Gravity Reduce the gravity

Classical Mechanics Level 3

The height, vertically above, at which the acceleration due to gravity becomes 1% as compared to that on the surface of the Earth is _ _ _ _ _ _ _ _ ?

DETAILS AND ASSUMPTIONS:

1) Provide the answer in terms of R, "NOT" in kilometers.

2) R here means the "RADIUS OF THE EARTH".

For more such problems, try my set Gravity

-10 10 -9 9

Sravanth C. by Sravanth C. , 21, India

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2 solutions

Suppose

g ′ g' is the gravitational acceleration at h h above the surface of the earth,

g g is the gravitational acceleration at the surface of the earth.

It is given that:

g ′ g = 0.01 \frac{g'}{g} = 0.01

We know that,

g ′ = G M ( R + h ) 2 g' = \frac{GM}{(R+h)^{2}}

g = G M R 2 g = \frac{GM}{R^2}

So,

g ′ g = ( R R + h ) 2 \frac{g'}{g} = (\frac{R}{R+h})^{2}

( R R + h ) 2 = 0.01 (\frac{R}{R + h})^{2} = 0.01

R R + h = 0.1 \frac{R}{R+h} = 0.1

R + h R = 10 \frac{R + h}{R} = 10

R + h = 10 R R+h=10R

h = 9 R h = \boxed{9R}

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Rishabh Tiwari
May 10, 2016

We know g=GM/R^2....(1), & g'=GM/(R+h)^2..... (2) Divide eq. (1) by (2) So g/g' = (R+h)^2/R^2.... Now since given g/g' = 100. Hence we simplify our answer as →(R+h)/R=10 → h=9R. Ans.

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