A Level Physics 13.4 Gravitational Potential MCQs – 30 Practice Questions with Answers (Cambridge 9702)
This page covers Section 13.4 of the Cambridge AS & A Level Physics 9702 syllabus, an A Level (A2) topic on gravitational potential. Work through 30 original multiple-choice questions on potential, potential energy changes and escape velocity, each with a fully worked answer.
Formula recap: V = −GM/r | Ep = mV = −GMm/r | ΔEp = mΔV | vesc = √(2GM/r) = √(2|V|) | g = −dV/dr. Full notes and more formulas are on our A Level Physics Formula Sheet.
Gravitational Potential
Q1. Which statement correctly defines the gravitational potential at a point in a gravitational field?
- (A) The force acting on a unit mass placed at the point
- (B) The work done per unit mass in bringing a mass from infinity to the point
- (C) The total work done in bringing a mass from infinity to the point
- (D) The potential energy stored in a unit mass at infinity
Show Answer
Answer: B – Potential is work done per unit mass, V = W/m, for a mass brought from infinity. Option (A) defines gravitational field strength g, not potential.
Q2. The gravitational potential at every point near a planet is negative. This is because
- (A) the zero of potential is chosen at the surface of the planet
- (B) the potential decreases as the distance from the planet increases
- (C) the zero of potential is taken at infinity, and the field does work on a mass as it moves in from infinity, so energy must be supplied to take a mass back out to infinity
- (D) gravitational forces are too weak to produce a positive potential
Show Answer
Answer: C – With V = 0 at infinity, a mass falling inwards gains kinetic energy as the field does positive work, so its potential (and potential energy) drops below zero. Energy input is then needed to escape, which is why V stays negative everywhere near a mass.
Q3. The unit of gravitational potential is
- (A) J kg−1
- (B) N kg−1
- (C) J
- (D) kg J−1
Show Answer
Answer: A – Potential is energy per unit mass, so its unit is joules per kilogram, J kg−1. N kg−1 is the unit of field strength g.
Q4. At a point P, the potential due to planet 1 alone is −4.0 × 107 J kg−1 and the potential due to planet 2 alone is −1.5 × 107 J kg−1. The total potential at P is
- (A) −2.5 × 107 J kg−1
- (B) +5.5 × 107 J kg−1
- (C) −4.0 × 107 J kg−1
- (D) −5.5 × 107 J kg−1
Show Answer
Answer: D – Potential is a scalar, so the two contributions simply add algebraically: −4.0 × 107 + (−1.5 × 107) = −5.5 × 107 J kg−1. There is no direction to consider, unlike field strength.
Q5. For the Earth, take G = 6.67 × 10−11 N m2 kg−2, mass M = 6.0 × 1024 kg and radius R = 6.4 × 106 m. The gravitational potential at the Earth’s surface is approximately
- (A) −6.3 × 106 J kg−1
- (B) −6.3 × 107 J kg−1
- (C) +6.3 × 107 J kg−1
- (D) −9.8 J kg−1
Show Answer
Answer: B – V = −GM/r = −(6.67 × 10−11 × 6.0 × 1024) / (6.4 × 106) = −(4.0 × 1014) / (6.4 × 106) = −6.25 × 107 ≈ −6.3 × 107 J kg−1. (−9.8 is the value of g, in N kg−1, not V.)
Q6. The potential at the Earth’s surface is −6.3 × 107 J kg−1. A point at a distance of 2R from the Earth’s centre, where R is the Earth’s radius, has potential
- (A) −3.1 × 107 J kg−1
- (B) −1.6 × 107 J kg−1
- (C) −1.26 × 108 J kg−1
- (D) −6.3 × 107 J kg−1
Show Answer
Answer: A – V = −GM/r, so V is proportional to 1/r, not 1/r2. Doubling r halves the magnitude: −6.3 × 107 / 2 = −3.1 × 107 J kg−1. Quartering it (option B) is the behaviour of g, not of V.
Q7. A point is at a distance of 1.92 × 107 m from the centre of the Earth (three Earth radii). Using GM = 4.0 × 1014 m3 s−2 for the Earth, the potential at this point is
- (A) −6.3 × 107 J kg−1
- (B) −7.0 × 106 J kg−1
- (C) −2.1 × 107 J kg−1
- (D) +2.1 × 107 J kg−1
Show Answer
Answer: C – V = −GM/r = −(4.0 × 1014) / (1.92 × 107) = −2.08 × 107 ≈ −2.1 × 107 J kg−1. This is one third of the surface value, because r is three times larger and V ∝ 1/r.
