In short

The gravitational potential energy of two point masses M and m separated by a distance r is U = -\dfrac{GMm}{r}, with the zero of potential energy chosen at infinite separation. The gravitational potential is the potential energy per unit test mass, V = -\dfrac{GM}{r}, measured in joules per kilogram. The negative sign is not cosmetic — it says the system is bound: you would have to do positive work to tear the two masses apart and send them to infinity. The familiar formula U = mgh is what -GMm/r looks like near Earth's surface, and only there. Equipotential surfaces are concentric spheres around a point mass, and they are everywhere perpendicular to the gravitational field lines.

A rocket sitting on the pad at Sriharikota is gravitationally bound to Earth. A satellite 36,000 km up is also bound — more loosely, but still bound. A photon of starlight that left Andromeda two million years ago and is arriving at your telescope tonight has completely escaped. Three stories, three different relationships to Earth's gravity — and one number decides all of them.

That number is the gravitational potential energy: how much work you would have to do against gravity to lift something out to infinity. A small number (in magnitude) means loosely bound; a large negative number means deeply bound; zero means "just free." Sort the world by this one quantity and every orbital mechanics question — from Chandrayaan-3 to the proton inside a hydrogen atom — becomes a problem about a height on a landscape.

This article builds that landscape. It starts from Newton's law F = GMm/r^2, integrates once to get potential energy, and then shows why the clean separation between "how the mass acts" and "how it feels the field" leads to the idea of gravitational potential — a property of space itself, not of any test mass.

From force to potential energy

You already know, from work and energy, that for any conservative force \vec{F} there is a potential energy U such that

\vec{F} = -\frac{dU}{dr}\,\hat{r}.

Gravity is conservative — the work it does between two points depends only on the endpoints, not on the path. So gravity has a potential energy. The only question is: what is it?

Setting the zero of U at infinity

Potential energy is defined up to a constant — only differences in U are physical. So you get to pick where U = 0. For gravity, the natural choice is:

U \to 0 \quad \text{as} \quad r \to \infty.

Why: infinity is the one place where the two masses exert no force on each other, so a system of two masses infinitely far apart has no stored interaction energy to worry about. Choose the zero to live at that natural reference, and all other configurations get a definite number.

With that choice, the potential energy at a finite separation r is the negative of the work done by gravity as the two masses were brought from infinity to their current separation. Equivalently, U(r) equals the work you would have to do against gravity to lift the mass m from r back out to infinity.

The integration

Take mass M fixed at the origin. Bring a test mass m from infinity inward along a straight radial path to a distance r. Gravity pulls m inward; as m moves from some radius r' to r' - dr', gravity does positive work F\,dr'.

The work done by gravity over the full path from \infty to r is

W_{\text{grav}} = \int_{\infty}^{r} -\frac{GMm}{{r'}^{\,2}}\,dr' \cdot (-\hat{r})

Why: the gravitational force on m points toward M (negative \hat{r}), and the displacement along the inward path is also in the negative-\hat{r} direction, so force and displacement are parallel — the dot product is positive. Written as a scalar integral, it is cleaner to just take both as magnitudes and the sign comes from the r-direction limits.

Reduce to the scalar form using \vec{F}\cdot d\vec{s} = +\dfrac{GMm}{r'^2}\,dr' along the inward path:

W_{\text{grav}} = \int_{\infty}^{r} \frac{GMm}{{r'}^{\,2}}\,dr' \cdot (-1)

Here the extra (-1) comes from dr' running from \infty to r (decreasing), while the displacement magnitude is positive inward. It is easier and less error-prone to do the potential-energy calculation directly as the line integral of the force component along the displacement. Let r' decrease from \infty to r. The force along the radial inward direction has magnitude GMm/r'^2. Work equals force times displacement with the same direction:

W_{\text{grav}} = \int_{\infty}^{r} \frac{GMm}{{r'}^2}\,(-dr')

Now the minus sign is absorbed into the variable: -dr' is the infinitesimal inward displacement (positive in the direction of the force). Substitute u = -r', or just compute directly:

W_{\text{grav}} = -GMm \int_{\infty}^{r} \frac{dr'}{r'^2} = -GMm \left[-\frac{1}{r'}\right]_{\infty}^{r} = -GMm\left(-\frac{1}{r} - 0\right) = \frac{GMm}{r}.

