Gravitation - Result Question 38
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40. 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:
======= ####40. 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:
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/gravitation/gravitation—result-question-38.md (a) $8.56 km s^{-1}$
(b) $9.13 km s^{-1}$
(c) $6.67 km s^{-1}$
(d) $7.76 km s^{-1}$
[2015 RS]
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Answer:
Correct Answer: 40. (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, $v_0=$ ?
$v_0=\sqrt{\frac{GM}{(R+h)}}=\sqrt{\frac{GM}{R^{2}} \cdot \frac{R^{2}}{(R+h)}}$
$ \begin{matrix} =\sqrt{\frac{9.8 \times 6.38 \times 6.38}{6.63 \times 10^{6}}} \\ =7.76 km / s & {[\because \frac{GM}{R^{2}}=g]} \end{matrix} $
Escape velocity
$ \begin{aligned} & =\sqrt{\frac{2 GM}{R}}=c=\text{ speed of light } \\ & \Rightarrow R=\frac{2 GM}{c^{2}} \\ & =\frac{2 \times 6.6 \times 10^{-11} \times 5.98 \times 10^{24}}{(3 \times 10^{8})^{2}} m \\ & =10^{-2} m \end{aligned} $
