Question

A body is orbiting around earth at a mean radius which is two times as greater as the parking orbit of a satellite, the period of body is

• A. 4 days
• B. 16 days
• C. $2\sqrt{2}$ days
• D. 64 days

Answer 1

Answer: (C)

Parking orbit is a geostationary satellites orbit.
From Keplers third law of planetary motion, the square of period of revolution (T) is directly proportional to cube of semi-major axis of its elliptical orbit (a), i.e..
$T^2\propto a^3$
Given, $T_1=1$ day (geostationary)
$a_1=a,a_2=2a$
$\therefore \frac{T_1^2}{T_2^2}=\frac{a_1^3}{a_2^3}$
$\Rightarrow T_2^2=\frac{a_2^3}{a_1^3}{T}_1^2=\frac{(2a{)}^3}{a^3}\times 1=8$
$\Rightarrow T_2=2\sqrt{2}$ days