Explain why the total mechanical energy of a satellite in a circular orbit around a planet is equal to half the gravitational potential energy at that orbit.

IB DP Physics Higher Level (2023 syllabus) — D.1 Gravitational fields · Explain · 4 marks · View as Markdown

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

A satellite of mass m orbits a planet of mass M in a circular orbit of radius r. The gravitational potential energy of the satellite at this orbit is U = −GMm/r.

Model answer (4 marks)

For a circular orbit the gravitational force supplies the centripetal force:
GMm/r² = mv²/r.
Thus v² = GM/r and the kinetic energy is EK = ½mv² = ½m(GM/r) = GMm/2r.
The gravitational potential energy at radius r is EP = –GMm/r.
Hence EK = –EP/2 = –½(–GMm/r).
The total mechanical energy is E = EK + EP = GMm/2r – GMm/r = –GMm/2r.
Therefore E = EP/2, i.e. the total energy equals half the gravitational potential energy (negative, showing a bound orbit).

Examiner tips

  • Show the force balance to get v², then calculate EK; link EK to EP; sum to find E; state the final relation E = EP/2.
  • Use the negative sign of EP to explain why E is negative and the orbit is bound.

Common mistakes

  • Confusing EP with +GMm/r instead of –GMm/r.
  • Forgetting to divide EK by 2 when relating it to EP; or mis‑applying the sign when adding EK and EP.

Mark scheme (4 marks)

  1. For a circular orbit, the gravitational force provides the centripetal force, so GMm/r² = mv²/r, giving the kinetic energy EK = ½mv² = GMm/2r.
  2. The gravitational potential energy at radius r is EP = −GMm/r, so the kinetic energy EK = −EP/2 = −½(−GMm/r).
  3. The total mechanical energy is E = EK + EP = GMm/2r − GMm/r = −GMm/2r.
  4. Therefore E = EP/2 (i.e. the total energy equals half the gravitational potential energy), and the negative value confirms the satellite is in a bound orbit.

Key terms in this question

gravitational potential energy · total mechanical energy · circular orbit

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