Physics · Gravitation

Gravitational potential energy in a central field

Q: What does gravitational potential energy in a central field describe?

With $U(\infty)=0$, −$\frac{GMm}r$ describes interaction energy of two point masses or nonoverlapping spherical mass distributions. Measure distance between their centers. Negative U alone does not imply a bound orbit: use total energy in the appropriate frame. For a light test body in a fixed central field that is K+U; for two freely moving masses it is relative energy in the center-of-mass frame. Increasing r raises U toward zero while gravity does negative work. Changing the zero reference changes neither force nor energy differences. At small heights $mg\Delta h$ may approximate the difference, but h never replaces center distance r.

Q: How do I check units and scale for U=−GMm/r?

$\frac{GMm}r$ has units kg·m²/s²=J. The parameter $\mu=GM$ has m³/s², whereas $\Phi=\frac Um$ has J/kg. Convert kilometers to meters before substitution and distinguish μ from reduced mass measured in kilograms. Using the retained inputs G=6.67430×10⁻¹¹ N·m²/kg², M=5.97×10²⁴ kg, m=1000 kg and r=7.0×10⁶ m gives U≈−5.6922×10¹⁰ J. The value −5.6922×10¹⁰ J is a calculated value; strictly applying the two significant figures of r=7.0×10⁶ m gives −5.7×10¹⁰ J. Extra displayed digits do not imply extra measurement accuracy. Doubling test mass doubles U; doubling distance halves its magnitude. Force magnitude instead falls to one quarter when distance doubles. These different powers provide a useful independent check. The negative sign belongs to the chosen energy reference, rather than a force direction.

Q: When should I combine gravitational potential energy with other relations?

A speed problem requires the combined balance $E=\frac12mv^2-\frac{GMm}r$ for a light test body in a fixed central field. Potential difference alone does not determine final kinetic energy. For a circular orbit add $v^2=\frac{GM}r$; at the escape threshold $E=0$ gives $\sqrt{\frac{2GM}r}$. Two freely moving bodies of comparable mass have the same pair potential, but relative kinetic energy uses $m_{\mathrm{red}}=\frac{Mm}{M+m}$. Arbitrary center-of-mass translation must not determine their mutual binding classification. When lifting from R to R+h, the relative error of $mg_0h$ against the exact change is $\frac hR$. Choose the approximation using the actual height interval.

Using $U(\infty)=0$ , the pair interaction energy is $U=-\frac{GMm}{r}$ . r is center distance; binding is determined by appropriate total energy, not by U alone.

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Formula

Calculate

U (r) = - \frac{GM m}{r}

U in J; G in $\mathrm{N\,m^2\,kg^{-2}}$ ; M,m in kg; r in m.

Variables & units

Variables & units
Symbol	Meaning	Unit
$U$	Gravitational potential energy relative to infinity	J
$G$	Gravitational constant	N·m²/kg²
$M$	Central mass	kg
$m$	Object mass	kg
$r$	Distance between centers	m

LaTeX

U(r)=-\frac{GMm}{r}

QUANTA STUDYquanta-study.de

Normalised energy increase versus h/R from zero to 0.6. Exact curve x/(1+x) lies below the local approximation line x. At x = 0.25, values 0.20 and 0.25 are marked. — Own normalised energy curve: at h/R = 0.25, the exact lift is ΔU/(μm/R) = 0.20. Constant-g₀ approximation gives 0.25, overestimating the energy increase by 25%. The models agree only near h = 0.

Gravitational potential energy calculator

Find the potential energy of two masses with zero at infinity, or the energy change when the centre-to-centre distance changes.

Calculation model

Find

Gravitational parameter μ = GM

μ = G · M of the central body, 398,600 km³/s² for the Earth.

Second body mass m

Mass of the second body, e.g. 1000 kg for a small satellite.

Centre-to-centre distance r

Distance between the centres, not the height: Earth radius 6371 km plus height, e.g. 7000 km.

Filled with example values. Type your own and the result updates straight away.

Result: Potential energy U

U = -56 940 MJ

With μ = 398 600 km³/s², m = 1000 kg and r = 7000 km, U = -56 940 MJ.

