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A rocket is launched normal to the surface of earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is $3 \times 10^5$ times heavier than the Earth and is at a distance $2.5 \times 10^4$ times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is $\mathrm{ve}_{\mathrm{e}}=11.2 \mathrm{~km} \mathrm{~s} \mathrm{~s}^{-1}$. The minimum initial $\left(\mathrm{v}_{\mathrm{s}}\right)$ required for the rocket to be able to leave the Sun-earth system is closest to
(Ignore the rotation and revolution of the Earth and the presence of any other planet)
$\mathrm{V}_5=22 \mathrm{~km} \mathrm{~s}^{-1}$
$\mathrm{v}_5=42 \mathrm{~km} \mathrm{~s}^{-1}$
$\mathrm{v}_5=62 \mathrm{~km} \mathrm{~s}^{-1}$
$\mathrm{v}_5=72 \mathrm{~km} \mathrm{~s}^{-1}$
Solution
$\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^2-\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{\mathrm{G}\left(3 \times 10^5 \mathrm{M}\right) \times \mathrm{m}}{2.5 \times 10^4 \mathrm{R}}=0$
$\Rightarrow \mathrm{V}_{\mathrm{e} 5}=\sqrt{13} \quad \mathrm{ve}^2=40.3 \mathrm{~km} / \mathrm{s}$
