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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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Period of revolution of an earth satellite close to surface of earth is 90 minutes. The time period of another satellite in an orbit at a distance three times the radius of earth from its surface will be

\[(a)\;720\;min \quad (b)\;360\;min \quad (c)\;90 \sqrt 8 min \quad (d)\;270\;min \]
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Distance of satellite from the center of earth is 4 times
$T^2 \; \alpha \; r^3$
$T \; \alpha \; r^{\large\frac{3}{2}}$
Time period of second satellite will become $(4)^{3/2}$ ie 8 times
Hence $ T'= 8T=8 \times 90 =720\;min$
Hence a is the correct answer.


answered Aug 29, 2013 by meena.p
edited Feb 18, 2014 by meena.p

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