The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three times its previous value, then approximate new time period of the satellite will be

Solution

Physics sir · 3 steps

Kepler's third law: the period grows as radius to the power 3/2, so multiply 7 hours by 3√3.

1. Kepler's third law. For satellites going around the same planet, T² ∝ r³, so T ∝ r^(3/2). The mass of the Earth and G are the same for both orbits, so they cancel when you take a ratio.

   T₂/T₁ = (r₂/r₁)^(3/2) = 3^(3/2) = 3√3

2. Put in numbers. 3√3 = 3 × 1.732 = 5.196. So T₂ = 7 × 5.196 = 36.37 hours.
3. So. The new period is about 36 hours, which is option (b).

The idea to keep. Whenever two orbits share the same central body, compare them with T ∝ r^(3/2). Tripling the radius multiplies the period by 3√3 ≈ 5.2, not by 3.

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