Q8. When the distance from a planet’s centre is doubled, the field strength g falls to one quarter of its original value. The potential V becomes
- (A) one quarter of its original value
- (B) one eighth of its original value
- (C) unchanged
- (D) one half of its original (negative) value
Show Answer
Answer: D – g ∝ 1/r2 but V ∝ 1/r. Doubling r therefore halves V: for example from −6.3 × 107 to −3.1 × 107 J kg−1. Note V becomes less negative, i.e. it increases towards zero.
Q9. A point P is 2.0 × 107 m from the centre of a mass of 8.0 × 1023 kg and 1.0 × 107 m from the centre of a mass of 4.0 × 1023 kg. Taking G = 6.67 × 10−11 N m2 kg−2, the potential at P is
- (A) −2.7 × 106 J kg−1
- (B) −5.3 × 106 J kg−1
- (C) 0
- (D) −1.1 × 107 J kg−1
Show Answer
Answer: B – Potentials add as scalars. V1 = −(6.67 × 10−11 × 8.0 × 1023) / (2.0 × 107) = −2.7 × 106 J kg−1. V2 = −(6.67 × 10−11 × 4.0 × 1023) / (1.0 × 107) = −2.7 × 106 J kg−1. Total V = −2.7 × 106 + (−2.7 × 106) = −5.3 × 106 J kg−1.
Q10. A mass of 4.0 kg is moved from infinity to a point where the gravitational potential is −5.0 × 107 J kg−1. The work done by the gravitational field during this move is
- (A) +2.0 × 108 J
- (B) −2.0 × 108 J
- (C) +5.0 × 107 J
- (D) zero
Show Answer
Answer: A – The change in potential energy is ΔEp = mV = 4.0 × (−5.0 × 107) = −2.0 × 108 J. The work done by the field is −ΔEp = +2.0 × 108 J: the field does positive work as it pulls the mass inwards.
Potential Energy & Moving Between Points
Q11. A mass of 2.0 kg rests on the Earth’s surface, where the gravitational potential is −6.25 × 107 J kg−1. Its gravitational potential energy is
- (A) −6.25 × 107 J
- (B) −1.25 × 108 J
- (C) +1.25 × 108 J
- (D) −3.1 × 107 J
Show Answer
Answer: B – Ep = mV = 2.0 × (−6.25 × 107) = −1.25 × 108 J. Potential energy is negative because it is measured relative to zero at infinity.
Q12. The gravitational potential energy of a mass m at a distance r from the centre of a planet of mass M is given by
- (A) Ep = +GMm/r
- (B) Ep = GMm/r2
- (C) Ep = −GM/r
- (D) Ep = −GMm/r
Show Answer
Answer: D – Ep = mV = m(−GM/r) = −GMm/r. Option (C) is the potential V itself (per unit mass), and option (B) is related to the force law, not the energy.
Q13. A satellite of mass 200 kg is moved from an orbit where the potential is −3.13 × 107 J kg−1 to a higher orbit where the potential is −1.56 × 107 J kg−1. The change in its gravitational potential energy is
- (A) −3.1 × 109 J
- (B) +1.57 × 107 J
- (C) +3.1 × 109 J
- (D) zero, because potential energy does not change in orbit
Show Answer
Answer: C – ΔEp = m(V2 − V1) = 200 × [−1.56 × 107 − (−3.13 × 107)] = 200 × (+1.57 × 107) = +3.1 × 109 J. Moving outwards, V becomes less negative, so Ep increases: the change is positive.
Q14. How much energy must be supplied to lift a 50 kg mass from the Earth’s surface (V = −6.25 × 107 J kg−1) to a point at a distance of 2R from the Earth’s centre (V = −3.13 × 107 J kg−1), ignoring any change in kinetic energy?
- (A) 1.6 × 109 J
- (B) 3.1 × 107 J
- (C) 4.7 × 109 J
- (D) 1.6 × 108 J
Show Answer
Answer: A – Energy supplied = ΔEp = mΔV = 50 × [−3.13 × 107 − (−6.25 × 107)] = 50 × (3.12 × 107) = 1.56 × 109 ≈ 1.6 × 109 J.
Q15. A small mass falls from rest at a distance of 2R from the Earth’s centre towards the surface at distance R. The potential changes from −3.13 × 107 J kg−1 to −6.25 × 107 J kg−1. Ignoring the atmosphere, the speed of the mass when it reaches the surface is approximately
- (A) 5.6 × 103 m s−1
- (B) 1.1 × 104 m s−1
- (C) 3.1 × 103 m s−1
- (D) 7.9 × 103 m s−1
Show Answer
Answer: D – Per kilogram, the loss of potential energy is 6.25 × 107 − 3.13 × 107 = 3.12 × 107 J kg−1, and this becomes kinetic energy: ½v2 = 3.12 × 107, so v = √(2 × 3.12 × 107) = √(6.24 × 107) ≈ 7.9 × 103 m s−1. (Option B is the full escape velocity from the surface.)
