Gravitational potential energy

Gravity on a mass m points straight at the centre of M, and its size GMm/r² depends only on the distance. Break any path into small steps: a step around the centre, at constant r, is perpendicular to the force and does no work, so only the radial steps count. The work from r₁ to r₂ is therefore a one-variable integral, W_grav = ∫ −GMm/r² dr from r₁ to r₂ = GMm(1/r₂ − 1/r₁), the same for a straight climb, a spiral or a detour. Raise a 500 kg satellite from r₁ = 7.0 × 10⁶ m to r₂ = 4.2 × 10⁷ m and W_grav = 3.986 × 10¹⁴ × 500 × (1/(4.2 × 10⁷) − 1/(7.0 × 10⁶)) = −2.37 × 10¹⁰ J. Gravity does negative work on anything moving outward. That independence of route is exactly what allows a potential energy: define ΔU = −W_grav, so this climb has ΔU = +2.37 × 10¹⁰ J whichever path the satellite took.

ΔU fixes differences only, so some position must be declared U = 0. The force fades to nothing as r grows, and a zero at infinite separation belongs to no particular planet, so the standard choice is U(∞) = 0. Integrating in from infinity then gives U(r) = −GMm/r, negative at every finite distance: the pair has less energy than it would have fully separated, because gravity does positive work as they come together. Nothing physical hangs on that choice. Put the zero at Earth's surface instead and Uₛ(r) = GMm(1/R − 1/r), positive above the ground. Per kilogram at r = 3R the two conventions read −20.9 MJ kg⁻¹ and +41.7 MJ kg⁻¹, yet from R to 3R both give a rise of 41.7 MJ kg⁻¹, because they differ only by the constant GM/R = 62.6 MJ kg⁻¹. The surface form also exposes a ceiling that mgh hides: however far a kilogram is carried, from rest to rest, the bill never reaches 62.6 MJ.

Energy and force carry the same information. Differentiating, Fᵣ = −dU/dr = −GMm/r²: U rises with r, so its slope is positive and the radial force is negative, pointing inward. The graph shows how differently the two fall off. At r = 2R the well is half as deep as at the surface, 31.3 MJ kg⁻¹ instead of 62.6, but the slope, which is the pull on each kilogram, has dropped to a quarter: 2.46 N kg⁻¹ instead of 9.82. Replacing the curve by its tangent at the surface means holding the force at mg₀, which is exactly ΔU = mg₀h with h = r − R. Because −1/r bends downward, every tangent lies above the curve, so a constant-force estimate always overstates the energy of a climb: from R to 1.5R it gives 31.3 MJ kg⁻¹ against the true 20.9. The tangent at any radius r₁ reaches U = 0 at 2r₁. A force that kept its starting strength would empty the well within one more r₁ of climbing; the real force weakens, and the curve reaches zero only at infinity.

The distance from a point on the curve up to the zero line is the binding energy, E_bind = GMm/r: the least energy that takes the pair from rest at r to rest at infinite separation. Run it backwards and it is the kinetic energy a body gains falling in from rest far away. Per kilogram, the depth at a body's surface is GM/R, which grows with mass and with compactness: 2.82 MJ kg⁻¹ at the Moon, 62.6 MJ kg⁻¹ at Earth, and 1.91 × 10⁵ MJ kg⁻¹ at the Sun's surface. That factor of 22 between the Moon and Earth is why lifting cargo off the Moon is so much cheaper. The energy belongs to the pair, not to m alone: for Earth and the Moon, 3.844 × 10⁸ m apart, U = −GM_E M_M/r = −7.62 × 10²⁸ J, so pulling them apart, if both were held at rest, would take 7.62 × 10²⁸ J whichever body you picture moving. A body that is already moving needs less than E_bind, because its kinetic energy counts toward the bill; how much less belongs to orbital energy and escape.

Potential energy belongs to pairs, so a group of bodies gets one term per pair: U = −Σ G mᵢmⱼ/rᵢⱼ over i < j. The reason is the order of assembly. Bring the bodies in from infinity one at a time: the first costs nothing, the second adds one pair term, the third adds two more, and every pair is counted exactly once. Three 12.0 kg lead spheres at the corners of an equilateral triangle of side s = 0.250 m therefore have U = −3Gm²/s = −3 × 6.674 × 10⁻¹¹ × 12.0² ÷ 0.250 = −1.15 × 10⁻⁷ J, a reminder of how feeble gravity is between laboratory masses. Pulling one sphere far away while the other two are held in place removes the two pairs it belongs to and costs 2Gm²/s = 7.69 × 10⁻⁸ J, two-thirds of the total. The classic slip is to add up each sphere's energy in the field of the other two: every pair then appears twice, and the answer doubles to −2.31 × 10⁻⁷ J.

The formula rests on the shell theorem, which lets a spherical body act as a point at its centre only for points outside it. Inside a uniform planet only the mass within radius r pulls, the force falls in proportion to r, and integrating that force inward from the surface gives U(r) = −GMm(3R² − r²)/(2R³). At the centre that is −1.5 GMm/R, or −93.8 MJ kg⁻¹ for a uniform Earth: deeper than at the surface, but finite, where −GMm/r would plunge without limit. The real Earth, with its dense core, is somewhat deeper still. Near a body that is not spherical, shape enters: Earth's equatorial bulge changes U near the surface by about one part in a thousand, and beside an elongated asteroid the point-mass form can be badly wrong. Finally, −GMm/r is the weak-field limit of general relativity, reliable while GM/(rc²) ≪ 1. That ratio is 7.0 × 10⁻¹⁰ at Earth's surface and 2.1 × 10⁻⁶ at the Sun's, but about 0.17 at the surface of a 1.4-solar-mass neutron star 12 km in radius, where Newton's potential is no longer a safe guide.