mgh or −GMm/r: Which Potential Energy Formula Applies

Two formulas for one quantity, and exams punish the wrong choice. Tell me how far you moved and how big the body is, and this page tells you which form to use and by exactly how much the easy one is wrong — because the fractional error in mgh is precisely h/R, which is a closed result and not an estimate.

Last updated 11 Nov 2026

Which potential energy formula your problem needs

The error in mgh is exactly h/R

How far the body moves, verticallyThe height change, not an altitude above sea level: if a satellite goes from 300 km to 700 km, this is 400. The unit is the next field. Zero is allowed and gives the obvious answer. Negative values are not accepted here because a descent is the same judgement with the sign flipped — the fractional error depends only on how far you moved relative to the body’s radius, not on which way.

…measured inThe third option is there because the judgement on this page depends only on the ratio of the height to the radius, so for a far-field problem it is often the natural way to state it: “two Earth radii out” is a complete specification and needs no kilometres at all.

Radius of the body you are moving away fromIn kilometres, from the centre to the surface — or to wherever the motion starts, if that is not the surface. The Earth’s nominal equatorial radius is 6378.1 km and its polar radius 6356.8 km (IAU 2015), which makes no difference to any judgement this page offers. The Moon is about 1737 km and Mars about 3390; those two are not defaults here because this batch did not source them.

How accurate the answer has to beThis is the input that makes the page a judgement rather than a formula. “Does mgh work here?” has no answer until you say how wrong you are willing to be, and the honest version of the rule of thumb is therefore conditional. Most school questions are decided at the one per cent level; a question that quotes its data to four significant figures has implicitly asked for the tighter one.

Use −GMm/r. At this height mgh is out by more than one per cent, and the error grows in direct proportion to how far you go.Example

A satellite raised 400 km above the Earth, 6378.1 km radius, 1% accuracy wanted

Why the error in mgh is exactly h/R, with no approximation in the statement

ΔE_P = GMm(1/R − 1/(R+h)) · mgh = (GM/R²)mh · mgh / ΔE_P = 1 + h/R

the exact form: ΔE_P = GMm(1/r₁ − 1/r₂), the difference of two −GMm/r terms. Always positive going up, always with the radii measured from the centre. Cambridge 9702 gives the potential-energy form at outcome 13.4.3 and the potential at 13.4.2.
the easy form: mgΔh, with g = GM/R² the field strength at the surface. The syllabus writes it with the delta and with the qualifier “in a uniform gravitational field” attached at outcomes 5.2.1 and 5.2.2, and both the delta and the qualifier are load-bearing.
the algebra: put x = h/R. The exact change is (GMm/R) x/(1+x); mgh is (GMm/R) x. Divide one by the other and everything cancels except 1 + x. No series expansion, no small-angle step, no dropped terms — this is an identity, which is why the page can state the error as a number rather than as a bound.
the direction of the error: mgh is always the larger of the two, because it charges the full surface g over the whole climb while gravity has been weakening throughout. So if a calculation has gone wrong in this way the answer is too big, and that is a usable check on your own work.
the sign, and the zero: mgh takes the starting point as its zero and gives a positive number. −GMm/r takes infinity as its zero and is therefore negative everywhere, reaching zero only at infinite separation. They disagree wildly about absolute values and agree exactly about differences — and only differences mean anything, which is why the two conventions can coexist without contradiction.

