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

Here, h=1000 km=1000×103m=106m

r=R+h=6.38×106+106=7.38×106m

Orbital speed,

v0=R+hGM=7.38×1066.67×10−11×6×1024=7364m s−1

Time period,

T=v02πr=7.364×1032×(22/7)×(7.38×106)=6297s.

hope it helps u.....✌✌