A Level Physics 13.2 Gravitational Force Between Point Masses MCQs – 30 Practice Questions with Answers (Cambridge 9702)

This page covers Section 13.2 of the Cambridge International AS & A Level Physics (9702) syllabus, an A Level topic: Newton’s law of gravitation and the inverse-square force between point masses. Use G = 6.67 × 10−11 N m2 kg−2 throughout.

Formula recap: F = G M m / r2, where r is the separation of the centres of the two masses. See the full A Level Physics Formula Sheet for every equation used on this page.

Newton’s Law of Gravitation

Q1. Which statement correctly describes the gravitational force F between two point masses M and m separated by a distance r?

  • (A) F is proportional to the product of the masses and proportional to r2
  • (B) F is proportional to the product of the masses and inversely proportional to r2
  • (C) F is inversely proportional to the product of the masses and proportional to r2
  • (D) F is proportional to the sum of the masses and inversely proportional to r
Show Answer

Answer: B – Newton’s law of gravitation states F α M m / r2: double the product of the masses and the force doubles, but double the separation and the force falls to one quarter.

Q2. Which expression gives the gravitational force between two point masses M and m whose centres are a distance r apart?

  • (A) F = G M m / r2
  • (B) F = G M m / r
  • (C) F = G (M + m) / r2
  • (D) F = G M m r2
Show Answer

Answer: A – F = G M m / r2, where G = 6.67 × 10−11 N m2 kg−2 is the gravitational constant. The separation r appears squared in the denominator: this is an inverse-square law.

Q3. In the equation F = G M m / r2, applied to two solid spheres, the distance r is measured between

  • (A) the surfaces of the two spheres
  • (B) the centre of the larger sphere and the surface of the smaller sphere
  • (C) the centres of the two spheres
  • (D) the centres of the spheres, minus their radii
Show Answer

Answer: C – A uniform sphere attracts as if all its mass were concentrated at its centre, so r is always the centre-to-centre separation, never the gap between the surfaces.

Q4. The Earth pulls the Moon with a gravitational force of magnitude F. The size of the gravitational force with which the Moon pulls the Earth is

  • (A) smaller than F, because the Moon has less mass
  • (B) larger than F, because the Earth has more mass
  • (C) zero, because only the larger body attracts
  • (D) also F, acting in the opposite direction
Show Answer

Answer: D – The two forces form a Newton’s third law pair: equal in size, opposite in direction, acting on different bodies. The same equation F = G M m / r2 gives both forces, so they must be equal even though the masses differ.

Q5. The gravitational constant G has the value and unit

  • (A) 6.67 × 10−11 N kg2 m−2
  • (B) 6.67 × 10−11 N m2 kg−2
  • (C) 9.81 N kg−1
  • (D) 6.67 × 1011 N m2 kg−2
Show Answer

Answer: B – Rearranging F = G M m / r2 gives G = F r2 / (M m), so the unit of G is N m2 kg−2, and its value is 6.67 × 10−11 in these units.

Q6. Two students each of mass 50 kg stand 1.0 m apart. They do not slide towards each other mainly because

  • (A) the attractive force is extremely small, about 1.7 × 10−7 N, because G is so tiny
  • (B) gravitational force only acts between objects in space
  • (C) the two forces cancel because the students have equal masses
  • (D) air resistance holds them apart
Show Answer

Answer: A – F = G M m / r2 = 6.67 × 10−11 × 50 × 50 / 1.02 = 1.7 × 10−7 N. Gravity acts between all masses, but with G ≈ 10−11 the force between everyday objects is far too small to notice; friction easily prevents any movement.

Q7. When calculating the gravitational force between the Earth and a satellite, the Earth may be treated as a point mass located at

  • (A) the point on its surface nearest the satellite
  • (B) every point throughout its volume equally
  • (C) half a radius below its surface
  • (D) its centre
Show Answer

Answer: D – Outside a uniform (or spherically symmetric) sphere, the gravitational force is the same as if the whole mass were concentrated at the centre. That is why the satellite’s orbital radius is measured from the Earth’s centre, not from its surface.

