Gravitational fields: A Level Physics questions with answers
Free Gravitational fields questions for A Level Physics (9702), part of Fields & motion. Each question has the answer and the marking points an examiner would look for.
The Earth has mass 5.98 × 1024 kg and radius 6.37 × 106 m. A satellite is in a geostationary orbit.
(a)
State two features of a geostationary orbit, other than its period.
[2]
(b)
Show that the radius of a geostationary orbit is about 4.2 × 107 m.
[3]
(c)
A satellite of mass 2000 kg is moved from a low orbit of radius 6.97 × 106 m to geostationary orbit. Calculate the change in gravitational potential energy.
[3]
(d)
Explain why the satellite’s kinetic energy is less in the geostationary orbit.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
equatorial orbit; moves west to east (same direction as Earth’s rotation)
above the equator B1 same direction of rotation as the Earth B1
2
(b)
r3 = GMT2 ÷ 4π2; r = 4.23 × 107 m
GMm/r2 = mω2r C1 ω = 2π ÷ 86 400 C1 r = 4.23 × 107 m A1
v2 = GM/r so speed decreases with radius; KE = GMm/2r is smaller
v2 = GM ÷ r B1 larger r, smaller speed and KE B1
2
Question 2Structured · 8 marksA* challenge
A satellite of mass 500 kg is placed in a synchronous orbit around Mars. Mars has mass 6.42 × 1023 kg, radius 3.39 × 106 m and rotates once in 88600 s.
(a)
State two features of a synchronous (geostationary) orbit.
[2]
(b)
Show that the radius of the orbit is about 2.0 × 107 m.
[3]
(c)
Calculate the gravitational potential at the orbit.
[1]
(d)
Calculate the change in gravitational potential energy of the satellite when it is moved from the surface to the orbit.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
period = rotation period; above equator, same direction
period equal to the planet’s rotation period B1 orbit above the equator moving in the same direction of rotation B1
2
(b)
2.04 × 107
GMm/r2 = mω2r with ω = 2π/T C1 r3 = GMT2/4π2C1 r = 2.04 × 107 m A1
3
(c)
−2.10 × 106 J kg−1
φ = −GM/r = −2.10 × 106 J kg−1B1
1
(d)
5.27 × 109 J
ΔE = GMm(1/R − 1/r) C1 = 5.27 × 109 J A1
2
Question 3Structured · 6 marks
A satellite orbits a planet of mass 6.0 × 1024 kg in a circular orbit of radius 7.4 × 106 m.
(a)
Show that T2 = 4π2r3GM.
[2]
(b)
Calculate the orbital period.
[2]
(c)
State two features of a geostationary orbit.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
gravitational force = centripetal force
GMm/r2 = mrω2M1 ω = 2π/T leads to result A1
2
(b)
6320 s
substitution C1 6320 s A1
2
(c)
period 24 h; equatorial; west to east
Any 2 of: period 24 hours; above the equator; moves west to east (same direction as Earth rotates) B2
2
Question 4Structured · 10 marks
A geostationary satellite orbits the Earth. The mass of the Earth is 5.97 × 1024 kg and its radius is 6.37 × 106 m.
(a)
State three features of a geostationary orbit.
[3]
(b)
Define gravitational potential at a point.
[2]
(c)
Calculate the minimum energy needed to move a 1200 kg satellite from the Earth’s surface to the geostationary orbit, ignoring the Earth’s rotation and the satellite’s orbital kinetic energy.
[3]
(d)
Explain why gravitational potential is always negative.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
period 24 h; equatorial; same direction as Earth rotates
period 24 hours (one day) B1 orbit above the equator B1 moves west to east (same direction as Earth’s rotation) B1
3
(b)
work done per unit mass bringing a small mass from infinity to the point
work done per unit mass B1 in bringing a small test mass from infinity to the point B1
zero at infinity; work done by field as mass approaches, so energy decreases
potential is defined as zero at infinity B1 the gravitational force is attractive so work is done by the field as a mass moves in, so potential decreases below zero B1
2
Question 5Structured · 9 marks
Mercury has mass 3.30 × 1023 kg and radius 2.44 × 106 m.
(a)
State Newton’s law of gravitation.
[2]
(b)
Calculate the gravitational field strength at the surface.
[2]
(c)
Show that the escape speed from the surface is given by v = √(2GM/R), and calculate it.
[3]
(d)
Sketch a graph showing how gravitational potential varies with distance from the centre, from the surface outwards.
Fig. 5.1
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
force proportional to product of masses, inversely to square of separation
the gravitational force between two point masses is proportional to the product of their masses B1 and inversely proportional to the square of their separation B1
2
(b)
3.70 N kg−1
g = GM/r2C1 3.70 N kg−1A1
2
(c)
4250 m s−1
½mv2 = GMm/R (KE equals gain in potential energy to infinity) M1 v = √(2GM/R) A1 4250 m s−1A1
3
(d)
negative, increasing towards zero as 1/r
curve with all values negative B1 increasing (becoming less negative) towards zero as distance increases, ∝ −1/r B1
2
Question 6Structured · 9 marks
A satellite orbits the Earth in a circular orbit at a height of 800 km above the surface. (Earth: mass 5.97 × 1024 kg, radius 6.37 × 106 m)
(a)
Explain why the satellite does not need an engine to stay in orbit.
[2]
(b)
Show that the orbital speed is v = √(GM/r) and calculate it.
[3]
(c)
Use your answer to (b) to calculate the period of the orbit in minutes.
[2]
(d)
Calculate the gravitational potential energy of a 500 kg satellite in this orbit.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
gravity provides centripetal force; no resistance in space
the gravitational force provides the centripetal force needed for circular motion B1 there is (almost) no air resistance to slow it B1
2
(b)
7450 m s−1
GMm/r2 = mv2/r M1 r = 7.17 × 106 m C1 7450 m s−1A1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in