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7.Gravitation
normal
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
