gravitational energy gained by the load” is a line worth writing before you divide anything, and it is the line that stops a marker guessing what you meant. A common student slip is to divide by the output instead of the input, which gives a figure above 1 and ought to stop you on the spot.

The conversions themselves — which store of energy becomes which, and by what route — are a separate subject with its own page. The guide to energy conversion examples walks those and then hands over exactly this question: how much of the input arrives in the form you wanted.

Chains: Why Stage Efficiencies Multiply

Stage efficiencies multiply because each stage is handed only what the stage in front of it managed to pass on, so a chain can never beat its worst single link and almost always comes out below it.

Start with three stages, given as 40 %, 92 % and 85 %. The product is 0.400 times 0.920 times 0.850, which is 0.3128, or 31.28 % overall. That is below the 40 % stage, and the two good stages that follow the bad one did not rescue it — each simply took a further bite.

Now the scene that defeats every wrong intuition at once. Four stages, given as 38 %, 96 %, 92 % and 88 %: multiply the fractions and you get 0.29534208, which is 29.53 %.

Look at what that figure is not. It is not the first stage, it is not the average of the four, and it is nowhere near the three strong stages that follow the weak one.

One bad link sets the scale of the whole chain, and everything downstream can only shave what is left of it.

Say it in units rather than per cent and it stops being abstract. Put 1.000 MJ into that chain — 1000000 J — and 295342.08 J arrives while 704657.92 J does not. The percentage tells you how good the chain is; the second of those two numbers tells you what it costs.

One more property is worth noticing, because it is the one that makes chains easy to check: order is irrelevant. Multiplication does not care which stage you write first, so the same four numbers in any sequence give 0.29534208. A stage left at exactly 100 % multiplies by 1 and changes nothing, which is how you tell the arithmetic that a stage is not there.

Two Wrong Ways to Combine a Chain: Adding and Averaging

Adding stage efficiencies and averaging them are both wrong, and a two-stage case shows how badly: given 0.800 and 0.900, the overall figure is 72.00 %, not 170.00 % and not 85.00 %.

Adding fails first and fails loudly. 0.800 plus 0.900 is 1.700, and an efficiency above 1 would mean more came out than went in. Any answer above 100 % is the arithmetic telling you that you combined the stages wrongly, not that you have found a remarkable machine.

Averaging is the dangerous one, because it looks reasonable and lands in the right region. The mean of 0.800 and 0.900 is 0.8500, or 85.00 %, which is above the correct 72.00 % and — tellingly — above the weaker of the two stages. An overall efficiency can never be above the smaller of the two, and 72.00 % sits below it while 85.00 % sits above it.

There is a quick sanity test here that costs nothing. Your overall figure must not be larger than any stage you multiplied, and if it is, you have added, averaged or dropped a stage. Check it before you check anything else.

Why Efficiency Can Never Pass 100 %

No device can exceed 100 %, and the reason is conservation of energy rather than anything subtler: you cannot draw more out of a supply than was put into it.

Exactly 100 % is a different matter — it is allowed, and it describes the ideal lossless device in the table above, where every joule supplied arrives as the useful output. Real devices fall short of it. The point is that the ceiling is arithmetic, not engineering: it would still be there in a universe with no friction.

So an efficiency above 1 is not a result but a signal. It means the useful output and the total input have been swapped, or they are not measuring the same thing, or a unit slipped by a factor of a thousand. The calculator for this page declines that combination in words instead of printing a figure, and the simulator cannot even be driven into that state, because its slider stops at the input.

Two neighbouring ratios wear the same word and are not this one. One belongs to mechanical advantage, where it compares the advantage a machine actually gives with the advantage its geometry promises, and it is not computed here. A heat engine carries a second and lower ceiling fixed by the two temperatures it works between — that is the Carnot efficiency, a separate calculation, and nothing on this page performs it.

And the apparent counterexample, the heat pump quoted above 100 %, is not a violation either. What gets quoted there compares heat moved against work supplied — two different quantities, so a different ratio with its own name, and no energy appearing from nowhere.

Where Rounding Bites

Multiplying percentages you have already printed shifts the result: two thirds of something at 0.850 comes to 56.67 %, while the same pair run through a stage printed first as 66.7 % comes to 56.70 %.

The gap is 0.0283 percentage points. It hides completely in the first decimal place and is plainly wrong in the second, and it grows with every extra stage you chain the same way. The rule that fixes it is short: carry the fractions, round at the very end, once.

A second trap runs the other way. A device at 0.99949 prints as 99.95 % to two decimal places, which is honest, but as 99.9 to three significant figures — and a reader may well file that as lossless. It is not: 0.00051 of the input is still going somewhere. That is why nothing on this page quotes an efficiency to three significant figures.

Thirds deserve a mention because exam questions are full of them. One third is 33.33 % and two thirds is 66.67 %, both correctly rounded, and the two printed figures add to 100.00 % only because the rounding happens to go that way.

None of this is a rule about how many figures an answer deserves — that is a different question with its own answer. The question here is purely when in a chain you are allowed to round, and the answer is once, at the end.

Worked Problems

Frequently Asked Questions

Dimensional analysis physics is the method of writing every quantity as a triple of exponents over the base dimensions mass, length and time: [M], [L] and [T]. An equation is admissible only if those exponents agree on both sides, and with three ingredients the same rule derives the shape of a law, never its constant.

Make a pendulum four times as long and its small swings take exactly twice as long. That factor of two is not measured and not looked up: it falls out of the exponents of mass, length and time, and nothing else about the pendulum gets a say.

This is the cheapest tool in physics and the most underrated one. It will tell you that a pendulum’s period cannot depend on the mass of the bob, that an orbital period has to go as the three-halves power of the radius, and that the third term in your working is in the wrong place — all before you solve anything. It also has a hard limit, and this page measures that limit rather than asserting it.

What Is Dimensional Analysis in Physics?

Dimensional analysis is the arithmetic of a quantity’s exponents: every mechanical quantity is mass to some power times length to some power times time to some power, and physics only lets you equate quantities whose three powers agree.

A dimension is not a unit, and the difference is the reason the method works. Measure a speed in metres per second, in miles per hour or in furlongs per fortnight and the number changes every time, while the exponents — one of length, minus one of time — never budge. The three dimensions do get their numbers from the SI base units, and the guide to SI units is where that scaffolding is set up.

Neither the idea nor the notation is new: Joseph Fourier introduced dimensional analysis and the dimension of a quantity in 1822.

Everything below rests on three claims, and each section is the evidence for one of them.

• Claim 1. A dimensional formula is a vector of three exponents, and the algebra is addition. Multiplying two quantities adds their vectors; raising one to a power scales its vector.
• Claim 2. Homogeneity does more than audit an equation. Three base dimensions give three equations, so three ingredients usually pin down exactly one set of exponents — which makes this a derivation, not a check.
• Claim 3. The method is necessary and nowhere near sufficient. A dimensionless factor is invisible to it, a dimensionally perfect equation can still be false, and when the ingredients are dimensionally dependent it cannot even choose between three different answers.

Claim 3 is the one most accounts leave out. It is also the most valuable of the three, because knowing precisely what a check cannot see is what stops you trusting it further than it goes.

The Dimensional Formula: [M], [L] and [T]

A quantity’s dimensional formula is its exponent triple — the powers a, b and c below, one for mass, one for length and one for time.

• a, b, c — the dimensional exponents of the quantity Q: the powers of mass, length and time in its dimensional formula. Pure numbers, with no unit of their own.
• [M], [L], [T] — the three base dimensions this page works in: mass, length and time. Their SI base units are the kilogram (kg), the metre (m) and the second (s).
• p, q, r — the trial exponents: the powers of the three ingredients in a trial law. Pure numbers, and what the method solves for.
• k — the dimensionless constant a law carries. A pure number, and never a dimensional result.
• residual — one number per base dimension: the trial’s exponent in that row minus the target’s. All three zero means homogeneous.
• n — a scaling exponent, the single power in Q₂ / Q₁ = (X₂ / X₁)ⁿ. Dimensionless.

Two rules do all of the work, and both are arithmetic on the triple.

Force is mass times acceleration, so add the vectors: (1, 0, 0) + (0, 1, -2) = (1, 1, -2). Force is [M^1.00 L^1.00 T^-2.00], and no table was consulted to get there.

Energy is force times length, which adds (0, 1, 0) and lands on (1, 2, -2). Reach it the other way, as mass times the square of a speed, and the second rule gives (1, 0, 0) + 2 × (0, 1, -1) — the identical triple. Two routes, one answer: that is the first sign this is algebra rather than bookkeeping.

Pressure is energy per unit volume, and division subtracts: (1, 2, -2) − (0, 3, 0) = (1, -1, -2). Every quantity below was built this way.

Here is the whole mechanical vocabulary this page needs. The unit in the third column is a consequence of the exponents rather than an input to them, and if you want the finished laws these quantities appear in, the physics formulas reference is the list.

Read any row backwards and you have a sanity check worth having. A quantity with a mass exponent of zero cannot depend on a mass, however the algebra is dressed up.

Dimensional Homogeneity: Why a Balanced Equation Can Still Be Wrong

An equation is dimensionally homogeneous when every exponent on the left matches the exponent on the right — and homogeneity is necessary for a correct equation without being anywhere near sufficient.

The rule itself is as strong as rules in physics get: any physically meaningful equation must carry the same dimensions on both sides, which is what dimensional homogeneity means, and checking it is usually described as a plausibility test on a derived result. Two quantities of the same dimension can be compared even when they are written in different units; two of different dimensions cannot be compared at all, in any units whatever.

Try it on energy = mass × speed². The left side is (1, 2, -2). On the right, (1, 0, 0) + 2 × (0, 1, -1) = (1, 2, -2), so the residual is zero in all three rows and the equation passes.

Now the part that matters. Write 2 m v² instead of 1/2 m v² and the check passes exactly as happily, because a pure number has no exponents to contribute. At m = 2.000 kg and v = 3.000 m/s, 1/2 m v² gives 9.000 J while 2 m v² gives 36.00 J: one of them is four times too big and dimensionally perfect. The mechanics formulas guide states that conclusion about the unit check; this is the page that measures it.

Three separate things are wrong with reading a successful check as a proof, and they are independent of each other.

• A dimensionless factor is invisible. One half, two, pi, and even a function of an angle all pass through unchanged, because none of them has a dimension to contribute.
• A balanced equation can simply be false. Energy = m v² balances in every row and is wrong by a factor of two. No amount of checking will find that.
• Sometimes the check cannot even choose. When the ingredients are dimensionally dependent, three different exponent triples can all balance perfectly — the last section walks through exactly that case.

So the honest reading of a passed check is narrow: the equation is still in the running, and nothing more has been established.

How to Derive a Formula From Dimensions Alone

Write the quantity you want as a constant times three ingredients raised to unknown powers, demand that the exponents agree in every base dimension, and you have three linear equations for three unknowns.

Take a pendulum. The quantity wanted is a period, so the target triple is (0, 0, 1), and the three ingredients offered are the length L, the gravitational acceleration g and the mass m of the bob.

One row per base dimension, and each row is an equation:

• M row. Of the three ingredients only the mass carries a mass dimension, so r = 0.
• L row. Length and gravity each carry one power of length, so p + q = 0.
• T row. Only gravity carries a time, and it carries two inverse powers of it, so −2q = 1.

