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

According to question, a satellite is launched into a circular orbit of radius R around the Earth, while a second satellite is launched into a circular orbit of radius 1.02 R.

As we know, according to Kepler’s law of planetary motion,

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

where, T = time period

and R = length of the semi-major axis of planet’s orbit.

$\Rightarrow \mathrm{T}\propto {\mathrm{R}}^{3/2}$

Differentiating above equation, we get

$\frac{\mathrm{dT}}{\mathrm{T}}=\frac{3}{2}\frac{\mathrm{dR}}{\mathrm{R}}$

Here, dR = 1.02R - R = 0.02R

$\Rightarrow \frac{\mathrm{dT}}{\mathrm{T}}=\frac{3}{2}\times \frac{0.02\mathrm{R}}{\mathrm{R}}=0.03$

$\therefore$ % difference in time periods of two satellites is

$\frac{\mathrm{dT}}{\mathrm{T}}\times 100=0.03\times 100=3\mathrm{\%}$