Q8. The gravitational force between two masses is always

  • (A) repulsive when the masses are small
  • (B) attractive only when at least one mass is a planet
  • (C) attractive
  • (D) attractive at large separations and repulsive at small separations
Show Answer

Answer: C – Masses always attract each other. Unlike electric charges, there is no negative mass, so the gravitational force can never be repulsive.

Q9. Two point masses attract each other. Each force acts along

  • (A) a line perpendicular to the line joining the masses
  • (B) the line joining the two masses
  • (C) the vertical, towards the centre of the Earth
  • (D) the direction in which the masses are moving
Show Answer

Answer: B – Each mass is pulled directly towards the other, so both forces lie along the line joining the two centres.

Q10. Which single change would increase the gravitational force between two point masses?

  • (A) Decreasing the separation between them
  • (B) Decreasing one of the masses
  • (C) Increasing the separation between them
  • (D) Moving them to a region where G is smaller
Show Answer

Answer: A – Since F α 1 / r2, a smaller separation gives a larger force. G is a universal constant: it has the same value everywhere in the Universe.

Inverse-Square Calculations

Q11. Two point masses of 1.0 kg each are placed with their centres 1.0 m apart. The gravitational force between them is

  • (A) 1.3 × 10−10 N
  • (B) 6.67 N
  • (C) 6.67 × 10−11 N
  • (D) 6.67 × 10−9 N
Show Answer

Answer: C – F = G M m / r2 = 6.67 × 10−11 × 1.0 × 1.0 / 1.02 = 6.67 × 10−11 N. This tiny value for kilogram masses one metre apart shows how weak gravity is on a human scale.

Q12. Two point masses of 2.0 kg each are placed with their centres 1.0 m apart. The gravitational force between them is

  • (A) 6.67 × 10−11 N
  • (B) 1.33 × 10−10 N
  • (C) 5.34 × 10−10 N
  • (D) 2.67 × 10−10 N
Show Answer

Answer: D – F = G M m / r2 = 6.67 × 10−11 × 2.0 × 2.0 / 1.02 = 4 × 6.67 × 10−11 = 2.67 × 10−10 N.

Q13. A point mass of 4.0 kg and a point mass of 5.0 kg are separated by 2.0 m between their centres. The gravitational force between them is

  • (A) 3.34 × 10−10 N
  • (B) 1.33 × 10−9 N
  • (C) 8.34 × 10−11 N
  • (D) 6.67 × 10−10 N
Show Answer

Answer: A – F = G M m / r2 = 6.67 × 10−11 × 4.0 × 5.0 / 2.02 = 6.67 × 10−11 × 20 / 4 = 5 × 6.67 × 10−11 = 3.34 × 10−10 N. Option B is the common slip of forgetting to divide by r2.

Q14. The gravitational force between two point masses is F. One of the masses is doubled while the other mass and the separation stay the same. The new force is

  • (A) F / 2
  • (B) 2F
  • (C) 4F
  • (D) F
Show Answer

Answer: B – F α M m, so doubling one mass doubles the product M m and the force becomes 2F.

Q15. The gravitational force between two point masses is F. Both masses are doubled and the separation is unchanged. The new force is

  • (A) 2F
  • (B) F
  • (C) 8F
  • (D) 4F
Show Answer

Answer: D – The product M m becomes (2M)(2m) = 4Mm, so the force is multiplied by 4, giving 4F.

Q16. The gravitational force between two point masses is F. The separation between them is doubled, with the masses unchanged. The new force is

  • (A) F / 4
  • (B) F / 2
  • (C) 4F
  • (D) F / 16
Show Answer

Answer: A – F α 1 / r2: replacing r by 2r multiplies the force by 1 / 22 = 1/4, so the new force is F / 4. This is the inverse-square behaviour.

Q17. Two point masses attract each other with force F when their separation is r. The force when the separation is increased to 3r is

  • (A) F / 3
  • (B) F / 6
  • (C) F / 9
  • (D) F / 27
Show Answer

Answer: C – The force is multiplied by 1 / 32 = 1/9, giving F / 9.

