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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover 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. Use G = 6.67 × 10⁻¹¹ N m² kg⁻², Earth mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m, Moon mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m. Revise first with the study guide or the revision notes. The course pages are the OCR A Level Physics hub and the checklist.

Questions

1. Define gravitational potential at a point in a gravitational field. [2]

2. Explain why the gravitational potential is negative at every point near a planet. [2]

3. Calculate the gravitational potential at the surface of the Earth, and at a point whose distance from the centre of the Earth is three times the Earth’s radius. [3]

4. Calculate the work that must be done to move a 4.0 kg mass from the surface of the Earth to a point at a distance of three Earth radii from the centre of the Earth. [3]

5. Europa, a moon of Jupiter, has mass 4.80 × 10²² kg and radius 1.56 × 10⁶ m.

(a) Calculate the gravitational potential at its surface. [2] (b) Calculate the escape velocity from its surface. [2]

6. A 650 kg satellite is moved from a circular path of radius 1.2 × 10⁷ m to one of radius 4.2 × 10⁷ m, both measured from the centre of the Earth.

(a) Calculate the change in the gravitational potential energy of the satellite. [3] (b) State whether the satellite gains or loses gravitational potential energy. [1]

7. A 50 kg object moves directly away from the centre of the Earth, from the surface to a distance of two Earth radii from the centre.

(a) Describe the shape of the graph of gravitational force against distance from the centre of the Earth for this journey. [2] (b) Calculate the area under this graph between the two distances and state what it represents. [3]

8. A planet has an escape velocity of 7.4 × 10³ m s⁻¹ and a radius of 3.1 × 10⁶ m. Calculate the mass of the planet. [3]

9. A small projectile is launched vertically from the surface of the Moon at 2.0 × 10³ m s⁻¹. Ignore any atmosphere.

(a) Calculate the greatest height above the surface that it reaches. [5] (b) A student uses h = v²/2g with g = 1.62 N kg⁻¹ and obtains 1.23 × 10⁶ m. Explain why this is not the same as your answer to (a). [2]

10. Planet Y has mass 3.3 × 10²³ kg and radius 2.44 × 10⁶ m. Two gases are present near its surface. Their atoms have mean speeds of 5.1 × 10² m s⁻¹ (gas A) and 3.6 × 10³ m s⁻¹ (gas B).

(a) Calculate the escape velocity from the surface of planet Y. [3] (b) Discuss which gas the planet is more likely to retain, and explain why some of the other gas could still escape. [2]

11. A body is released from rest at a distance of two Earth radii from the centre of the Earth. Ignore air resistance and the rotation of the Earth.

(a) Calculate its speed when it reaches the Earth’s surface. [4] (b) Explain why a real object released from this height would arrive more slowly. [1]

12. A spherical asteroid has radius 2.0 × 10⁵ m and a uniform density of 2.5 × 10³ kg m⁻³. The volume of a sphere is 4πr³/3.

(a) Show that its mass is about 8.4 × 10¹⁹ kg. [2] (b) Calculate the escape velocity from its surface. [2] (c) An astronaut can throw a stone at 12 m s⁻¹. State, with a reason, whether the stone could escape. [1] (d) Calculate the energy needed to take a 0.50 kg stone from the surface to infinity. [3]

Answers

1. The work done per unit mass [1] in bringing a small (test) mass from infinity to that point [1]. [2] Examiner insight: Both ideas are needed: “per unit mass” and “from infinity”. “Energy at a point” alone describes potential energy and does not gain credit.

2. Gravity is attractive, so the field does positive work as a mass moves in from infinity [1]. Potential is defined as zero at infinity, so the work done by an external agent is negative, and every nearer point is below zero [1]. [2] Examiner insight: Linking the sign to the zero at infinity and the attractive force is what earns the second mark; “because it is an attractive force” on its own is incomplete.

3. Vg = -GM/r [1]. At the surface: -3.982 × 10¹⁴ / 6.37 × 10⁶ = -6.25 × 10⁷ J kg⁻¹ [1]. At 3R: one third of this, -2.08 × 10⁷ J kg⁻¹ [1]. [3] Examiner insight: Show the substitution into Vg = -GM/r and keep the minus sign; a positive value is wrong even if the size is right.

4. W = mΔVg [1]. ΔVg = -2.084 × 10⁷ - (-6.251 × 10⁷) = +4.167 × 10⁷ J kg⁻¹ [1]. W = 4.0 × 4.167 × 10⁷ = 1.67 × 10⁸ J [1]. [3] Examiner insight: Take final minus initial. Subtracting the other way gives a negative energy, which suggests the field does the work, contradicting a mass being moved away.

5. (a) Vg = -GM/r = -(6.67 × 10⁻¹¹ × 4.80 × 10²²) / 1.56 × 10⁶ [1] = -2.05 × 10⁶ J kg⁻¹ [1] (b) v = √(2GM/r) = √(2 × 3.202 × 10¹² / 1.56 × 10⁶) [1] = 2.03 × 10³ m s⁻¹ [1] Examiner insight: If you have found Vg in part (a), v = √(2 × |Vg|) saves a step and an arithmetic slip.