Working · rearrange for U, substitute, evaluate

1. Rearrange for U
$U=-\frac{\mu m}{r}$
2. Substitute values (in SI units)
- μ = 398 600 km³/s² = 3.986 · 10¹⁴ m³/s²
- r = 7000 km = 7 000 000 m
$-\frac{398600000000000\cdot 1000}{7000000}=-56942857142.9$
3. Result
U = -56 940 MJ

Quick tips

Twice the centre distance, half the size of U: U is proportional to 1/r. That is why each extra kilometre of height costs less energy than the one before.
Near the surface ΔU ≈ m · g · h with g = μ/R² is enough. The error is about h/R: at 64 km above the Earth, m · g · h comes out about 1 % too large.

Assumptions and limits

Newtonian point-mass model or nonoverlapping spherical bodies in the exterior field; μ=GM>0, m>0, r>0. Reference U(∞)=0. μ denotes one central mass, not reduced mass; interaction energy does not require small test mass. Negative U alone does not establish binding.

Derivation & background

For point masses or a nonoverlapping spherical exterior field, $F_r=-\frac{dU}{dr}=-\frac{GMm}{r^2}$ gives attractive force. Two states have $\Delta U=GMm\left(\frac1{r_1}-\frac1{r_2}\right)$ , while gravity does $W_{\mathrm{grav}}=-\Delta U$ . For altitude h above R, exactly $\Delta U=\frac{mg_0h}{1+\frac hR}$ , approximately $mg_0h$ at small relative altitude. The potential function itself does not require a light test body; the simplified motion balance treats the central body as fixed. For comparable freely moving masses, relative mechanical energy in the centre-of-mass frame determines mutual binding.

Exam blueprint

Validity range

Newtonian point masses or a nonoverlapping spherical exterior field; r>0 is center distance and $U(\infty)=0$ . Interior fields, other masses and relativistic effects need expanded potentials. m≪M is required only for the fixed-central-body motion approximation.

Derivation steps

Integrate conservative inverse-square force with a specified zero reference.

1Choose the positive outward radial direction and one reference $U(\infty)=0$ .
2Radial gravitational force is $F_r=-\frac{GMm}{r^2}$ .
3The conservative interaction gives $dU=-F_r\,dr=\frac{GMm}{r^2}\,dr$ .
4Integrate from infinity to the specified separation; the antiderivative is $-\frac{GMm}{r}$ .
5The boundary term at infinity vanishes, leaving $U(r)=-\frac{GMm}{r}$ .
6Two states give $\Delta U=GMm\left(\frac1{r_1}-\frac1{r_2}\right)$ . Add kinetic energy in the appropriate frame before classifying binding.

Rearrangements

Center distance from U

r=-\frac{GMm}{U}

With positive masses and $U(\infty)=0$ , $U<0$ ; this sign alone does not establish a bound orbit.

Task variant

Using $G=6.67430\cdot10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$ , $M=5.97\cdot10^{24}\,\mathrm{kg}$ , $m=1000\,\mathrm{kg}$ and center distance $r=7.0\cdot10^{6}\,\mathrm{m}$ , find U with $U(\infty)=0$ .

$U=-\frac{GMm}{r}\approx-5.6922\cdot10^{10}\,\mathrm{J}$ ≈−5.69×10¹⁰ J.

Common mistakes

Substituting height h for center distance r, or treating negative U alone as proof of binding.

Use $r=R+h$ . Check $E=K+U$ for a light test body; use relative center-of-mass-frame energy for two freely moving masses.

Exam context

Calculate lifting potential changes, satellite energy balances and escape thresholds. Account separately for final kinetic energy, propulsion and losses.

Formula cluster

Gravitation

Related relations: energie-potentiell, gravitationsgesetz, kepler-drittes-gesetz.

