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A geostationary satellite is orbiting the Earth at a height of 5R above that surface of the Earth, R being the radius of the Earth. The time period of another satellite in hours at a height of 2R from the surface of the Earth is:

  • Option 1)

    5

  • Option 2)

    10

  • Option 3)

    6\sqrt2

  • Option 4)

    \frac{6}{\sqrt2}

 

Answers (1)

best_answer

From Kepur's law T^{2}\alpha r^{3}

r is distance from earth centre

\therefore T_{1}^{2}=k.\left ( 6R \right )^{3}

[ for geostationary statellite r=bR]

\therefore T_{2}^{2}=k.\left ( 6R \right )^{3}

or \left ( \frac{T_{2}}{T_{1}} \right )^{2}=\frac{1}{8} or  T_{2}^{2}=\frac{T_{1}^{2}}{8}

T_{2}=\frac{T_{1}}{2\sqrt{2}}=\frac{24}{2\sqrt{2}}hr

=6\sqrt{2} hrs


Option 1)

5

Option 2)

10

Option 3)

6\sqrt2

Option 4)

\frac{6}{\sqrt2}

Posted by

prateek

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