Why: the antiderivative of 1/r'^2 is -1/r'. Evaluate at the upper limit r and subtract the value at the lower limit \infty. The \infty term vanishes (1/∞ = 0), leaving +GMm/r. Gravity did positive work — it helped, by pulling m inward — so W_{\text{grav}} is a positive number.

The potential energy is defined so that U(r) - U(\infty) = -W_{\text{grav}} (the work you would have to do against gravity to undo this motion). With U(\infty) = 0:

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

This is the potential energy of the system of two masses — not of mass m alone. Both masses share this energy; if you pull them apart, you are pulling on the system. It is a property of the pair.

Reading the formula

Every piece is doing something specific. Break it apart:

Gravitational potential energy U as a function of r Graph of U = -1/r against r, showing the curve diving toward negative infinity as r approaches zero and asymptotically approaching zero from below as r grows large. The horizontal axis labels mark r = 1, 2, 3, 5; the vertical axis labels mark U = 0 and the characteristic shape of the 1/r well. r → U 0 1 2 3 5 U = −GMm/r U → 0⁻ as r → ∞ the well gets deeper as r shrinks
The potential energy of two masses versus their separation. At large $r$, $U$ approaches zero from below. As $r$ shrinks, $U$ dives to $-\infty$. A particle in a $-1/r$ well is bound whenever its total energy is negative (below the $U = 0$ line) — and free the moment its total energy reaches zero or above.

Why the negative sign is not cosmetic

A sign is not just a sign. Here is what the minus in U = -GMm/r tells you physically.

Two masses at finite separation have less energy than two masses at infinite separation. That sentence is the negative sign, translated. To pull the two masses apart — to lift them from r out to \infty — you must add energy to the system. You are climbing out of a well.

That is what "bound" means: the system sits at the bottom of an energy well, and any attempt to separate the pair costs energy. A ball resting in a valley is bound to the valley — you must climb up to remove it. A proton and electron in a hydrogen atom sit in an electrostatic well, and 13.6 eV of work must be done to ionise the atom. A satellite in Earth orbit sits in a gravitational well, and to send it to interstellar space you must supply extra kinetic energy. Same idea, same sign.

The energy budget:

If E < 0, the system is bound — the two masses can never reach infinity, because at r = \infty you would need KE = E < 0, which is impossible.

If E \geq 0, the system is unbound — the two masses can reach infinity, arriving there with KE = E \geq 0.

The whole framework of escape velocity is "find the speed that makes E = 0," which you will develop in full in Escape Velocity and Orbital Velocity.

Near Earth: recovering U = mgh

The -GMm/r formula is true at every distance from Earth — whether you are at the surface, in low Earth orbit, or at the Moon. But at school you met U = mgh, where h is the height above some reference level. How do they relate?

Take M = mass of Earth, R = radius of Earth, and consider a point at height h above the surface. The distance from the centre is r = R + h. The potential energy is

U(R+h) = -\frac{GMm}{R+h}.

This is a big, negative number. To get the usable mgh form, you care about differences in U, not its absolute value. Compare U at height h with U at the surface:

\Delta U = U(R+h) - U(R) = -\frac{GMm}{R+h} + \frac{GMm}{R}.

Combine the fractions over a common denominator:

\Delta U = GMm\left(\frac{1}{R} - \frac{1}{R+h}\right) = GMm\left(\frac{(R+h) - R}{R(R+h)}\right) = \frac{GMm\,h}{R(R+h)}.

Why: pull GMm out, use a single denominator R(R+h), and the numerator telescopes to h.

Now apply the near-surface approximation h \ll R. Earth's radius is R \approx 6.37 \times 10^6 m; even a jetliner at 10 km is only h/R \approx 1.6 \times 10^{-3}. So to excellent accuracy, R + h \approx R for anything near the surface:

\Delta U \approx \frac{GMm\,h}{R \cdot R} = \frac{GM}{R^2}\,mh.

And you already know, from Gravitational Field and Acceleration Due to Gravity, that g = GM/R^2 \approx 9.8 \text{ m/s}^2. Substitute:

\Delta U \approx mgh.

