Deriving Kepler's third law for a circular orbit

Overview

Recall questions on equating GMm/r² with mrω², substituting ω = 2π/T, rearranging to T² = 4π²r³/GM, why the constant depends only on the central mass, and what is plotted to test the law as a straight line.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   State the two physical relationships needed to derive Kepler's third law for a circular orbit.

   Model answer

   Newton's law of gravitation, F = GMm/r², and the expression for centripetal acceleration, a = v²/r, are the two relationships required.

2. 2

   Derive Kepler's third law for a circular orbit using the orbital speed.

   Model answer

   Equating the gravitational force to the centripetal force gives GMm/r² = mv²/r, so v² = GM/r. Substituting v = 2πr/T gives 4π²r²/T² = GM/r, and rearranging gives T² = 4π²r³/GM.

3. 3

   Derive Kepler's third law for a circular orbit using angular velocity.

   Model answer

   The centripetal acceleration may be written as ω²r, so GMm/r² = mω²r gives ω² = GM/r³. Substituting ω = 2π/T gives 4π²/T² = GM/r³, so T² = 4π²r³/GM.

4. 4

   State the constant of proportionality in Kepler's third law for bodies orbiting a mass M.

   Model answer

   The constant is 4π²/GM, so that T² = (4π²/GM)r³.

5. 5

   Why does Kepler's third law hold whatever the mass of the orbiting body?

   Model answer

   The orbiting body's mass appears on both sides of GMm/r² = mv²/r and cancels, so the relationship between period and radius involves only the central mass.

6. 6

   Show that the expression T² = 4π²r³/GM has consistent units.

   Model answer

   The unit of r³/GM is m³ divided by N m² kg⁻² × kg, giving m³ ÷ (N m² kg⁻¹). Since 1 N = 1 kg m s⁻², that becomes m³ ÷ (m³ s⁻²), leaving s².

Other ways to work this

The same content, four other ways.

What is inside Levo

Six ways to work the same material.