Given radius of Earth ?R? and length of a day ?T? the height of a geostationary satellite is [G?Gravitational Constant, M?Mass of Earth] [MP PMT 2002]

\[{{\left( \frac{GM{{T}^{2}}}{4{{\pi }^{2}}} \right)}^{1/3}}+R\]

\[{{\left( \frac{4{{\pi }^{2}}GM}{{{T}^{2}}} \right)}^{1/3}}\]

\[{{\left( \frac{4\pi GM}{{{R}^{2}}} \right)}^{1/3}}-R\]

\[{{\left( \frac{GM{{T}^{2}}}{4{{\pi }^{2}}} \right)}^{1/3}}-R\]

Given radius of Earth ?R? and length of a day ?T? the height of a geos

1.	Given radius of Earth ?R? and length of a day ?T? the height of a geostationary satellite is [G?Gravitational Constant, M?Mass of Earth] [MP PMT 2002]
A.	\[{{\left( \frac{4{{\pi }^{2}}GM}{{{T}^{2}}} \right)}^{1/3}}\]
B.	\[{{\left( \frac{4\pi GM}{{{R}^{2}}} \right)}^{1/3}}-R\]
C.	\[{{\left( \frac{GM{{T}^{2}}}{4{{\pi }^{2}}} \right)}^{1/3}}-R\]
D.	\[{{\left( \frac{GM{{T}^{2}}}{4{{\pi }^{2}}} \right)}^{1/3}}+R\]
Answer» D. \[{{\left( \frac{GM{{T}^{2}}}{4{{\pi }^{2}}} \right)}^{1/3}}+R\]