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Gravitational potential questions
Gravitational potential attaches a number to every location in a field, the work per kilogram needed to arrive there from infinity. Every value comes out negative, the zero sits at infinity, and one pair of graphs converts between potential and field strength in both directions.
18 original questions · 46 marks · the gravitational potential notes · Gravitational fields
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Define the gravitational potential at a point in a gravitational field.
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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.
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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.
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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.
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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 the planet, 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.
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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.
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.
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.
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.
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.
A satellite is boosted from an orbit at r = 7.0 × 106 m to a higher orbit at r = 1.4 × 107 m. Calculate the change in gravitational potential between the two orbits.
Describe how gravitational field strength and gravitational potential are related graphically for a radial field.
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.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.
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.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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