A remote-sensing satellite of earth revolves in a circular orbit at a height of m above the surface of earth. If earth’s radius is m and , then the orbital speed of the satellite is: [2015]

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Solution

Q.
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]

1 $9.13$ $km$ $s^{- 1}$

2 $6.67$ $km$ $s^{- 1}$

3 $7.76$ $km$ $s^{- 1}$

4 $8.56$ $km$ $s^{- 1}$

Ans.
(3)

The orbital speed of the satellite is, $v_{0} = R \sqrt{\frac{g}{(R + h)}}$

where $R$ is the earth’s radius, $g$ is the acceleration due to gravity on earth’s surface, and $h$ is the height above the surface of Earth.

Here, $R = 6.38 \times 10^{6} m, g = 9.8 {ms}^{- 2}, h = 0.25 \times 10^{6} m$

$∴$ $v_{0} =$ $(6.38 \times 10^{6} m)$ $\sqrt{\frac{(9.8 {ms}^{- 2})}{(6.38 \times 10^{6} m + 0.25 \times 10^{6} m)}}$

$= 7.76 \times 10^{3} {ms}^{- 1} = 7.76 km$ $s^{- 1}$

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