Solve from the bottom up. The T row gives q = −0.50; the L row then gives p = 0.50; and the M row has already sent r to 0.00. Substituting those back reproduces the target exactly: 0.50 × (0, 1, 0) − 0.50 × (0, 1, -2) = (0, 0, 1).

The mass has dropped out of the problem altogether, and that is the method saying something physical. A heavy bob and a light one of the same length keep the same period.

What the working has earned is this, and dimensions stop here.

Those exponents are exact. No constant is available anywhere in that working: at small amplitude it happens to be 2 pi, which comes out of solving the equation of motion, and the pendulum period formula is where that constant and its conditions belong.

Three ingredients and three base dimensions make a three-by-three system, which is why it usually has exactly one answer. The word “usually” is carrying real weight there: when the ingredient matrix is singular, the system has either no solution or infinitely many, and both of those turn up further down this page.

The lab below is the same ledger with the exponents under your control. Each row is one base dimension, the bars are what the three ingredients contribute to it, and the gap between where they finish and where the target sits is the residual — the number that tells you which ingredient is wrong rather than merely that something is. Underneath, a log-log panel sets what your trial law predicts against what the exact physics does.

Why Dimensions Can Never Give You the Constant

Dimensions fix every exponent in a law and say nothing whatever about the dimensionless number in front of it.

The cleanest demonstration is three problems that come out with the identical exponent triple and have constants that are nothing like each other. A pendulum’s period from length and gravity, the time to fall from rest through a height, and the speed of a wave on a stretched string: all three land on (0.50, −0.50, 0.00).

Read the last column. Across three problems whose dimensional answer is identical, the constants span a factor of 6.283 — so calling an exponent triple “the formula” can leave you wrong by more than six times.

The third row is the honest counterweight. For the string, k is exactly 1, so the dimensional answer happens to be the whole answer. The reach of this method is not uniform, and it would be just as misleading to claim it never gets you all the way.

There is a sharper version of the same blindness. A real pendulum’s period depends on how far back you pull it, and that dependence is invisible here, because an angle is a pure number: T = k sqrt(L / g) multiplied by any function of the angle you like is dimensionally identical. Work the exact series and you can see the size of what is being missed — at 30 degrees the true period runs 1.741 per cent above the small-amplitude value, and at 5 degrees it runs 0.04762 per cent above it.

That is why every appearance of 2 pi on this page carries the words “small amplitude”. The usual statement of the condition is an angle much less than one radian, often quoted as under 0.1 radians or about 6 degrees, and what goes as the cube of that angle is the error in the small-angle step itself, where sin theta is replaced by theta in radians. The period error that step produces goes as the square instead, which is what the two percentages above already show: cutting the amplitude sixfold divides the error by roughly six squared, not six cubed.

Where Scaling Works Without Any Formula at All

Because the constant cancels when you divide one instance of a power law by another, an exponent on its own predicts exactly how a quantity changes.

Divide Q = k Xⁿ at two values of X and the k disappears — not approximately, and not only for small changes. What is left is a ratio raised to the exponent dimensions handed you, which you can do in your head or hand to the Dimensional Scaling Calculator, which also runs the relation backwards and returns the exponent itself.

The pendulum again, this time with no constant anywhere in sight. Four times the length is a length ratio of 4.000, and 4.000^0.50 = 2.000, so the period doubles. Evaluate the exact small-amplitude formula separately at each length, with g at its defined standard value of 9.80665 m/s², and it returns 1.419 s at 0.5000 m and 2.837 s at 2.000 m — two figures each rounded once from full precision, with an underlying ratio of 2.000.

Notice what you never had to do: convert anything. Both ratios are pure numbers, so whatever unit you measured in divides out of its own ratio and leaves no trace, and seconds with metres gives the identical factor to milliseconds with centimetres. The SI prefix converter is the tool for the conversion job proper; this calculation simply does not have one.

The same relation runs the other way, and that direction needs no theory at all. Take two satellites around one planet: the first at a radius of 7.000e+06 m with a period of 5829 s, the second at 4.200e+07 m with a period of 8.566e+04 s. Their periods stand in the ratio 14.70 and their radii in the ratio 6.000, so n = ln(14.70) / ln(6.000) = 2.688 / 1.792 = 1.500 — the exponent dimensions predict for an orbit, recovered from two observations with the constant cancelling out of the arithmetic.

Square that statement and you have the familiar one: the square of the period goes as the cube of the radius, which is Kepler’s third law. Dimensions got the three-halves out of a radius and a gravitational parameter with no theory of gravitation in sight; the constant is what they could not get, and the Kepler third law calculator is where that lives.

A wrong exponent does not shift a prediction up or down — it tilts the whole trend. Normalise two curves at one point and the constant is gone from both, which leaves the slope on a log-log plot as the only thing left to disagree about.

When the Method Refuses, and When It Cannot Choose

Two things can go wrong with those three equations, and neither is a failure of the method: they can have no solution, which means an ingredient is missing, or infinitely many, which means the ingredients are not independent.

Ask for a period from a length, a mass and an area. The M row sends the mass exponent to zero, the L row ties the length and area exponents together, and then the T row reads 0 = 1 — not one of the three ingredients carries a time, and the target is a time.

A brute-force sweep settles it more convincingly than the algebra does. Scan every exponent triple on a quarter-step grid from −4 to 4, which is 35,937 of them, and the smallest the T residual ever gets is exactly 1 in size, never 0. Nothing has broken: the method is telling you that an ingredient carrying a time is missing, and if you put gravity back the pendulum answer appears at once.

The second case is subtler and rather more interesting. Ask for a speed from a length, a gravitational acceleration and a time, and three different exponent triples all balance perfectly:

• (1.00, 0.00, −1.00), which reads v = L / t
• (0.50, 0.50, 0.00), which reads v = sqrt(L g)
• (0.00, 1.00, 1.00), which reads v = g t

All three are dimensionally perfect, and they are three physically different laws. The reason is that these ingredients are dimensionally dependent — an acceleration is already a length over a time squared — so the three equations are not independent and the system has no single answer. Homogeneity has not been violated here; it has run out of information.

The residual is therefore worth reading one dimension at a time, rather than as a single pass or fail. Nudge the pendulum’s mass exponent up to 0.25 and the M row breaks by 0.25 while the L and T rows stay at exactly zero. Nudge the gravity exponent to −0.25 instead and two rows break at once: the L row by 0.25 and the T row by −0.50. Which rows fail is what tells you which ingredient to look at.

Finally, the honest edge of the page. Three base dimensions are a choice rather than the whole system, and several neighbouring subjects are deliberately not here.

• Four more base dimensions exist. The International System of Quantities has seven base quantities — length, mass, time, electric current, thermodynamic temperature, amount of substance and luminous intensity — and a quantity’s dimension is a product of all seven raised to positive, negative or zero exponents. Charge, current, voltage and resistance need the electric-current one; entropy and specific heat need temperature; molar mass needs amount of substance. The same elimination runs in seven dimensions completely unchanged.
• Counting dimensionless groups is a different subject. That is the Buckingham pi theorem, which counts how many independent dimensionless groups a problem has, and nothing on this page computes or implies any part of it. For those groups in practice, and for dynamic similarity, the Reynolds number guide is the place to go.
• Nothing here is about natural units. Planck or geometrised units, directed dimensions and dimensional regularisation are all real physics and all out of scope; no figure above depends on any of them.

Worked Problems

Frequently Asked Questions

Black hole physics is the study of a mass sealed inside its own event horizon, the one-way surface at radius R_s = 2GM/c^2. That radius depends on the mass alone: 2.953 km per solar mass. A static clock at distance r ticks sqrt(1 – R_s/r) times for each far-away tick, and at the horizon it stops.

Almost nothing in the universe is this easy to describe. Choose a mass and three numbers follow at once: how wide the horizon is, how slowly a clock runs at a given distance from it, and how hard a body held there would be pulled apart. No composition, no temperature, no chemistry, no age — the mass does all of it.

That is also why the arithmetic is worth doing by hand. Ten Suns’ worth of mass hides behind a horizon of radius 29.53 km, and a clock held at the innermost stable orbit around it runs at 0.8165 of the rate yours does. Change the mass and the horizon follows in exact step.

What Is a Black Hole?

A black hole is a region of spacetime wrapped in an event horizon: a surface that matter and light cross inwards and never cross back out. Everything else about it — how big it is, how slowly clocks run near it, how violently it pulls a body apart — follows from one number, its mass.

Start with the geometry rather than the force. General relativity describes the empty space outside a non-rotating, uncharged mass with the Schwarzschild solution, and that solution contains a factor, 1 – R_s/r, which shrinks as you approach the centre and reaches exactly zero at one particular radius. That radius is the horizon. It is not a wall and not an object; it is the place where the geometry stops allowing an outward path.

The speed of light is what sets the scale of the whole thing. Because c is the limit on how fast anything can move or signal, the radius at which the geometry closes over is built out of c, and the ideas underpinning that limit belong to special relativity. Relativity puts a ceiling on speed; a horizon is what happens when a mass is packed tightly enough for that ceiling to matter.

Everything on this page reduces to three claims. The sections below are the evidence for each.

• Claim 1. The horizon radius is exactly proportional to the mass, and depends on nothing else — not on your distance, not on what the hole is made of, not on how it formed.
• Claim 2. A static clock near the horizon runs slow by a factor sqrt(1 – R_s/r), and at the horizon it stops. That factor depends only on your distance measured in horizon radii, never on the mass.
• Claim 3. The tidal stretch at a hole’s own horizon falls as one over the mass squared. A stellar-mass hole would shred a human body at its horizon; a supermassive one would barely tug at it.

Most accounts stop after the first. The second and third are what decide whether the picture in your head is right — and the third turns upside down if you compare two black holes carelessly.

The Schwarzschild Radius Formula

The radius of the event horizon is twice the mass times the gravitational constant, divided by the speed of light squared. Put a mass in kilograms on the right and a length in metres comes out.

• R_s — the Schwarzschild radius, the radius of the event horizon. Unit: metre (m), usually quoted in kilometres.
• M — the mass of the black hole. Unit: kilogram (kg), usually quoted in solar masses.
• G — the gravitational constant, 6.67430 x 10^-11. Unit: m³ kg^-1 s^-2.
• c — the speed of light, 299,792,458 m/s exactly, by the definition of the metre. Unit: metre per second (m/s).
• r — your distance from the centre, which does not appear in the formula at all. Unit: metre (m).
• x — that same distance counted in horizon radii, r / R_s, so x = 1 is the horizon itself. Dimensionless, and never less than 1.

Because mass is the only variable on the right-hand side, the relation is pure proportionality, and it is worth memorising in the units astronomers actually use. One solar mass corresponds to 2.953 km of horizon radius. Ten solar masses give 29.53 km, a hundred give 295.3 km, and so on for ever — or you can hand the mass straight to our Schwarzschild radius calculator, which also runs the relation backwards from a horizon radius to the mass behind it.

Notice what is not in the formula. The product GM is an input, taken from whatever mass you were given, so none of the Newtonian machinery of the law of gravitation is being derived here. Nor does r appear: move closer or further away and the horizon radius you compute does not budge by a millimetre.

The constant c carries all of the smallness. Dividing by c² — about 9 x 10¹⁶ in SI units — is what turns a stellar mass into a city-sized length, and the speed of light is where that figure comes from. It also contributes no uncertainty, because the metre is defined from it.

