First slide

Keplers 3 Laws

Question

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

Moderate

A

$9.13 km s^{- 1}$
B

$6.67 km s^{- 1}$
C

$7.76 km s^{- 1}$
D

$8.56 km s^{- 1}$

Solution

The orbital speed of the satellite is
$v_{o} = R \sqrt{\frac{g}{(R + h)}}$
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.
$Here, R = 6.38 \times 10^{6} m, g = 9.8 m s^{- 2} and$ $h = 0.25 \times 10^{6} m$
$∴ v_{o} = (6.38 \times 10^{6} m) \sqrt{\frac{(9.8 m s^{- 2})}{(6.38 \times 10^{6} m + 0.25 \times 10^{6} m)}}$
$= 7.76 \times 10^{3} m s^{- 1} = 7.76 km s^{- 1}$
$(∵ 1 km = 10^{3} m)$

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