Question

At what height the acceleration due to gravity will be reduced by 36 % its value o the surface of the earth?

Answer 1

Given,

Radius of earth, $R=6400km$

Gravity reduced 36 %, $g^{\prime }=(1-\frac{36}{100})g=0.64g$

Gravity acceleration above height $h$ from the earth surface.

$g^{\prime }=g{\left[\frac{R}{R+h}\right]}^2$

$0.64g=g{\left[\frac{R}{R+h}\right]}^2$

$0.8(R+h)=R$

$h=\frac{R}{4}=\frac{6400}{4}=1600km$

Hence, height is one fourth of earth radius, $1600km$