Question 302: Gravitation

Question

A body (mass $\mathrm{m}$ ) starts its motion from rest from a point distant $R_0\left(R_0>R\right)$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of earth will be ( $\mathrm{G}=$ universal constant of gravitation, $\mathrm{M}=$ mass of earth, $\mathrm{R}$ = radius of earth)

Accepted Answer

According to law of conservation of energy,
$\begin{aligned}
& \frac{1}{2} m v^2=-\frac{G M m}{R_0}-\left(\frac{G M m}{R}ight)=G M m\left(\frac{1}{R}-\frac{1}{R_0}ight) \
	herefore \quad & v^2=2 G M\left(\frac{1}{R}-\frac{1}{R_0}ight)
\end{aligned}$
$	herefore \quad$ The velocity acquired by the body when it reaches the surface of earth is:
$\mathrm{V}=\left[2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}ight)ight]^{\frac{1}{2}}$
Amongst the given options, only option (B) dimensionally equates to velocity.

Answer

$2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)$

Answer

$\left[2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\right]^{\frac{1}{2}}$

Answer

$\mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)$

Answer

$2 \mathrm{GM}\left[\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\right]^{\frac{1}{2}}$

Gravitation

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