JEE MAIN - Physics (2022 - 27th July Evening Shift - No. 6)
A body of mass $$\mathrm{m}$$ is projected with velocity $$\lambda \,v_{\mathrm{e}}$$ in vertically upward direction from the surface of the earth into space. It is given that $$v_{\mathrm{e}}$$ is escape velocity and $$\lambda<1$$. If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be :
(R : radius of earth)
$$\frac{\mathrm{R}}{1+\lambda^{2}}$$
$$\frac{R}{1-\lambda^{2}}$$
$$\frac{R}{1-\lambda}$$
$$\frac{\lambda^{2} \mathrm{R}}{1-\lambda^{2}}$$
Explanation
Using energy conservation
$$ - {{G{M_e}m} \over {{R_e}}} + {1 \over 2}m{\left( {\lambda \sqrt {{{2G{M_e}} \over {{R_e}}}} } \right)^2} = - {{G{M_e}m} \over r}$$
$${{G{M_e}m} \over r} = {{G{M_e}m} \over {{R_e}}} - {{G{M_e}m} \over {{R_e}}}{\lambda ^2}$$
$$r = {{{R_e}} \over {1 - {\lambda ^2}}}$$
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