Epot = mgh

Potential Energy (Gravitational)

F = G·m₁m₂/r²

Newton's Law of Universal Gravitation

T²/a³ = 4π²/[G(M+m)]

Kepler Third Law

Worked example

Using $G=6.67430\cdot10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$ , $M=5.97\cdot10^{24}\,\mathrm{kg}$ , $m=1000\,\mathrm{kg}$ and $r=7.0\cdot10^6\,\mathrm m$ gives $U\approx-5.6922\cdot10^{10}\,\mathrm J$ . This is about −5.69·10¹⁰ J as a rounded model value; displayed digits do not imply extra measurement accuracy. Distance is measured from Earth’s centre. Potential per test mass is $\Phi=\frac Um\approx-5.69\cdot10^7\,\mathrm{J\,kg^{-1}}$ .

Applications

Calculate lifting potential changes, satellite energy balances and escape thresholds. Account separately for final kinetic energy, propulsion and losses.

How do you calculate with Gravitational potential energy in a central field?

Here is how to work through a typical Gravitational potential energy in a central field (U = −GMm/r) task step by step:

1
Task
Using $G=6.67430\cdot10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$ , $M=5.97\cdot10^{24}\,\mathrm{kg}$ , $m=1000\,\mathrm{kg}$ and center distance $r=7.0\cdot10^{6}\,\mathrm{m}$ , find U with $U(\infty)=0$ .
Solution path
$U=-\frac{GMm}{r}\approx-5.6922\cdot10^{10}\,\mathrm{J}$ ≈−5.69×10¹⁰ J.

Understand and apply the equation

Central attraction varies with distance. Integrating the force gives $U=-\frac{GMm}{r}$ . Binding and escape depend on the balance with the appropriate kinetic energy.

How much energy does greater separation require?

A body can be lifted without increasing its speed. Over small heights, the familiar gravitational-energy approximation often suffices. Over large distances gravity becomes weaker, so an equal additional height requires less energy. First identify distance and the energy reference; then follow the force integral, two state balances and the limits of the near-surface approximation.[1]

Prior knowledge: work, kinetic energy and differences. The force integral explains the formula; the two-body extension can initially be skipped.

An interaction energy for two masses

U belongs to the gravitational interaction of the pair. In a fixed central field it is often described as the energy of the test body m, but the expression contains both masses. r is center-to-center distance, not height above a surface. At height h above a body of radius R, use $r=R+h$ .[1]

\begin{aligned}U(r)&=-\frac{GMm}{r}=-\frac{\mu m}{r}\\\mu&=GM\end{aligned}

With $U(\infty)=0$ : for point masses, a point test body outside a spherical mass distribution, or two non-overlapping spherical bodies. r is always centre-to-centre separation.[1][3]

m cancels from many speed calculations, but remains in an energy balance measured in joules. Doubling the test mass doubles its potential-energy change over the same path. $\mu=GM$ has units $\mathrm{m^3\,s^{-2}}$ ; this gravitational parameter is different from reduced mass.

Potential rises during lifting

QUANTA STUDYquanta-study.de

Original example: with

\mu=3.20\cdot10^{14}\,\mathrm{m^3\,s^{-2}}

and

m=1200\,\mathrm{kg}

, U rises from −48 to −38.4 GJ as r increases from 8 to 10 million metres. The zero is at infinity, outside the displayed domain; the positive change is 9.6 GJ.[1]

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From force to potential

Choose outward as the positive radial direction. Attraction then has $F_r=-\frac{GMm}{r^2}$ . A conservative force obeys $dU=-F_r\,dr$ : during slow lifting U rises although gravity points inward. A sideways displacement at constant r does no work against the central field.[1]

U(r)-U(\infty)=\int_{\infty}^{r}\frac{GMm}{s^2}\,ds=\left[-\frac{GMm}{s}\right]_{\infty}^{r}=-\frac{GMm}{r}

The integral runs from the chosen zero reference to finite distance; its negative result is not a sign error.

\begin{aligned}\Delta U&=GMm\left(\frac1{r_1}-\frac1{r_2}\right)\\W_{\mathrm{grav}}&=-\Delta U\end{aligned}

For $r_2$ > $r_1$ , $\Delta U$ is positive and gravitational work negative.

