how far away from the surface of earth does the acceleration due to gravity becomes 4% of its value on surface of earth?
Answers
Answered by
77
Acceleration due to gravity at height h is given as
g’ = g [R / (R + h)]^2
4g/100 = g [R / (R + h)]^2
0.04 = [R / (R + h)]^2
0.2 = R / (R + h)
0.2R + 0.2h = R
0.2h = 0.8R
h = 4R
It will be 4% that on surface of earth at height equal to 4R (where R is radius of Earth)
g’ = g [R / (R + h)]^2
4g/100 = g [R / (R + h)]^2
0.04 = [R / (R + h)]^2
0.2 = R / (R + h)
0.2R + 0.2h = R
0.2h = 0.8R
h = 4R
It will be 4% that on surface of earth at height equal to 4R (where R is radius of Earth)
Answered by
1
Answer:
Acceleration due to gravity g
′
=
100
4
g
s
Using g
′
=g
s
[
R+h
R
]
2
Or
100
4
g
s
=g
s
[
R+h
R
]
2
Or
R+h
R
=
10
2
⟹ h=4R=4×6400=25600 km
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