Question

An artificial satellite of the earth is launched in a circular orbit in the equatorial plane of the earth and the satellite is moving from west to east. With respect to a person on the equator, the satellite is completing one round trip in 24 h. Mass of the earth is $M=6\times {10}^{24}kg$. For this situation, the orbital radius of the satellite is

• A. $2.66\times {10}^{4}km$
• B. 6400 km
• C. 36,000 km
• D. 29,600 km

Answer 1

Answer: (A)

$2.66\times {10}^{4}km$

Here time period of the satellite w.r.t. an observer on the equator is 24 h and the satellite is moving from west to east, so angular velocity of the satellite w.r.t. earth's axis of rotation (considered as fixed) is $\omega =\frac{2\pi }{T_s}+\frac{2\pi }{T_e}$, where $T_s$ and $T_e$ are time periods of satellite and earth, respectively.
$\omega =\frac{\pi }{6}\left(h^{-1}\right)=1.45\times {10}^{-4}rad{s}^{-1}$
From, $v=\sqrt{\frac{GM}{r}}$
$r\omega =\sqrt{\frac{GM}{r}}$
$r^{\frac{3}{2}}=\sqrt{\frac{GM}{r}}$
$r=2.66\times {10}^{7}m=2.66\times {10}^{4}km$.