One last thing about this expression, and then the article leaves it alone: a purely Newtonian calculation about escape velocity arrives at the identical algebra, which is a coincidence of arithmetic rather than a derivation, since a horizon is a property of the geometry and not a speed a projectile fails to reach.

What the Event Horizon Actually Is

The honest signature of a horizon is not a size but a stopped clock. Take a clock and hold it at a fixed distance r from the centre, well outside the horizon, and compare it with an identical clock far away: the near one runs slow, by a factor that has R_s in it and nothing else.

Write the distance in horizon radii and the mass disappears from the problem entirely. At x = 3, the innermost stable orbit, the rate is 0.8165. On the photon sphere at x = 1.5 it is 0.5774. At x = 2 it is 0.7071, and at the horizon itself, x = 1, the bracket is exactly zero and so is the rate: 0.0000.

That last number is the point. A one-solar-mass hole and a ten-billion-solar-mass hole have horizons differing in size by a factor of ten billion, and both stop a static clock completely. The radius is a length and lengths scale with mass; the clock rate at a given multiple of the radius does not scale at all, which is what makes the horizon a feature of the geometry instead of a measurement of bigness.

Be careful about which kind of slowing this is. The effect here comes from where the clock sits in the geometry, not from how fast it is travelling; a clock that is simply moving quickly also runs slow, and that velocity effect is time dilation in the special-relativistic sense, with a different cause and a different formula that this page never uses.

Two more cautions, both of which are easy to get wrong. The rate never actually reaches 1, however far out you stand — it is still below 1 at ten thousand horizon radii — so the correct phrasing is always “far enough away that the difference does not matter”, never “normal”. And the rate applies to a clock held still at that radius, so nothing on this page describes what a falling clock would read.

The simulator below carries the same three relations. The upper panel draws one hole’s horizon, photon sphere and innermost stable orbit against a labelled scale bar whose own value in kilometres is printed beside it; the lower panel plots the clock rate against distance, and that curve is identical at every mass, which is claim 2 made visible.

Three Radii That Matter: Horizon, Photon Sphere, ISCO

A non-rotating black hole has three special radii, and all three are fixed multiples of the horizon: the horizon at 1 R_s, the photon sphere at 1.5 R_s, and the innermost stable circular orbit at 3 R_s. Because they are multiples, a single drawing serves every mass.

The photon sphere, at 1.5 R_s = 3GM/c², is the radius at which a ray of light can travel in a circle around the hole. It is a real orbit and a useless one: the orbit is unstable, so the faintest nudge inward sends the ray spiralling in and the faintest nudge outward sends it away for good. Nothing can linger there.

The innermost stable circular orbit, or ISCO, sits at 3 R_s = 6GM/c². Outside it, matter can hold a stable circular path; inside it, no circular orbit survives a small disturbance. It is quoted here as a radius of the same geometry, not worked out from an orbital period, which is a different calculation belonging to a different page.

Put real numbers on them and the scale of the thing lands. For the ten-solar-mass hole of the hook, the horizon is 29.53 km in radius, the photon sphere 44.30 km and the ISCO 88.60 km — three surfaces inside a region you could cycle across in an afternoon, if a bicycle were any use there.

Why a Bigger Black Hole Is Gentler at Its Own Horizon

Compare two black holes each at its own horizon, and the more massive one is the gentler of the two — that qualifier, “at its own horizon”, is the whole claim, and dropping it makes the sentence false. What does the pulling is the tidal stretch: the difference in gravitational pull between your head and your feet, which is what tears a body apart rather than the pull itself.

Here h is the length of the body being stretched; every figure below is quoted across h = 1.80 m, roughly a person. The first form is the general one. The second is what you get by substituting r = R_s into it, and the mass has moved into the denominator and been squared.

So at its own horizon, ten times the mass means one hundredth of the stretch. Run that down the ladder and the numbers are startling.

Read the last column downwards. A 1.80 m body at the horizon of the ten-solar-mass hole is stretched at nearly nineteen million g; raise the mass to 4.350e+4 solar masses and the stretch at that hole’s horizon is 0.9996 g, a head-to-foot difference no bigger than one g; raise it to 4.300e+6 and the stretch is 1.023e-4 g, a ten-thousandth of that. Multiply the mass by 4.300e+5 and the stretch falls by its square, 1.849e+11.

Now the trap, and it is the single most likely way this claim gets repeated wrongly. Hold the distance fixed in kilometres instead, and the comparison flips completely. At a fixed 1.000e+5 km from the centre, a 1.80 m body feels 4.872e-4 g of stretch from the ten-solar-mass hole and 209.5 g from the 4.300e+6-solar-mass one: the bigger hole is 4.300e+5 times stronger, which is exactly the mass ratio, because at fixed r the stretch is simply proportional to M.

Both statements are true, and they are answers to different questions. “Gentler” belongs to the comparison made at each hole’s own horizon, and nowhere else.

How Big Are Real Black Holes?

Real black holes come in two well-established size ranges: stellar-mass holes of a few to a few tens of solar masses, and supermassive holes of millions to billions. Their horizons therefore run from a few tens of kilometres to tens of billions.

There is a floor, and it is an observational one rather than an arithmetic one. Collapse theory and the measured masses of compact objects together give the rule that an object heavier than three solar masses can only be a black hole, which has become the observational definition in use. At three solar masses, the smallest well-established horizon is 8.860 km in radius.

Stellar-mass holes are the ones the formula makes most vivid. The ten-solar-mass example used throughout this article has a horizon 29.53 km in radius, and ten Suns’ worth of mass sits inside a sphere of exactly that size. The size itself gives no hint of what it contains.

At the other extreme sits the black hole at the centre of our own galaxy, Sagittarius A*, whose mass is given as about 4.3 million solar masses. If that is its mass, its horizon is about 1.3 x 10⁷ km in radius, some eighteen times the radius of the Sun. That figure is a consequence of the stated mass and not a measured property of the object; the two sources consulted for this article agree on its mass only to one significant figure, so two figures are all its horizon gets here and only the leading one is firm.

Larger ones exist still. A ten-billion-solar-mass hole, of the kind found at the centres of the biggest galaxies, would have a horizon 2.953e+10 km in radius, and the stretch across a 1.80 m body at that horizon would be 1.891e-11 g — utterly imperceptible. The largest black holes are the mildest places to stand.

Where This Model Breaks Down: 4 Common Mistakes

Everything above describes a Schwarzschild black hole: non-rotating, uncharged, isolated, with nothing falling into it. The arithmetic inside that model is exact, and four mistakes account for most of the wrong statements made with it.

Mistake 1: comparing two black holes at the wrong distance

“Bigger black holes are gentler” is true at each hole’s own horizon and false at a fixed distance in kilometres, where the stretch is proportional to the mass and the bigger hole wins. The two comparisons differ by a factor of the mass ratio cubed, so getting this wrong is not a rounding error. Always say which comparison you mean.

Mistake 2: reading the clock rate as what a falling observer sees

The factor sqrt(1 – R_s/r) is the rate of a clock held at rest at that radius, measured against one far away. It is not what a falling clock reads, and it is not a Doppler shift. Nothing in this model describes infall at all, so any sentence beginning “as you fall in, your clock…” is outside what has been computed here.

Mistake 3: using these radii for a real black hole

Real black holes spin, and spin changes every radius above. The source consulted for this article gives the rotational parameter of the hole in M87 as 0.90 plus or minus 0.05, which is close to the theoretical maximum — so the Schwarzschild figures are an idealisation, not the right numbers for that object. Treat every radius here as the non-rotating case, which is what it is.

Mistake 4: letting the formula decide what exists

R_s = 2GM/c² will happily take any mass at all, including a third of a solar mass, and hand back a radius. Problem 3 below runs that backwards — a 1.000 km horizon needs 0.3386 solar masses — which is well under the three-solar-mass observational floor. The arithmetic is correct; the object is not one anybody has seen.

And a longer list of things this model simply does not describe, for which no figure appears anywhere on this page:

• Hawking radiation, evaporation times and black-hole entropy.
• Rotation: the ergosphere and frame dragging.
• The interior, and whatever the singularity means.
• Accretion-disc temperatures, jets and the geometry of lensed images.
• Gravitational waves from mergers.

Those are real physics and most are active research. They are simply not what three static relations about one isolated mass can tell you.

Worked Problems

Frequently Asked Questions

Resonance physics is what happens when a system is driven at its own natural frequency and responds with its largest motion. For a mass on a spring that frequency is f0 = (1/2pi) sqrt(k/m), set by the spring and the mass alone. At resonance the amplitude is Q times the static deflection, and Q is fixed by the damping.

You already know how to do this. Push a child on a swing and you do not shove harder and harder; you give one small push per swing, timed to go with the motion rather than against it, and the arc grows on its own. Mistime those pushes and the same effort does almost nothing.

That is the whole of resonance, and the physics only sharpens it. Every object with springiness and mass has one frequency it answers to, and driving it at that frequency turns a gentle push into a large motion. The interesting question is not whether the motion grows but how far it grows before something stops it — and the answer to that has nothing to do with how hard you pushed.

What Is Resonance in Physics?

Resonance is the large response a system gives when it is driven at or near its own natural frequency. The natural frequency is the rate at which the system would swing if you displaced it once and let go. Drive it slower or faster and the response shrinks; drive it in step and the motion builds, cycle after cycle.

The system this page works with is the simplest one that shows the whole effect: a block of mass m on a spring of stiffness k, with a damper that resists motion in proportion to speed, pushed back and forth by a force of fixed size. Everything that follows is measured on that oscillator. Its own free motion — the displacement, the speed, the acceleration cycle by cycle — belongs to simple harmonic motion, and this page takes it as read.

Three rules carry the whole topic, and the rest of this article is the evidence for them.

• Rule 1. The natural frequency depends on the stiffness and the mass, and on nothing else — not on how hard you push, not on the damping, not on the frequency you drive at.
• Rule 2. How big the response gets at the natural frequency is decided by the damping alone. It is exactly Q times the deflection the same force would produce if you simply leaned on the spring.
• Rule 3. The frequency of the largest response is not quite the natural frequency. Damping pulls the peak below it, and past a certain amount of damping there is no peak at all.

Rule 1 is the one every textbook states. Rules 2 and 3 are the ones that decide whether a real machine survives, and they are where most of the confusion lives.

The Resonance Frequency Formula

The natural frequency of a mass on a spring is one square root and one division by 2π. Write the stiffness in newtons per metre and the mass in kilograms and the answer comes out in hertz.

The same relation in angular form drops the 2π, and the two are connected by the notation w = 2πf that runs through all oscillation work.

Four more quantities turn that frequency into a prediction about size. Two of them are dimensionless, which is why they travel between a spring, a tuning fork and a tuned circuit without changing form.

• f₀ — the natural frequency, the rate the system swings at when nothing is driving it. Unit: hertz (Hz).
• w₀ — the same thing in angular measure, 2π times f₀. Unit: radian per second (rad/s).
• k — the spring constant, the force needed per metre of stretch. Unit: newton per metre (N/m).
• m — the mass that moves. Unit: kilogram (kg).
• c — the damping coefficient, the resisting force per unit of speed. Unit: newton second per metre (N s/m).
• zeta — the damping ratio, usually printed as the Greek letter zeta. Dimensionless, and the only thing that decides how tall resonance gets.
• Q — the quality factor, one over twice the damping ratio. Dimensionless.
• f and r — the frequency you drive at, in hertz, and the ratio of that to the natural frequency. r is dimensionless, and r = 1 means you are driving exactly on resonance.
• M — the magnification: the steady amplitude divided by the static deflection F₀/k. Dimensionless.

