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7.Gravitation
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The rotation period of an earth satellite close to the surface of the earth is $83$ minutes. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be .......... $\min$
A
$83$
B
$83 \times \sqrt 8$
C
$664$
D
$249$
Solution
(c) For first satellite ${r_1} = R$ and $T_1 = 83\,minutes$
For second satellite ${r_2} = 4R$
${T_2} = {T_1}{\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^{3/2}} = {T_1}{(4)^{3/2}} $
$= 8{T_1} = 8 \times 83$
$=664 \,minutes$
Standard 11
Physics
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