Differentiating gives $\frac{dU}{dr}=\frac{GMm}{r^2}$ and hence $-\frac{dU}{dr}=F_r$ . This checks both the power and the sign. The derivative has units $\mathrm{J\,m^{-1}}=\mathrm N$ . Integrating force therefore changes its radial dependence; the $r^2$ denominator cannot simply be copied into the potential.

Why is potential energy negative and what determines binding?

\begin{aligned}\Phi&=\frac{U}{m}=-\frac{\mu}{r}\\E&=K+U\end{aligned}

$\Phi$ has units $\mathrm{J\,kg^{-1}}$ = $\mathrm{m^2\,s^{-2}}$ ; E and U are measured in joules.

Changing the zero reference shifts U and the associated mechanical energy by the same constant; forces and energy differences remain unchanged. With $U(\infty)=0$ , $U<0$ initially means only finite separation. In a fixed central field, binding is governed by $E=\frac12mv^2+U$ . For two freely moving masses, instead use $E_{\mathrm{rel}}=\frac12m_{\mathrm{red}}v_{\mathrm{rel}}^2+U$ , the energy of relative motion in the centre-of-mass frame. Shared motion of the entire pair through the laboratory is excluded from this binding energy. Thus negative U alone does not establish binding: sufficiently large relative kinetic energy gives $E_{\mathrm{rel}}>0$ .[4]

The shorthand “negative energy means bound” requires the stated reference and the appropriate relative or fixed-central-field energy. It applies within the isolated Newtonian model. For extended bodies, the trajectory must additionally be compatible with their radii: even an energetically unbound body can hit a surface. Positive relative mechanical energy therefore does not guarantee a collision-free path.[1]

Two complete energy balances

Lifting in a specified model field

Given: Hypothetical spherical central body with $\mu=3.20\cdot10^{14}\,\mathrm{m^3\,s^{-2}}$ and $R=8.00\cdot10^{6}\,\mathrm{m}$ . A test body $m=1200\,\mathrm{kg}$ , modelled as a point mass, is lifted from $r_1$ = R to $h=2.00\cdot10^{6}\,\mathrm{m}$ . The central body is treated as stationary, neglecting the test body's back reaction. Initial and final speeds are zero; other fields are excluded.
Find: Find the energy change and compare it with constant surface g.

Approach: Measure both radii from the center; lifting adds no further kinetic energy. $\Delta U=\mu m\left(\frac1{r_1}-\frac1{r_2}\right)$ . Specify center distances.
$\begin{aligned}r_1&=8.00\times10^6\,\mathrm m\\r_2&=1.00\times10^7\,\mathrm m\end{aligned}$
Evaluate both energies with the same reference.
$\begin{aligned}U_1&=-4.80\times10^{10}\,\mathrm J\\U_2&=-3.84\times10^{10}\,\mathrm J\end{aligned}$
Take the difference and check the linear approximation.
$\begin{aligned}\Delta U&=9.60\times10^9\,\mathrm J\\{} g_0&=\frac{\mu}{R^2}=5.00\,\mathrm{m/s^2}\\{} mg_0h&=1.20\times10^{10}\,\mathrm J\end{aligned}$

Result: $\Delta U=9.60\,\mathrm{GJ}$ . Constant g overestimates the difference by 2.40 GJ, or 25% relative to the exact model value.

Check: The difference is positive. The mean $\frac{\Delta U}{mh}=4.00\,\mathrm{m\,s^{-2}}$ lies between local accelerations of 3.20 and 5.00 m/s². These parameters are model inputs, not measurements of a real planet.

Reaching altitude is not yet a circular orbit

Given: The same point test body in the same fixed model field, with target radius $r_2=1.00\cdot10^{7}\,\mathrm{m}$ . Initially it rests at the nonrotating surface; the final state is a circular orbit. Test-body mass remains constant in this state balance; propellant mass and losses are excluded.
Find: Calculate the required net change in mechanical energy.

