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Question:

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⁻¹

vs=72kms⁻¹

vs=62kms⁻¹

vs=22kms⁻¹

Solution:

Given:
Ms = 3×10⁵ Me
d = 2.5×10⁴ Re
Also Ve = 11.2Km/s = √(2GMe/Re)
Now for the sun earth system,
(1/2)mVs² - GMe m/Re - GMsm/(Re+d) = 0
Assuming Re << d and taking the value of GMe/R = 11.2²
Vs ≈ 42 Km/s