Question

A geosynchronous orbit is one in which the satellite makes one revolution around the Earth in 24 hrs.
How far above the surface of the Earth does this satellite have to orbit?
Assume the radius of the Earth is $6.37\times {10}^{6}m$, and the mass of the Earth is $5.98\times {10}^{24}kg$.

• A. $3.59\times {10}^{7}m$
• B. $4.23\times {10}^{7}m$
• C. $2.76\times {10}^{6}m$
• D. $6.37\times {10}^{6}m$
• E. $1.27\times {10}^{7}m$

Answer 1

Answer: (A)

$3.59\times {10}^{7}m$

Given : $T=24$ hrs $=84400$ s $M_e=5.98\times {10}^{24}$ kg $R_e=0.637\times {10}^7$ m

Time period of satellite moving in orbit of radius $r$ $T=2\pi \sqrt{\frac{r^3}{G{M}_e}}$

$\therefore$ $86400=2\pi \sqrt{\frac{r^3}{(6.67\times {10}^{-11})\times (5.98\times {10}^{24})}}$ $⟹r=4.23\times {10}^7$ m

Thus height of satellite above the surface $h=r-R_e=(4.23-0.637)\times {10}^7\approx 3.59\times {10}^7$ m