GATE-2018 (24) :Wyllie – Time average equation

UPSC-2014

1. The earth rotates once per sidereal day of 23 hours 56 minutes 4.09 seconds. Find the radius of the geostationary earth orbit satellite.

Solution:

The relation between the Time period and orbital radius is

$T^{2} = \frac{4*\pi^{2}*a^{3}}{\mu}$--(1)

Rearranging the equation the orbital radius ‘a’ is given by

$a^{3} = \frac{T^2 \mu}{4 *\pi^2}$

For one sidereal day, T= 86,164.09 s

$a^{3} = \frac{(86,164.09)^2 *(3.986*10^5)}{4 *(3.14)^2}$

$a^{3} = 7.4960 * 10^{13} km^3$

a= 42,164.17km

This is the orbital radius for a geostationary satellite.

( μ = G M

M= mass of the Earth =$ 5.9 * 10^{24} kg$

G= Gravitational constant =$ 6.674 * 10^{-11} m^3/ (kg* s^2) )$