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 $$6.38\times 10^6m$$ and $$g = 9.8 ms^{-2}$$, then the orbital speed of the satellite is:

A
6.67kms1
B
7.76kms1
C
8.56kms1
D
9.13kms1

Solution

## The correct option is D $$7.76 \,km \,s^{-1}$$Orbital speed: $$V = \sqrt{\frac{GM}{r}}= \sqrt{ \frac{GM}{R^2}\frac{R^2}{r}}=\sqrt{ \frac{gR^2}{r} }= \sqrt{ \frac{9.8 \times ( 6.38 \times 10^6)^2}{ 0.25 \times 10^6}}=7.76\: km/s$$Physics

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