Two planets A and B having masses m₁ and m₂ move around the sun in circular orbits of r₁ and r₂ radii respectively. If angular momentum of A is L and that of B is 3 L, the ratio of time period (T_A/T_B) is:

Solution

Physics sir · 4 steps

The options carry masses, not radii, so write the period in terms of L and m and let r drop out.

1. Orbital speed. Gravity supplies the centripetal force: GMm/r² = mv²/r, so v = √(GM/r), where M is the mass of the sun. The angular momentum is then L = mvr = m√GMr.
2. Remove the radius. Square and rearrange to get r = L²/(m² GM). Put this into T = (2π r)/v = (2π r^(3/2))/√GM.

   T = (2π L³)/(m³ (GM)²) ⇒ T ∝ (L/m)³

3. Take the ratio. T_A/T_B = (L/3L)³ (m₂/m₁)³ = 1/27(m₂/m₁)³.
4. So. The ratio of time periods is 1/27(m₂/m₁)³, which is option (d).

The idea to keep. For a circular orbit, L = m√GMr links angular momentum to radius. Use it to swap r for L/m, and then T ∝ (L/m)³.

Answer from the NTA final answer key. Solution by PrepWiser.