Why: mgh falls straight out of the exact formula -GMm/r once you take h \ll R. It is not a separate law — it is the first term in a Taylor expansion of the true potential energy, valid near Earth's surface. The approximation gets worse as h grows: at 100 km (h/R \approx 1.6\%), mgh overestimates by about 1.6%; at 1000 km, by 16%; at the Moon's distance, the mgh form is wrong by orders of magnitude.

So U = mgh is not wrong — it is a local approximation of the universal formula. The moment you leave the surface and go to orbital or interplanetary scales, you must use U = -GMm/r. Both answers must agree where they overlap, and they do.

Gravitational potential — per-test-mass

Now here is a useful conceptual move. The formula U = -GMm/r has two masses in it: the source M, and the test mass m. But the landscape of gravity — the shape of the well, the places where gravity is strong versus weak — should be a property of the source alone. Why should I care which test mass I put in it?

So define the gravitational potential V at a point in space as the potential energy per unit test mass:

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

V is a property of the field, not of the test mass. It has units of joules per kilogram (J/kg), which is the same as m²/s² (though you will rarely see it written that way in textbooks). For a point source of mass M:

\boxed{\; V(r) = -\frac{GM}{r} \;}

To use V: place a test mass m at some point, and its potential energy is just U = mV. The potential V does the mass-independent work of describing the well; multiplying by m gives the energy of a specific test body.

This is the same move you make everywhere in physics. The electric field E = F/q strips the test-charge dependence out of force; the gravitational potential V = U/m strips the test-mass dependence out of potential energy. Once you do this, the field (or potential) is a property of the source — and you can talk about "the potential at this point in space" without mentioning what test body is there.

Superposition: potentials just add

A deep consequence: gravitational potentials from multiple sources add as scalars. If there are several masses M_1, M_2, \ldots, M_n at distances r_1, r_2, \ldots, r_n from a given point, the total potential there is

V_{\text{total}} = -G\sum_i \frac{M_i}{r_i}.

This is much easier than adding forces, because forces are vectors (you have to add them component-by-component) but potentials are scalars (you just add numbers). For any configuration of sources, you can compute the potential everywhere simply by summing -GM_i/r_i from each source, and then — if needed — get the force by taking the gradient.

Why: this is the single biggest reason working with potentials is usually easier than working with forces. For a complicated source distribution (a galaxy, a planet with bumps, a charge configuration), you add scalars instead of vectors, and the field follows.

Equipotential surfaces

An equipotential surface is the set of all points at which the gravitational potential takes the same value. For a point source, V(r) = -GM/r depends only on r — so the equipotentials are concentric spheres centred on the mass.

Two key properties of equipotential surfaces make them powerful tools:

  1. No work is done moving along an equipotential. If two points A and B lie on the same equipotential, then V(A) = V(B), so U(A) = mV(A) = mV(B) = U(B), so \Delta U = 0. The work done by gravity is -\Delta U = 0. Conclusion: gravity does no work when a test mass slides along an equipotential — the walk is free of gravitational cost.

  2. The gravitational field is perpendicular to the equipotential. Suppose the field had a component tangent to the equipotential. Then moving along the equipotential in that direction, you would do work against that component — but you just said the work is zero. Contradiction. So the field must be purely perpendicular.

Combine the two: equipotentials and field lines form a perpendicular grid in space, the way lines of latitude are perpendicular to lines of longitude on a globe.

Equipotential surfaces and field lines around a point mass A central mass with three concentric dashed circles representing equipotential surfaces. Radial arrows pointing inward toward the mass represent gravitational field lines, perpendicular to the equipotentials at every intersection. M V = −GM/r field line equipotential
Around a point mass $M$, equipotential surfaces (dashed circles) are concentric spheres. Gravitational field lines (solid arrows) point inward along radii — perpendicular to every equipotential at the point of crossing. Moving along any dashed circle takes no work; moving along any arrow is a "downhill" descent through the gravitational well.

On a topographic map, the contour lines are the equipotentials of Earth's gravity near the surface. Water flowing across a landscape moves perpendicular to the contour lines — which is why rivers run "downhill" in exactly the direction the gravitational field points. The entire geography of drainage basins in the Himalayas is the equipotential-surface picture of gravity made visible.

