Answer
The altitude $h$ of the orbit above the earth's surface is
$h \simeq 35872$ km
Work Step by Step
We have the period of the orbit $T = 24$ hours $ = 86400$ s.
According to Kepler's third law:
${T^2} = \left( {\frac{{4{\pi ^2}}}{{GM}}} \right){a^3}$,
where $M$ is the mass of the earth and $a$ is the radius of the orbit.
Since the orbit of the satellite is circular geosynchronous, so $a$ is the distance from the center of the earth. Thus,
$a = {\left( {\frac{{GM}}{{4{\pi ^2}}}{T^2}} \right)^{1/3}}$
$a \simeq {\left( {\frac{{6.673 \times {{10}^{ - 11}} \times 5.974 \times {{10}^{24}}}}{{4{\pi ^2}}}{{86400}^2}} \right)^{1/3}} \simeq 4.22426 \times {10^7}$ m
$a \simeq 42243$ km
Thus, the altitude $h$ of the orbit above the earth's surface is
$h \simeq 42243 - 6371 \simeq 35872$ km
