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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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A geostationary satellite is orbiting the earth at a height of 6R where R is radius of earth. The time period of another satellite at a height of $2.5 R$ from the surface of earth is

\[(a)\;10\;hours \quad (b)\;\frac{6}{\sqrt 2}\;hours \quad (c)\;6\;hours \quad (d)\;6 \sqrt 2 \;hours \]
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Distance of satellite from center of earth are 7R and 3.5 R respectively
Time period of geostationary satellite is 24 hours
$\large\frac{T_2}{T_1}=\bigg(\large\frac{R_2}{R_1}\bigg)^{\frac{3}{2}}$
$T_2 =24 \bigg(\large\frac{3.5 R}{7R}\bigg)^{3/2}$
$\qquad= 6 \sqrt 2\;hours$
answered Aug 24, 2013 by meena.p
 

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