Question

Calculate the distance from the surface of the earth at which the acceleration due to gravity is the same below and above the surface of the earth.

Answer 1

Gravitaitonal field inside the earth varties as

$g^{\prime }=\frac{GM}{r^2}\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3}=\frac{GMr}{R^3}=g\frac{r}{R}$

Above the surface, gravitational field varies as

$g^{\prime }=\frac{GM}{r^2}=g\frac{R^2}{r^2}$

Let the required height be h. Then,

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

$(R-h){(R+h)}^2=R^3$

One solution to the equation is h=0.

Required solution is $h=\frac{\sqrt{5}-1}{2}R$

Answer is $\frac{\sqrt{5}-1}{2}R$.