At what height above the surface of the earth will the acceleration due to gravity becomes 1 percent of its value at the earth surface given radius of earth is 6400km
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Answered by
2
GM/(R+h)²=(1/100)GM/R²
1/(R+h)²=1/100R²
(R+h)²=100R²
square root both the sides:
R+h= 10R
h= 10R-R
h=9R
therefore the height must be 9 times the radius of the earth above the surface of earth.
9×6400= 57600km
1/(R+h)²=1/100R²
(R+h)²=100R²
square root both the sides:
R+h= 10R
h= 10R-R
h=9R
therefore the height must be 9 times the radius of the earth above the surface of earth.
9×6400= 57600km
Answered by
0
Given : d=1600km R=6400km
Acceleration due to gravity at a depth d, g
d
=g
s
(1−
R
d
)
where g
s
=9.8ms
−2
is acceleration due to gravity at earth's surface.
∴ g
d
=9.8(1−
6400
1600
)=7.35ms
−2
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