Gravitation Question 258

Question: 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]

Options:

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{GMT^{2}}{4{{\pi }^{2}}} \right)}^{1/3}}-R] $

D) $ [{{\left( \frac{GMT^{2}}{4{{\pi }^{2}}} \right)}^{1/3}}+R] $

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Answer:

Correct Answer: C

Solution:

$ [T=2\pi \sqrt{\frac{r^{3}}{GM}},\Rightarrow ,T^{2}=\frac{4{{\pi }^{2}}}{GM}{{(R+h)}^{3}}] $

$ [\Rightarrow R+h={{\left[ \frac{GMT^{2}}{4{{\pi }^{2}}} \right]}^{1/3}}\Rightarrow ,h={{\left[ \frac{GMT^{2}}{4{{\pi }^{2}}} \right]}^{\frac{1}{3}}}-R] $



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