A Level Physics (A2) chapter 2: Gravitational fields MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
96 exam-style questions on Gravitational fields, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Gravitational force between point masses24 questions
Gravitational field of a point mass24 questions
Gravitational potential24 questions
Ten Gravitational fields questions with answers
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. Mars has 0.11 times the mass of the Earth and 0.53 times its radius. The field strength at the Earth’s surface is 9.81 N kg−1. What is the weight of a 60 kg rover on the surface of Mars?
A
65 N
B
120 N
C
230 N
D
1500 N
Show answer
Answer: C. gMars = 9.81 × 0.11/0.532 = 3.8 N kg−1, so W = 60 × 3.84 = 230 N. 120 N forgets to square the radius ratio, and 65 N leaves out the radius altogether.
2. Point P lies on the line between the Earth and the Moon at the place where the resultant gravitational field strength is zero. Which statement about the gravitational potential at P is correct?
A
it is zero, because the two fields cancel at P
B
it is equal to the potential due to the Earth alone, because the Moon’s contribution cancels
C
it is positive, because work must be done to move a mass from P to either body
D
it is negative, because the potentials due to the Earth and the Moon are both negative and add as scalars
Show answer
Answer: D. Potential is a scalar; each body gives a negative contribution −GM/r, so the total at P is negative. Zero field means zero potential gradient at P, not zero potential.
3. A satellite orbits the Earth at a height of 1.363 × 107 m above the surface. The Earth has mass 5.97 × 1024 kg and radius 6.37 × 106 m. What is the orbital speed?
A
4.5 × 103 m s−1
B
5.4 × 103 m s−1
C
6.3 × 103 m s−1
D
2.0 × 107 m s−1
Show answer
Answer: A. r = 1.363 × 107 + 6.37 × 106 = 2.0 × 107 m and v = √(GM/r) = 4.5 × 103 m s−1. 5.4 × 103 uses the height as the orbital radius, and 2.0 × 107 forgets the square root.
4. Why can g be treated as constant for an object that rises a few hundred metres above the Earth's surface?
A
the mass of the Earth is concentrated in a thin layer close to its surface
B
the field lines become closer together with height, which cancels the effect of distance
C
the rise is tiny compared with the Earth's radius, so r, and so GM/r2, barely changes
D
g depends only on the mass of the object, which does not change as it rises
Show answer
Answer: C. In g = GM/r2, r is measured from the Earth's centre, about 6.4 × 106 m away. A rise of a few hundred metres changes r by only about 0.01%, so the change in g is negligible and the field lines there are almost parallel.
5. The radius of the Earth is 6.37 × 106 m. By what percentage is g at the top of a mountain 8.8 km high less than g at sea level?
A
0.14%
B
2.8%
C
28%
D
0.28%
Show answer
Answer: D. g ∝ 1/r2: gtop/gsea = (6.370/6.3788)2 = 0.9972, a fall of 0.28%. Taking g ∝ 1/r would give half this. Such a small change is why g is treated as constant near the surface.
6. To derive the field strength at distance r from a point mass M, a small test mass m is placed there. Which expression is the force on m, before it is divided by m?
A
GM/r
B
Gm/r2
C
GMm/r
D
GMm/r2
Show answer
Answer: D. Newton’s law gives F = GMm/r2; dividing by m gives g = GM/r2. GMm/r is the size of the potential energy, not a force.
7. Two stars, of masses M and 4M, have their centres a distance d apart. What is the resultant gravitational field strength at the point half-way between them?
A
3GM/d2 towards the star of mass 4M
B
12GM/d2 towards the star of mass M
C
12GM/d2 towards the star of mass 4M
D
20GM/d2 towards the star of mass 4M
Show answer
Answer: C. Each star is d/2 away. The fields are opposite: G(4M)/(d/2)2 − GM/(d/2)2 = 16GM/d2 − 4GM/d2 = 12GM/d2, towards the larger mass. 20GM/d2 adds the fields, and 3GM/d2 uses d instead of d/2.
8. A mass m moves from a distance r1 to a greater distance r2 from the centre of a planet of mass M. What is the change in its gravitational potential energy?
A
a decrease of GMm(1/r1 − 1/r2)
B
an increase of GMm(1/r1 − 1/r2)
C
an increase of GMm(1/r12 − 1/r22)
D
an increase of GMm/(r2 − r1)
Show answer
Answer: B. EP = −GMm/r, so ΔEP = −GMm/r2 − (−GMm/r1) = GMm(1/r1 − 1/r2), which is positive: moving away from a planet raises the potential energy. The 1/r2 form belongs to force, not energy.
9. A space station orbits 400 km above the Earth, whose radius is 6.37 × 106 m. The field strength at the surface is 9.81 N kg−1. What is the field strength at the station?
A
8.7 N kg−1
B
9.2 N kg−1
C
9.8 N kg−1
D
2.5 × 103 N kg−1
Show answer
Answer: A. g = 9.81 × (6.37/6.77)2 = 8.7 N kg−1, using r = 6.37 × 106 + 4.0 × 105 m. 9.2 forgets to square the ratio, and 2.5 × 103 uses the 400 km height as r in GM/r2.
10. The gravitational field strength at the surface of a planet of radius 3.0 × 106 m is 4.5 N kg−1. What is the mass of the planet?
A
6.1 × 1023 kg
B
2.0 × 1017 kg
C
2.4 × 1024 kg
D
1.5 × 1023 kg
Show answer
Answer: A. M = gr2/G = 4.5 × (3.0 × 106)2/6.67 × 10−11 = 6.1 × 1023 kg. Not squaring r gives 2.0 × 1017 kg; using the diameter for r gives 2.4 × 1024 kg.
96, each with the answer and a short explanation, across 4 topics: Gravitational field, Gravitational force between point masses, Gravitational field of a point mass, Gravitational potential. Every chapter test draws ten of them, so you can take it many times.
Is the A Level Physics (A2) chapter 2 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
