A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25×106m above the surface of earth. If earth's radius is 6.38×106 m and g=9.8 ms-2, then the orbital speed of the satellite is :

For the satellite revolving around earthv0=GMe(Re+h)=GMeRe1+hRe=gRe1+hReSubstituting the valuesv0=60×106 m/s v0=7.76×103 m/s =7.76 km/s

A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25×106m above the surface of earth. If earth's radius is 6.38×106 m and g=9.8 ms-2, then the orbital speed of the satellite is :

For the satellite revolving around earthv0=GMe(Re+h)=GMeRe1+hRe=gRe1+hReSubstituting the valuesv0=60×106 m/s v0=7.76×103 m/s =7.76 km/s

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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 $g = 9.8 m s^{- 2}$ , then the orbital speed of the satellite is :

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$9.13 k m s^{- 1}$

$6.67 k m s^{- 1}$

$8.56 k m s^{- 1}$

$7.76 k m s^{- 1}$

answer is B.

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Detailed Solution

For the satellite revolving around earth

$v_{0} = \sqrt{\frac{G M_{e}}{(R_{e} + h)}} = \sqrt{\frac{G M_{e}}{R_{e} (1 + \frac{h}{R_{e}})}} = \sqrt{\frac{g R_{e}}{1 + \frac{h}{R_{e}}}}$

Substituting the values

$v_{0} = \sqrt{60 \times 10^{6}} m / s v_{0} = 7.76 \times 10^{3} m / s = 7.76 k m / s$

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