The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is :

Solution

Physics sir · 3 steps

Kepler's third law: square the period, cube the radius, and 9 cubed under a root gives 27.

1. Origin of the law. Gravity supplies the centripetal force: GMm/r² = mω² r with ω = 2π/T. Rearranging gives T² = 4π²/GM r³, so T² ∝ r³ for satellites of the same planet.

   T₂/T₁ = (r₂/r₁)^(3/2)

2. Put in numbers. Here r₂/r₁ = 9R/R = 9. So T₂/T = 9^(3/2) = (√9)³ = 3³ = 27.
3. So. The second satellite has period T₂ = 27T, which is option (b).

The idea to keep. For anything orbiting the same central body, T ∝ r^(3/2). Take the square root of the radius ratio first, then cube it. Perfect squares like 4, 9 and 16 make this quick.

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