Worked example

A satellite raised 400 km above the Earth, 6378.1 km radius, 1% accuracy wanted
THE ONLY RATIO THAT MATTERS. x = h/R = 400 / 6378.1 = 0.062 715. Nothing else about the problem enters the judgement — not the mass, not GM, not g. Two bodies with the same h/R get the same answer.
SO THE ERROR IN mgh IS 6.27 PER CENT, exactly. Not approximately: mgh divided by the true change is 1 + x = 1.062 715, so mgh is 6.2715 per cent too large. Against the one per cent you asked for, that is six times too much, and the verdict is to use the exact form.
WHAT THE TWO FORMULAS ACTUALLY GIVE, per kilogram. Exact: GM(1/R − 1/(R+h)) = 3.986 004 × 10¹⁴ × (1/6.378 1 × 10⁶ − 1/6.778 1 × 10⁶) = 3.688 1 MJ/kg. With mgh, taking g = GM/R² = 9.798 4 m/s²: 9.798 4 × 400 000 = 3.919 4 MJ/kg. The ratio of those two numbers is 1.062 715, which is 1 + x to every figure printed — the identity, confirmed on real numbers.
AND THE DIRECTION, WHICH IS ALWAYS THE SAME. mgh came out larger. It always does, because it charges the surface value of g for the whole 400 km while g has actually fallen to 8.676 m/s² by the top — 11.5 per cent weaker. Using the average of the two field strengths, 9.237, gives 3.695 MJ/kg, which is within 0.2 per cent of the exact answer: that is the mid-height trick, and it is why it works.
THE SAME JUDGEMENT AT OTHER HEIGHTS, to calibrate your instinct. A 10 m drop in a laboratory: 0.000 157 per cent — mgh is effectively exact. The summit of Everest at 8.849 km: 0.139 per cent, so mgh is good to three significant figures and not four. An airliner at 11 km: 0.172 per cent. The Kármán line at 100 km: 1.57 per cent. Geostationary at 35 786 km: 561 per cent, where mgh is not an approximation but a different and wrong model.

The error in mgh at every height anyone asks about

Every number in the third column is just the second column as a percentage, because the error is h/R — exactly, with no approximation anywhere in the derivation. That is what makes this table a result rather than a table of examples. Read the Everest row as the practical boundary of the easy formula: at the top of the highest mountain on the planet, mgh is wrong by 0.14 per cent, so it is good to three significant figures and not to four. Everything a school mechanics question happens inside — buildings, slopes, lifts, pendulums, pile drivers, roller coasters — is in the top five rows, which is the honest reason mgh is taught first and used without apology. And read the 400 km row as the trap: a satellite in low orbit is in the range where mgh is wrong by six per cent, which is far too much to ignore and nowhere near enough to look wrong on the page. The last row is the limiting case and gives a result worth memorising on its own: let the final radius go to infinity and the exact formula collapses to GMm/R, which is the energy needed to escape, and dividing it by the mass and solving for speed gives √(2GM/R).
Height change on the Earth	h/R	Error in mgh	Good to how many significant figures	Verdict at the 1% level
10 m — a laboratory drop	0.000 001 57	0.000 157%	5 and beyond	mgh
100 m — a tall building	0.000 015 7	0.001 57%	4 comfortably	mgh
1 km	0.000 157	0.0157%	4	mgh
8.849 km — the summit of Everest	0.001 387	0.1387%	3, not 4	mgh
11 km — an airliner’s cruise	0.001 725	0.1725%	3	mgh
100 km — the Kármán line	0.015 679	1.568%	1, barely 2	−GMm/r
400 km — the Space Station	0.062 715	6.271%	1	−GMm/r
2000 km	0.313 573	31.36%	none	−GMm/r, and not nearly the surface
35 786 km — geostationary	5.610 762	561.1%	none	−GMm/r; mgh is the wrong model
infinity — escape	infinite	infinite	none	−GMm/r, and the answer is GMm/R

The two formulas side by side

The row that resolves most of the confusion is the first. These two formulas measure energy from different places, so their absolute values are not comparable and were never meant to be: mgh says a book on the floor has zero gravitational potential energy, while −GMm/r says it has about −6.26 × 10⁷ joules per kilogram. Neither is wrong, because an absolute potential energy has no physical meaning — only a *change* does, and on changes the two agree to within the h/R of the row below. The negative sign in the second form is not a quirk either: it is what makes “bound” a meaningful word. An object whose total energy is below zero can never reach infinity and is therefore in orbit or on the ground; one whose total energy is zero or above escapes.
	mgh (strictly mgΔh)	−GMm/r
Where the zero of energy is	wherever you choose — usually the floor, the ground or the lowest point	at infinite separation, always
Sign of the energy	positive above your zero, negative below it	negative everywhere, approaching zero from below
What the distance is measured from	your chosen zero level	the centre of the body
Assumes	g is constant over the height — a uniform field	nothing; it is the general form
Fractional error	exactly h/R	none
Right for	anything that happens in a building, a mine, a slope or a mountain range	satellites, orbits, escape, any motion comparable with a radius
The syllabus wording	CIE 5.2.2, ΔE_P = mgΔh, in a uniform gravitational field	CIE 13.4.3, E_P = −GMm/r