The field is (minus) the gradient of the potential

One more piece of machinery, and you are done with the scaffolding. The relation between field and potential is:

\vec{g} = -\frac{dV}{dr}\,\hat{r}.

For V = -GM/r:

\frac{dV}{dr} = -GM \cdot \left(-\frac{1}{r^2}\right) = \frac{GM}{r^2}.

So

\vec{g} = -\frac{GM}{r^2}\,\hat{r}.

Why: the minus sign says the field points "downhill" — in the direction of decreasing potential. Gravity points toward the source, from regions of higher V (farther from the source, V closer to zero) to regions of lower V (closer to the source, V deeply negative). The magnitude GM/r^2 matches the familiar inverse-square field strength.

If you know the potential everywhere, you know the field everywhere — take the gradient. And vice versa. The potential contains the same information as the field, packaged as a single scalar instead of three components of a vector.

Worked examples

Example 1: Lifting a satellite — exact vs approximate

A 500 kg satellite is to be lifted from the surface of the Earth (altitude 0 km) to a geostationary orbit at altitude 35,800 km. How much gravitational potential energy does it gain? Compare (a) the exact -GMm/r calculation with (b) the near-surface mgh estimate. Use GM = 3.986 \times 10^{14} m³/s² and R_\oplus = 6.371 \times 10^6 m.

Satellite lifted from the surface to geostationary orbit Earth drawn as a circle at the left, with a satellite starting at the surface and ending at geostationary altitude, 35800 km above. Radial distances R and R+h are labeled. Earth start GEO h = 35,800 km R + h ≈ 42,171 km R ≈ 6,371 km
The satellite starts at the surface (distance $R$ from Earth's centre) and ends at geostationary altitude (distance $R+h$).

Step 1. Compute the starting and ending distances from Earth's centre.

r_1 = R = 6.371 \times 10^6 m

r_2 = R + h = 6.371 \times 10^6 + 3.58 \times 10^7 = 4.217 \times 10^7 m

Step 2. Exact calculation using U = -GMm/r.

U_1 = -\frac{GMm}{r_1} = -\frac{(3.986 \times 10^{14})(500)}{6.371 \times 10^6} = -\frac{1.993 \times 10^{17}}{6.371 \times 10^6} \approx -3.129 \times 10^{10} \text{ J}
U_2 = -\frac{GMm}{r_2} = -\frac{(3.986 \times 10^{14})(500)}{4.217 \times 10^7} \approx -4.726 \times 10^9 \text{ J}
\Delta U_{\text{exact}} = U_2 - U_1 = -4.726 \times 10^9 - (-3.129 \times 10^{10}) \approx +2.656 \times 10^{10} \text{ J}

Why: both U_1 and U_2 are negative, but U_2 is less negative (shallower well at greater distance). The difference U_2 - U_1 is positive — the satellite has gained potential energy, which is what "lifting" means.

Step 3. Approximate calculation using \Delta U \approx mgh.

\Delta U_{mgh} = mgh = 500 \times 9.8 \times 3.58 \times 10^7 = 1.754 \times 10^{11} \text{ J}

Step 4. Compare.

\frac{\Delta U_{mgh}}{\Delta U_{\text{exact}}} = \frac{1.754 \times 10^{11}}{2.656 \times 10^{10}} \approx 6.6

Why: mgh overestimates by a factor of 6.6 — it is wildly wrong at this altitude. The reason: mgh assumes g stays at its surface value throughout the climb. In reality, g falls off as 1/r^2, so at geostationary altitude it has dropped to g_{\text{GEO}} = GM/(R+h)^2 \approx 0.22 m/s² — about 2% of its surface value. Using the surface value for the whole climb badly overshoots.

Result: The exact gain in potential energy is about 2.66 \times 10^{10} J (26.6 GJ). The mgh approximation gives 1.75 \times 10^{11} J (175 GJ), an overestimate by a factor of 6.6.

What this shows: U = mgh is only a near-surface approximation. For any problem where h is a non-negligible fraction of R — which includes every orbital problem — you must use the exact formula U = -GMm/r.

