Why SL's mgh is not enough: Near the ground the field is roughly constant, so SL uses Ep = mgh. But over astronomical distances g changes with height (g = GM/r²), so that shortcut breaks down. HL replaces it with two exact field quantities:
- gravitational potential Vg — the PE per kilogram at a point - gravitational potential energy Ep — the PE of an actual mass m at that point
Where is 'zero'?: We set the zero of potential at infinity — infinitely far from every mass. Bring a mass in from infinity and it loses PE, so everywhere closer the value is negative.
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The gravitational potential Vg at a point is the work done per unit mass to bring a small test mass from infinity to that point. Because gravity does the work for you on the way in, that work is negative — so Vg is negative and tends to zero at infinity.
- gravitational potential (J kg⁻¹)
- gravitational constant, 6.67×10⁻¹¹ N m² kg⁻²
- mass producing the field (kg)
- distance from the centre of M (m)
Worked example — potential at Earth's surface
Earth has mass M = 6.0×10²⁴ kg and radius r = 6.4×10⁶ m. Find the gravitational potential at its surface. (G = 6.67×10⁻¹¹ N m² kg⁻²)
Solution
- Write the given formula first:
- Substitute the values:
- Work it out — keep the unit:
Final answer
Vg = −6.3×10⁷ J kg⁻¹. The minus sign shows you sit in a potential well.
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From potential to energy: Potential is per kilogram; multiply by the actual mass m to get the energy. So Ep = m × Vg, which gives the full-field expression below. Like Vg it is negative and zero at infinity.
- gravitational potential energy (J)
- mass placed in the field (kg)
- mass producing the field (kg)
- separation of the centres (m)
Worked example — PE of a satellite at the surface
Find the gravitational potential energy of a 500 kg satellite sitting at Earth's surface, where Vg = −6.3×10⁷ J kg⁻¹.
Solution
- Use the given relation between energy and potential:
- Substitute m = 500 kg and Vg = −6.3×10⁷ J kg⁻¹:
- Work it out — keep the unit:
Final answer
Ep = −3.1×10¹⁰ J (the same as −GMm/r). It is negative: energy must be added to lift the satellite away.
Surfaces of equal potential: An equipotential surface links all points at the same Vg. Around a point mass these are concentric spheres. They are always perpendicular to the field lines, and no work is done moving a mass along one (ΔVg = 0).
Field lines
- Point radially inward toward the mass
- Closer together where the field is stronger
- Show the direction of the gravitational force
Equipotentials
- Spheres of constant potential (rings in 2D)
- Always perpendicular to the field lines
- Moving along one does no work (ΔVg = 0)
To move a mass between equipotentials you do work equal to the mass times the change in potential. This is the HL way to find energy changes when g is not constant.
- work done on the mass (J)
- mass moved (kg)
- change in gravitational potential, V_final − V_initial (J kg⁻¹)
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Where it shows up: Gravitational potential is HL only (D.1):
- Paper 1A — 'why is the potential negative?', 'which point has the larger Vg?', or 'how do equipotentials relate to field lines?'. - Paper 2 — determine Vg or Ep at a distance, or find the work (= mΔVg) to move a probe between two points.
Three easy marks: (1) Keep the minus sign — Vg and Ep are negative below infinity. (2) Use the separation of centres for r, not the height above the surface. (3) Work between points = m × (Vfinal − Vinitial), so a rise in potential gives positive work.
IB-style question — work to lift a probe
At a planet's surface the gravitational potential is −4.0×10⁷ J kg⁻¹; at the orbit of a probe it is −1.5×10⁷ J kg⁻¹. Determine the work needed to raise a 200 kg probe from the surface to that orbit.
Solution
- Use the given work formula:
- Find ΔVg = Vfinal − Vinitial:
- Multiply by the mass — keep the unit:
Final answer
W = 5.0×10⁹ J. Positive, because the probe climbs to a higher (less negative) potential.