Two formulas, one quantity, and a rule of thumb that turns out to be exact

The question this page answers is a judgement, and nothing on the web arbitrates it: given a problem, do you use mgh or −GMm/r? Most sources tell you that mgh is “for small heights” and leave you to guess what small means. It can be made precise, and the result is unusually clean. If g is taken as the surface value GM/R², then mgh divided by the true change in potential energy is exactly 1 + h/R. So the fractional error in using the easy formula is exactly h/R — the height you moved, divided by the radius of the body you moved away from. No series expansion, no small-angle step, no neglected terms: it is an identity.

The derivation is three lines and worth seeing, because it is what makes the number trustworthy. The exact change in going from r = R to r = R + h is GMm(1/R − 1/(R+h)), which rearranges to (GMm/R) × h/(R+h). Write x = h/R and that is (GMm/R) × x/(1+x). Meanwhile mgh with g = GM/R² is (GM/R²)mh = (GMm/R) × x. Divide the second by the first and everything cancels but 1 + x. That is the whole proof, and it explains the direction of the error as well as its size: mgh is always the larger, because it charges you the full surface strength of gravity for the entire climb when gravity has been getting weaker the whole way up.

What that means in practice is that mgh is better than its reputation and then fails suddenly. The summit of Everest is an error of 0.14 per cent, so mgh is good to three significant figures at the top of the highest mountain on Earth, and every mechanics question that happens in a building, a mine, on a slope, on a roller coaster or at the end of a pendulum is well inside that. And then the Kármán line at 100 km is 1.6 per cent, the Space Station at 400 km is 6.3 per cent, and geostationary orbit is 561 per cent. The satellite range is the dangerous one, because a six per cent error is far too large to ignore and nowhere near large enough to look wrong. The table on this page has the whole span.

The two formulas also measure from different places, and that is where the rest of the confusion lives. mgh takes its zero wherever you put it — the floor, the ground, the lowest point of the swing — and gives a positive number above it. −GMm/r takes its zero at infinite separation and is therefore negative everywhere. They disagree completely about absolute values: mgh says a book on the floor has zero gravitational potential energy, and −GMm/r says it has about −6.26 × 10⁷ J per kilogram. Both are right, because an absolute potential energy means nothing; only a change does. The reason for putting the zero at infinity is that it makes “bound” a usable criterion: a body with total energy below zero can never reach infinity, and one with zero or more escapes. That is why the escape-speed derivation on the escape velocity page is a single line.

And where the boards sit on this, because it is not quite the same place. Cambridge 9702 writes the near-surface form as ΔE_P = mgΔh and attaches the phrase “in a uniform gravitational field” to it both times it appears, at outcomes 5.2.1 and 5.2.2 — the qualifier is part of the outcome, not a footnote. It then asks separately, at 13.3.3, that candidates “understand why g is approximately constant for small changes in height near the Earth’s surface”, which is this page’s question asked as bookwork. The CBSE Class XI syllabus lists “gravitational potential energy and gravitational potential” in its gravitation unit with no qualifier recorded in the curriculum document, and expects the −GMm/r form alongside escape speed and satellite energy. Either way the examinable skill is the same one: noticing which regime the question is in before writing anything down.

Frequently asked questions

When exactly can I use mgh instead of −GMm/r?

Whenever h/R is smaller than the accuracy you need, because the fractional error in mgh is exactly h/R — the height change divided by the radius of the body. At the one per cent level that means heights up to about 64 km on the Earth, which covers every ordinary mechanics question. For a four-figure answer you need h/R below 0.000 1, which is about 640 m. The reason the rule can be stated exactly rather than as a bound is that mgh divided by the true change is identically 1 + h/R, with no approximation in the derivation.

Why is gravitational potential energy negative?