Approach: The circular orbit requires orbital speed as well as the increase in height. $v_c=\sqrt{\frac{\mu}{r_2}},\quad K_2=\frac{\mu m}{2r_2}$ . The circular-orbit condition gives speed and kinetic energy.
$\begin{aligned}v_c&=\sqrt{\frac{\mu}{r_2}}=5656.854\,\mathrm{m/s}\\{} K_2&=\frac{\mu m}{2r_2}=1.92\times10^{10}\,\mathrm J\end{aligned}$
Combine potential and kinetic energy.
$\begin{aligned}E_2&=K_2+U_2=-1.92\times10^{10}\,\mathrm J\\E_1&=U_1=-4.80\times10^{10}\,\mathrm J\end{aligned}$
The total state difference gives net work.
$W_{\mathrm{ext}}=E_2-E_1=2.88\times10^{10}\,\mathrm J$

Result: Required net work on the test body: 28.8 GJ, comprising 9.6 GJ potential change and 19.2 GJ final kinetic energy.

Check: Check: $\Delta U+\Delta K=(9.6+19.2)\,\mathrm{GJ}=28.8\,\mathrm{GJ}$ . Negative final energy identifies a bound circular orbit. This is a mechanical state balance rather than fuel demand or a specified transfer plan.

When is near-surface energy in approximately constant gravity sufficient?

\begin{aligned}\Delta U&=\frac{\mu m h}{R(R+h)}=\frac{mg_0h}{1+\frac{h}{R}}\\{} \frac{mg_0h-\Delta U}{\Delta U}&=\frac hR\end{aligned}

For lifting from R to R+h with h>0 in the same spherical exterior field.

This exact algebraic rearrangement makes the approximation error explicit. For $\frac hR=0.001$ , $mg_0h$ exceeds the exact difference by 0.1%; for 0.25, by 25%. This is relative error in the energy change, not the entire negative potential energy. Small-height problems may use $mg\Delta h$ ; large-height problems must account for $g(r)$ . Gravitational energy in approximately constant gravity here describes only the small height change with its own energy reference.

The tangent to U at R has slope $mg_0$ . Its separation from the curved potential grows with the height interval. Never insert h instead of r into the central-field expression; even $h=0$ has finite surface value $U(R)$ .

Explore further: Comparable masses and circular orbits

$U=-\frac{GMm}{r}$ itself does not require a tiny second mass. That assumption enters when treating the central body as stationary and retaining only $K=\frac12mv^2$ . Two freely moving masses require both kinetic energies. In the center-of-mass frame this can be written in terms of relative motion.[4]

\begin{aligned}E_{\mathrm{rel}}&=\frac12m_{\mathrm{red}}v_{\mathrm{rel}}^2-\frac{GMm}{r}\\m_{\mathrm{red}}&=\frac{Mm}{M+m}\end{aligned}

Reduced mass $m_{\mathrm{red}}$ has units kg; gravitational parameter $\mu=GM$ still has $\mathrm{m^3\,s^{-2}}$ .[4]

In the fixed-central-body approximation a circular orbit has $K=\frac{\mu m}{2r}$ , $U=-\frac{\mu m}{r}$ and $E=-\frac{\mu m}{2r}$ . The factor two between $|U|$ and K belongs to that circular orbit. It cannot be applied to arbitrary initial speeds. Elliptical motion uses the same potential but different local speed. Speed on a circular orbit is therefore an additional dynamical condition, not a consequence of destination radius alone.[2]

Models that require additional terms

The point-mass expression applies outside a spherical mass distribution. Inside an extended body, its entire mass does not simply contribute $\frac{GM}{r^2}$ to local gravitational acceleration; density distribution and interior potential are needed. Nonspherical bodies introduce spatial field structure. Specifying center distance is necessary, but does not by itself validate the model.[3]

Atmospheric drag, propulsion work and other bodies alter the mechanical balance. An accelerating vehicle that loses mass also requires an expanded system model. Strong relativistic fields cannot be described completely by a Newtonian calculation. State the system and excluded contributions alongside $\mu$ and r in each numerical problem.