Example 2: Falling from the Moon's distance

A 1 kg mass is released from rest at the Moon's orbital distance from Earth's centre (r_1 = 3.84 \times 10^8 m), far from any other body. Assuming only Earth's gravity, what speed does it reach just before hitting Earth's surface (r_2 = R = 6.37 \times 10^6 m)? Use GM = 3.986 \times 10^{14} m³/s².

Mass falling from Moon's distance to Earth's surface Earth shown as a small sphere at the right, the Moon's orbit path as a dashed circle around it, and a mass released from rest at the orbital distance falling radially inward toward Earth. Earth m, v = 0 falls radially inward r₁ = 3.84 × 10⁸ m (Moon distance) r₂ = R = 6.37 × 10⁶ m
Released from rest at the Moon's distance, the mass falls inward under Earth's gravity alone. All potential-energy change goes into kinetic energy.

Step 1. Set up conservation of mechanical energy.

KE_1 + U_1 = KE_2 + U_2
0 + \left(-\frac{GMm}{r_1}\right) = \tfrac{1}{2}mv^2 + \left(-\frac{GMm}{r_2}\right)

Step 2. Solve for v^2.

\tfrac{1}{2}mv^2 = \frac{GMm}{r_2} - \frac{GMm}{r_1} = GMm\!\left(\frac{1}{r_2} - \frac{1}{r_1}\right)
v^2 = 2GM\!\left(\frac{1}{r_2} - \frac{1}{r_1}\right)

Why: mass m cancels, as it always does in free gravitational motion. The result depends only on the source GM and the two radii. Note that r_2 < r_1, so 1/r_2 > 1/r_1 and the bracket is positive.

Step 3. Plug in numbers.

\frac{1}{r_2} = \frac{1}{6.37 \times 10^6} = 1.570 \times 10^{-7} \text{ m}^{-1}
\frac{1}{r_1} = \frac{1}{3.84 \times 10^8} = 2.604 \times 10^{-9} \text{ m}^{-1}
\frac{1}{r_2} - \frac{1}{r_1} = 1.570 \times 10^{-7} - 0.026 \times 10^{-7} = 1.544 \times 10^{-7} \text{ m}^{-1}
v^2 = 2 \times (3.986 \times 10^{14}) \times (1.544 \times 10^{-7}) \approx 1.231 \times 10^{8} \text{ m}^2/\text{s}^2
v \approx \sqrt{1.231 \times 10^8} \approx 1.109 \times 10^4 \text{ m/s} \approx 11.1 \text{ km/s}

Step 4. Compare to escape velocity.

The escape velocity from Earth's surface — the speed that would just barely carry the mass to infinity with zero kinetic energy left over — is v_{\text{esc}} = \sqrt{2GM/R} \approx 11.2 km/s.

Why: the speed you calculated (11.1 km/s) is slightly less than the escape velocity (11.2 km/s) because the mass started at the Moon's distance rather than from infinity — it had a tiny nonzero amount of potential energy to begin with, so less of the fall was available to convert to kinetic energy. The two numbers are close because the Moon's distance, while vast to us, is not infinity — and the 1/r potential is already nearly at zero by then.

Result: The mass arrives at Earth's surface at about 11.1 km/s, just shy of escape velocity. This is why "falling from the Moon's orbit" is, to excellent accuracy, the same energy calculation as "arriving from infinity" — the 1/r potential is almost flat out there.

What this shows: Energy conservation — write down the total at two instants, equate them, solve — is almost always the fastest way to handle gravitational problems. You never need to know the detailed trajectory; only the endpoints matter.

Common confusions

If you came here to compute gravitational potential energies and solve orbital-problem-style energy questions, you have everything you need. What follows is for readers who want the deeper structure: the potential from an extended mass distribution (shell theorem again, this time for potential), the internal gravitational potential of a solid sphere, and the connection between V, \vec{g}, and Poisson's equation that underlies all of gravitational theory.

Potential of a uniform spherical shell

You already know, from Newton's Law of Universal Gravitation, that a uniform spherical shell of mass M and radius R attracts an external point mass as though all its mass were concentrated at the centre. The same result applies to the potential:

V_{\text{outside}}(r) = -\frac{GM}{r} \quad (r \geq R).

