OCR A Level Physics A: Gravitational potential, potential energy and escape velocity (H556) – Revision Notes

Condensed notes for section 5.4.4 (Gravitational potential and energy), statements (a) to (e), of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. For full explanations and worked examples, use the study guide, then test yourself with the practice questions. The OCR A Level Physics hub, the checklist and the free diagnostics sit alongside these notes.

Definitions

Term	Meaning
Gravitational potential, Vg (5.4.4(a))	Work done per unit mass in bringing a small mass from infinity to the point. Unit J kg⁻¹. Scalar.
Zero of potential	Defined to be zero at infinity, so every finite point has a negative value
Gravitational potential energy, E (5.4.4(d))	E = mVg: the work done in bringing mass m from infinity to the point. Unit J.
Escape velocity (5.4.4(e))	Minimum speed at which an object must be launched so that it just reaches infinity with no further propulsion

Key equations

Quantity	Equation	Notes
Potential	Vg = -GM/r	r from the centre of M, which may be a point or a sphere
Change in potential	ΔVg = Vg(final) - Vg(initial)	Positive when moving away from the mass
Potential energy	E = mVg = -GMm/r	Zero at infinity, negative elsewhere
Change in energy	ΔE = mΔVg = GMm(1/r₁ - 1/r₂)	Energy needed to move from r₁ out to r₂
Force	F = GMm/r²	Magnitude; plotted on the force-distance graph
Work from graph	W = area under F-r graph	Equals mΔVg between the two distances
Escape velocity	v = √(2GM/r)	Also v = √(2gR) at a planet’s surface
Surface link	GM = gR²	Use to convert between g and GM

Method in steps

Potential and energy changes:

1. Convert heights to distances from the centre, r = R + h, in metres.
2. Calculate Vg at both positions, keeping an extra digit.
3. Subtract: final minus initial.
4. Multiply by m for the energy change. Positive means energy must be supplied.

Escape velocity:

1. State that the total energy at infinity is zero for the minimum case.
2. Write ½mv² - GMm/r = 0.
3. Cancel m and rearrange to v = √(2GM/r).
4. Substitute, using r in metres from the centre.

Energy conservation in a fall or a throw (no other forces):

1. Write the total energy at the start: ½mv² - GMm/r.
2. Equate with the total energy at the end.
3. Cancel m and solve for the unknown speed or distance.

Graph features to know

Vg against r (5.4.4(b)):

• Starts steeply negative at the surface radius and rises towards zero as r increases.
• Never crosses the axis. Vg is zero only at infinity.
• Doubling r halves the size of Vg.

F against r (5.4.4(c)):

• Falls steeply, then flattens; the inverse square law.
• Area under the graph between two distances is the work done, equal to mΔVg.
• Counting squares or trapezia: trapezia overestimate the area under this curve because it is bowed in towards the axes.

Small worked reminders

Potential at the surface of the Earth, using GM = 3.982 × 10¹⁴ m³ s⁻² and R = 6.37 × 10⁶ m: Vg = -3.982 × 10¹⁴ / 6.37 × 10⁶ = -6.25 × 10⁷ J kg⁻¹.

Energy to take a 2.0 kg mass from the surface to infinity: 2.0 × 6.25 × 10⁷ = 1.25 × 10⁸ J.

Escape speed from a planet with g = 6.0 N kg⁻¹ at the surface and R = 4.0 × 10⁶ m: v = √(2 × 6.0 × 4.0 × 10⁶) = 6.9 × 10³ m s⁻¹.

Maximum distance reached: a probe leaves the surface of the Earth at 8.0 × 10³ m s⁻¹, below the escape speed, and moves with no engine and no air resistance. Energy per kilogram at launch: ½v² + Vg = 3.20 × 10⁷ - 6.25 × 10⁷ = -3.05 × 10⁷ J kg⁻¹. At the highest point the speed is zero, so Vg = -GM/r = -3.05 × 10⁷, giving r = 3.982 × 10¹⁴ / 3.05 × 10⁷ = 1.31 × 10⁷ m from the centre, which is 6.68 × 10⁶ m above the surface. Notice that a negative total energy means the probe is bound and must return.