Five solved exercises

Calculate · Foundation

$\mu=6.00\cdot10^{12}\,\mathrm{m^3\,s^{-2}}$ , $m=5.00\,\mathrm{kg}$ , $r=3.00\cdot10^{6}\,\mathrm{m}$ . Find U and $\Phi$ using $U(\infty)=0$ .

Show hint

$\Phi$ is energy per mass.

Learning goal: Distinguish potential from potential energy in the exterior field.

Show solution and calculation

Approach: Use $U=-\frac{\mu m}{r}$ and $\Phi=\frac Um$ with one reference.
Substitution gives $U=-1.00\cdot10^7\,\mathrm J$ and $\Phi=-2.00\cdot10^6\,\mathrm{J\,kg^{-1}}$ .

Check: $-\frac{Ur}{m}=6.00\cdot10^{12}\,\mathrm{m^3\,s^{-2}}$ reconstructs the given gravitational parameter.

\begin{aligned}U&=-10^7\,\mathrm J\\\Phi&=-2\times10^6\,\mathrm{J/kg}\end{aligned}

Interpret · Application

A body is lifted from R to 1.01 R in a spherical exterior field. By what percentage does $mg_0h$ exceed the exact energy change?

Show hint

Use exact energy change as the error denominator.

Learning goal: Evaluate relative error with an explicit reference quantity.

Show solution and calculation

Approach: Compare $mg_0h$ with $\Delta U=\frac{\mu mh}{R(R+h)}$ , relative to the exact result.
$\frac{h}{R}=0.01$ gives 1%.
Relative to the approximation rather than exact value, the difference would be $\frac{0.01}{1.01}\approx0.9901\,\%$ ; name the denominator.

\frac{mg_0h-\Delta U}{\Delta U}=0.01=1\%

Transfer · Application

A 500 kg test body has $K=10.0\,\mathrm{MJ}$ and $U=-8.00\,\mathrm{MJ}$ in an isolated central field with $U(\infty)=0$ . Is it energetically bound, and what speed remains at infinity on a collision-free outward trajectory?

Show hint

Use $E=K+U$ and $U(\infty)=0$ .

Learning goal: Determine escape status and residual speed from total energy.

Show solution and calculation

Approach: First calculate $E=K+U$ ; along the assumed collision-free outward path, $\frac12mv_\infty^2=E$ .
This gives $E=+2.00\,\mathrm{MJ}$ , an energetically unbound state, and $v_\infty=\sqrt{\frac{2E}{m}}=89.443\,\mathrm{m\,s^{-1}}$ .

Check: kinetic energy at that limiting speed is 2.00 MJ again. Negative potential alone does not establish binding.

\begin{aligned}E&=2\times10^6\,\mathrm J\\v_\infty&=\sqrt{8000}\,\mathrm{m/s}\end{aligned}

Interpret · Application

For a nonrelativistic test body in an isolated Newtonian central field with $M\gg m$ , the energy zero is shifted by +100 J. The diagram then shows $U_\infty$ =100 J and total $E=20\,\mathrm{J}$ . Does $E>0$ imply escape? Interpret with constant m and no further propulsion.

Show hint

Reaching infinity requires nonnegative $K_\infty=E-U_\infty$ .

Learning goal: Assess binding relative to the infinity level rather than an arbitrary zero.

Show solution and calculation

Approach: $E-U_\infty=-80\,\mathrm J$ , so infinity is energetically inaccessible despite positive displayed E.
The usual $E<0$ binding criterion assumes $U_\infty=0$ .

Check: shifting back by 100 J gives $E=-80\,\mathrm{J}$ and $U_\infty=0$ with identical physics. Bound energy alone does not guarantee a collision-free orbit without trajectory information.

K_\infty=E-U_\infty=20\,\mathrm J-100\,\mathrm J=-80\,\mathrm J<0

[1][2]

Error analysis · Application

A test body outside a spherical body of radius R reaches height h=R above its surface. A solution uses $U=-\frac{\mu m}{h}$ with $U_\infty=0$ and claims infinite surface potential at $h=0$ . Correct both statements.

Show hint

Exterior-field radius is $r=R+h$ .

Learning goal: Avoid substituting surface height for centre separation.

