A satellite is launched into a circular orbit of radius R around earth, while a second satellite is launched into a circular orbit of radius 1.02 R. The percentage difference in the time periods of the two satellites is :

Solution

Physics sir · 4 steps

Kepler's third law gives T ∝ R^(3/2), so a 2% change in radius becomes a 3% change in period.

1. Period of an orbit. For a circular orbit, gravity supplies the centripetal force, which gives T = 2π√(R³/GM). So T ∝ R^(3/2), Kepler's third law.

   (Δ T)/T = 3/2 (Δ R)/R

2. Put in numbers. The radius grows from R to 1.02R, so (Δ R)/R = 2%. Then (Δ T)/T = 3/2 × 2% = 3%.
3. Check exactly. Without the approximation, T₂/T₁ = (1.02)^(3/2) ≈ 1.0301. That is a 3.01% increase, so the small-change estimate holds.
4. So. The time periods differ by about 3%, option (d).

The idea to keep. When y ∝ xⁿ and x changes by a small percentage, y changes by n times that percentage. For orbits, n = 3/2 for the period.

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