Escape velocity and atmospheres

• Total energy at infinity is zero for the minimum case, so a launch at exactly the escape speed arrives with zero kinetic energy.
• A projectile with total energy below zero is bound. It rises to a maximum distance and falls back.
• A launch above the escape speed leaves the field with some kinetic energy remaining.
• To predict the atmosphere of a planet, calculate the escape speed from its mass and radius and compare it with the speeds of the gas atoms, which are given as data. Light atoms move faster than heavy ones at the same temperature, so they are the first to be lost.
• Explain the result in words: the atoms with the highest speeds in the range can escape, so the atmosphere is lost over time even if the mean speed is lower than the escape speed.
• A larger mass or a smaller radius raises the escape speed, so a planet is more likely to hold on to its gases.

Must-know distinctions

• Potential is per kilogram (J kg⁻¹); potential energy is for a mass (J).
• Potential is a scalar, so there is no direction to worry about. A negative sign means lower than infinity, not “pointing” anywhere.
• Escape velocity is about reaching infinity. It is not the speed needed to orbit.
• The mgh formula is a uniform-field approximation. Vg = -GM/r is general for a point or spherical mass.
• Atoms escape from an atmosphere when their speeds are comparable to or above the escape speed. A mean speed below the escape speed does not guarantee that none escape, since speeds are spread over a range.

Quick self-test

1. Define gravitational potential at a point.
2. State the unit of gravitational potential and say whether it is a scalar or a vector.
3. Calculate the gravitational potential at the surface of the Earth (GM = 3.982 × 10¹⁴ m³ s⁻², R = 6.37 × 10⁶ m).
4. How much energy is needed to take a 2.0 kg mass from the Earth’s surface to infinity?
5. Calculate the potential at r = 2.0 × 10⁷ m from the centre of the Earth.
6. By what factor does the magnitude of Vg change when r is doubled?
7. What does the area under a force-distance graph represent?
8. Explain why the escape velocity does not depend on the mass of the projectile.
9. A planet has g = 6.0 N kg⁻¹ at its surface and radius 4.0 × 10⁶ m. Calculate its escape velocity.
10. Calculate the gravitational potential energy of a 3000 kg satellite at r = 4.2 × 10⁷ m from the centre of the Earth.
11. A gas has atoms with mean speed 70% of a planet’s escape speed. Can you say that none will escape?

Answers

1. The work done per unit mass in bringing a small mass from infinity to that point.
2. J kg⁻¹; a scalar.
3. Vg = -3.982 × 10¹⁴ / 6.37 × 10⁶ = -6.25 × 10⁷ J kg⁻¹.
4. 2.0 × 6.25 × 10⁷ = 1.25 × 10⁸ J.
5. Vg = -3.982 × 10¹⁴ / 2.0 × 10⁷ = -1.99 × 10⁷ J kg⁻¹.
6. It halves (it is multiplied by 1/2).
7. The work done in moving between the two distances, equal to mΔVg.
8. Both the kinetic energy needed (½mv²) and the potential energy to be overcome (GMm/r) are proportional to m, so m cancels.
9. v = √(2gR) = √(2 × 6.0 × 4.0 × 10⁶) = 6.9 × 10³ m s⁻¹.
10. E = -GMm/r = -3.982 × 10¹⁴ × 3000 / 4.2 × 10⁷ = -2.84 × 10¹⁰ J.
11. No. Speeds are spread over a range, so a fraction of the atoms will be above the mean and some could exceed the escape speed.

Where marks are usually lost

• Leaving out the minus sign in Vg or E, or giving a positive answer for energy needed from infinity.
• Using the height above the surface instead of the distance from the centre.
• Subtracting potentials the wrong way round, so the sign of ΔVg is wrong.
• Rounding the potentials early and then subtracting two close numbers.
• Applying mgh over a height that is a sizeable fraction of the planet’s radius.
• Missing the square root in the escape velocity, or not using metres.
• Defining potential without “per unit mass”, or without “from infinity”.
• Explaining the escape-velocity result without saying that mass cancels.
• Saying that the area under a force-distance graph is “energy” without naming the work done by the force or against it.

Official syllabus

OCR Advanced GCE in Physics A (H556), Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Section 5.4.4, Gravitational potential and energy.