Show solution and calculation

Approach: at the target height $r=2R$ , giving $U=-\frac{\mu m}{2R}$ .
At the surface $r=R$ and $U=-\frac{\mu m}{R}$ is finite for $R>0$ .
The error confuses surface height with centre separation and doubles the target magnitude.

Check: lifting increases energy by $+\frac{\mu m}{2R}$ . The exterior model using the whole M also cannot be continued to the centre of an extended body.

\begin{aligned}U(R+h)&=-\frac{\mu m}{R+h}\\U(2R)&=-\frac{\mu m}{2R}\\\Delta U&=U(2R)-U(R)=\frac{\mu m}{2R}>0\end{aligned}

[1][3]

What to remember

Choose the energy reference and measure both separations from the centre. Lifting increases potential energy while gravity does negative work. A completed orbit also requires kinetic energy. Binding depends on total energy relative to the infinity level; potential energy alone is insufficient.[1]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Refined question headings on energy reference, binding and the near-surface approximation; added relevant paragraph connections to gravitational height energy and circular-orbit speed.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an individual introduction and conclusion, checked existing calculations, clarified notation and model references, and added an original learning figure.

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Quanta. Gravitational potential energy in a central field. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/gravitations-potentielle-energie

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Frequently asked questions about Gravitational potential energy in a central field

What does gravitational potential energy in a central field describe?+

With $U(\infty)=0$ , − $\frac{GMm}r$ describes interaction energy of two point masses or nonoverlapping spherical mass distributions. Measure distance between their centers. Negative U alone does not imply a bound orbit: use total energy in the appropriate frame. For a light test body in a fixed central field that is K+U; for two freely moving masses it is relative energy in the center-of-mass frame. Increasing r raises U toward zero while gravity does negative work. Changing the zero reference changes neither force nor energy differences. At small heights $mg\Delta h$ may approximate the difference, but h never replaces center distance r.

How can I check a gravitational-potential-energy calculation?+

Evaluate both states with one zero reference, then subtract U₂−U₁. In the original model field μ=3.20×10¹⁴ m³/s², m=1200 kg, r₁=8.00×10⁶ m and r₂=1.00×10⁷ m, the energies are −48.0 GJ and −38.4 GJ, giving ΔU=+9.60 GJ. Gravitational work is −9.60 GJ. Independently, $\frac{\Delta U}{m\Delta r}=4.00\,\mathrm{m\,s^{-2}}$ lies between local accelerations 5.00 and 3.20 m/s². Positive change at greater separation confirms the sign. A final speed requires additional ΔK; reaching a circular orbit at the destination requires 28.8 GJ rather than 9.6 GJ. Both calculations exclude drive losses and describe net mechanical changes rather than fuel consumption.

What assumptions and limits apply to U=−GMm/r?+

The point-mass expression applies in the Newtonian model at positive finite separation. Extended spherical bodies can be represented by central point masses in the nonoverlapping exterior region. Interior fields and nonspherical distributions require other potentials. The energy function itself does not assume m≪M; that assumption belongs to the simplified motion balance with a fixed central body. For two moving bodies, reduced mass accounts for relative kinetic energy. Other fields, air resistance or propulsion alter the mechanical balance. Use the same zero reference for both endpoints. Strong relativistic fields require another model. Positive relative mechanical energy does not ensure a collision-free path for finite bodies.

How do I check units and scale for U=−GMm/r?+

$\frac{GMm}r$ has units kg·m²/s²=J. The parameter $\mu=GM$ has m³/s², whereas $\Phi=\frac Um$ has J/kg. Convert kilometers to meters before substitution and distinguish μ from reduced mass measured in kilograms. Using the retained inputs G=6.67430×10⁻¹¹ N·m²/kg², M=5.97×10²⁴ kg, m=1000 kg and r=7.0×10⁶ m gives U≈−5.6922×10¹⁰ J. The value −5.6922×10¹⁰ J is a calculated value; strictly applying the two significant figures of r=7.0×10⁶ m gives −5.7×10¹⁰ J. Extra displayed digits do not imply extra measurement accuracy. Doubling test mass doubles U; doubling distance halves its magnitude. Force magnitude instead falls to one quarter when distance doubles. These different powers provide a useful independent check. The negative sign belongs to the chosen energy reference, rather than a force direction.

