how much below the surface of the earth does the value of g become 75% of it's value on the surface? take the radus of the earth =6400km
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Explanation:
Let the depth at which acceleration due to gravity(g) becomes 70% or 7g/10 =d
We know that
g
d
=g(1−
R
d
)
Given
g
d
=
10
7g
Also, R=6400km (Radius of earth)
10
7g
=g(1−
6400
d
)
10
7
=1−
6400
d
6400
d
=1−
10
7
6400
d
=
10
3
d=
10
3
×6400
d=1920km
Therefore,
1920km below the surface of earth does the acceleration due to gravity become 70% of its value at the surface of earth.
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