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  • An artificial satellite is describing an equatorial orbit at above the earth's

An artificial satellite is describing an equatorial orbit at 3600 \mathrm{~km} above the earth's surface. Calculate its orbital speed.
 

Option: 1

\mathrm{6.335 \mathrm{~km} / \mathrm{sec}}

 


Option: 2

\mathrm{\quad 7.335 \mathrm{~km} / \mathrm{sec}}
 


Option: 3

\mathrm{8.335 \mathrm{~km} / \mathrm{sec}}
 


Option: 4

\mathrm{9.335 \mathrm{~km} / \mathrm{sec}}


Answers (1)

best_answer

The time period of satellite is given by

\mathrm{T^2 =\frac{4 \pi^2}{\mathrm{GM}}(\mathrm{R}+\mathrm{h})^3=\frac{4 ^2}{\mathrm{~g}} \frac{(\mathrm{R}+\mathrm{h})^3}{\mathrm{R}^2} }

\mathrm{\mathrm{~T} =\frac{2 }{\mathrm{R}} \sqrt{\frac{(\mathrm{R}+\mathrm{h})^3}{\mathrm{~g}}}=\frac{2 \pi}{6400 \times 10^3} \sqrt{\frac{10^{21}}{9.8}}}\mathrm{\left[\text { as } \mathrm{R}+\mathrm{h}=(6400+3600) \times 10^3 \mathrm{~m}=10^7 \mathrm{~m}\right] }

\mathrm{ =31360.78 \mathrm{sec}=8.71 \mathrm{hrs}}

Orbital speed is

\mathrm{\mathrm{V}_{\mathrm{o}} =\sqrt{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}}=\sqrt{\frac{\mathrm{gR}^2}{(\mathrm{R}+\mathrm{h})}}=\frac{9.8 \times\left(6400 \times 10^3\right)^2}{10^7} \mathrm{~m} / \mathrm{s}=6335 \mathrm{~m} / \mathrm{s} }

\mathrm{=6.335 \mathrm{~km} / \mathrm{sec}}

Hence option 1 is correct.



 

Posted by

SANGALDEEP SINGH

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