Put numbers through it once and the shape of the answer sticks. A 2.0 kg block on a 200 N/m spring gives k/m = 200 / 2.0 = 100.0 1/s², so w₀ = 10.00 rad/s and f₀ = 1.592 Hz. If you would rather not run the square root by hand for every spring, our Resonance Frequency Calculator solves this relation in any direction — frequency from k and m, or the stiffness or mass you would need for a frequency you want — and prints the damping ratio, Q and the response at your drive frequency beside it.

The spring constant is an input here, never a result. If you do not have a figure for the spring you are holding, the spring constant calculator gets it from a load and the extension that load produces, and then this formula takes over. Confuse the two and you will solve the wrong problem entirely.

One comparison worth making and then leaving alone: a simple pendulum obeys a formula of the same shape, with a length and the strength of gravity in place of a stiffness and a mass, which the pendulum period formula works through. Same algebra, different physical ingredients. Nothing in this article transfers to it unchanged.

How Resonance Builds Up: Push in Time With the Swing

Resonance builds because a well-timed force adds energy on every single cycle instead of taking some of it back. A force does positive work when it points the same way the object is already moving. Keep that true for the whole cycle and the energy in the oscillator rises steadily.

Here is the timing that makes it work, and it is sharper than “push when it is at the top”. The driving force has to lead the motion by a quarter of a cycle, so that the force is at its largest exactly when the block is moving fastest through the middle. Then the force and the velocity point the same way the whole time, and no part of the cycle is spent cancelling what another part achieved.

Drive it at the wrong frequency and the timing drifts. The force now spends part of every cycle opposing the motion, so most of what it puts in comes straight back out, and the oscillation settles at a much smaller size. That is why the response falls away so sharply on either side of the natural frequency when the damping is light.

What stops the growth is the damper. Over one cycle the force adds an amount of energy proportional to the amplitude, while the damper removes an amount proportional to the amplitude squared. The two match at exactly one amplitude, and that is where the motion settles. Make the damper weaker and that balance point moves higher, which is Rule 2 in one sentence.

The simulator below is the same oscillator with the sliders exposed. Set the mass and the spring, choose a drive frequency, and the drawing answers in two stacked panels: the block is already at its steady swing in the upper one, and the response curve underneath shows what every other drive frequency would have given.

How Big Does It Get? Damping Decides

At the natural frequency the magnification is exactly Q, which means the amplitude is Q times the deflection the same force would produce if it were simply held steady. Set r = 1 in the magnification formula and the first bracket vanishes; what is left is 1/(2 zeta), and that is the definition of Q. No force appears in the result at all.

Read that again, because it is the sentence this whole topic turns on. The size of the push sets the static deflection, and the damping sets the number you multiply it by. Double the force and you double both; halve the damping and the response doubles with the same push.

Take the block from the formula section — 2.0 kg on a 200 N/m spring — with a damper of 2.0 N s/m, pushed by a force whose peak size is 2.00 N. Lean on that spring with 2.00 N and it moves 10.00 mm. Push it back and forth at 1.592 Hz instead and it swings 100.0 mm, because the damping ratio is 0.05000 and Q is therefore 10.00.

No table can make the point those curves make. The two lightly damped ones lie on top of each other everywhere except close to the dashed line, where the 0.01 curve reaches 50.00 and the 0.05 curve reaches 10.00 — their two quality factors. Heavy damping barely lifts its curve at all, and past a ratio of about 1.8 all three fall away together.

Seven settings of that same oscillator are tabulated below, and the pair to compare is the second and third rows. Same spring, same mass, same natural frequency, same drive — only the damper changes, and the amplitude goes from 500.0 mm to 10.00 mm, a factor of 50.00.

Every amplitude in that table comes from the same 2.00 N push, and six of the seven rows are driven at their own natural frequency. The exception is Driven well below f₀, which has the same spring, mass and damper as the first row but is driven at half the natural frequency, 0.7958 Hz rather than 1.592 Hz — and that alone takes the swing from 100.0 mm down to 13.30 mm. The last row is Rule 1 on its own: four times the stiffness, the same mass, and the natural frequency doubles from 1.592 Hz to 3.183 Hz while nothing about the damper has been touched.

In practice Q is also the number engineers quote, because it has a second meaning that falls out of the same algebra. A lightly damped system is sharp as well as tall: for light damping the width of the response at its half-power points is about f₀/Q. High Q buys you a big response and a narrow band in one purchase, which is exactly what a radio tuner wants and exactly what a bridge does not.

Where the Peak Actually Sits

The largest response is not at the natural frequency once there is any damping at all. It sits at f₀ times sqrt(1 – 2 zeta²), which is always below f₀, and the gap widens as the damping grows. At a damping ratio of 0.05000 the shift is too small to see; by 0.3162 the peak has moved 10.56% below the natural frequency.

Push the damping further and something more dramatic happens than the peak moving. It dies. Once the damping ratio reaches 1/sqrt(2) = 0.7071 the response has no maximum anywhere: it simply falls away from its static value as the drive speeds up.

The approach to that threshold is a lesson in reading formulae honestly. At a damping ratio of 0.7000 the algebra still reports a peak, at 0.2251 Hz for our block — but the magnification there is 1.000 to four figures, the same as the static response. By the time the damping is that heavy the peak is already gone in any measurable sense; at 0.7071 it formally disappears.

So if the peak wanders, what is the exact signature of the natural frequency? The phase. However light or heavy the damping, the motion runs exactly a quarter of a cycle — 90.00 degrees — behind the driving force when the drive sits on f₀, and nowhere else. That is the timing described in the building-up section, and it is exact, with no approximation anywhere in it.

A lightly damped system turns that corner violently: almost exactly in step with the force below the natural frequency, half a cycle behind it above, with the whole transition squeezed into a narrow band. Heavy damping spreads the same transition across the entire range. The crossing point never moves.

Everyday Examples of Resonance

Resonance is everywhere once you know what to look for, and the examples divide neatly into the ones we want and the ones we design against.

• A playground swing. The classic. One small push per cycle, timed with the motion, and the arc grows until air resistance and friction at the pivot take as much as each push delivers.
• A wine glass answering a voice. The rim has a natural frequency of its own. Sing that note and the rim flexes further on every cycle, and because the glass is lightly damped the magnification is large.
• A radio or a phone picking one station. The same mathematics describes a tuned electrical circuit, where a capacitance and an inductance take the roles of the mass and the spring. Tuning means moving the circuit’s natural frequency onto the one you want and letting its sharpness reject the rest.
• Magnetic resonance imaging. Nuclei in a magnetic field respond strongly to radio waves at one particular frequency and ignore everything else. The scanner is built around that selectivity.
• A washing machine on its spin cycle. The drum passes through the frame’s natural frequency on the way up to speed, which is why a machine can shake alarmingly for a few seconds and then settle once it is spinning faster than the resonance.

Notice what the useful cases have in common. Light damping is the asset when you want selectivity, and the liability when something structural has to survive the response. Most engineering around resonance is a choice about where to sit on that trade.

Clamp a string at both ends instead and it answers at a whole series of frequencies rather than one, which is the subject of standing waves. Our oscillator has a single mode, and that is the only case this page measures.

Where Resonance Breaks Down: 4 Common Mistakes

Four beliefs about resonance survive a surprising distance into physics courses, and each one is corrected by something already on this page.

Mistake 1: resonance means the amplitude grows without limit. It does not. At the natural frequency the magnification is Q, a finite number, for every real system — 10.00 for the block above, 50.00 when the damper is weakened. The formula only diverges at a damping ratio of exactly zero, which no real object has.

Mistake 2: a bigger push is what makes resonance dramatic. Push twice as hard and you get twice the swing, exactly as you would by leaning on the spring. The drama comes from the multiplier, and the multiplier is the damping’s business. A 2.00 N push producing a 100.0 mm swing on a spring that the same 2.00 N bends by 10.00 mm is the whole topic in one line.

Mistake 3: the resonance frequency is where the response peaks. Those are two different frequencies whenever the damping is not zero, as the table above sets out. The natural frequency is where the phase lag is 90.00 degrees; the peak sits slightly below it and moves as the damping changes.

Mistake 4: the 1940 Tacoma Narrows bridge collapse is the textbook case of resonance. It is the textbook case of the textbooks getting it wrong. As Wikipedia’s account of the 1940 Tacoma Narrows Bridge sets out, the event is presented in many physics textbooks as elementary forced mechanical resonance, but the collapse is attributed to aeroelastic flutter: a self-exciting oscillation driven by the structure’s own motion rather than by a driving force at a fixed frequency. The correction is usually credited to a 1991 American Journal of Physics paper by Billah and Scanlan, titled “Resonance, Tacoma Narrows Bridge Failure, and Undergraduate Physics Textbooks”.

There is a fifth caveat, and it is about the model rather than about anyone’s misunderstanding. Everything here assumes one mass, one spring, one mode, and a damping force proportional to speed.

Real objects break every one of those assumptions. A bridge deck or a turbine blade has many natural frequencies rather than one; springs stiffen or soften at large deflections, so the natural frequency starts to depend on the amplitude; and real losses — friction, air drag, internal heating — are rarely proportional to speed. Treat the single-mode result as the first and clearest case, not as the last word.

One more boundary worth stating plainly: every number on this page is a steady-state figure. How long an oscillator takes to build up to its steady swing is a different calculation, and nothing here makes it. Amplitude itself, as a quantity with its own energy rule, belongs to amplitude in physics, and the hertz-and-seconds bookkeeping behind f and T belongs to the frequency formula.

Worked Problems

Frequently Asked Questions

Newtons third law examples all share one shape: two bodies, one interaction, and two forces equal in magnitude and opposite in direction, written F_12 = -F_21. The pair acts on different bodies, one force on each, so the two can never cancel. Equal forces on unequal masses give unequal accelerations.

Lean on a wall and the wall leans back on you, with exactly the force you used. Lean harder and it pushes harder. That is why nothing dramatic happens to either of you.

Now push a shopping trolley instead. The trolley shoves your hands backwards just as hard as you shove it forwards, and yet off it goes. Both halves of that are true at once, and the moment you see why, the third law stops being a slogan you recite and becomes a tool for naming forces you would otherwise walk straight past.

What Is Newton’s Third Law?

Newton’s third law says that when two bodies interact, each exerts a force on the other that is equal in magnitude and opposite in direction. One interaction, two forces, two bodies. The two forces together are called a third-law pair, or an action-reaction pair.

It is the third of the three laws Newton’s laws of motion sets out, and the only one most students can quote word for word and still misapply. “Equal and opposite” is half a sentence. The half that does the work is the rest of it: equal and opposite, on different bodies.

Name every force “on X by Y”

Here is the habit that turns the law into a method. Never write “the force of friction” or “the tension”; write the force on the body that feels it, by the body that applies it. One push, named twice, from two points of view.

