A rocket is launched normal to the surface of the 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 $v_{e} = 11.2 k m / s$ . The minimum initial velocity $(v_{s})$ 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 $)$

$v_{s} = 62 k m / s$

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$v_{s} = 42 k m / s$

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v_{s} = 72 k m / s

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v_{s} = 22 k m / s

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Solution

The correct option is B
$v_{s} = 42 k m / s$

For earth, the escape velocity $v_{e} = \sqrt{\frac{2 G M}{R}}$
and for the system (Sun +Earth),
potential energy $= (- \frac{G M}{R} - \frac{3 \times 10^{5}}{2.5 \times 10^{4}} \frac{G M}{R}) m$
$= - \frac{G M}{R} (1 + 12) = - \frac{13 G M}{R}$
$∴$ $\frac{1}{2} m v^{2} = \frac{13 G M m}{R}$
$v = \sqrt{\frac{2 G M}{R}} . \sqrt{13} = v_{e}$
$∴$ $v_{e} = 40.4 k m / s \approx 42 k m / s .$

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The correct option is B vs=42 km/s For earth, the escape velocity ve=√2GMR and for the system (Sun +Earth), potential energy =(−GMR−3×1052.5×104GMR)m =−GMR(1+12)=−13GMR ∴ 12mv2=13GMmR v=√2GMR.√13=ve ∴ ve=40.4 km/s≈42 km/s.