An Earth satellite is in circular orbit at a distance of 384,000 km from the Earth's surface. What is its period of revolution in days? Take mass of the Earth
M= 6.0×10²⁴ and its Radius= 6400km.
Note: I have added a picture of solution from my book. But when I put the values of orbital velocity as in picture in the calculator the answer is wrong. Pls Help
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Answer:
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.
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