Q16. The total energy (kinetic + potential) of a satellite in a closed orbit around the Earth is negative. This means that
- (A) the satellite is continually losing energy and must fall to Earth
- (B) the satellite is bound to the Earth, and extra energy must be supplied before it can escape to infinity
- (C) the kinetic energy of the satellite is negative
- (D) the potential energy of the satellite is positive
Show Answer
Answer: B – Kinetic energy is always positive, but Ep is negative and larger in magnitude. A negative total energy means the satellite cannot reach infinity (where the total would need to be at least zero) unless energy is supplied: it is a bound system.
Q17. A satellite of mass 100 kg is in a circular orbit of radius 1.28 × 107 m around the Earth, for which GM = 4.0 × 1014 m3 s−2. The total energy of the satellite is approximately
- (A) −1.6 × 109 J
- (B) −3.1 × 109 J
- (C) +1.6 × 109 J
- (D) zero
Show Answer
Answer: A – Ep = −GMm/r = −(4.0 × 1014 × 100) / (1.28 × 107) = −3.1 × 109 J. For a circular orbit, GMm/r2 = mv2/r, so KE = +GMm/(2r) = +1.6 × 109 J. Total E = −3.1 × 109 + 1.6 × 109 = −1.6 × 109 J, exactly half of Ep.
Q18. A satellite is raised from a low circular orbit to a higher circular orbit. In the higher orbit it moves more slowly, yet its total energy has
- (A) decreased, because its kinetic energy has decreased
- (B) stayed exactly the same
- (C) increased, because the gain in potential energy is twice the loss in kinetic energy
- (D) become positive for the first time
Show Answer
Answer: C – Total E = −GMm/(2r). As r increases, E becomes less negative, i.e. it increases. The potential energy rises by GMm(1/r1 − 1/r2), while the kinetic energy falls by only half of that amount, so energy must be supplied by the satellite’s motors.
Q19. Less fuel is needed to launch a rocket into orbit when it is launched eastwards from a site near the equator, mainly because
- (A) the value of g is zero at the equator
- (B) the Earth’s magnetic field gives the rocket an extra push
- (C) there is no atmosphere above the equator
- (D) the Earth’s rotation already gives the rocket a large eastward speed (about 470 m s−1 at the equator), so the rocket has to supply less additional kinetic energy
Show Answer
Answer: D – The launch site itself is moving eastwards with the Earth’s surface. At the equator this free speed is 2πR/T = 2π × 6.4 × 106 / 86400 ≈ 470 m s−1, and launching eastwards adds the rocket’s speed to it, saving fuel.
Q20. A space probe of mass 400 kg moves from a point where the gravitational potential is −4.0 × 107 J kg−1 to a point where it is −2.5 × 107 J kg−1. The increase in its potential energy is
- (A) 1.5 × 107 J
- (B) 6.0 × 109 J
- (C) −6.0 × 109 J
- (D) 1.0 × 1010 J
Show Answer
Answer: B – ΔEp = mΔV = 400 × [−2.5 × 107 − (−4.0 × 107)] = 400 × (1.5 × 107) = 6.0 × 109 J. The probe ends up further from the attracting mass, so its potential energy increases.
Escape Velocity & Potential Calculations
Q21. The escape velocity from the surface of a planet is
- (A) the minimum speed at which an object must be projected from the surface so that it can just reach infinity without any further propulsion
- (B) the speed of a satellite in the lowest possible orbit around the planet
- (C) the speed at which the weight of the object becomes zero
- (D) the maximum speed that a rocket engine can produce
Show Answer
Answer: A – At the escape velocity the object’s initial kinetic energy is just enough to overcome the negative potential energy, so it reaches infinity with (ideally) zero speed left. It is a minimum launch speed, and in practice air resistance is ignored.
Q22. By applying conservation of energy to an object of mass m projected from the surface of a planet of mass M and radius R, the escape velocity v is given by
- (A) v = √(GM/R)
- (B) v = GM/R2
- (C) v = √(2GM/R)
- (D) v = 2GM/R
Show Answer
Answer: C – The object just escapes if its kinetic energy equals the magnitude of its potential energy: ½mv2 = GMm/R, so v2 = 2GM/R and v = √(2GM/R). (Option A is the speed of a satellite in a low circular orbit.)
