What is gravitational potential energy?
Potential energy is energy stored because of position. Gravity is a conservative force: the work it does depends only on where you start and where you end, not on the path. So we can give every position a potential energy U.
The gravitational potential energy of a mass at a point is the work done by an outside agent to bring it slowly from infinity to that point (with no change in kinetic energy).
Deriving U = −GMm/r
- At distance x from Earth's centre, gravity on m is GMm/x², pointing inward.
- To bring m in slowly, we pull outward with the same size of force, GMm/x². Our pull is outward but the mass moves inward, so our work is negative.
- Each small step of length dx gives work −(GMm/x²) dx. Add all the steps between r and ∞: ∫ dx/x² from r to ∞ = 1/r. So W = −GMm/r.
- So U(r) = −GMm/r, with U(∞) = 0.
Why is it negative?
Gravity pulls the mass in by itself, so the outside agent does negative work while bringing it in. Said simply: the mass is stuck in a well, and you need to give it energy (GMm/r) to set it free at infinity. So its energy starts below zero.
For a system of masses
For two masses m₁ and m₂ a distance r apart, U = −Gm₁m₂/r. For many masses, add U for every pair.
Link with mgh
Lift m from the surface (r = R) to height h (r = R + h). The rise in U is
ΔU = −GMm/(R + h) − (−GMm/R) = GMm h / [R(R + h)]
If h ≪ R, then R(R + h) ≈ R² and GM/R² = g, so ΔU ≈ mgh.
So mgh is the near-ground version. It only measures a change from a chosen floor. The full formula −GMm/r measures from infinity. For big heights, always use the full formula; mgh gives too large an answer.
Neat form: ΔU = mgh / (1 + h/R).
Gravitational potential V
Gravitational potential at a point = potential energy per unit mass = work done to bring 1 kg from infinity to that point.
V = U/m = −GM/r. Unit: J/kg. Dimensions [L² T⁻²]. It is a scalar.
- At Earth's surface: V = −GM/R = −gR ≈ −6.3 × 10⁷ J/kg.
- Many masses: V = −G(m₁/r₁ + m₂/r₂ + …). Just add numbers, no arrows.
- Work to move m from point A to point B = m (V_B − V_A).
- Link with the field (g): the field points toward lower V, and its size is how fast V changes with distance, g = −dV/dr.
Inside a hollow shell, V is the same everywhere (−GM/R), so the field there is zero.
Try it: a funnel well
Make a cone from chart paper (or use a kitchen funnel) and roll a marble around its inside. It circles fast near the narrow bottom and slowly near the wide top, just like a satellite in Earth's energy well. Push it harder and it climbs higher. In the 3D, slide the ball out to 6 R: U gets closer and closer to 0 but never becomes positive.
Key formulas and definitions
- U = −GMm/r (U = 0 at infinity)
- U = −Gm₁m₂/r for two masses
- ΔU (R → R + h) = GMm h/[R(R + h)] = mgh/(1 + h/R)
- ΔU ≈ mgh when h ≪ R
- V = −GM/r (J/kg)
- V at surface = −gR
- W(A → B) = m(V_B − V_A)
- g = −dV/dr
Worked examples
1. Find the gravitational PE of a 100 kg satellite on Earth's surface. (GM = 4 × 10¹⁴ m³/s², R = 6.4 × 10⁶ m)
U = −GMm/R = −4 × 10¹⁴ × 100 / 6.4 × 10⁶ = −6.25 × 10⁹ J.
2. Find the gravitational potential at Earth's surface (g = 9.8 m/s², R = 6.4 × 10⁶ m).
V = −gR = −9.8 × 6.4 × 10⁶ ≈ −6.27 × 10⁷ J/kg.
3. How much work is needed to lift a 10 kg body from the surface to a height h = R? (g = 9.8, R = 6.4 × 10⁶ m)
W = mgh/(1 + h/R) = mgR/2 = 10 × 9.8 × 6.4 × 10⁶ / 2 = 3.14 × 10⁸ J.
4. A 2 kg body is lifted 5 m near the ground. Find the change in PE. Why is mgh fine here?
ΔU = mgh = 2 × 9.8 × 5 = 98 J. 5 m is tiny compared with 6400 km, so h ≪ R.
5. Find the PE of the Earth-Moon pair. (M = 6 × 10²⁴ kg, m = 7.4 × 10²² kg, r = 3.84 × 10⁸ m)
U = −GMm/r = −6.67 × 10⁻¹¹ × 4.44 × 10⁴⁷ / 3.84 × 10⁸ ≈ −7.7 × 10²⁸ J.
6. Find the potential at the midpoint of two 10 kg masses kept 2 m apart.
Each is 1 m away. V = −G(10/1) − G(10/1) = −20G = −1.33 × 10⁻⁹ J/kg.
7. Three equal masses m sit at the corners of an equilateral triangle of side a. Find the PE of the system.
There are 3 pairs, each −Gm²/a. U = −3Gm²/a.
8. The potential at a point is −5 × 10⁷ J/kg. How much work is needed to take a 20 kg body from there to infinity?
W = m(V_∞ − V) = 20 × (0 − (−5 × 10⁷)) = 1 × 10⁹ J.
Common mistakes
- Using mgh for large heights (like h = R). mgh only works when h is tiny compared with R.
- Thinking a negative PE means 'no energy'. It means the body is bound; you must add energy to free it.
- Adding potentials as vectors. Potential is a scalar: just add the numbers (with signs).
- Mixing up U and V. U = −GMm/r is energy (J); V = −GM/r is energy per kg (J/kg).