A Level Physics 13.3 Gravitational Field of a Point Mass MCQs – 30 Practice Questions with Answers (Cambridge 9702)

This page covers Cambridge AS & A Level Physics (9702) Section 13.3, Gravitational Field of a Point Mass – an A Level (A2) topic. Practise using g = GM/r2 to find field strengths, work out g for planets and satellites, and predict how g changes with mass and distance.

Formula recap: g = GM/r2, where r is measured from the centre of the mass M. Rearranged: M = gr2/G. So g is proportional to M and proportional to 1/r2. At the surface of a uniform sphere, g = (4/3)πGρR. See the full A Level Physics formula sheet for every equation used on this page.

Field Strength of a Point Mass

Q1. The force between a point mass M and a test mass m is F = GMm/r2. Using g = F/m, which expression gives the gravitational field strength at distance r from M?

  • (A) g = GMm/r2
  • (B) g = GM/r2
  • (C) g = Gm/r2
  • (D) g = GM/r
Show Answer

Answer: B – g = F/m = (GMm/r2)/m. The test mass m cancels, giving g = GM/r2. Option A is the force F, not the field strength.

Q2. In the equation g = GM/r2, the distance r must be measured from

  • (A) the surface of the mass M
  • (B) the centre of the test mass
  • (C) the centre of the mass M
  • (D) the nearest point of the mass M
Show Answer

Answer: C – r is the separation of the centres of the two masses. For a spherical planet, r is measured from the planet’s centre, so r = (planet radius + height above the surface).

Q3. A point mass has a mass of 1.0 × 1020 kg. What is the gravitational field strength at a point 1.0 × 105 m from its centre? (G = 6.67 × 10-11 N m2 kg-2)

  • (A) 0.67 N kg-1
  • (B) 6.7 N kg-1
  • (C) 0.067 N kg-1
  • (D) 6.7 × 105 N kg-1
Show Answer

Answer: A – g = GM/r2 = (6.67 × 10-11 × 1.0 × 1020) / (1.0 × 105)2 = 6.67 × 109 / 1.0 × 1010 = 0.67 N kg-1.

Q4. A planet is a uniform sphere. For a point outside the planet, its gravitational field is the same as that of

  • (A) a ring of the planet’s mass around its equator
  • (B) a point mass spread evenly throughout its volume
  • (C) a point mass placed at the point of its surface nearest to the point considered
  • (D) a point mass equal to the planet’s mass placed at its centre
Show Answer

Answer: D – Outside a spherical mass, the field is identical to that of a point mass of the same mass at the sphere’s centre, which is why r in GM/r2 is measured from the centre.

Q5. Which statement about the direction of the gravitational field strength g is correct?

  • (A) g points away from the mass creating the field
  • (B) g points towards the centre of the mass creating the field
  • (C) g points in the direction in which a mass placed there is already moving
  • (D) g has no direction because it is a scalar quantity
Show Answer

Answer: B – Gravity is always attractive, so the force on a test mass, and therefore g, points towards the centre of the mass creating the field. g is a vector quantity.

Q6. The gravitational field strength at a point does NOT depend on

  • (A) the mass creating the field
  • (B) the distance from the centre of that mass
  • (C) the mass of the object placed at the point
  • (D) the gravitational constant G
Show Answer

Answer: C – g = GM/r2 contains only G, M and r. A larger test mass experiences a proportionally larger force F, so the ratio F/m, and hence g, is unchanged.

Q7. A point mass of 4.0 × 1024 kg creates a gravitational field. What is the field strength at a distance of 2.0 × 106 m from its centre?

  • (A) 6.7 N kg-1
  • (B) 670 N kg-1
  • (C) 0.67 N kg-1
  • (D) 67 N kg-1
Show Answer

Answer: D – g = GM/r2 = (6.67 × 10-11 × 4.0 × 1024) / (2.0 × 106)2 = 2.668 × 1014 / 4.0 × 1012 = 66.7, so about 67 N kg-1.

