Physics › Practise › Questions › Gravitational potential
Questions on gravitational potential
Gravitational potential is the work done per kilogram to bring a mass in from infinity. Gravity only attracts, so V = −GM/r is always negative. No work is done moving along an equipotential, which is why a circular orbit needs no thrust. Field strength is minus the gradient of V, and the area under g against r is ΔV.
18 original questions · 46 marks · the gravitational potential notes · Gravitational fields
To print: the questions (PDF, 79 KB) · the worked answers (PDF, 75 KB)
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening a mark scheme: the schemes award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Define the gravitational potential at a point in a gravitational field.
Mark scheme
Work done per unit mass (1); in bringing a small test mass from infinity to the point (where the potential at infinity is zero) (1).A planet has a mass of 6.0 × 1024 kg and a radius of 6.4 × 106 m. Calculate the gravitational potential at its surface.
G = 6.67 × 10−11 N m2 kg−2Mark scheme
V = −GM/r = −(6.67 × 10−11 × 6.0 × 1024)/(6.4 × 106) (1)
V = −6.25 × 107 J kg−1 (1)Explain why gravitational potential always has a negative value.
Mark scheme
The potential is defined as zero at infinity and the gravitational force is attractive (1); work must be done against the field to move a mass to infinity, so at every point in the field the potential is below zero (1).A satellite moves along an equipotential surface of the Earth's gravitational field. State the work done against gravity during this motion and give a reason.
Mark scheme
Zero (1); the potential does not change along an equipotential, so ΔW = mΔV = 0 (1).State the angle between an equipotential surface and the gravitational field lines that cross it.
Mark scheme
90° (they always cross at right angles) (1).Calculate the gravitational potential at a distance of 1.0 × 107 m from the centre of a planet of mass 6.0 × 1024 kg.
G = 6.67 × 10−11 N m2 kg−2Mark scheme
V = −GM/r = −(6.67 × 10−11 × 6.0 × 1024)/(1.0 × 107) (1)
V = −4.00 × 107 J kg−1 (1)A space vehicle of mass 500 kg is moved from the surface of a planet of mass 6.0 × 1024 kg, at r = 6.4 × 106 m, to a point at r = 1.28 × 107 m from its centre. Calculate the work done against the gravitational field.
Mark scheme
Vsurface = −GM/r = −6.25 × 107 J kg−1 (1)
Vfinal = −3.13 × 107 J kg−1 (1)
ΔV = 3.13 × 107 J kg−1 (1)
W = mΔV = 500 × 3.13 × 107 = 1.56 × 1010 J (1)Calculate the escape velocity from the surface of the planet of mass 6.0 × 1024 kg and radius 6.4 × 106 m.
G = 6.67 × 10−11 N m2 kg−2Mark scheme
½mv2 = GMm/r, so v = √(2GM/r) (1)
v = √(2 × 6.67 × 10−11 × 6.0 × 1024/(6.4 × 106)) (1)
v = 1.12 × 104 m s−1 (about 11.2 km s−1) (1)Near a certain point in the field, the gravitational potential changes by 1.0 × 106 J kg−1 over a radial distance of 1.0 × 105 m. Estimate the gravitational field strength at this point.
Mark scheme
g = ΔV/Δr = (1.0 × 106)/(1.0 × 105) (1)
g = 10 N kg−1 (1)The area under a graph of gravitational field strength against distance, between two distances from a planet's centre, is 2.1 × 106 J kg−1. State what this area represents, and calculate the energy needed to move an 800 kg probe between the two distances.
Mark scheme
The area is the (magnitude of the) change in gravitational potential ΔV between the two distances (1)
W = mΔV = 800 × 2.1 × 106 (1)
W = 1.7 × 109 J (1)Starting from the condition for a mass launched from the surface of a planet to just escape its gravitational field, show that the escape velocity does not depend on the mass of the escaping object.
Mark scheme
The kinetic energy supplied must equal the energy needed to reach infinity: ½mv2 = GMm/r (1); m cancels from both sides, giving v = √(2GM/r), which is independent of the escaping mass (1).A probe falls freely from a high orbit towards a planet's surface. State the sign of ΔV for the fall, and describe the energy transfer taking place.
Mark scheme
ΔV is negative: the probe moves to a point of lower (more negative) potential (1). Gravitational potential energy is transferred to kinetic energy, the field doing work on the probe (1).A spacecraft of mass 1000 kg is launched from the surface of the planet, where the gravitational potential is −6.25 × 107 J kg−1. Determine the minimum energy needed for the spacecraft to escape from the planet's gravitational field.
Mark scheme
To escape, the spacecraft must reach a point where V = 0 (1)
ΔV = 0 − (−6.25 × 107) = 6.25 × 107 J kg−1 (1)
E = mΔV = 1000 × 6.25 × 107 = 6.25 × 1010 J (1)A satellite is boosted from an orbit at r = 7.0 × 106 m to a higher orbit at r = 1.4 × 107 m around a planet of mass 6.0 × 1024 kg. Calculate the change in gravitational potential between the two orbits.
Mark scheme
V1 = −GM/r1 = −5.72 × 107 J kg−1 (1)
V2 = −2.86 × 107 J kg−1 (1)
ΔV = V2 − V1 = 2.86 × 107 J kg−1 (an increase) (1)Describe how gravitational field strength and gravitational potential are related graphically for a radial field.
Mark scheme
The field strength equals the negative of the gradient of the potential-distance graph, g = −ΔV/Δr (1). The change in potential between two distances equals the area under the field strength-distance graph between them (1). Both g and the magnitude of V decrease with distance, g as 1/r2 and V as 1/r (1).A 1500 kg satellite sits on the surface of the planet, where the gravitational potential is −6.25 × 107 J kg−1. Its launch vehicle can transfer at most 2.0 × 1010 J to the satellite. The table gives the gravitational potential at three possible operating orbits.
orbit A: −5.00 × 107 J kg−1
orbit B: −4.50 × 107 J kg−1
orbit C: −5.50 × 107 J kg−1
Ignoring the kinetic energy of the satellite in orbit, deduce which of the orbits the launch vehicle can reach.Mark scheme
W = mΔV, with ΔV measured from the surface value (1)
Orbit A: W = 1500 × 1.25 × 107 = 1.88 × 1010 J; orbit C: W = 1.13 × 1010 J, both within 2.0 × 1010 J (1)
Orbit B: W = 1500 × 1.75 × 107 = 2.63 × 1010 J, which exceeds the limit (1)
The vehicle can reach orbits A and C but not orbit B (1).At a distance r from the centre of a planet, the gravitational field strength is 6.0 N kg−1 and the gravitational potential is −3.0 × 107 J kg−1. State the values of both quantities at a distance 2r from the centre, and explain why they change by different factors.
Mark scheme
g = 6.0/4 = 1.5 N kg−1 (1)
V = (−3.0 × 107)/2 = −1.5 × 107 J kg−1 (1)
g is proportional to 1/r2 but V is proportional to 1/r, so doubling r quarters g while only halving the magnitude of V (1).The gravitational potential at the surface of an airless moon is −2.8 × 106 J kg−1. Show that the escape velocity from its surface is about 2.4 km s−1.
Mark scheme
To just escape, ½mv2 = m × (0 − V), so ½v2 = 2.8 × 106 J kg−1 (1)
v = √(2 × 2.8 × 106) (1)
v = 2.37 × 103 m s−1 ≈ 2.4 km s−1 (1)
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise gravitational potential one question at a time
The player marks nothing for you. It shows one question, waits, then shows the scheme so you can mark yourself, and brings a question back sooner when it went badly.