Question

A geostationary satellite is orbiting the earth at a height 5 R above the surface of earth, where R is radius of earth. Find the time period of another satellite at a height of 2 R from the the surface of earth.answer rhe question only of u know...
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Answer 1

$\huge \mathfrak \color {teal} \underline {\underline {Question}}$

A geostationary satellite is orbiting the earth at a height 5 R above the surface of earth, where R is radius of earth. Find the time period of another satellite at a height of 2 R from the the surface of earth.

$\huge \mathfrak \color {green} \underline {\underline {Answer}}$

The answer is $\Large 6 \sqrt {2}$ hours

$\Large \mathfrak \color {Magenta} \underline {\underline {Detailed\:Explanation}}$

For Geostationary satellite

$\Large {r_1} = R + 5R = 6R$

For 2nd Satellite

$\Large {r_2} = R + 2R = 3R$

As we know, Time period of satelite is

️ $2\pi\sqrt{ \frac{ {r}^{3} }{GM} } \\ \frac{T_2}{T_1} = \sqrt{\frac {{r_2}^{3}} {{r_1}^{3}}}$

T1 is the time period of earth = 24, So

$\frac{T_2}{24} = \sqrt{ \frac {{(3R)}^{3}} {{(6R)}^{3}}} \\ = > \frac{T_2}{24} = \sqrt{ \frac{ {3}^{3} {R}^{3} }{ {6}^{3} {R}^{3} } } \\ = > \frac{T_2}{24} = \sqrt{ \frac{27}{216} } \\ = > \frac{T_2}{24} = \sqrt{ \frac{1}{8} } \\ = > \frac{T_2}{24} = \frac{ \sqrt{2} }{4} \\ = > {T_2} = \frac{ \sqrt{2} \times 24}{4} \\ = > {T_2} = 6 \sqrt{2} \: hours$

Hope it helps,

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