Q8. Gravitational field strength at a point is best defined as

  • (A) the force per unit mass on a small mass placed at the point
  • (B) the force on any mass placed at the point
  • (C) the mass per unit volume at the point
  • (D) the work done per unit mass at the point
Show Answer

Answer: A – g = F/m by definition: the force experienced per unit mass. Option D describes gravitational potential, not field strength.

Q9. A small moon has a mass of 2.7 × 1023 kg. What is the gravitational field strength at a point 3.0 × 106 m from its centre?

  • (A) 0.20 N kg-1
  • (B) 20 N kg-1
  • (C) 2.0 N kg-1
  • (D) 6.0 N kg-1
Show Answer

Answer: C – g = GM/r2 = (6.67 × 10-11 × 2.7 × 1023) / (3.0 × 106)2 = 1.80 × 1013 / 9.0 × 1012 = 2.0 N kg-1.

Q10. At a point in the field of a planet, the field strength is 2.5 N kg-1. What is the gravitational force on a 40 kg rock placed at that point?

  • (A) 16 N
  • (B) 100 N
  • (C) 0.063 N
  • (D) 1000 N
Show Answer

Answer: B – Since g = F/m, the force is F = mg = 40 × 2.5 = 100 N.

Calculating g for Planets & Satellites

Q11. Earth has a mass of 6.0 × 1024 kg and a radius of 6.4 × 106 m. The gravitational field strength at Earth’s surface is approximately

  • (A) 9.8 N kg-1
  • (B) 0.98 N kg-1
  • (C) 98 N kg-1
  • (D) 1.6 N kg-1
Show Answer

Answer: A – g = GM/R2 = (6.67 × 10-11 × 6.0 × 1024) / (6.4 × 106)2 = 4.00 × 1014 / 4.10 × 1013 = 9.8 N kg-1.

Q12. The field strength at Earth’s surface is 9.81 N kg-1. A satellite is at a height above the surface equal to Earth’s radius R. What is the field strength at the satellite?

  • (A) 9.81 N kg-1
  • (B) 4.9 N kg-1
  • (C) 2.45 N kg-1
  • (D) 1.23 N kg-1
Show Answer

Answer: C – The distance from Earth’s centre is r = R + h = R + R = 2R. Since g follows an inverse-square law, g = 9.81 / 22 = 9.81 / 4 = 2.45 N kg-1. (Simply halving, option B, is the common trap.)

Q13. The surface field strength of a planet of radius R is 9.81 N kg-1. A probe is at a height of 2R above the surface. What is the field strength at the probe?

  • (A) 3.3 N kg-1
  • (B) 2.45 N kg-1
  • (C) 4.9 N kg-1
  • (D) 1.1 N kg-1
Show Answer

Answer: D – The distance from the centre is r = R + 2R = 3R, so g = 9.81 / 32 = 9.81 / 9 = 1.09, about 1.1 N kg-1. Dividing 9.81 by 3 (option A) forgets to square.

Q14. A planet has a mass of 6.4 × 1023 kg and a radius of 3.4 × 106 m. What is the field strength at its surface?

  • (A) 37 N kg-1
  • (B) 3.7 N kg-1
  • (C) 0.37 N kg-1
  • (D) 13 N kg-1
Show Answer

Answer: B – g = GM/R2 = (6.67 × 10-11 × 6.4 × 1023) / (3.4 × 106)2 = 4.27 × 1013 / 1.156 × 1013 = 3.7 N kg-1.

Q15. The surface field strength of a planet is 10 N kg-1 and its radius is 5.0 × 106 m. The mass of the planet is approximately

  • (A) 3.7 × 1023 kg
  • (B) 3.7 × 1025 kg
  • (C) 7.5 × 1024 kg
  • (D) 3.7 × 1024 kg
Show Answer

Answer: D – Rearranging g = GM/R2 gives M = gR2/G = 10 × (5.0 × 106)2 / (6.67 × 10-11) = 2.5 × 1014 / 6.67 × 10-11 = 3.7 × 1024 kg.