Inside the shell, every piece of the shell still contributes to the potential — but now the contributions conspire to give a constant potential, equal to the value at the shell's inner surface:

V_{\text{inside}}(r) = -\frac{GM}{R} \quad (r < R).

Why constant? Because \vec{g} = 0 inside a uniform shell (shell theorem, force version). A vanishing field means a flat potential — no "slope" for gravity to pull along. Inside the shell, you could float weightlessly at any point, and the potential is the same everywhere. Only outside the shell does V start changing with distance.

This is the "inside a hollow planet" scenario of science fiction. The force is zero; the potential is a flat plateau at V = -GM/R.

Potential inside a uniform solid sphere

A uniform solid sphere of total mass M and radius R is built by stacking shells. Outside, V = -GM/r as before. Inside (at distance r < R from the centre), the potential at a point comes from two contributions:

  1. The shells of radius less than r (total mass M_{\text{in}} = M(r/R)^3) act like a point mass M_{\text{in}} at the centre, contributing -GM_{\text{in}}/r.
  2. The shells of radius between r and R contribute a constant. Each such shell of radius r' contributes -GM_{\text{shell}}(r')/r' (by the inside-shell result — you are inside every one of these shells).

You need to integrate over shell radii from r to R, accounting for the mass of each differential shell. After doing the integral, the result is

V_{\text{inside}}(r) = -\frac{GM}{2R^3}(3R^2 - r^2) \quad (r \leq R).

Two checks:

  • At r = R (the surface from inside): V = -\frac{GM}{2R^3}(3R^2 - R^2) = -\frac{GM}{2R^3}(2R^2) = -\frac{GM}{R}. Matches the outside value — the potential is continuous across the surface, as it must be.
  • At r = 0 (the centre): V = -\frac{GM}{2R^3}(3R^2) = -\frac{3GM}{2R}. The centre is the deepest point of the well — 50% deeper than the surface.

The field inside the sphere is \vec{g} = -dV/dr\,\hat{r} = -(GMr/R^3)\,\hat{r}, which is linear in r — exactly the result from the force version of the shell theorem.

The connection to Poisson's equation

Everything in this article is a special case of a single differential equation. For any gravitational source distribution \rho(\vec{r}) (mass per unit volume), the gravitational potential V(\vec{r}) satisfies

\nabla^2 V = 4\pi G \rho.

In regions of empty space (\rho = 0), this reduces to Laplace's equation \nabla^2 V = 0. Solutions of Laplace's equation have a beautiful property: they take their maximum and minimum values on the boundary of any region, never in the interior. This is the mathematical reason a satellite in free space cannot sit at a local potential minimum that is not also a mass location — "gravitational traps" in empty space do not exist.

For a spherically symmetric source, \nabla^2 V = \dfrac{1}{r^2}\dfrac{d}{dr}\!\left(r^2 \dfrac{dV}{dr}\right), and you can solve the one-dimensional equation directly. Outside the source, V = -GM/r satisfies it identically (you can verify: r^2 (GM/r^2) = GM, a constant, whose derivative is zero). Inside a uniform sphere, the quadratic form -(GM/2R^3)(3R^2 - r^2) satisfies the equation with constant \rho. The inside-outside matching at the surface is automatic.

This is a preview: the same equation, with \rho replaced by the charge density and G replaced by -1/(4\pi\epsilon_0), governs electrostatic potential. The mathematics of gravitation and electrostatics is, at bottom, the mathematics of the Laplacian — everything you learn here carries over to electric fields with a sign change on G.

A three-check summary

You should be able to write down all of the following in one line, without thinking:

  1. Force: F = GMm/r^2, always attractive.
  2. Potential energy: U = -GMm/r, chosen so U(\infty) = 0.
  3. Potential: V = U/m = -GM/r, a property of the source.
  4. Field: \vec{g} = -dV/dr\,\hat{r} = -GM/r^2\,\hat{r}, pointing toward the source.

Each is a single integration or differentiation away from the others. The whole of Newtonian gravitation — for point sources or for any spherically symmetric mass distribution viewed from outside — sits in those four lines.

Where this leads next