Question 128 of 223

A body of mass $m$ is projected with velocity $\lambda {v}_{e}$ in vertically upward direction from the surface of the earth into space. It is given that ${v}_{e}$ is escape velocity and $\lambda \lt 1$. If air resistance is considered to the negligible, then the maximum height from the centre of earth, to which the body can go, will be ($R$ : radius of earth)

A$\dfrac{R}{1+{\lambda }^{2}}$B$\dfrac{R}{1-{\lambda }^{2}}$C$\dfrac{R}{1-\lambda }$D$\dfrac{{\lambda }^{2}R}{1-{\lambda }^{2}}$

Show answer and solution▾

Answer: (B) $\dfrac{R}{1-{\lambda }^{2}}$

Escape velocity is given by, ${v}_{e}=\sqrt{\dfrac{2G{M}_{e}}{{R}_{e}}}$
If the body goes to the height $h$ from the surface of the earth, using mechanical energy conservation,
$\Rightarrow \dfrac{1}{2}m{\left(\lambda \sqrt{\dfrac{2G{M}_{e}}{{R}_{e}}}\right)}^{2}-\dfrac{G{M}_{e}m}{{R}_{e}}=-\dfrac{G{M}_{e}m}{{R}_{e}+h}$
$\Rightarrow {\lambda }^{2}\dfrac{G{M}_{e}m}{{R}_{e}}-\dfrac{G{M}_{e}m}{{R}_{e}}=-\dfrac{G{M}_{e}m}{{R}_{e}+h}$
$\Rightarrow \left({R}_{e}+h\right)=\dfrac{{R}_{e}}{1-{\lambda }^{2}}$