Gravitational Potential and Energy

The previous lesson gave us Newton's law of gravitation and the field strength g = GM/r². To understand energy conservation in gravitational problems — dropping probes onto Mars, launching satellites, computing escape velocities — we need the partner concept: gravitational potential. This lesson develops the idea, derives the formulae V_g = −GM/r and E_p = −GMm/r, and applies them to several exam-style problems.

This is OCR A-Level Physics A Module 5.4 (Gravitational fields).

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From Force to Energy

An inverse-square force field has a very special property: the work done in moving a test mass from one point to another depends only on the start and end points, not on the path taken. Such a field is called conservative, and we can define a potential function at each point such that the work done against the field between two points is the difference in potential energy.

For gravity, the potential is called gravitational potential and denoted V_g (sometimes just V, when context is clear).

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Gravitational Potential — Definition

The gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point.

In symbols:

V_g = W / m

Units: joules per kilogram (J kg⁻¹).

The reference point "infinity" is chosen because it is the only place where a test mass experiences zero gravitational attraction from the source. By convention, the gravitational potential at infinity is defined to be zero.

Exam Tip: Students often confuse gravitational potential V_g (in J kg⁻¹) with gravitational potential energy E_p (in J). The relationship is E_p = mV_g. If you are reading about "potential" alone, it is an energy per unit mass; potential energy is the total.

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Why Is Potential Negative?

Here is a subtlety that trips up almost every first-time learner. Consider bringing a test mass from infinity (where V_g = 0) towards the Earth. The gravitational field pulls the test mass inward, so the field does positive work on it. By energy conservation, its gravitational potential energy must decrease.

Since V_g starts at 0 at infinity and decreases as we approach the Earth, it must become negative everywhere close to the Earth. And indeed:

V_g = −GM / r

The minus sign is not a typo. It is a statement of the choice of reference. Any point closer than infinity has a lower potential, and because infinity is zero, all finite points have negative potential.

Derivation Outline

For those who like calculus, the potential is defined so that g = −dV_g/dr — the field is the negative gradient of the potential. Starting from g = GM/r² (inward, so g = −GM/r² when r is measured outward) and integrating from infinity inward gives

V_g(r) − V_g(∞) = −∫(from ∞ to r) g dr' = −∫(from ∞ to r) (−GM/r'²) dr' = [−GM/r'] from ∞ to r = −GM/r

Since V_g(∞) = 0, we have V_g(r) = −GM/r.

At A-Level you do not need to reproduce this derivation, but you must know the result.

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Gravitational Potential Energy

If a mass m is placed at a point where the gravitational potential is V_g, its gravitational potential energy is

E_p = m V_g = −GMm / r

This replaces the simple E_p = mgh formula you learned at GCSE. The reason is that mgh assumes g is constant — a good approximation near the Earth's surface, but badly wrong at astronomical scales.

Consistency with mgh

For heights h ≪ R_E (much smaller than the Earth's radius), the change in the full formula is

ΔE_p = −GMm/(R_E + h) − (−GMm/R_E) = GMm(1/R_E − 1/(R_E + h))
     = GMm × h/(R_E(R_E + h)) ≈ GMm × h/R_E² = mgh

So mgh is exactly the small-h limit of the exact formula. The two are not in disagreement — mgh is just a first-order approximation.

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Graphs of V_g and E_p versus r

Both V_g and E_p are negative everywhere outside the Earth and approach zero only in the limit r → ∞. A sketch of V_g against r looks like a −1/r curve, starting at a large negative number near the surface and rising asymptotically toward 0 at infinity.

graph LR
    A[r = R: V = −GM/R<br/>large negative] --> B[r = 2R: V = −GM/2R<br/>half as negative]
    B --> C[r = 10R: V = −GM/10R<br/>nearly zero]
    C --> D[r → ∞: V = 0]

A common OCR question asks you to add the graph of g against r (from the previous lesson — a 1/r² curve) and the graph of V_g against r (a −1/r curve) on adjacent axes and discuss how they are related. Since g = −dV_g/dr, the gradient of the potential curve gives the field strength.

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Equipotentials

An equipotential surface is a surface on which V_g is constant. No work is done in moving along an equipotential because ΔV_g = 0. For a spherical source, the equipotentials are concentric spheres around the source.

Field lines and equipotentials are perpendicular to each other everywhere. For the Earth, the equipotentials are horizontal (tangent to the Earth's surface) and the field lines are vertical.

Exam Tip: OCR sometimes asks you to sketch a field-line diagram and equipotentials on the same figure. Draw radial field lines converging inward, and draw concentric circles as the equipotentials. Make sure they are perpendicular where they cross.

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Worked Example 1 — Potential at the Surface of the Earth

Calculate the gravitational potential at the surface of the Earth. Take M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m.

V_g = −GM/R = −(6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / 6.37 × 10⁶
    = −3.98 × 10¹⁴ / 6.37 × 10⁶
    = −6.25 × 10⁷ J kg⁻¹