JEE MAIN - Physics (2023 - 15th April Morning Shift - No. 10)
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.)
(Given $g=$ acceleration due to gravity on the earth.)
$\sqrt{\frac{g R}{2}}$
$\sqrt{4 g R}$
$\sqrt{2 g R}$
$\sqrt{g R}$
Explanation
_15th_April_Morning_Shift_en_10_1.png)
By conservation of mechanical energy
$$ \begin{aligned} & \mathrm{U}_{\mathrm{i}}+\mathrm{K}_{\mathrm{i}}=\mathrm{U}_{\mathrm{f}}+\mathrm{K}_{\mathrm{i}} \\\\ & -\frac{\mathrm{GMm}}{2 \mathrm{R}}+0=-\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}^2 \\\\ & \frac{\mathrm{GMm}}{2 \mathrm{R}}=\frac{1}{2} \mathrm{mv}^2 \\\\ & \mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}=\sqrt{\mathrm{gR}} \end{aligned} $$
Comments (0)
