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

Answer:

Here, h=1000 km=1000×10

3

m=10

6

m

r=R+h=6.38×10

6

+10

6

=7.38×10

6

m

Orbital speed,

v

0

=

R+h

GM

=

7.38×10

6

6.67×10

−11

×6×10

24

=7364m s

−1

Time period,

T=

v

0

2πr

=

7.364×10

3

2×(22/7)×(7.38×10

6

)

=6297s.