Question

A geosynchronous satellite is a satellite that orbits the earth with an orbital period of 24 hours, thus matching the period of the earth's rotational motion. A special class of geosynchronous satellites is a geostationary satellite. A geostationary satellite orbits the earth in 24 hours along an orbital path that is parallel to an imaginary plane drawn through the Earth's equator. Such a satellite appears permanently fixed above the same location on the Earth. If a geostationary satellite wishes to orbit the earth in 24 hours (86400 s), then how high above the earth's surface must it be located? (Given: $M_{earth}=5.98\times {10}^{24}kg,R_{earth}=6.37\times {10}^{6}m$)

• A. 10.1 x 10⁷ m
• B. 6.38 x 10⁷ m
• C. 3.59 x 10⁷ m
• D. 4.56x 10⁷ m

Answer 1

The correct option is C

3.59 x 10⁷ m

T = 86400 s
$M_{earth}=5.98\times {10}^{24}kg$
$R_{earth}=6.37\times {10}^{6}m$
$G=6.673\times {10}^{-11}N{m}^2/k{g}^2$
$\frac{T^2}{R^3}=\frac{4\times {\pi }^2}{G\times M_{central}}$
$R^3=\frac{T^2\times G\times M_{central}}{4\times {\pi }^2}$
$R^3=\frac{\left(\right(86400s{)}^2\times (6.673\times {10}^{-11}N{m}^2/k{g}^2)\times (5.98\times 1024kg))}{4\times (3.14{)}^2}$
$R^3=7.54\times {10}^{22}{m}^3$
By taking the cube root of $7.54\times {10}^{22}{m}^3$, the radius can be determined to be
$R=4.23\times {10}^{7}m$
The radius of orbit indicates the distances that the satellite is from the centre of the earth. Now that the radius of orbit has been found, the height above the earth can be calculated. Since the earth's surface is $6.37\times {10}^{6}m$ from its center (that's the radius of the earth), the satellite must be a height of $4.23\times {10}^{7}m-6.37\times {10}^{6}m=3.59\times {10}^{7}m$, above the surface of the earth.
So the height for the satellite is $3.59\times {10}^{7}m$.