A body is projected vertically upwards from earth's surface with velocity $2 \mathrm{v}_{\mathrm{e}}$, where…
A body is projected vertically upwards from earth's surface with velocity $2 \mathrm{v}_{\mathrm{e}}$, where
$\mathrm{v}_{\mathrm{e}}$ is escape velocity from earth's surface. The velocity when body escapes the
gravitational pull is
$\sqrt{7} \mathrm{v}_{\mathrm{e}}$
$\sqrt{3} \mathrm{v}_{\mathrm{e}}$
$\sqrt{5} \mathrm{v}_{\mathrm{e}}$
$2 \mathrm{v}_{\mathrm{e}}$
Solution
$\begin{aligned} \text { The kinetic energy given to the body }=\mathrm{k} &=\frac{1}{2} \mathrm{~m}(2 \mathrm{Ve})^{2} \\ &=\frac{1}{2}\left(4 \mathrm{mV}_{\mathrm{o}}^{2}\right) \end{aligned}$
The kinetic energy required to escape from the earth's gravitational field $=\mathbf{K}^{\prime}=\frac{1}{2} \mathbf{m V}_{0}^{2}$
If its velocity is $\mathrm{V}$ after escaping from the earth's gravitational field then its kinetic energy will be
$\begin{aligned}
\frac{1}{2} \mathrm{mV}^{2} &=\frac{1}{2}\left(4 \mathrm{mV}_{\mathrm{e}}^{2}\right)-\frac{1}{2} \mathrm{mV}_{\mathrm{e}}^{2} \\
&=\frac{1}{2}\left(3 \mathrm{mV}_{\mathrm{e}}^{2}\right) \\
\therefore \mathrm{V}^{2} &=3 \mathrm{~V}_{\mathrm{e}}^{2} \\
\therefore \mathrm{V} &=\sqrt{3} \mathrm{~V}_{\mathrm{e}}
\end{aligned}$
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