Question 99: Gravitation

Question

A body is released from a height equal to the radius $(\mathrm{R})$ of the earth. The velocity of the body when it strikes the surface of the earth will be

(Given $g=$ acceleration due to gravity on the earth.)

Accepted Answer

By conservation of mechanical energy
$$
\begin{aligned}
& \mathrm{U}_{\mathrm{i}}+\mathrm{K}_{\mathrm{i}}=\mathrm{U}_{\mathrm{f}}+\mathrm{K}_{\mathrm{i}} \\
& -\frac{\mathrm{GMm}}{2 \mathrm{R}}+0=-\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}^2 \\
& \frac{\mathrm{GMm}}{2 \mathrm{R}}=\frac{1}{2} \mathrm{mv}^2 \\
& \mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}=\sqrt{\mathrm{gR}}
\end{aligned}
$$

Answer

$\sqrt{\frac{g R}{2}}$

Answer

$\sqrt{4 g R}$

Answer

$\sqrt{2 g R}$

Answer

$\sqrt{g R}$

Gravitation

A body is released from a height equal to the radius (\mathrm{R}) of the earth. The velocity of the body when it strikes the surface of the earth will be (Given g= acceleration due to gravity on the earth.)

Choose the correct answer: