Question 135 of 223

A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height $h$ is $S.$

A$\sqrt{\dfrac{{R}_{e}}{2g}}\left[{\left(1+\dfrac{h}{{R}_{e}}\right)}^{\dfrac{3}{2}}-1\right]$B$\sqrt{\dfrac{2{R}_{e}}{g}}\left[{\left(1+\dfrac{h}{{R}_{e}}\right)}^{\dfrac{3}{2}}-1\right]$C$\dfrac{1}{3}\sqrt{\dfrac{{R}_{e}}{2g}}\left[{\left(1+\dfrac{h}{{R}_{c}}\right)}^{\dfrac{3}{2}}-1\right]$D$\dfrac{1}{3}\sqrt{\dfrac{2{R}_{c}}{g}}\left[{\left(1+\dfrac{h}{{R}_{e}}\right)}^{\dfrac{3}{2}}-1\right]$

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Answer: (D) $\dfrac{1}{3}\sqrt{\dfrac{2{R}_{c}}{g}}\left[{\left(1+\dfrac{h}{{R}_{e}}\right)}^{\dfrac{3}{2}}-1\right]$

Applying energy conservation from ($\left(1\right)\text{to}\left(2\right)$,
$\dfrac{1}{2}m\cdot \left(\dfrac{2GM}{{R}_{e}}\right)-\dfrac{GMm}{{R}_{e}}=\dfrac{1}{2}m{v}^{2}-\dfrac{GMm}{R+r}$
$\Rightarrow \dfrac{1}{2}m{v}^{2}=\dfrac{GMm}{R+r}$
$\Rightarrow v=\sqrt{\dfrac{2GM}{R+r}}=\dfrac{dr}{dt}$
$\Rightarrow \sqrt{2GM}{\int }_{0}^{t}dt={\int }_{{R}_{e}}^{{R}_{e}+h}(\sqrt{R+r})dr$
$\sqrt{2GM}\cdot t=\dfrac{2}{3}{\left[(R+r{)}^{\dfrac{3}{2}}\right]}_{{R}_{e}}^{{R}_{e}+h}$
$t=\dfrac{2}{3}\sqrt{\dfrac{{R}_{e}^{3}}{2GM}}\left[{\left(1+\dfrac{h}{{R}_{e}}\right)}^{\dfrac{3}{2}}-1\right]$
$\dfrac{GM}{{R}_{e}^{2}}=g$
$t=\dfrac{1}{3}\sqrt{\dfrac{2{R}_{e}}{g}}\left[{\left(1+\dfrac{h}{{R}_{e}}\right)}^{\dfrac{3}{2}}-1\right]$

JEE Main Physics practice question on Gravitation (Gravitational potential and Potential Energy), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.