calculate the period of revolution of satellite revolving at a distance of 20 km above the surface of the earth?
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Answer:
Orbital velocity,
4v
0
=
R
e
+h
GM
e
6380+2620
(6.67×10
−11
)(6×10
24
)
=6.67km/sec
Timeperiodofrevolution,
T=2π
v
0
(R
e
+h)
=
6.67×10
3
2×3.14×(6380+2620)
=8474sec=2.35hours
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