This page fixes one convention and never departs from it: F_12 means the force on body 1 exerted by body 2. The first subscript is the body that feels the force. The second is the body doing the pushing or the pulling.

Textbooks genuinely split on which subscript comes first, so check any source before you trust a sign in somebody else’s working. Read the pair the other way round and you have F_21, the force on body 2 by body 1 – the same size, the opposite direction.

One sentence on the question every class asks, because it is answered properly elsewhere: the two halves of a pair never cancel, because cancelling is something forces do when they act on the same body, and these two never do – common physics misconceptions takes that apart with a figure.

The pair statement as written holds for contact forces and for Newtonian gravity, which covers every example on this page. For forces between moving charges it needs the momentum carried by the field itself counted in as well, and the general law is then the conservation of momentum rather than the force pair alone, a boundary set out in Reaction (physics) – Wikipedia.

The Newton’s Third Law Formula and What Each Symbol Means

The third law is an identity, not an equation you solve for an unknown. One magnitude, two directions, two bodies.

Read it aloud as a sentence: the force on body 1 by body 2 is equal in size and opposite in direction to the force on body 2 by body 1. The minus sign carries the direction and nothing else. Strip it away and what is left is a single magnitude, which this page calls F, felt by each body – never shared out between them and never doubled.

Every number on this topic then comes from the second law, applied to one body of the pair at a time. That is the only arithmetic there is, and it is the same arithmetic twice.

Divide those two lines by one another and the masses swap sides. Multiply them out instead and the two products come back to the one force you started with.

• F_12 – the force on body 1 exerted by body 2. Unit: newton (N).
• F_21 – the force on body 2 exerted by body 1, equal in magnitude to F_12 and opposite in direction. Unit: newton (N).
• F – the single magnitude both halves of the pair share, felt in full by each body. Unit: newton (N).
• m_1, m_2 – the masses of the two interacting bodies. Unit: kilogram (kg).
• a_1, a_2 – the acceleration the pair force alone gives each body. Unit: metre per second squared (m/s²).
• g – gravitational field strength at the Earth’s surface, taken as 9.81 m/s² throughout this page. Unit: metre per second squared (m/s²), which for a field strength is the same thing as the newton per kilogram (N/kg).

Because step two is just the second law one body at a time, our Newton’s Second Law calculator finishes it: feed it the pair force with the first mass, read off a_1, then feed it the same force with the second mass and read off a_2. Run it twice, once per body – never once for “the pair”, which has no mass of its own.

Put numbers on that. A 40 N pair force between a 2.0 kg trolley and an 8.0 kg trolley gives 40 / 2.0 = 20.00 m/s² for the light one and 40 / 8.0 = 5.000 m/s² for the heavy one. Multiply each back and the force reappears: 2.0 × 20.00 = 40.00 N and 8.0 × 5.000 = 40.00 N. One force, two accelerations, one product.

How to Find the Pair: Name Both Bodies

To find a third-law pair, name the two bodies in the interaction, then write the two forces as “on X by Y” and “on Y by X”. If you cannot name both bodies, you have not found a pair – you have found one force and guessed at its partner.

The three-condition test

1. The two forces arise from one interaction between exactly two bodies.
2. They act on different bodies – one force on each.
3. They are the same type of force: both gravitational, or both normal contact, or both friction, or both tension. If you are hazy on which type is which, types of forces sorts them out.

All three conditions are required, and failing any single one is enough to disqualify a candidate. Condition 3 is the quiet one. Two forces can be equal, opposite and on different bodies and still not be a pair, if they came from two different interactions.

One related habit worth picking up while you are here: a force’s third-law partner never belongs on the same diagram as the force itself, because the partner acts on the other body – free-body diagrams sets out which arrows do belong.

Newton’s Third Law Examples, With Both Bodies Named

Every everyday example of the third law is the same move twice: find the two bodies, then name the two forces. The table below does it for seven of them, and the column that usually surprises people is the last one.

Aerodynamics leans on this law constantly, and NASA Glenn Research Center’s page on Newton’s laws of motion gives three cases of its own. A jet engine throws exhaust gas out of the back and is pushed forwards; a spinning ball deflects air to one side and curves the other way; and a wing deflects air downwards and in reaction is pushed upwards.

Notice what every row has in common. The two forces are the same size, so the interesting question is never “which force is bigger” but “which body was easier to move”.

The Book on the Table: the Pair That Is Not One

A book’s weight and the table’s normal force are not a third-law pair. They act on the same body – the book – so they fail condition 2 outright, and they are not even the same type of force. They are equal here for a completely different reason.

Take a 2.0 kg book sitting still on a table, with g = 9.81 m/s². There are four forces in play and two pairs, and all four happen to have the same magnitude: 2.0 × 9.81 = 19.62 N.

• W, on the book by the Earth: 19.62 N down. Its partner is W’.
• W’, on the Earth by the book: 19.62 N up. Its partner is W.
• N, on the book by the table: 19.62 N up. Its partner is N’.
• N’, on the table by the book: 19.62 N down. Its partner is N.

Read the “on” column and the whole confusion dissolves. W and N both act on the book. W’ acts on the Earth and N’ acts on the table, which is exactly why they are the partners and the other two are not.

So why are W and N equal? Because the book is not accelerating, which is the first law talking, not the third. That distinction sounds academic until you move the table.

The lift settles it

Put the same book in a lift accelerating upwards at 3.00 m/s². The table in the lift now has to do more than hold the book up; it has to accelerate it, so N = m(g + a) = 2.0 × (9.81 + 3.00) = 2.0 × 12.81 = 25.62 N.

Nothing at all has happened to the weight: W = 2.0 × 9.81 = 19.62 N, because neither the mass nor g changed. The gap between the two is the net force doing the accelerating, N – W = 25.62 – 19.62 = 6.000 N, which is exactly m a = 2.0 × 3.00.

So W and N are not equal in the lift. A third-law pair is equal in every circumstance whatsoever, which is enough on its own to rule these two out – and the real contact pair obliges: N’ = N = 25.62 N, the book pressing on the table precisely as hard as the table presses on the book. In practice this is the quickest check a student can run on any suspected pair: imagine a situation that changes one of them, and see whether the other follows.

Equal Forces, Unequal Accelerations

Because the pair force is one magnitude felt by two different masses, the accelerations it produces are in the inverse ratio of those masses: a_1 / a_2 = m_2 / m_1. The force is symmetric. The result almost never is.

That is the whole reason the third law looks false in daily life. Lean on a parked van and nothing moves but you; lean on a supermarket trolley with the same effort and it sails off down the aisle. Each one returns exactly the force you gave it, and the only thing that changed was the mass on the other end.

The simulator below is built for exactly this. Set two masses and one pair force, and watch the two arrows stay the same length while the two accelerations pull apart – the Newton’s Third Law Simulator also has a page of its own if you would rather open it full-screen.

Here is the same identity at two very different scales. A 150 N shove between a 30 kg child and a 1200 kg car gives 150 / 30 = 5.000 m/s² and 150 / 1200 = 0.1250 m/s², a ratio of 1200 / 30 = 40.00. Stretch it to a 100 N pair between a 2.0 kg body and a 2000 kg one and you get 50.00 m/s² against 0.05000 m/s² – a ratio of 2000 / 2.0 = 1000, with the pair force not having budged a newton.

Scale it up once more and the arithmetic stops being polite. A car driving away on a 4800 N friction push from the road presses back on the road with the same 4800 N, which gives the 1200 kg car 4800 / 1200 = 4.000 m/s² and gives the Earth 4800 / (5.972 × 10²⁴) = 8.038 × 10^-22 m/s². The mass ratio 5.972 × 10²⁴ / 1200 = 4.977 × 10²¹ is the ratio of those two accelerations. Nobody has ever felt the Earth’s half, and the force is the same size all the same.

Tug of War, Horse and Cart: Where the Law Looks Wrong

The law looks wrong in a tug of war and in a horse and cart because people add the two halves of a pair together, and the two halves are on different bodies, so they never belong in the same sum. Add forces on one body at a time and both puzzles evaporate.

If the rope pulls both teams equally, how does anyone win?

The rope’s tension really is the same both ways – team A pulls the rope with 600 N and the rope pulls team A back with 600.0 N, and the same holds at the other end. The rope cannot possibly decide the contest.

Something else does. Look at team A alone and the rope is only one of two horizontal forces on it; the other is friction from the ground. Standing still or sliding at a steady pace, that friction matches the rope exactly at 600.0 N. To accelerate the 240 kg team forwards at 0.50 m/s², the ground has to push 240 × 0.50 = 120.0 N harder than the rope pulls, so it must supply 600 + 120.0 = 720.0 N.

A tug of war is won at the feet, not on the rope. That is the honest resolution, and it is why boots and body weight matter so much more than arm strength.

The horse and cart

The classic objection runs: if the cart pulls back on the horse as hard as the horse pulls the cart, how does the pair ever move? Because that pair is two forces on two bodies, and neither body’s own sum contains both of them.

Give the horse 500 kg and the cart 300 kg. The ground pushes the horse forwards with 1000 N and the cart drags with 200 N of resistance. Treat the pair as one 800 kg object and (1000 – 200) / (500 + 300) = 800 / 800 = 1.000 m/s².

Now split them. The cart needs T = 200 + 300 × 1.000 = 500.0 N of harness tension to accelerate at that rate, and the horse feels the same 500.0 N backwards, leaving it 1000 – 500.0 = 500.0 N of net force – which is 500 × 1.000, as it has to be. Both bodies accelerate at 1.000 m/s² and the harness pair never shows up twice in one calculation.

And the result that surprises everyone

Two blocks in contact, pushed along as one, raise a sharper question: how hard do they press on each other? Take a 3.0 kg block A and a 5.0 kg block B side by side on a smooth floor, with 24 N applied to the whole arrangement. The acceleration is 24 / (3.0 + 5.0) = 3.000 m/s² however you push.

Push on B so that B drives A, and the only force on A is the contact push, so it must be 3.0 × 3.000 = 9.000 N. Turn the whole thing round and push A instead: now the only force on B is the contact push, which must be 5.0 × 3.000 = 15.00 N.

Same applied force, same acceleration, and an internal pair that differs by 15.00 – 9.000 = 6.000 N depending on which end your hand is at. Problem 8 works both cases in full, because this is the single result on this page most worth keeping.

Worked Problems

Frequently Asked Questions

Relative velocity is the velocity one body has as measured by another, and you get it by subtracting the two velocity vectors rather than the two speeds: v_AB is v_A minus v_B, taken component by component. Both velocities have to be written down in one common frame first. Reverse one body and the answer grows.

Stand on a bridge over a motorway and the traffic below is doing 110 km/h. Sit in one of those cars and the one beside you in the next lane, also doing 110, might as well be parked. Neither reading is wrong.

The two of you are simply measuring against different things — the bridge and the car. Pick who is doing the measuring, subtract their velocity from the one you care about, and every answer on this page follows. Get the subtraction the wrong way round and you have a car closing on you at 220 km/h that is actually sitting on your wing mirror.

What Is Relative Velocity?

Relative velocity is the velocity one body has as measured by another, and it is the vector difference of the two velocities written down in a single common frame. No comparison of speeds will do it. The directions have to be subtracted along with the magnitudes.

That common frame is usually the ground, and choosing it is the first thing you do, not an afterthought. Write both velocities against the same reference, then subtract. The frame itself never appears in the answer, which is the quiet miracle of the whole subject.

