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Question

An artificial satellite revolves round the earth at a height of 1000km. The radius of the earth is 6.38×103km. Mass of the earth =6×1024kg; G=6.67×1011 Nm2 kg2. Find the orbital speed and period of revolution of the satellite.

A
7364m s1, 6297s.
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B
1064m s1, 6297s.
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C
9364m s1, 9297s.
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D
8364m s1, 7297s.
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Solution

The correct option is A 7364m s1, 6297s.
Here, h=1000 km=1000×103m=106m
r=R+h=6.38×106+106=7.38×106m
Orbital speed,
v0=GMR+h=6.67×1011×6×10247.38×106=7364m s1
Time period,
T=2πrv0=2×(22/7)×(7.38×106)7.364×103=6297s.

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