Question

Assuming that the earth has constant density, at what distance $\left(d\text{in km}\right)$ from the earth's surface the gravity above the earth is equal to that below the surface.

Answer 1

Formula used: $g_h=g_0\frac{R^2}{(R+h{)}^2}$,

$g_d=g_0(1-\frac{d}{R})$

If $g_0$ is the gravity at the earth's surface, $g_h$ at
height $h$ and $g_d$ at depth $d$, then

$g_h=g_0\frac{R^2}{(R+h{)}^2}...\left(i\right)$

$g_d=g_0(1-\frac{d}{R})...\left(ii\right)$

Since, $g_d=g_h$ at $h=d...\left(iii\right)$

From $\left(i\right)$ and $\left(ii\right)$, we get

$d^2+dR-R^2=0$

$d=\frac{(\sqrt{5}-1)}{2}R$

$=0.618R$

$=0.618\times 6400$

$=3955km$

Final Answer: $3955$