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# 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=11.2kms−1. The minimum 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) vs=42kms−1vs=62kms−1vs=72kms−1vs=22kms−1

A
vs=42kms1
B
vs=22kms1
C
vs=72kms1
D
vs=62kms1
Solution
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#### Given:Ms=3×105 Med=2.5×104 ReAlso Ve=11.2Km/s = √2GMeReNow for the sun earth system, 12mV2s−GMemRe−GMsmRe+d=0Assuming Re<<dand taking the value of GMeR=11.222Vs≈ 42 Km/s

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