Gravitation Ques 36
36. A remote - sensing satellite of earth revolves in a circular orbit at a height of $0.25 \times 10^{6} m$ above the surface of earth. If earth’s radius is $6.38 \times 10^{6} $ $m$ and $g=9.8$ $ ms^{-2}$, then the orbital speed of the satellite is:
[2015 RS]
(a) $8.56 $ $km $ $s^{-1}$
(b) $9.13 $ $km $ $s^{-1}$
(c) $6.67 $ $km $ $s^{-1}$
(d) $7.76 $ $km $ $s^{-1}$
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Answer:
Correct Answer: 36.(d)
Solution:
- (d) Given: Height of the satellite from the earth’s surface $h=0.25 \times 10^{6} $ $m$
Radius of the earth $R=6.38 \times 10^{6}$ $ m$
Acceleration due to gravity $g=9.8 $ $m / s^{2}$ Orbital velocity, $v_0=$ ?
$v_0=\sqrt{\frac{GM}{(R+h)}}=\sqrt{\frac{GM}{R^{2}} \cdot \frac{R^{2}}{(R+h)}}$
$=\sqrt{\frac{9.8 \times 6.38 \times 6.38}{6.63 \times 10^{6}}} $
$=7.76 $ $km / s \quad {[\because \frac{GM}{R^{2}}=g]}$
BETA
