Question

An artificial satellite revolves around the earth at a height of 1000 km .

The radius of the earth is 6.38 × 10 ^-11 Nm^2 kg^-2.

Find its orbital velocity and period of revolution

Answer 1

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Answer 2

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R = 6000 km

= 6×10^6 m

h = 1000 km

= 10^6 m

g = 10 m/s^2

Orbital velocity of satellite

v = √[g(R+h)]

v = √[10(6×10^6+10^6)]

v = √(7×10^7)

v = 8.36×10³

v = 8360m/s