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A body is projected from earth's with thrice the escape velocity from the surface of the earth. What will be its velocity when it will escape the gravitational pull?
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Verified Answer
The correct answer is:
$2 \sqrt{2} \mathrm{~V}_{\mathrm{e}}$
Energy required to escape the earth's gravitational field is $\frac{1}{2} \mathrm{mV}_e^2$
Energy given to the body is $=\frac{1}{2} \mathrm{~m}\left(3 \mathrm{~V}_{\mathrm{e}}\right)^2$
$=\frac{9}{2} \mathrm{mV}_{\mathrm{e}}^2$
$\therefore$ If $\mathrm{V}$ is velocity of the body when it has escaped from earth's gravitational field then
$$
\begin{aligned}
& \frac{1}{2} \mathrm{mV}^2=\frac{9}{2} \mathrm{mV}_{\mathrm{e}}^2-\frac{1}{2} \mathrm{mV}_{\mathrm{e}}^2 \\
& \therefore \frac{1}{2} \mathrm{mV}^2=4 \mathrm{mV}_{\mathrm{e}}^2 \\
& \therefore \mathrm{V}^2=8 \mathrm{~V}_{\mathrm{e}}^2 \\
& \mathrm{~V}=2 \sqrt{2} \mathrm{~V}_{\mathrm{e}}
\end{aligned}
$$
Energy given to the body is $=\frac{1}{2} \mathrm{~m}\left(3 \mathrm{~V}_{\mathrm{e}}\right)^2$
$=\frac{9}{2} \mathrm{mV}_{\mathrm{e}}^2$
$\therefore$ If $\mathrm{V}$ is velocity of the body when it has escaped from earth's gravitational field then
$$
\begin{aligned}
& \frac{1}{2} \mathrm{mV}^2=\frac{9}{2} \mathrm{mV}_{\mathrm{e}}^2-\frac{1}{2} \mathrm{mV}_{\mathrm{e}}^2 \\
& \therefore \frac{1}{2} \mathrm{mV}^2=4 \mathrm{mV}_{\mathrm{e}}^2 \\
& \therefore \mathrm{V}^2=8 \mathrm{~V}_{\mathrm{e}}^2 \\
& \mathrm{~V}=2 \sqrt{2} \mathrm{~V}_{\mathrm{e}}
\end{aligned}
$$
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