Q16. Planet X has half the mass of Earth and half the radius of Earth. Earth’s surface field strength is 9.8 N kg-1. What is the surface field strength of planet X?

  • (A) 20 N kg-1
  • (B) 9.8 N kg-1
  • (C) 4.9 N kg-1
  • (D) 39 N kg-1
Show Answer

Answer: A – Using ratios, G cancels: gX/gE = (MX/ME) / (RX/RE)2 = 0.5 / 0.25 = 2, so gX = 2 × 9.8 = 19.6, about 20 N kg-1.

Q17. At the surface of a planet, the field strength is g0. At a distance of 4R from the planet’s centre, where R is the planet’s radius, the field strength is

  • (A) g0/4
  • (B) g0/8
  • (C) g0/16
  • (D) g0/2
Show Answer

Answer: C – The distance from the centre has increased by a factor of 4, so by the inverse-square law g falls by a factor of 42 = 16, giving g0/16.

Q18. A satellite is in orbit at a height of 3.6 × 107 m above Earth’s surface. Earth has a mass of 6.0 × 1024 kg and a radius of 6.4 × 106 m. What is the field strength at the satellite?

  • (A) 0.31 N kg-1
  • (B) 0.22 N kg-1
  • (C) 2.2 N kg-1
  • (D) 0.022 N kg-1
Show Answer

Answer: B – The distance from Earth’s centre is r = 3.6 × 107 + 6.4 × 106 = 4.24 × 107 m. Then g = GM/r2 = 4.00 × 1014 / (4.24 × 107)2 = 4.00 × 1014 / 1.80 × 1015 = 0.22 N kg-1. Option A results from wrongly using the height alone as r.

Q19. A planet has a mass of 1.0 × 1024 kg and a radius of 5.0 × 106 m. What is the field strength at its surface?

  • (A) 0.27 N kg-1
  • (B) 27 N kg-1
  • (C) 270 N kg-1
  • (D) 2.7 N kg-1
Show Answer

Answer: D – g = GM/R2 = (6.67 × 10-11 × 1.0 × 1024) / (5.0 × 106)2 = 6.67 × 1013 / 2.5 × 1013 = 2.7 N kg-1.

Q20. Planet A has a surface field strength of 5.0 N kg-1. Planet B has 8 times the mass of planet A and twice the radius of planet A. What is the surface field strength of planet B?

  • (A) 2.5 N kg-1
  • (B) 40 N kg-1
  • (C) 10 N kg-1
  • (D) 20 N kg-1
Show Answer

Answer: C – By the ratio method (G cancels): gB/gA = 8 / 22 = 8/4 = 2, so gB = 2 × 5.0 = 10 N kg-1.


How g Varies

Q21. A spacecraft moves from a distance r to a distance 2r from the centre of a planet. The gravitational field strength

  • (A) becomes one quarter of its original value
  • (B) becomes one half of its original value
  • (C) becomes twice its original value
  • (D) becomes one eighth of its original value
Show Answer

Answer: A – g is proportional to 1/r2. Doubling r multiplies g by 1/22 = 1/4.

Q22. The distance from the centre of a mass is increased by a factor of 3. The field strength becomes

  • (A) one third of its original value
  • (B) one ninth of its original value
  • (C) one twenty-seventh of its original value
  • (D) one sixth of its original value
Show Answer

Answer: B – By the inverse-square law, multiplying r by 3 divides g by 32 = 9.

Q23. The mass of a planet is doubled while its radius stays the same. The surface field strength

  • (A) stays the same
  • (B) becomes one quarter of its original value
  • (C) becomes one half of its original value
  • (D) becomes twice its original value
Show Answer

Answer: D – g = GM/R2 is directly proportional to M when R is fixed, so doubling the mass doubles g.

