Calculate the value of theacceleration due to gravity at a place 3200 above the earth surface
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Shubusingh58:
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Answered by
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Hey bro!!!
I think you're question is incomplete...
I know this question so I can answer...
The acceleration due to gravity near earth surface is -
g=GMR2g=GMR2
At height hh, above earths surface, the gravity is
gh=GM(R+h)2gh=GM(R+h)2
Taking the ratio of the two
ggh=(R+h)2R2ggh=(R+h)2R2
Substitute the given value gh=4g9gh=4g9
(R+h)2R2=94(R+h)2R2=94
(R+h)R=32(R+h)R=32
1+hR=321+hR=32
hR=12hR=12
R=2hR=2h
R=2(3200)R=2(3200) km
R=6400R=6400 km
hope it helps
I think you're question is incomplete...
I know this question so I can answer...
The acceleration due to gravity near earth surface is -
g=GMR2g=GMR2
At height hh, above earths surface, the gravity is
gh=GM(R+h)2gh=GM(R+h)2
Taking the ratio of the two
ggh=(R+h)2R2ggh=(R+h)2R2
Substitute the given value gh=4g9gh=4g9
(R+h)2R2=94(R+h)2R2=94
(R+h)R=32(R+h)R=32
1+hR=321+hR=32
hR=12hR=12
R=2hR=2h
R=2(3200)R=2(3200) km
R=6400R=6400 km
hope it helps
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