Gravitational fields practice questions.

Planet Y moves in a circular orbit with four times the orbital radius of planet X around the same star. Determine the ratio of their orbital periods T(Y) : T(X).

• 2 : 1
• 4 : 1
• 8 : 1
• 16 : 1

Answer: 8 : 1

By Kepler's third law T² ∝ r³, so T(Y)/T(X) = 4^(3/2) = √(4³) = 8.

The gravitational field strength at the Earth's surface is 9.8 N kg⁻¹. Calculate the field strength at a height above the surface equal to one Earth radius.

• 4.9 N kg⁻¹
• 2.45 N kg⁻¹
• 9.8 N kg⁻¹
• 1.2 N kg⁻¹

Answer: 2.45 N kg⁻¹

At height R the distance from Earth's centre doubles to 2R; since g ∝ 1/r², the field strength falls to 9.8/4 = 2.45 N kg⁻¹.

The gravitational force between two point masses is F. Both masses are doubled and their separation is also doubled. State the new gravitational force between them.

• F/4 (one quarter of the original F)
• F/2
• 2F
• F (unchanged)

Answer: F (unchanged)

F ∝ m₁m₂/r²: doubling both masses multiplies the force by 4, while doubling the separation divides it by 4, so the force is unchanged.

Which statement best describes the gravitational field over a small region close to the Earth's surface?

• Approximately uniform, with parallel field lines directed vertically downward
• Radial, with all the field lines directed straight away from the centre of the Earth
• Circular, with field lines forming loops around the Earth
• Zero, because the ground supports objects resting on it

Answer: Approximately uniform, with parallel field lines directed vertically downward

Over a region small compared with the Earth's radius the field lines are effectively parallel and equally spaced, so the field is treated as uniform and directed downward.

A satellite in a circular orbit of radius r about a planet has orbital speed v. Determine its orbital speed if instead it orbits the same planet at radius 4r.

• v/4
• v/2
• 2v
• v (unchanged)

Answer: v/2

For a circular orbit v = √(GM/r), so v ∝ 1/√r; quadrupling the radius halves the orbital speed.

Two identical uniform spheres, each of mass M, have their centres a distance 2d apart. Determine the resultant gravitational field strength at the midpoint of the line joining the centres.

• 2GM/d², because the two field strengths at that point have equal magnitudes and add together.
• Zero, because the two fields there are equal in magnitude and opposite in direction.
• GM/d², because the field of one sphere is cancelled by half of the field of the other.
• GM/2d², because each sphere halves the field that the other one produces at that point.

Answer: Zero, because the two fields there are equal in magnitude and opposite in direction.

Gravitational field strength is a vector: each sphere contributes GM/d² at the midpoint, but the two point in opposite directions and cancel. Adding the magnitudes is the classic scalar treatment of a vector quantity — note that a mass placed there still has gravitational potential energy, it simply feels no resultant force.