Which way round the subscripts read

The notation carries the answer to a question students ask constantly: who is watching whom? On this page v_AB means the velocity of A as measured by B, so A is the body being watched and B is the observer doing the watching.

Textbooks genuinely disagree about which subscript goes first, and that is worth knowing before you trust a sign in someone else’s working. Read the pair the other way and you get v_BA, which has the same relative speed on exactly the opposite bearing. The two are mirror images: v_BA is minus v_AB, always.

A velocity and a speed are different animals, and the difference is the reason this is a subtraction rather than an arithmetic comparison — velocity versus speed is where that distinction is set out properly. Here it is simply assumed.

The Relative Velocity Formula

The relative velocity formula subtracts the two velocities component by component, then recombines the components into a speed and a direction. One line does all of it.

In one dimension that is the whole calculation: pick a positive direction, give each velocity its sign, subtract. In two dimensions you resolve both velocities onto a pair of axes first, subtract axis by axis, then put the two differences back together.

Resolving onto axes is the only piece of vector machinery this page needs, and it is borrowed rather than built here — scalar and vector quantities teaches the component method, and the vector addition calculator does the arithmetic the other way round, adding where this page subtracts.

The quickest way to feel what the formula does is to put numbers through it and change one of them at a time. Our relative velocity calculator takes two speeds and two bearings, hands back the relative speed with its components and its reciprocal, and prices the difference of the two speeds beside the real answer so you can watch the two part company. The simulator below does the same job visually, and it also has a page of its own if you would rather open it full-screen.

One Dimension: Why Head-On Speeds Add

Head-on speeds add because reversing one body flips the sign of the velocity you subtract, and subtracting a negative number is addition. Nothing else changes. The slower body’s speed is untouched; only its sign moves.

Take two cars on a straight road and call east positive. A does 100 km/h east, B does 80 km/h east, so v_AB is 100 – 80 = 20 km/h. From inside B, A pulls ahead at the pace of an unhurried cyclist, which is 5.556 m/s.

Now turn B round without changing its speedometer. Its velocity becomes minus 80 km/h, and v_AB is 100 – (-80) = 180 km/h. The two cars are identical in every respect except the direction of one of them, and the answer has moved by 180 – 20 = 160 km/h — exactly twice the 80.

That factor of two is worth holding on to, because it is where the danger on a road actually lives. Overtaking on a two-lane road is a 180 km/h problem dressed up as a 100 km/h one.

The overtaking train

Two trains run the same way on parallel tracks at 120 and 90 km/h. The express sees the goods train drifting backwards at 30 km/h, which is 8.333 m/s — exactly 25/3 of a metre per second. If it needs 400 m of relative movement to clear the goods train completely, that takes 400 x 3 / 25 = 48.00 s.

For two bodies strung out along one line, the relative speed and the rate the gap closes are the same number, which is why a single division finishes the problem. Hold that thought until the closing-speed section below, where it stops being true.

Walking on a moving train

A passenger strolls along the aisle at 1.50 m/s while the train does 20.0 m/s over the ground. Forwards, the ground sees 21.5 m/s; back towards the rear, 18.5 m/s. The two differ by 3.00 m/s, which is twice the walking speed and the same factor of two in a gentler setting.

One sentence on a neighbouring idea so it does not confuse you later: in a collision the quantity that matters is the relative velocity of the two bodies just before and just after impact, and elastic and inelastic collisions is where that identity is worked out.

Two Dimensions: Crossing a River

A boat crossing a river moves with its own velocity through the water plus the water’s velocity over the ground, and the crossing time depends only on the component that points across. The current contributes nothing across. It contributes everything downstream.

Take a river 120 m wide with a 3.00 m/s current, and a boat that makes 4.00 m/s through still water. Point it straight at the far bank. Over the ground it travels sqrt(3.00 x 3.00 + 4.00 x 4.00) = sqrt(25.00) = 5.00 m/s, on a track atan(0.7500) = 36.87° downstream of the straight line across. It is the 3-4-5 triangle, hiding in a boat.

Now the part almost everyone gets wrong. The crossing takes 120 divided by 4.00 m/s, which is 30.00 s, because the only velocity carrying the boat across the river is that 4.00 m/s. Dividing the width by the 5.00 m/s over the ground gives 24.00 s instead — 6.00 s short, a 20.0 per cent error, and the single commonest mistake on this problem.

In 30.00 s the current has carried the boat 3.00 x 30.00 = 90.00 m downstream, so it lands 90.00 m along the far bank from the point directly opposite. Spin the rudder as you like: aimed straight across, the time is fixed and the drift is the price.

Landing directly opposite

To kill the drift you have to aim upstream far enough that the boat’s upstream component cancels the current exactly: 4.00 x sin(theta) = 3.00, so sin(theta) = 0.7500 and theta = 48.59°. That is a long way round — nearly half a right angle into the flow.

What is left to carry you across is sqrt(4.00 x 4.00 – 3.00 x 3.00) = sqrt(7.000) = 2.646 m/s, and the crossing now takes 120 divided by sqrt(7.000) m/s = 45.36 s. The drift is zero. The extra 15.36 s is what the zero drift cost, and it is a real trade: you pay for position in time.

Carry the root rather than the rounded figure into that division. Dividing 120 by the printed 2.646 returns 45.35 s, one in the last place out, which is the kind of discrepancy that makes a reader think the page is wrong when it is only rounded.

Flying in a Crosswind: Drift Angle and Crab Angle

A crosswind hands a pilot two different angles, and they are not the same number. The drift angle is how far off course the wind pushes you if you do nothing; the crab angle is how far you must turn into the wind to hold the course you wanted.

An aircraft holds an airspeed of 250 km/h due north while a 60.0 km/h wind blows from the west. Airspeed is the aircraft’s velocity relative to the air, so the ground sees the airspeed vector plus the wind vector: 60.0 km/h east and 250 km/h north. NASA Glenn states the same relation the other way round — airspeed is the vector difference between the ground speed and the wind speed, which is v_AB = v_A – v_B with the air as the observer.

Over the ground that comes to sqrt(60.0 x 60.0 + 250 x 250) = sqrt(66100) = 257.1 km/h, and the track runs atan(0.2400) = 13.50° east of north. Notice what has not changed: the northward progress is still 250 km/h, because the wind has no northward part to contribute. The aircraft is not flying north any slower. It is simply also flying east.

To track due north instead, the aircraft has to point west of north by enough that its own eastward component cancels the wind: 250 x sin(crab) = 60.0, so sin(crab) = 0.2400 and the crab angle is asin(0.2400) = 13.89°.

Those two angles come from the same ratio of 0.2400 and they are still 13.89 – 13.50 = 0.39° apart, because one is an arctangent and the other an arcsine. Using one for the other is a real error, not a rounding quibble, and at four significant figures there is nowhere for it to hide.

Holding the track also costs ground speed. With the crab applied the aircraft makes sqrt(250 x 250 – 60.0 x 60.0) = sqrt(58900) = 242.7 km/h, so the correction is worth sqrt(66100) – sqrt(58900) = 14.41 km/h of progress. Pilots call that turn the wind correction angle; the FAA publishes its whole Pilot’s Handbook of Aeronautical Knowledge as free chapter PDFs, navigation among them, if you want the aviation treatment rather than the physics one.

This is the arithmetic behind a sentence you may already have met: a plane’s airspeed indicator reads one steady number while a crosswind nudges its actual track sideways. Velocity versus speed raises that as an illustration of why velocity needs a direction. This is the page that puts the angles on it.

Relative Speed Is Not Closing Speed

Relative speed is the magnitude of v_AB. Closing speed is only the part of v_AB that points along the line joining the two bodies, which makes it a projection — and a projection can never be bigger than the thing it is a projection of.

Here is the case that separates them. Aircraft A flies north at 200 m/s. Aircraft B is 1000 m due east of A and also flying north, at 150 m/s. The relative speed is 200 – 150 = 50.0 m/s, so in B’s frame A is sliding steadily northwards at 50.0 m/s.

And yet the gap between them is, at that instant, not changing at all. The relative velocity points due north; the line joining the two aircraft points due east; the two are at right angles, so none of the 50.0 m/s does any closing. The rate of change of the separation is 0 m/s.

That is an instantaneous rate, though, and not a standing promise — which is exactly the sort of claim worth testing rather than trusting. Ten seconds later A has gone 500 m further north than B, and the separation is sqrt(1000 x 1000 + 500 x 500) = sqrt(1250000) = 1118 m. The gap was momentarily frozen and then began to open.

Put the same two aircraft in line instead, one 1000 m ahead of the other on the same track, and the two quantities coincide: the relative speed is 50.0 m/s and the gap really does shrink at 50.0 m/s. That coincidence is why the terms get muddled. It holds only for motion along the line of sight.

Collision-avoidance radar lives on this distinction, and so does one other place you have met: a siren’s pitch depends on the relative motion of source and observer along the line between them and on nothing else, which is what the Doppler effect is built from.

Every result on this page in one place

Here is every case worked above, with its inputs and its answers side by side. Each figure is rounded once, at four significant figures, so the table and the prose agree digit for digit.

Where the Galilean Rule Breaks Down

Subtracting velocities is exact for anything you will ever drive, row or fly, and it fails only when the speeds involved become a measurable fraction of the speed of light. Everywhere else it is not an approximation at all. It is the answer.

Put a number on that. For the two cars meeting head-on at 100 and 80 km/h, the relativistic correction to the 180 km/h answer is a fractional 6.9 x 10^-15 — about seven parts in a thousand million million. The four-significant-figure ceiling this page works to is about ten orders of magnitude coarser than that, so there is nothing to correct and no apology to make.

Now push it. Two craft approach each other, each doing 0.800c in the ground frame. Subtracting gives 1.600c, which is not a velocity at all — nothing carries information faster than light, so the rule has handed back an answer outside its own domain without the slightest complaint.

The correct composition returns 0.9756c, which is fast and still under the limit, and special relativity is where that composition law belongs. The same framework is what makes moving clocks run slow, which time dilation handles; neither is derived here.

Worth knowing where the crossover sits, because it is so absurdly far away. A simple subtraction survives the four-figure ceiling until each body passes roughly 0.007c, about 2,120 km/s — getting on for two hundred times the speed needed to escape the Earth’s gravity. Below that, use v_AB = v_A – v_B and do not look back.

Common Misconceptions About Relative Velocity

Four mistakes account for almost every wrong answer on this topic, and the first one is in a class of its own because it is so nearly right.

“Relative speed is just the difference of the two speeds”

It is, in exactly one case: when the two bodies move along the same line in the same sense. Outside that case it can be wildly wrong, because subtracting two numbers throws away the directions they came with.

Rain falling at 8.00 m/s and a cyclist riding at 6.00 m/s are the cleanest demonstration. The difference of speeds is 8.00 – 6.00 = 2.00 m/s. The relative speed, with the two motions at right angles, is 10.00 m/s — five times larger, and the rain arrives from in front rather than from above.

Quieter, but the crosswind case says it too. The aircraft and the wind differ in speed by 250 – 60.0 = 190.0 km/h, while the ground speed the subtraction actually produces is 257.1 km/h. In one case the naive answer is a fifth of the truth; in the other it is too small by more than a third.

