Question

How much the surface of earth does the acceleration due to gravity reduce by $36\%$ of its value on the surface of earth? Radius of earth $=6400$km.

Answer 1

Since the acceleration due to gravity reduces by $36\%$, the value of acceleration due to gravity there is $100-36=64\%$
It means, $g^{\prime }=\frac{64}{100}$g. if h is the height of location above the surface of earth, then
$g^{\prime }=g\frac{R^2}{(R+h{)}^2}$
or $\frac{64}{100}g=g\frac{R^2}{(R+h{)}^2}$
$\frac{8}{10}=\frac{R}{R+h}$ or $8R+8h=10R$
$\Rightarrow h=\frac{2R}{8}=\frac{R}{4}=\frac{6.4\times {10}^6}{4}=1.6\times {10}^6$m.