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Q.
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
Bihar CECEBihar CECE 2007Gravitation
Solution:
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)
$a_{1}=a, a_{2}=2 a$
$\therefore \frac{T_{1}^{2}}{T_{2}^{2}}=\frac{a_{1}^{3}}{a_{2}^{3}}$
$\Rightarrow T_{2}^{2}=\frac{a_{2}^{3}}{a_{1}^{3}} T_{1}^{2}=\frac{(2 a)^{3}}{a^{3}} \times 1=8$
$\Rightarrow T_{2}=2 \sqrt{2}$ days