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A spherical planet has a mass $M_p$ and diameter $D_p$. A particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity, equal to
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Verified Answer
The correct answer is:
$4 G M_p / D_p^2$
Force,
$F=\frac{G M_e m}{R^2}$
Here, $F=\frac{G M_P m}{\left(D_P / 2\right)^2}$
$\begin{aligned}
& F=\frac{4 G M_P m}{D_P^2} \\
& F=m a
\end{aligned}$
Acceleration due to gravity
$a=\frac{F}{m}=\frac{4 G M_P}{D_P^2}$
$F=\frac{G M_e m}{R^2}$
Here, $F=\frac{G M_P m}{\left(D_P / 2\right)^2}$
$\begin{aligned}
& F=\frac{4 G M_P m}{D_P^2} \\
& F=m a
\end{aligned}$
Acceleration due to gravity
$a=\frac{F}{m}=\frac{4 G M_P}{D_P^2}$
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