Gravitation Ques 40
40. A geostationary satellite is orbiting the earth at a height of $5 $ $R$ 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 $2 $ $R$ from the surface of the earth is :
[2012]
$5$
(b) $10$
(c) $6 \sqrt{2}$
(d) $\frac{6}{\sqrt{2}}$
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Answer:
Correct Answer: 40.(c)
Solution:
- (c) According to Kepler’s law of period $T^{2} \propto R^{3}$
$\frac{T_1^{2}}{T_2^{2}}=\frac{R_1^{3}}{R_2^{3}}=\frac{(6 R)^{3}}{(3 R)^{3}}=8$
$\frac{24 \times 24}{T_2^{2}}=8$
$T_2^{2}=\frac{24 \times 24}{8}=72=36 \times 2$
$T_2=6 \sqrt{2}$
BETA