Q18. Two point masses attract each other with force F. Their separation is halved. The new force is

  • (A) 2F
  • (B) 4F
  • (C) F / 4
  • (D) 16F
Show Answer

Answer: B – Replacing r by r / 2 multiplies the force by 1 / (1/2)2 = 4, giving 4F.

Q19. Two point masses, separation d, attract each other with force F. At what separation would the force between them be F / 16?

  • (A) 2d
  • (B) 16d
  • (C) d / 4
  • (D) 4d
Show Answer

Answer: D – The force must fall by a factor of 16 = 42, so by the inverse-square law the separation must increase by a factor of 4, to 4d.

Q20. A satellite of mass 500 kg orbits with its centre 7.0 × 106 m from the centre of the Earth, mass 6.0 × 1024 kg. The gravitational force between the Earth and the satellite is about

  • (A) 4.1 × 101 N
  • (B) 4.1 × 105 N
  • (C) 4.1 × 103 N
  • (D) 2.0 × 1017 N
Show Answer

Answer: C – F = G M m / r2 = 6.67 × 10−11 × 6.0 × 1024 × 500 / (7.0 × 106)2 = 2.0 × 1017 / 4.9 × 1013 = 4.1 × 103 N. Option D is the numerator before dividing by r2 – the classic slip in this calculation.


Comparing & Combining Gravitational Forces

Q21. A satellite is pulled towards a planet of mass M by a force F when its distance from the planet’s centre is r. The force on the same satellite at distance 2r from the centre of a planet of mass 2M would be

  • (A) F / 2
  • (B) F
  • (C) 2F
  • (D) F / 4
Show Answer

Answer: A – Work with ratios so G and the satellite mass cancel: F’ / F = (2M / M) × (r / 2r)2 = 2 × 1/4 = 1/2, so F’ = F / 2.

Q22. Two point masses attract each other with force F. Both masses are doubled AND the separation between them is halved. The new force is

  • (A) 4F
  • (B) 16F
  • (C) 8F
  • (D) F
Show Answer

Answer: B – Doubling both masses multiplies F by 4; halving the separation multiplies F by 1 / (1/2)2 = 4. Combined change: 4 × 4 = 16, so the new force is 16F.

Q23. A mass of 8.0 kg and a mass of 2.0 kg are fixed 4.0 m apart (centre to centre). A 1.0 kg mass is placed midway between them. The resultant gravitational force on the 1.0 kg mass is

  • (A) 1.67 × 10−10 N towards the 2.0 kg mass
  • (B) 1.33 × 10−10 N towards the 8.0 kg mass
  • (C) 1.00 × 10−10 N towards the 2.0 kg mass
  • (D) 1.00 × 10−10 N towards the 8.0 kg mass
Show Answer

Answer: D – The 1.0 kg mass is 2.0 m from each fixed mass. Pull towards 8.0 kg: G × 8.0 × 1.0 / 2.02 = 2G. Pull towards 2.0 kg: G × 2.0 × 1.0 / 2.02 = 0.5G. The forces act in opposite directions, so the resultant is 2G − 0.5G = 1.5G = 1.5 × 6.67 × 10−11 = 1.00 × 10−10 N, towards the larger (8.0 kg) mass.

Q24. Two point masses, 9M and M, are fixed a distance 4.0 m apart. At the null point between them the resultant gravitational force on a test mass is zero. The null point is at a distance from the mass 9M of

  • (A) 1.0 m
  • (B) 2.0 m
  • (C) 3.0 m
  • (D) 3.6 m
Show Answer

Answer: C – At the null point the two forces balance: G × 9M / x2 = G × M / (4.0 − x)2. Taking square roots: 3 / x = 1 / (4.0 − x), so 3(4.0 − x) = x, giving x = 3.0 m. The null point lies closer to the smaller mass, as it must: the smaller mass needs a smaller distance to pull equally hard.

