JEE Main 2025 — Gravitation Question with Solution

$$T\propto R^{\frac{3}{2}}$$

For a satellite in a circular orbit with radius $$R$$, the time period is given by

$$T=2\pi \sqrt{\frac{R^3}{GM}}.$$

If the radius is increased to $$1.03R$$, the new time period $$T^{\prime }$$ becomes

$$T^{\prime }=2\pi \sqrt{\frac{(1.03R{)}^3}{GM}}=T\times (1.03{)}^{\frac{3}{2}}.$$

For a small change, we can approximate

$$(1.03{)}^{\frac{3}{2}}\approx 1+\frac{3}{2}\times 0.03=1+0.045=\mathrm{1.045.}$$

This means the new time period is approximately 4.5% larger than the original time period.

Thus, the time period of the second satellite is larger by approximately 4.5%.