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Q.

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  3×105 times heavier than the Earth and is at a distance 2.5×104 times larger than the radius of the Earth. The escape velocity from Earth’s gravitational field is ve=11.2 km s−1. The maximum initial velocity vs 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)

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a

vS=72 km s−1

b

vS=22 km s−1

c

vS=42 km s−1

d

vS=62 km s−1

answer is C.

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Detailed Solution

mV22−GMMERE−GMMSx+RE=0V022−GMERE−GMSx+RE=0 V0=2GMERE+2GMSx+RE=Ve2+2GME×3×105RE2.5×104=Ve2+V023×102×25=Ve13×10=11.2×3.6V0=40.38kms−1
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