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JEE Advanced · Physics · 9. Gravitation

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 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)

  1. A vs=22 km s-1
  2. B vs=72 km s-1
  3. C vs=42 km s-1
  4. D vs=62 km s-1
Verified Solution

Answer & Solution

Correct Answer

(C) vs=42 km s-1

Step-by-step Solution

Detailed explanation

Given ve=11.2km/sec= 2GMeRe 12mvs2-GMsmr-GMemRe=0-0      where r = distance of rocket from Sun
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