“The current makes the crossing take longer”

Aimed straight across, it makes no difference whatsoever. The current runs along the bank, so it has no component across the river and cannot touch the crossing time: 120 m divided by a 4.00 m/s across-component is 30.00 s in a 3.00 m/s current, in a 1.00 m/s current and in dead still water.

What the current changes is where you arrive, 90.00 m downstream here. It only starts costing you time once you aim upstream to cancel the drift, and then it costs 15.36 s.

“Drift angle and crab angle are the same angle”

They are two different inverse functions of the same ratio. With 60.0 km/h of wind on 250 km/h of airspeed the ratio is 0.2400 both times, but the drift is atan(0.2400) = 13.50° and the crab is asin(0.2400) = 13.89°.

Half a degree sounds like nothing until you notice that a four-figure answer has no room for it. The two angles answer different questions: one is where the wind takes you, the other is what you must do about it.

“There must be a frame in which things really move”

There is not, and the formula says so in one line. Add the same velocity to both bodies and the subtraction cancels it exactly, so v_AB comes out identical in every frame you could choose.

Two passengers facing each other down a carriage agree about each other’s motion whether the train is standing in a siding or doing 300 km/h. Neither of them is “really” moving. The 300 km/h belongs to the choice of frame, not to the passengers.

One honest consequence, and it is a valid answer rather than an error: if two bodies have identical velocities, the relative velocity is exactly zero. A zero vector has no direction, so there is no bearing to report — and a tool that printed “due north” there would be inventing one.

Worked Problems

Frequently Asked Questions

Binding energy is the energy needed to pull a nucleus completely apart into free protons and neutrons, and the same energy that was released when those nucleons came together. Divide it by the mass number A and you get binding energy per nucleon, which peaks just below 8.8 MeV in the iron and nickel region.

Snap a stick, split a log, smash a plate. Each of those costs energy, and in every case the bill is a rounding error beside the mass of the thing you broke. Nuclei are not like that at all.

Take one iron-56 nucleus fully to pieces — 26 free protons, 30 free neutrons, nothing left bound — and the bill comes to 492.3 MeV. Chemistry works in single electronvolts. The nucleus works in millions of them, and that gap is the whole reason nuclear energy is worth the trouble.

What Is Binding Energy?

Binding energy is the energy you would have to supply to separate a nucleus into its individual protons and neutrons, with every one of them at rest and far apart. The United States Nuclear Regulatory Commission puts it the same way in its glossary definition of binding energy: the minimum energy required to separate a nucleus into its component neutrons and protons.

Read that sentence backwards and you have the other half of the idea. The energy you would need to spend taking the nucleus apart is exactly the energy that came out when it was assembled. One number, two directions.

So a bound nucleus is not a store of energy waiting to be released. It is a deficit. Separated nucleons sit higher; the bound nucleus sits lower by the binding energy, and that is precisely why it stays together.

The word “binding” carries no sense of strain here. Nothing in the nucleus is under tension, waiting to snap. The deficit is simply how much work the strong nuclear force has already done for you.

The Binding Energy Formula

The binding energy of a nucleus shows up as missing mass. A bound nucleus weighs less than the sum of its separated parts, and the shortfall, the mass defect, converts straight into the binding energy:

Spread that over the nucleons and you get the quantity the rest of this page lives on:

• BE — the binding energy of the whole nucleus, in MeV (1 MeV = 1.602177 x 10^-13 J)
• Δm — the mass defect: how much lighter the bound nucleus is than its separated nucleons, in unified atomic mass units (u)
• c — the speed of light, 299,792,458 m/s
• A — the mass number, the total count of nucleons
• Z — the proton number; N — the neutron number
• BE/A — binding energy per nucleon, in MeV per nucleon

Only one constant is needed to get from a mass in u to an energy in MeV: one unified atomic mass unit is worth 931.494 MeV, a figure NIST publishes as the atomic mass constant energy equivalent. Why a mass difference becomes an energy at all is the subject of E = mc² explained, and the E = mc2 calculator handles that conversion for any mass you like; this page takes it as given.

Assembling the mass defect by hand is fiddly, because the tabulated mass of a nuclide is the mass of its neutral atom rather than its bare nucleus. Our nuclear binding energy calculator takes Z, N and that measured atomic mass and returns both the total and the per-nucleon figure, with the electron bookkeeping already handled.

The lab below plots binding energy per nucleon against mass number and lets you drag along the curve. Watch the readout: it barely moves across the middle of the chart and falls away hard at both ends.

One caution about any sim of this kind, including this one. It reads binding energy per nucleon off the mass number alone, so it draws the shape of the curve beautifully but cannot tell iron-56 from iron-58 — for that you need each nuclide’s own measured mass, which is what the table further down uses. The fission and fusion simulator page explains the controls in detail.

Total Binding Energy vs Binding Energy per Nucleon

These two quantities answer different questions, and mixing them up is the single most common error on this topic. The total says how much energy the whole nucleus represents. The per-nucleon figure says how tightly the average nucleon in it is held.

Uranium-238 makes the point loudly. Its total binding energy is 1802 MeV against iron-56’s 492.3 MeV, so by that measure uranium wins by a mile. Per nucleon it loses: 7.570 MeV against 8.790 MeV.

The reason is pure bookkeeping. Uranium-238 holds 3.660 times iron-56’s total binding energy, but it has 4.250 times as many nucleons to spread it over, so each one ends up with less.

Which one you want depends on the question. Asking how much energy a reaction can release means comparing totals. Asking which nucleus is the more stable means comparing per-nucleon values, and stability is what the famous curve is about.

The Binding Energy Curve, Nuclide by Nuclide

Here is the curve as numbers rather than a sketch. Every entry was computed from that nuclide’s own measured atomic mass, so the table and the graph below it come from the same arithmetic and agree to the last digit shown.

Energies are quoted to four significant figures throughout, which is coarser than the data behind them. Every per-nucleon value here carries a published uncertainty below 0.04 keV, and even the loosest of the 23, barium-141’s, is pinned to about one part in 220,000. Four figures is a choice made for readability, not a limit of the source.

Binding energies computed from the IAEA Nuclear Data Services ground-state atomic mass table, fetched 21 October 2026. The three highlighted rows are the top of the curve.

Three features deserve naming. The rise on the left is brutal — hydrogen-2 manages 1.112 MeV per nucleon, helium-4 reaches 7.074, and after oxygen-16 at 7.976 the curve has essentially arrived.

The middle is a long, nearly level stretch. From oxygen-16 to krypton-86 — 70 nucleons apart — every value in the table lies between 7.976 and 8.795 MeV, a spread of under 0.9 MeV.

After that the curve slides slowly downhill, reaching plutonium-239 at 7.560 MeV per nucleon, which is lower than carbon-12 manages at 7.680.

Why the Binding Energy Curve Has That Shape

Two effects, pulling opposite ways, produce the whole shape. Neither needs a fitted formula to see.

The strong nuclear force has a very short range — roughly the width of a nucleon or two. A nucleon deep inside a large nucleus therefore feels only its immediate neighbours, not the hundreds of other nucleons further away. Once a nucleus is big enough that most nucleons are fully surrounded, adding more cannot buy any extra binding per nucleon. That saturation is the plateau.

Small nuclei fall short for the opposite reason: almost every nucleon in them is on the surface, with neighbours on one side only. Hydrogen-2 has two nucleons and no interior at all, which is why 1.112 MeV per nucleon is the worst value in the table.

Going the other way costs money. Every proton repels every other proton electrically, and that repulsion has unlimited range, so it grows roughly with the square of the proton number while the binding grows only with the nucleon count. Past the iron region the electrical price starts to win, and the curve tips over.

That downhill slope on both sides is why nuclei at the ends of the table can release energy by moving towards the middle, a consequence worked out in full in nuclear fission vs fusion.

Why Helium-4 Sits Above Its Neighbours

Look again at the left-hand end of the curve and you will find one point refusing to behave. Helium-4 reaches 7.074 MeV per nucleon, while lithium-6 next door manages only 5.332 and lithium-7 only 5.606. Helium-4 also towers over both three-nucleon systems, hydrogen-3 at 2.827 and helium-3 at 2.573.

Two protons and two neutrons is a closed configuration: every nucleon is paired, and there is no half-filled level left over. The result is a package so tightly bound that heavy nuclei frequently shed it whole rather than one nucleon at a time, which is exactly what alpha decay is.

Where Is the Peak of the Binding Energy Curve?

Three different questions hide inside that one, and they have three different answers: nickel-62 has the highest binding energy per nucleon, iron-56 has the lowest mass per nucleon, and iron-56 is the nuclide everyone names.

Start with what the measurements say. Ranked strictly by binding energy per nucleon, the order is nickel-62 at 8.795 MeV, then iron-58 at 8.792, then iron-56 at 8.790. Nickel-62 leads iron-56 by 4.2 keV per nucleon.

Is that gap real, or is it noise? The combined uncertainty on those two published figures is 0.0117 keV, so the gap is roughly 359 times the uncertainty. It is resolved data, not a wobble in the last digit.

Now the second question. Rank the same nuclides by mass per nucleon — atomic mass divided by A — and iron-56 wins at 0.998838135 u, ahead of nickel-62 at 0.998844270 u. The two rankings disagree because binding energy per nucleon and mass per nucleon are not the same quantity: the proton-to-neutron ratio differs between the two nuclides, and protons and neutrons do not weigh the same.

So the lab above marks “Fe-56 peak”, and so does almost every textbook diagram, and they are not wrong. They are answering the mass question, which is the one that governs what stellar burning can reach, and iron-56 is also vastly more abundant than nickel-62.

The honest statement is that the top of this curve is a plateau, flat to a few keV per nucleon across the iron and nickel region. “The iron peak” names a neighbourhood, not a winner. Which nuclide takes first place depends entirely on which question you asked.

Nuclear Binding Energy vs Atomic Binding Energy

Two completely different quantities share this name, and the scale between them is the easiest way to keep them apart. Hydrogen’s single electron is bound to its proton by 13.6 eV. A nucleon in iron-56 is bound by 8.790 MeV.

That is 8.790 million electronvolts against 13.6, a factor of about 646,000. Chemistry rearranges the first kind and never touches the second, which is why a tonne of coal and a kilogram of nuclear fuel are not in the same conversation.

The electron side belongs to atomic physics: the Bohr model is where that 13.6 eV comes from, and atomic structure covers how the electrons are arranged around the nucleus in the first place. This page is about the nucleus itself.

One practical consequence. Because published nuclide masses are atomic masses, electrons are already in the numbers, and getting the nuclear answer out means handling them deliberately rather than ignoring them.

Common Misconceptions About Binding Energy

“Binding energy is energy stored in the nucleus”

It is the opposite. A bound nucleus has less energy than its separated nucleons, by exactly the binding energy, and you have to put that energy in to take it apart. The tempting mental picture of a compressed spring inside the nucleus gets the sign wrong.

“Iron-56 is the most stable nucleus”

Only with a qualifier attached. Iron-56 has the lowest mass per nucleon of any nuclide, which is one perfectly good definition of most stable. By binding energy per nucleon, nickel-62 is higher, and iron-58 sits between them.

“A heavier nucleus is more tightly bound”

Heavier nuclei have larger total binding energies simply because they contain more nucleons. Plutonium-239’s total reaches 1807 MeV where iron-56’s stops at 492.3, and that surplus buys its nucleons nothing: each one is bound by 7.560 MeV, below iron-56’s 8.790.