Q24. Planet Y has twice the mass and twice the radius of planet Z. Compared with the surface field strength of planet Z, the surface field strength of planet Y is

  • (A) the same
  • (B) twice as large
  • (C) half as large
  • (D) one quarter as large
Show Answer

Answer: C – g is proportional to M/R2, so the ratio is 2 / 22 = 2/4 = 1/2. The doubled radius has the bigger effect because it is squared, so g is halved – a common examination trap.

Q25. Which change to a planet leaves its surface field strength unchanged?

  • (A) mass × 4 and radius × 2
  • (B) mass × 2 and radius × 2
  • (C) mass × 4 and radius × 4
  • (D) mass × 2 and radius × 4
Show Answer

Answer: A – The multiplying factor is (mass factor)/(radius factor)2 = 4 / 22 = 4/4 = 1, so g is unchanged. For comparison: B gives 2/4 = 1/2, C gives 4/16 = 1/4 and D gives 2/16 = 1/8.

Q26. For a uniform sphere, the surface field strength can be written as g = (4/3)πGρR, where ρ is the density of the sphere. Two planets have the same density, and planet Q has twice the radius of planet P. The surface field strength of Q compared with P is

  • (A) the same
  • (B) twice as large
  • (C) four times as large
  • (D) half as large
Show Answer

Answer: B – At fixed density, g is proportional to R (this follows from M = (4/3)πR3ρ substituted into g = GM/R2). Doubling R therefore doubles the surface g.

Q27. A uniform planet has a density of 5.0 × 103 kg m-3 and a radius of 6.0 × 106 m. Using g = (4/3)πGρR, the surface field strength is approximately

  • (A) 0.84 N kg-1
  • (B) 84 N kg-1
  • (C) 4.2 N kg-1
  • (D) 8.4 N kg-1
Show Answer

Answer: D – g = (4/3) × π × 6.67 × 10-11 × 5.0 × 103 × 6.0 × 106 = 4.19 × 6.67 × 10-11 × 3.0 × 1010 = 8.4 N kg-1.

Q28. A graph of field strength g against distance r is plotted for points outside a planet, starting at its surface. Which description of the graph is correct?

  • (A) g decreases in a straight line to zero
  • (B) g stays constant and then falls suddenly to zero
  • (C) g decreases in proportion to 1/r2, steeply at first and then more gradually
  • (D) g increases in proportion to r2
Show Answer

Answer: C – Outside the planet g = GM/r2, an inverse-square curve: it falls quickly just above the surface and levels off towards (but never below) zero at large distances.

Q29. Inside a planet of uniform density, the field strength g varies with distance r from the centre as follows:

  • (A) g increases linearly from zero at the centre to a maximum at the surface
  • (B) g is constant throughout the planet
  • (C) g decreases following 1/r2 all the way to the centre
  • (D) g is zero everywhere inside the planet
Show Answer

Answer: A – At depth, only the mass inside radius r contributes, and that mass is proportional to r3, so g is proportional to r3/r2 = r. Hence g rises linearly from zero at the centre to its maximum at the surface, where it joins the 1/r2 curve outside.

Q30. Earth has a mass of 6.0 × 1024 kg. At what distance from Earth’s centre is the field strength 0.20 N kg-1 (roughly the value at geostationary orbit height)?

  • (A) 4.5 × 106 m
  • (B) 4.5 × 107 m
  • (C) 2.0 × 1015 m
  • (D) 4.5 × 108 m
Show Answer

Answer: B – Rearranging g = GM/r2 gives r = √(GM/g) = √[(6.67 × 10-11 × 6.0 × 1024) / 0.20] = √(2.0 × 1015) = 4.5 × 107 m. Option C is r2 before taking the square root.

Keep practising: find more topics on the Physics MCQs hub, and keep the A Level Physics formula sheet open while you revise gravitation.

These are original practice questions written for revision. They are not Cambridge past-paper questions, and this page is not affiliated with Cambridge Assessment International Education.