Explain why the work done by the gravitational force on a satellite moving in a circular orbit around the Earth is zero, and state what this implies about the satellite's kinetic energy.

IB DP Physics Higher Level (2023 syllabus) — A.3 Work, energy and power · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (4 marks)

The gravitational force on a satellite is directed radially inward, i.e. along the radius of the orbit. In a circular orbit the satellite’s velocity is always tangential to the orbit, so the angle between the force and the displacement is 90°. Work is W=Fs cosθ; with θ=90° cosθ=0, therefore W=0. Since no work is done, the work–energy theorem gives ΔK=0, so the satellite’s kinetic energy (and speed) remains constant.

Examiner tips

  • Show the force is radial and velocity tangential → θ=90°
  • Use W=Fs cosθ to get W=0
  • State ΔK=0 and kinetic energy is constant

Common mistakes

  • Confusing the direction of the force with the direction of motion
  • Using the wrong angle (e.g. 0°) instead of 90°
  • Failing to link zero work to constant kinetic energy

Mark scheme (4 marks)

  1. The gravitational force acts along the radius (centripetally), directed towards the centre of the Earth.
  2. For a circular orbit, the velocity (displacement) of the satellite is always tangential, i.e. perpendicular to the gravitational force.
  3. Work done W = Fs cosθ, and since θ = 90°, cos 90° = 0, so W = 0.
  4. Since no work is done on the satellite, by the work–energy theorem its kinetic energy (and therefore its speed) remains constant.

Key terms in this question

work done · gravitational force · circular orbit · kinetic energy

Related

More Work, energy and power questions

▶ Try answering this question with AI marking (free) →