Question

At which height from earth surface acceleration due to gravity is decreased by 75% of its value at earth surface?

Answer 1

Since, the acceleration due to gravity reduces by 75%, then the value of acceleration due to gravity there is = 100 - 75= 25%.

It means, g' = 25/100 g.

If h is the height of location above the surface of earth(here h is not very less than R, therefore), then

$g^{\prime }=\frac{g}{{(1+\frac{h}{R})}^2}$

$\frac{g^{\prime }}{g}=\frac{1}{{(1+\frac{h}{R})}^2}$

$⟹\frac{25}{100}=\frac{1}{{(1+\frac{h}{R})}^2}$

$⟹1+\frac{h}{R}=2$

$⟹R+h=2R$

$⟹h=R$

Therefore at R = 6400 km from earth surface acceleration due to gravity is decreased by 75% of its value at earth surface.