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] $
Show Answer
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] $
