Question

The variation of acceleration due to gravity $g$ with distance $d$ from the center of the earth is best represented by ($R=$ Earth’s radius):

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

As we know that acceleration due to gravity on the surface of the earth is $g_s=\frac{GM}{R^2}$
Acceleration due to gravity inside the earth (at depth $d$) : -
$\because g_s=\frac{GM}{R^2}$ ... (1)
$\because \rho =\frac{M}{V}=\frac{M}{\frac{4}{3}\pi R^3}$
$\Rightarrow M=\rho \left(\frac{4}{3}\pi R^3\right)$
$\therefore$ From equation (1)
$g_s=\frac{G}{R^2}\left(\rho \frac{4}{3}\pi R^3\right)$
$g_s=\frac{4}{3}G\rho \pi R$
at depth $d:g^{\prime }=\frac{4}{3}\pi \rho G(R-d)$
$\therefore$linear relation
Acceleration due to gravity above the surface of the earth:
$\because g_s=\frac{GM}{R^2}$
at height $h$
$g^{\prime }=\frac{GM}{(R+h{)}^2}$
$g^{\prime }=\frac{GM}{R^2{(1+\frac{h}{R})}^2}$
$\therefore$ relation is parabola
$\therefore$ The correct option is (A).