A satellite is launched into a circular orbit of radius ' R ' around the earth. A second satellite is launched into an orbit of radius 1.03 R . The time period of revolution of the second satellite is larger than the first one approximately by

Solution

Physics sir · 4 steps

Kepler's third law makes T go as R^(3/2), so a small 3% change in radius becomes 1.5 times that in period.

1. Kepler's third law. For a circular orbit, T² ∝ R³, so T ∝ R^(3/2). The mass of the earth is the same for both satellites, so only the radius matters.

   (Δ T)/T ≈ 3/2 (Δ R)/R

2. Put in numbers. The radius goes up by (Δ R)/R = 0.03, that is 3%. So the period goes up by 3/2 × 3% = 4.5%.
3. Exact check. T₂/T₁ = (1.03)^(3/2) = 1.03 × √1.03 ≈ 1.045. That is about a 4.5% increase, which agrees.
4. So. The period of the second satellite is larger by about 4.5%, option (c).

The idea to keep. If y ∝ xⁿ, a small percentage change in x becomes n times that percentage change in y. Find the power first, then multiply.

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