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

Gravitation PYQ — JEE Main 2025 · 24 Jan · Shift 1 · Q56 with Solution

View written solutionFree

Correct answer: C

For a satellite orbiting Earth,

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

So,

$T \propto r^{3/2}$

Let the first satellite have orbit radius $R$ and time period $T_1$ .

The second satellite has orbit radius

$r_2 = 1.03R$

Hence,

$\frac{T_2}{T_1} = \left(\frac{r_2}{R}\right)^{3/2} = (1.03)^{3/2}$

For small changes,

$(1+x)^n \approx 1+nx$

Here, $x=0.03$ and $n=\frac{3}{2}$ , so

$\frac{T_2}{T_1} \approx 1 + \frac{3}{2}(0.03) = 1 + 0.045 = 1.045$

Thus the increase in time period is approximately

$0.045 \times 100\% = 4.5\%$

The second satellite's time period is larger by approximately

$\boxed{4.5\%}$

So the correct option is C.