Question
Prove that from the surface of earth :
(i) accleration due to the gravity at a height $h$$
g_h=g\left(1-\frac{2 h}{ R }\right)
$(ii) Acceleration due to gravity at depth $h ^{\prime}$$
g_h^{\prime}=g\left(1-\frac{h^{\prime}}{ R }\right)
$
Where g is the gravitational acceleration on earth's surface and $R$ is the radius of the Earth. (Here assume $\rho$ and $h \ll R)$
(i) accleration due to the gravity at a height $h$$
g_h=g\left(1-\frac{2 h}{ R }\right)
$(ii) Acceleration due to gravity at depth $h ^{\prime}$$
g_h^{\prime}=g\left(1-\frac{h^{\prime}}{ R }\right)
$
Where g is the gravitational acceleration on earth's surface and $R$ is the radius of the Earth. (Here assume $\rho$ and $h \ll R)$