“Binding energy per nucleon is the cost of removing one nucleon”

It is an average over the whole nucleus, not a price list. The energy to remove one particular nucleon is called its separation energy, and it depends sharply on which nucleon you pick and on whether the proton and neutron counts are even or odd.

Worked Problems

Answers are given to four significant figures. Two of these problems quote a figure to five, and both say why where it bites: round an input before you subtract or multiply it and the digit you were after is already gone.

Frequently Asked Questions

Orbital velocity is the speed a body needs to hold a circular orbit at a given radius, and it is fixed by that radius alone: v equals the square root of G times the central mass divided by the orbital radius. Raise the orbit and the required speed falls.

On a clear evening you can sometimes watch the International Space Station cross the sky as a steady white point that never blinks. It is 400 km up, and the whole time you are watching it, it is falling — it simply keeps missing the ground.

Missing the ground takes one very particular speed, and the station does not get to choose it. Gravity does. The surprise is which way that speed goes when you climb: a television satellite holding its place high above the equator is travelling at less than half the station’s speed.

What Is Orbital Velocity?

Orbital velocity is the speed an object must travel to stay on a circular orbit of a given radius, with gravity alone bending its path into the circle. Nothing is holding it up. It is falling continuously, and the ground is curving away beneath it at exactly the rate it falls.

Throw a stone horizontally and it lands a few metres away. Throw it harder and it lands further off, and the Earth’s surface has curved a little in the meantime. Keep increasing the throw and there comes a speed at which the curve of the fall matches the curve of the planet — and the stone never lands at all.

That speed is what this page is about. It is set by two things and two things only: how much mass you are orbiting, and how far from the centre of that mass you are. The thing doing the orbiting contributes nothing — not its mass, not its shape, not what it is made of.

One point of language, because it trips people up in exams. On a circular orbit the speed never changes but the velocity changes every instant, because the direction is always turning. “Orbital velocity” in everyday use means the orbital speed, and that is the quantity the formula below returns.

The Orbital Velocity Formula

The orbital velocity formula gives the speed directly from the central mass and the orbital radius, with no other input at all.

One lap of that circle takes the circumference divided by the speed, which is the second equation you need and the one examiners pair with the first.

Every Earth constant this page feeds into the formula is the value NASA publishes. Its Earth fact sheet gives the mass as 5.9722e24 kg, the volumetric mean radius as 6371.000 km, the equatorial radius as 6378.137 km and the mean surface gravity as 9.820 m/s² — and this page uses the mean radius throughout, which matters once in the table below.

Watch the units. G is quoted in metres, kilograms and seconds, so r has to go in as metres before anything is squared or rooted: 6,771 km becomes 6.771e6 m. Our orbital velocity calculator does that conversion for you and will also run the formula backwards, solving for the central mass or for the radius when the speed is the thing you measured.

One neighbouring quantity is worth naming here so you do not confuse the two: at the same radius the speed needed to leave altogether, rather than circle, is larger by exactly a factor of sqrt(2) — 10.85 km/s where a circular orbit needs 7.673 km/s — and escape velocity is where that quantity is worked out properly.

Here is the same formula with the controls handed over. Raise the orbit and watch the speed readout fall while the circle grows.

Where the Formula Comes From: Gravity Is the Centripetal Force

The formula comes from setting one force equal to another: the gravitational pull on the orbiting body is exactly the centre-pointing force a circle of that radius and that speed demands. Write both expressions down, set them equal, and the speed drops out in two lines.

The pull comes from the law of gravitation, which gives G M m / r² for a body of mass m at radius r. The demand comes from the geometry of going round in a circle: travelling a circle of radius r at speed v needs a centripetal force of m v² / r, pointing at the centre.

There is no third force in the picture, and in particular nothing is pushing outward. A circular orbit is gravity being precisely the right size to bend a straight line into that circle — no more and no less.

That cancellation is the most satisfying line in the derivation, and it has a real consequence. A spanner that drifts away from a space station keeps station with it: same radius, same speed, however far apart the two masses are.

It is also why the formula needs no information about the satellite at all. Size, shape, payload, fuel — none of it enters. Only the mass you are orbiting and the distance from its centre.

Why a Higher Orbit Is a Slower Orbit

A higher orbit is slower because the speed falls as one over the square root of the radius. Multiply the radius by four and the required speed halves, exactly, for any central body. That is the single most counter-intuitive thing about this formula.

Most people expect the opposite, and the instinct is understandable: getting higher plainly takes more work, so surely it takes more speed? It takes more energy. It does not take more speed. Those are different quantities, and this is the formula that separates them.

Look at why. Gravity weakens with the square of the distance, so a wider circle needs a gentler centre-pointing force; a gentler force bends the path more slowly, and a slowly bending path is a slow one. The station at 6,771 km does 7.673 km/s. Out at 27,084 km, four times as far, a satellite does 3.836 km/s.

Figure 1 above is that statement as a picture. Five real orbits sit on one falling curve, and the whole ladder from the surface to the Moon is a single line with no kinks, no thresholds and nothing special happening anywhere along it.

The climb still has to be paid for, in fuel and in energy, and nothing here says otherwise. What the formula says is that the circle waiting for you at the top is a slower one than the circle you left.

Real Orbits From a Surface Skim to the Moon

Every orbit below comes out of the same formula with the same two Earth constants, and nothing changes from row to row except the radius. The speeds fall, the periods grow, and the local gravity falls away faster than either.

Two rows need a word of explanation. The surface skim is the formula’s limit as the altitude goes to zero, not a place to put a satellite — at that height there is atmosphere, and plenty of it. The geostationary altitude of 35,793 km in that table is measured from the mean radius of 6,371 km, which is the radius every row uses; subtract the equatorial radius of 6,378.137 km from the same 42,164 km orbit instead and you get the more familiar 35,786 km.

Radius, speed and period do not depend on that choice at all, and 23.93 h is one sidereal day — which is the whole reason the orbit is useful. Why a sidereal day rather than 24 hours is the geostationary orbit‘s own story.

Notice how little the speed moves across the whole of low Earth orbit — 7.789 km/s at 200 km against 7.597 km/s at 535 km — while the lap time moves by nearly seven minutes. That is why a low orbit is quoted by its period rather than its speed.

Change the central body and only M and R change. The formula itself does not notice.

Jupiter has no surface, so its 69,911 km radius is the level where the pressure reaches one bar, and the 1,000 km in that row is measured from there. The last row is the Earth’s own orbit about the Sun, and it is a useful check on the whole method: 29.79 km/s from this formula against the 29.78 km/s NASA publishes as the Earth’s mean orbital velocity.

The lunar row and the Martian row are worth putting side by side for a different reason. Their periods come out within a quarter of a minute of each other — 117.8 min and 118.0 min — from two bodies with nothing in common, which is a coincidence rather than a pattern.

How Much Gravity Is There at the ISS?

There is plenty of gravity at the altitude of the space station: 8.694 m/s², about 89 per cent of the 9.820 m/s² at the surface. Astronauts float because they are falling, not because gravity has run out.

The same force balance that gave the speed gives this number too, because the acceleration a circle demands and the gravitational field at that radius are the same thing.

Put the ISS numbers in and you get 8.694 m/s². Drop a pencil in the station and it falls at very nearly the rate it would fall in your kitchen — the difference is about one part in nine. It only looks weightless because the station, the pencil and the astronaut watching are all falling together.

That is free fall, and it is the whole explanation. Weightlessness is a statement about what you can feel, not about how much gravity is present.

Climb to geostationary radius and gravity really has thinned out: 0.2242 m/s², 2.3 per cent of the surface value. Even there it is not zero, and it is still the only thing holding the satellite on its circle.

Common Misconceptions About Orbital Velocity

Four mistakes account for almost every wrong answer on this formula, and the first one is in a class of its own.

“The r in the formula is the altitude”

It is not. The r is measured from the centre of the body, so an altitude has to have the body’s own radius added to it first: r = R + h. For a 400 km orbit about the Earth that is 6,371 + 400 = 6,771 km, and 6,371 of those kilometres are underground.

Get this wrong and the error is enormous rather than marginal. Putting 400 km in as the radius gives G M / r = 3.98603e14 / 4.00e5 = 9.96508e8 m²/s² and a speed of sqrt(9.96508e8 m²/s²) = 31570 m/s, which is 31.57 km/s — more than four times the real answer, for a circle that would lie deep inside the planet.

“A heavier satellite must orbit faster”

The orbiting mass cancelled out of the derivation, so it cannot come back in. A one-tonne satellite and a ten-tonne satellite at 6,771 km both travel at 7.673 km/s, and if they are side by side they stay side by side.

The pull on the heavier one really is bigger — ten times bigger, by the law of gravitation — but the centripetal force its extra mass demands is ten times bigger as well, and the two grow together exactly. That is what cancelling means here, in physical terms rather than algebraic ones.

What the heavier one needs is more fuel to get there. That is a question about energy and thrust, not about the speed the orbit requires once you arrive.

“You have to go faster to reach a higher orbit”

You have to do more work, which is not the same thing. Every orbit above the one you are on is a slower orbit, and a satellite that fires its engine does not simply speed up where it is — it moves onto a bigger circle and ends up going more slowly than before.

“A satellite needs its engine running to stay up”

Nothing is running. Gravity does the whole job, and an orbit is the path a body takes when no force but gravity acts on it. Engines come into it only where the air is thick enough to matter, which is why a satellite 200 km up needs occasional help and one at 20,000 km needs none.

Where the Circular-Orbit Model Breaks Down

Every number on this page assumes a perfect circle around a single point mass, and four things eventually break that assumption. The arithmetic never complains, which is exactly the danger.

Real orbits are ellipses

A circular orbit has one speed. An elliptical one does not: the body runs fastest at its closest approach and slowest at its furthest, and no single figure describes it. For a nearly circular orbit this formula at the mean radius is a good approximation, which is all it is; Kepler’s laws are where the elliptical case is handled.

The Moon shows the size of the error. NASA’s Moon fact sheet gives its orbit an eccentricity of 0.0549 and a mean orbital velocity of 1.022 km/s, against the 1.018 km/s this formula returns at 384,400 km — about four parts in a thousand out, and problem 5 takes that gap apart.

There is an atmosphere, and the formula has never heard of it

Below about 150 to 200 km a circular orbit decays in days to weeks, so the surface-skim row of the table is a textbook limit and not a destination. Those altitudes are order-of-magnitude statements rather than measured thresholds, and no drag term appears anywhere in v = sqrt(GM/r).

There is only one central body in the formula

Far enough out, the body you are orbiting stops being the only thing that matters, and the number the formula returns stops describing anything real. The Earth keeps gravitational control out to roughly 1.5 million km; the Moon only to about 60,000 km. Both figures are orders of magnitude rather than published constants.

Four significant figures, and the fifth is not available

G is known to about five significant figures and the masses here to four or five, so their product cannot support more than four. Every speed on this page therefore stops at four: 7.673 km/s is honest, and a fifth digit on it would be noise rather than precision.

That ceiling has a sharp edge. Run the formula backwards — feed a four-figure speed in and ask for the radius or the mass — and the rounding is amplified, because both inverses go as the square of the speed. Problems 6 and 7 show exactly how much it costs.

Worked Problems

Frequently Asked Questions