A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is :
(Given = Radius of geo-stationary orbit for earth is 4.2 × 10⁴ km)

Solution

Physics sir · 3 steps

Kepler's third law says r³ ∝ MT², so compare the planet's satellite directly with Earth's.

1. Orbit condition. Gravity supplies the centripetal force, so GMm/r² = m(2π/T)² r. Rearranging gives r³ = GMT²/4π², so r³ ∝ MT².

   rₚ³/rₑ³ = Mₚ/Mₑ(Tₚ/Tₑ)²

2. Put in ratios. Mₚ/Mₑ = 1/4, and Tₚ/Tₑ = 6/24 = 1/4. So rₚ³/rₑ³ = 1/4 × 1/16 = 1/64, which gives rₚ/rₑ = 1/4.
3. So. rₚ = (4.2 × 10⁴)/4 = 1.05 × 10⁴ km, which is option (d).

The idea to keep. For any orbit, T² = (4π² r³)/GM. When two orbits are compared, set up the ratio and take the cube root only at the very end.

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