Question 278: Gravitation

Question

A body is projected in vertically upward direction from the surface of the earth of radius ' $R$ ' into space with velocity ' $n V_{\mathrm{e}}$ ' $(\mathrm{n}<1)$. The maximum height from the surface of earth to which a body can reach is

Accepted Answer

As per the law of conservation of energy,
$(\mathrm{K} . \mathrm{E}+\mathrm{P} . \mathrm{E})_{	ext {surface }}=(\mathrm{K} . \mathrm{E}+\mathrm{P} . \mathrm{E})_{	ext {max height }}$
$$\begin{aligned}
& 	herefore \quad \frac{-G M m}{R}+\frac{1}{2} m\left(\mathrm{nV}_{\mathrm{e}}ight)^2=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}+\frac{1}{2} \mathrm{~m}(0)^2 \
& 	herefore \quad \frac{1}{2} \mathrm{~m}\left(\mathrm{nV}_{\mathrm{e}}ight)^2=\mathrm{GMm}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}ight) \
& 	herefore \quad\left(\mathrm{nV}_{\mathrm{e}}ight)^2=2 \mathrm{GM}\left(\frac{\mathrm{R}+\mathrm{h}-\mathrm{R}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}ight) \
& 	herefore \quad\left(\mathrm{nV}_{\mathrm{e}}ight)^2=\mathrm{V}_{\mathrm{e}}^2\left(\frac{\mathrm{~h}}{(\mathrm{R}+\mathrm{h})}ight) \
& 	herefore\quad \frac{1}{\mathrm{n}^2}=\frac{\mathrm{R}}{\mathrm{~h}}+1 \
& 	herefore \quad \frac{\mathrm{R}}{\mathrm{~h}}=\frac{1-\mathrm{n}^2}{\mathrm{n}^2} \
& 	herefore\quad \mathrm{h}=\frac{\mathrm{n}^2 \mathrm{R}}{\left(1-\mathrm{n}^2ight)}
\end{aligned}
$$

Answer

$\frac{\mathrm{n}^2 \mathrm{R}}{\left(1-\mathrm{n}^2\right)}$

Answer

$\frac{\mathrm{n}^2 \mathrm{R}^2}{(1-\mathrm{n})}$

Answer

$\frac{n R^2}{\left(1+n^2\right)}$

Answer

$\frac{\mathrm{n}^2 \mathrm{R}^2}{(1+\mathrm{n})}$

Gravitation

A body is projected in vertically upward direction from the surface of the earth of radius ' R ' into space with velocity ' n V_{\mathrm{e}} ' (\mathrm{n}<1). The maximum height from the surface of earth to which a body can reach is

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