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  • 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 is__________S.Option: 1 <img alt="\sqrt{\fra

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.
Option: 1 \sqrt{\frac{\mathrm{R}_{\mathrm{e}}}{2 \mathrm{~g}}}\left[\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{3 / 2}-1\right]
Option: 2 \sqrt{\frac{2 \mathrm{R}_{\mathrm{e}}}{\mathrm{g}}}\left[\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{3 / 2}-1\right]
Option: 3 \frac{1}{3} \sqrt{\frac{\mathrm{R}_{\mathrm{e}}}{2 \mathrm{~g}}}\left[\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{3 / 2}-1\right]
Option: 4 \frac{1}{3} \sqrt{\frac{2 \mathrm{R}_{\mathrm{e}}}{\mathrm{g}}}\left[\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{3 / 2}-1\right]

Answers (1)

best_answer

\begin{gathered} V_{i}=V_{e}=\sqrt{\frac{2 G M}{R}} \\ \end{gathered}

Speed of projection  \Rightarrow \begin{gathered} V_{i}=\sqrt{2 g R} \end{gathered}

 

At height x above the (pt. p) above the Earth surface, the gravitational force acting is F=\frac{G M m}{(R+x)^{2}}

\begin{gathered} a=\frac{G M}{(R+x)^{2}} \\ \end{gathered}

\begin{gathered} \because a=\frac{d v}{d t} \ \& \ \frac{d x}{d t}=v \\ \end{gathered}

\begin{gathered} T E_{\text {surface }}=T E p \\ \end{gathered}

\begin{gathered} -\frac{G M m}{R}+\frac{1}{2} m v_{i}^{2}=-\frac{G M m}{(R+x)}+\frac{1}{2} m v^{2} \\ \end{gathered}

\begin{gathered} 0 =\frac{-G M m}{(R+x)}+\frac{1}{2} m v^{2} \\ \end{gathered}

\begin{gathered} \frac{d x}{d t}=v=\sqrt{\frac{2 G M}{(R+x)}} \\ \end{gathered}

\begin{gathered} \sqrt{2GM} \int dt=\int_{0}^{h} \sqrt{R+x} \: d x \\ \end{gathered}

\begin{aligned} \sqrt{2 G M} &t=\left[\frac{(R+x)^{3 / 2}}{3 / 2}\right]_{0}^{h} \\ t &=\left[\frac{2}{3}\left[(R+h)^{3 / 2}-R^{3 / 2}\right]\right] \times \frac{1}{\sqrt{2 G M}} \end{aligned}

\begin{aligned} &t=\frac{2}{3} \frac{R^{3 / 2}}{\sqrt{2 G M}}\left[\left(1+\frac{h}{R}\right)^{3 / 2}-1\right] \\ &t=\frac{2}{3} \sqrt{\frac{R^{3}}{2 G M}}\left[\left(1+\frac{h}{R}\right)^{3 / 2}-1\right] \\ &t=\frac{2}{3} \sqrt{\frac{R}{2 g}}\left[\left(1+\frac{h}{R}\right)^{3 / 2}-1\right] \\ &\because g=\frac{G M}{R^{2}} \end{aligned}

The correct option is (4)

Posted by

vishal kumar

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