Q25. Three point masses lie in a straight line: 1.0 kg, then 2.0 kg at a distance of 1.0 m from it, then 3.0 kg at a further 2.0 m from the 2.0 kg mass. The resultant gravitational force on the 2.0 kg mass is

  • (A) 3.3 × 10−11 N towards the 1.0 kg mass
  • (B) 3.3 × 10−11 N towards the 3.0 kg mass
  • (C) 2.3 × 10−10 N towards the 1.0 kg mass
  • (D) zero, because the two pulls cancel
Show Answer

Answer: A – Pull from the 1.0 kg mass: G × 2.0 × 1.0 / 1.02 = 2G towards the 1.0 kg mass. Pull from the 3.0 kg mass: G × 2.0 × 3.0 / 2.02 = 1.5G towards the 3.0 kg mass. Resultant = 2G − 1.5G = 0.5G = 0.5 × 6.67 × 10−11 = 3.3 × 10−11 N towards the 1.0 kg mass. Option C wrongly adds the two forces.

Q26. Two identical satellites orbit the same planet, one at a distance 2R from the planet’s centre and the other at 4R. The gravitational force on the nearer satellite is how many times the force on the farther one?

  • (A) 2 times
  • (B) 4 times
  • (C) 8 times
  • (D) 16 times
Show Answer

Answer: B – With equal satellite masses and the same planet, F α 1 / r2, so Fnear / Ffar = (4R / 2R)2 = 22 = 4.

Q27. A test mass is placed on the line joining two fixed masses of very different sizes. The point where the resultant gravitational force on the test mass is zero is

  • (A) always exactly midway between the two masses
  • (B) closer to the larger mass
  • (C) at the centre of the larger mass
  • (D) closer to the smaller mass
Show Answer

Answer: D – The pulls balance where G M1 / r12 = G M2 / r22. The smaller mass can only match the larger mass’s pull from a shorter distance, so the null point is always closer to the smaller mass (midway only when the masses are equal).

Q28. Three point masses lie in a straight line: 8.0 kg, then 2.0 kg at 1.0 m from it, then 18 kg at a further 2.0 m from the 2.0 kg mass. The resultant gravitational force on the 2.0 kg mass is

  • (A) 1.67 × 10−9 N towards the 8.0 kg mass
  • (B) 4.67 × 10−10 N towards the 18 kg mass
  • (C) 4.67 × 10−10 N towards the 8.0 kg mass
  • (D) 1.07 × 10−9 N towards the 8.0 kg mass
Show Answer

Answer: C – Pull from the 8.0 kg mass: G × 2.0 × 8.0 / 1.02 = 16G towards the 8.0 kg mass. Pull from the 18 kg mass: G × 2.0 × 18 / 2.02 = 36G / 4 = 9G towards the 18 kg mass. Resultant = 16G − 9G = 7G = 7 × 6.67 × 10−11 = 4.67 × 10−10 N towards the 8.0 kg mass.

Q29. A moon of a planet P, mass 3M, orbits at a distance 2d from the planet’s centre and is pulled by a force FP. A moon of a planet Q, mass M, orbits at distance d and is pulled by a force FQ. The two moons have the same mass. Then FP is equal to

  • (A) 0.75 FQ
  • (B) 1.5 FQ
  • (C) 3 FQ
  • (D) 6 FQ
Show Answer

Answer: A – G and the moon’s mass cancel in the ratio: FP / FQ = (3M / M) × (d / 2d)2 = 3 × 1/4 = 0.75, so FP = 0.75 FQ. Planet P’s triple mass does not compensate for the doubled distance.

Q30. Two equal point masses M, separated by a distance d, attract each other with force F. A third point mass M is now placed exactly midway between them. The resultant gravitational force on this third mass is

  • (A) 4F towards the nearer mass
  • (B) zero
  • (C) 2F towards one of the masses
  • (D) F / 2 towards one of the masses
Show Answer

Answer: B – The third mass is the same distance (d / 2) from two equal masses, so it is pulled equally hard in opposite directions – each pull is G M2 / (d / 2)2 = 4F. The two pulls cancel, so the resultant force is zero: the midpoint of two equal masses is a null point.

Keep practising: work through the full topic list on the Physics MCQs hub, and keep the A Level Physics Formula Sheet open while you revise – F = G M m / r2 and the rest of the gravitational fields equations are all collected there.

These are original practice questions written for QuaidianPoint; they are not Cambridge past-paper questions. QuaidianPoint is not affiliated with Cambridge Assessment International Education.