6. (a) ΔE = GMm(1/r₁ - 1/r₂) [1]. = 3.982 × 10¹⁴ × 650 × (1/1.2 × 10⁷ - 1/4.2 × 10⁷) [1]. = 1.54 × 10¹⁰ J [1] (b) The energy increases by this amount, so the satellite gains potential energy because it moves to a higher (less negative) potential [1]. Examiner insight: A state question carries a single mark for the correct direction. Quote the sign of the change as well; a final answer given as a loss is wrong even with correct working.

7. (a) The force falls with distance in a curve [1] that is steep at first and flattens, following an inverse-square law (the force at two radii is a quarter of the force at the surface) [1]. (b) The area under the graph is the work done in moving the object between the two distances [1]. Area = GMm(1/R - 1/2R) = 3.982 × 10¹⁴ × 50 / (2 × 6.37 × 10⁶) [1] = 1.56 × 10⁹ J [1] Examiner insight: When asked what an area represents, name the quantity and the process (“work done by the force in moving the object between these distances”) rather than just “energy”.

8. v² = 2GM/R, so M = v²R/2G [1]. M = (7.4 × 10³)² × 3.1 × 10⁶ / (2 × 6.67 × 10⁻¹¹) [1] = 1.27 × 10²⁴ kg [1]. [3] Examiner insight: Rearrange before substituting. Squaring 7.4 × 10³ as 7.4 × 10⁶ or squaring only the 7.4 are the common slips; check the power of ten.

9. (a) Energy conservation per kilogram: ½v² + Vg(surface) = Vg(highest point) [1]. Vg(surface) = -4.902 × 10¹² / 1.74 × 10⁶ = -2.8175 × 10⁶ J kg⁻¹ [1]. Total = 2.0 × 10⁶ - 2.8175 × 10⁶ = -8.175 × 10⁵ J kg⁻¹ [1]. r = GM/8.175 × 10⁵ = 5.997 × 10⁶ m from the centre [1]. Height = 5.997 × 10⁶ - 1.74 × 10⁶ = 4.26 × 10⁶ m [1] (b) The field strength decreases with height, so g is not uniform and 1.62 N kg⁻¹ applies only at the surface [1]. Using it throughout means the deceleration is overestimated, giving a height that is too small [1]. [7] Examiner insight: The final step is converting r back into a height. Giving 6.00 × 10⁶ m as “the height” is a common loss on this type of question.

10. (a) v = √(2GM/r) = √(2 × 6.67 × 10⁻¹¹ × 3.3 × 10²³ / 2.44 × 10⁶) [1]. = √(1.804 × 10⁷) [1] = 4.25 × 10³ m s⁻¹ [1] (b) Gas A, with mean speed about 12% of the escape velocity, is far more likely to be retained, whereas gas B has a mean speed about 85% of it [1]. Atoms have a range of speeds, so some of gas B (those in the fast tail of the distribution) can exceed the escape velocity and leave over time [1]. [5] Examiner insight: A good discussion compares numbers with the escape velocity and then makes a statement about the spread of speeds. Saying only “gas B is faster” is too thin.

11. (a) Energy conservation: ½v² = decrease in potential energy per kilogram = GM(1/R - 1/2R) [1]. = GM/2R = 3.982 × 10¹⁴ / (2 × 6.37 × 10⁶) = 3.126 × 10⁷ J kg⁻¹ [1]. v² = 2 × 3.126 × 10⁷ = 6.25 × 10⁷ [1]. v = 7.91 × 10³ m s⁻¹ [1] (b) Air resistance transfers kinetic energy to thermal energy, so the final speed is less [1]. [5] Examiner insight: Keep the energy balance per kilogram, since the mass cancels, and say which energy falls and which rises before substituting.

12. (a) M = ρ × 4πr³/3 = 2.5 × 10³ × 4π × (2.0 × 10⁵)³ / 3 [1] = 8.38 × 10¹⁹ kg, which rounds to about 8.4 × 10¹⁹ kg [1] (b) v = √(2GM/r) = √(2 × 6.67 × 10⁻¹¹ × 8.38 × 10¹⁹ / 2.0 × 10⁵) [1] = 2.4 × 10² m s⁻¹ [1] (c) No: 12 m s⁻¹ is far below the escape velocity (about 240 m s⁻¹) [1]. (d) Vg = -GM/r = -2.79 × 10⁴ J kg⁻¹ [1]. Energy = m × |Vg| [1] = 0.50 × 2.79 × 10⁴ = 1.40 × 10⁴ J [1] [8] Examiner insight: A “show that” needs the unrounded value and the stated result; give 8.38 × 10¹⁹ kg before rounding, then use the unrounded figure in later parts.

Where marks are usually lost

• Dropping the negative sign from Vg or E, or giving the energy to bring a mass in from infinity as positive.
• Using height above the surface rather than distance from the centre of the planet.
• Subtracting potentials in the wrong order, so the direction of the energy change is wrong.
• Rounding intermediate potentials and then subtracting two similar numbers.
• Using mgh for a large change in height, where the field is not uniform.
• Squaring or square-rooting incorrectly in v = √(2GM/r), or using g instead of GM.
• Stopping at the distance from the centre and not converting it to a height.
• Defining potential without “per unit mass” or without “from infinity”.

Next steps

• Recap with the revision notes or the study guide.
• Visit the OCR A Level Physics hub and tick off statements on the checklist.
• Take one of our free 10-minute diagnostics.
• Book a free trial class.

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.