Because of where the zero is put, and nothing more. The −GMm/r convention takes the energy to be zero at infinite separation, which is the only distance at which two masses genuinely do not interact. Bringing them closer releases energy, so every finite separation has less than zero. The reason for choosing that zero rather than the ground is that it makes a criterion out of the sign: a body whose total energy — kinetic plus potential — is negative can never reach infinity and is therefore bound, and one whose total is zero or positive escapes. With mgh’s convention you can choose the zero anywhere and the positive and negative signs tell you nothing in particular. Both conventions give the same answer for any change, which is the only thing that is measurable.

Does mgh overestimate or underestimate the energy?

It always overestimates, by exactly a factor of 1 + h/R. The reason is physical rather than algebraic: mgh charges the full surface value of g over the whole climb, while in reality g has been weakening the entire way up, so the true average pull was lower than the one you used. This is a genuinely useful check on your own work — if a mgh answer disagrees with an exact one, the mgh figure is the larger, every time. It also points at the standard fix: use g at the mid-height rather than at the surface and most of the error cancels. On a 400 km climb that trick gets you within 0.2 per cent of the exact answer.

What happens to −GMm/r when the object escapes?

The final radius goes to infinity, 1/r₂ goes to zero, and the exact change collapses to GMm/R — the full depth of the well you started in. Setting that equal to ½mv² gives the escape speed √(2GM/R), which is the cleanest derivation in the whole topic and is only available in this convention. It is also why mgh cannot be used for escape at all, at any level of tolerance: it would predict an infinite energy requirement for an infinite height, when the true answer is finite. That divergence — a finite energy buying an infinite distance — is the single sharpest reason to understand why the two formulas are not interchangeable.

Related calculators

References

Cambridge International AS & A Level Physics 9702, syllabus for use in 2025, 2026 and 2027, Version 1 (cambridgeinternational.org, read 11 November 2026). Quoted by outcome number so that the difference from the Indian syllabus is checkable rather than asserted. The two formulas are in two different topics and each carries its own condition, which is the whole subject of this page. 5.2.1 asks the candidate to “derive, using W = Fs, the formula ΔE_P = mgΔh for gravitational potential energy changes in a uniform gravitational field”, and 5.2.2 repeats the formula and the qualifier. 13.4.3 gives “E_P = −GMm/r” by way of the potential defined at 13.4.1 as “the work done per unit mass in bringing a small test mass from infinity to the point”. And 13.3.3 asks them to “understand why g is approximately constant for small changes in height near the Earth’s surface”, which is this page’s judgement set as bookwork. Searched for the word “escape” while reading: it does not appear in the syllabus at all.
PHYSICS (042), Class XI-XII, Central Board of Secondary Education senior secondary curriculum document for the session 2025-26 (cbseacademic.nic.in, read 11 November 2026). Cited for what the syllabus asks for, never as any kind of endorsement. Unit VI, Chapter 7 Gravitation, lists “Gravitational potential energy and gravitational potential, escape speed, orbital velocity of a satellite, energy of an orbiting satellite”, and Unit IV, Chapter 5, separately lists the “notion of potential energy”. The curriculum document records no uniform-field qualifier on the near-surface form — which is a difference from the Cambridge wording above, and is reported here as what the document says rather than as a claim about what any examiner expects.
IAU 2015 Resolution B3 on recommended nominal conversion constants for selected solar and planetary properties, XXIXth General Assembly of the International Astronomical Union (read 11 November 2026). Four numbers on this site come from its table: the nominal terrestrial equatorial radius 6.3781 × 10⁶ m, the nominal terrestrial polar radius 6.3568 × 10⁶ m, the nominal terrestrial mass parameter (GM) 3.986 004 × 10¹⁴ m³ s⁻², and the jovian and solar pairs. Two things about it are worth a student’s attention. The values are conversion factors, not measurements — they are fixed numbers close to the best estimates, so that everyone converting to SI gets the same answer. And the resolution gives no value for G: it notes that G is among the least precisely determined constants and asks that a publication needing a mass in kilograms state which estimate of G it used. That is exactly why this site tabulates GM and not M.