When should I combine gravitational potential energy with other relations?+

A speed problem requires the combined balance $E=\frac12mv^2-\frac{GMm}r$ for a light test body in a fixed central field. Potential difference alone does not determine final kinetic energy. For a circular orbit add $v^2=\frac{GM}r$ ; at the escape threshold $E=0$ gives $\sqrt{\frac{2GM}r}$ . Two freely moving bodies of comparable mass have the same pair potential, but relative kinetic energy uses $m_{\mathrm{red}}=\frac{Mm}{M+m}$ . Arbitrary center-of-mass translation must not determine their mutual binding classification. When lifting from R to R+h, the relative error of $mg_0h$ against the exact change is $\frac hR$ . Choose the approximation using the actual height interval.

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Curated exam set for "Gravitational potential energy in a central field":

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Which formula describes Gravitational potential energy in a central field?

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How do you rearrange Gravitational potential energy in a central field for Center distance from U?

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Which common mistake happens with Gravitational potential energy in a central field?

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[1]OpenStax, University Physics Vol. 1, §13.3: Gravitational Potential Energy and Total Energy

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Gravitational potential energy in a central field

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Gravitational potential energy calculator

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Gravitation

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How do you calculate with Gravitational potential energy in a central field?

How much energy does greater separation require?

An interaction energy for two masses

Potential rises during lifting

From force to potential

Why is potential energy negative and what determines binding?

Two complete energy balances

Lifting in a specified model field

Reaching altitude is not yet a circular orbit

When is near-surface energy in approximately constant gravity sufficient?

Explore further: Comparable masses and circular orbits

Models that require additional terms

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μ=6.00⋅1012 m3 s−2\mu=6.00\cdot10^{12}\,\mathrm{m^3\,s^{-2}}μ=6.00⋅1012m3s−2, m=5.00 kgm=5.00\,\mathrm{kg}m=5.00kg, r=3.00⋅106 mr=3.00\cdot10^{6}\,\mathrm{m}r=3.00⋅106m. Find U and Φ\PhiΦ using U(∞)=0U(\infty)=0U(∞)=0.

A body is lifted from R to 1.01 R in a spherical exterior field. By what percentage does mg0hmg_0hmg0​h exceed the exact energy change?

A 500 kg test body has K=10.0 MJK=10.0\,\mathrm{MJ}K=10.0MJ and U=−8.00 MJU=-8.00\,\mathrm{MJ}U=−8.00MJ in an isolated central field with U(∞)=0U(\infty)=0U(∞)=0. Is it energetically bound, and what speed remains at infinity on a collision-free outward trajectory?

A test body outside a spherical body of radius R reaches height h=R above its surface. A solution uses U=−μmhU=-\frac{\mu m}{h}U=−hμm​ with U∞=0U_\infty=0U∞​=0 and claims infinite surface potential at h=0h=0h=0. Correct both statements.

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More formulas: Mechanics

Retain Gravitational potential energy in a central field for exams

$\mu=6.00\cdot10^{12}\,\mathrm{m^3\,s^{-2}}$ , $m=5.00\,\mathrm{kg}$ , $r=3.00\cdot10^{6}\,\mathrm{m}$ . Find U and $\Phi$ using $U(\infty)=0$ .

A body is lifted from R to 1.01 R in a spherical exterior field. By what percentage does $mg_0h$ exceed the exact energy change?

A 500 kg test body has $K=10.0\,\mathrm{MJ}$ and $U=-8.00\,\mathrm{MJ}$ in an isolated central field with $U(\infty)=0$ . Is it energetically bound, and what speed remains at infinity on a collision-free outward trajectory?

A test body outside a spherical body of radius R reaches height h=R above its surface. A solution uses $U=-\frac{\mu m}{h}$ with $U_\infty=0$ and claims infinite surface potential at $h=0$ . Correct both statements.