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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 period of revolution?
 

Option: 1

8.71 \mathrm{hrs}

 


Option: 2

9.71 \mathrm{hrs}
 


Option: 3

10.71 \mathrm{hrs}
 


Option: 4

11.71 \mathrm{hrs}


Answers (1)

best_answer

The time period of satellite is given by

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

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

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

Hence option 1 is correct.


 

Posted by

Ritika Harsh

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