Question

An earth satellites $S$ has an orbit radius which is $4$ times that of communication satellite $C$. The period of revolution of $S$ will be:

• A. 32 day
• B. 18 day
• C. 8 day
• D. 9 day

Answer 1

Answer: (C)

Larger the distance of planet from the sun, larger will be its period of revolution around the sun.
From Kepler's third law of planetary motion, the square of the period of revolution of any planet around the sun is directly proportional to the cube of the semi-major axis of its elliptical orbit.
$\therefore T^2\propto R^3$
$\therefore \frac{T_s}{T_c}={\left(\frac{R_s}{R_c}\right)}^{3/2}$
Given, $R_s=4R_c$
$\therefore \frac{T_s}{T_c}={\left(\frac{4R_c}{R_c}\right)}^{3/2}=8$
For $T_c=1$ day
$T_s=8$ days.