A rocket is projected in vertically upwards direction with a velocity kve where vₑ is escape veloc

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7.Gravitation

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A rocket is projected in the vertically upwards direction with a velocity kve where $v_e$ is escape velocity and $k < 1$. The distance from the centre of earth upto which the rocket will reach, will be

A

$R_e\,(1 -k^2)$

B

$\frac{{\left( {1 - {k^2}} \right)}}{{{R_e}}}$

C

$\sqrt {R_e}\, (1 -k^2)$

D

$\frac{{{R_e}}}{{\left( {1 - {k^2}} \right)}}$

Solution

According to conservation of mechanical energy

$\mathrm{K}_{1}+\mathrm{u}_{1}=\mathrm{K}_{2}+\mathrm{u}_{2}$

$\frac{1}{2} m k^{2} v_{e}^{2}-\frac{G M m}{R}=0-\frac{G M m}{r}$

on solving $\mathrm{r}=\frac{\mathrm{R}}{1-\mathrm{k}^{2}}$

Standard 11

Physics

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