Q23. The gravitational potential at the Earth’s surface is −6.25 × 107 J kg−1. The escape velocity from the Earth’s surface is approximately
- (A) 7.9 × 103 m s−1
- (B) 1.1 × 104 m s−1
- (C) 6.3 × 103 m s−1
- (D) 2.2 × 104 m s−1
Show Answer
Answer: B – Each kilogram needs kinetic energy equal to |V| = 6.25 × 107 J to escape: ½v2 = 6.25 × 107, so v = √(2 × 6.25 × 107) = √(1.25 × 108) ≈ 1.1 × 104 m s−1 (about 11 km s−1).
Q24. A 1 kg object and a 1000 kg object are each projected from the Earth’s surface at their escape velocities. The escape velocity of the 1000 kg object is
- (A) 1000 times that of the 1 kg object
- (B) about 32 times that of the 1 kg object
- (C) smaller than that of the 1 kg object
- (D) exactly the same as that of the 1 kg object
Show Answer
Answer: D – In ½mv2 = GMm/R the object’s mass m cancels. Escape velocity depends only on the planet’s mass and radius, not on the mass of the escaping object (although the energy required, ½mv2, is of course much larger for the heavier object).
Q25. The escape velocity from a planet can also be written as v = √(2gR), where g is the surface field strength and R is the planet’s radius. For the Earth (g = 9.8 N kg−1, R = 6.4 × 106 m) this gives
- (A) 1.1 × 104 m s−1
- (B) 7.9 × 103 m s−1
- (C) 1.3 × 108 m s−1
- (D) 6.3 × 107 m s−1
Show Answer
Answer: A – v = √(2 × 9.8 × 6.4 × 106) = √(1.25 × 108) ≈ 1.1 × 104 m s−1. This agrees with √(2GM/R) because g = GM/R2, so 2gR = 2GM/R.
Q26. For the Moon, GM = 4.9 × 1012 m3 s−2 and the radius is 1.7 × 106 m. The escape velocity from the Moon’s surface is approximately
- (A) 1.7 × 103 m s−1
- (B) 2.9 × 106 m s−1
- (C) 2.4 × 103 m s−1
- (D) 5.8 × 106 m s−1
Show Answer
Answer: C – v = √(2GM/R) = √(2 × 4.9 × 1012 / 1.7 × 106) = √(5.8 × 106) ≈ 2.4 × 103 m s−1. This low escape velocity is why the Moon has no atmosphere: gas molecules can reach speeds comparable to it and escape.
Q27. At the surface of a planet the gravitational potential is −4.0 × 107 J kg−1. The escape velocity from this planet is approximately
- (A) 6.3 × 103 m s−1
- (B) 8.9 × 103 m s−1
- (C) 4.0 × 107 m s−1
- (D) 1.8 × 104 m s−1
Show Answer
Answer: B – vesc = √(2|V|) = √(2 × 4.0 × 107) = √(8.0 × 107) ≈ 8.9 × 103 m s−1. Forgetting the factor of 2 gives option A.
Q28. The field strength is related to the potential gradient by g = −dV/dr. The negative sign in this relation indicates that
- (A) gravitational potential is always a negative quantity
- (B) the field strength decreases as r increases
- (C) the potential gradient is always negative
- (D) the field acts in the direction in which the potential decreases, that is towards the attracting mass
Show Answer
Answer: D – As r increases, V increases (becomes less negative, heading for zero at infinity), so dV/dr is positive. The minus sign then gives g pointing towards smaller r, i.e. a mass is pulled in the direction of decreasing potential.
Q29. The gravitational potential changes from −6.25 × 107 J kg−1 at r = 6.4 × 106 m to −5.00 × 107 J kg−1 at r = 8.0 × 106 m from the Earth’s centre. The average field strength between these two points is approximately
- (A) 0.78 N kg−1
- (B) 78 N kg−1
- (C) 7.8 N kg−1
- (D) 9.8 N kg−1
Show Answer
Answer: C – |g| ≈ ΔV/Δr = (−5.00 × 107 + 6.25 × 107) / (8.0 × 106 − 6.4 × 106) = (1.25 × 107) / (1.6 × 106) ≈ 7.8 N kg−1. This is an average: g falls from 9.8 at the surface to about 6.3 at the outer point.
Q30. Planet X has twice the mass of planet Y but the same radius. The escape velocity from planet X is larger than that from planet Y by a factor of
- (A) √2
- (B) 2
- (C) 4
- (D) 1/√2
Show Answer
Answer: A – vesc = √(2GM/R) ∝ √M when R is fixed. Doubling M multiplies the escape velocity by √2 ≈ 1.41, not by 2.
Keep revising: Work through more A Level Physics practice on our Physics MCQs hub, and keep the key equations for gravitational fields and every other topic to hand on the A Level Physics Formula Sheet.
These are original practice questions written for revision purposes. They are not Cambridge past paper questions, and this page is not affiliated with or endorsed by Cambridge Assessment International Education.