Gravitation 2 Question 1

4. 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 Earth. The escape velocity from Earth’s gravitational field is ve=11.2kms1. 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)

(a) vs=72kms1

(b) vs=22kms1

(c) vs=42kms1

(d) vs=62kms1

(2017 Adv.)

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Answer:

Correct Answer: 4. (c)

Solution:

  1. Given, ve=11.2km/s=2GMeRe

From energy conservation,

Ki+Ui=Kf+Uf12mvs2GMsmrGMemRe=0+0

Here, r= distane of rocket from sun

vs=2GMeRe+2GMsr

Given, Ms=3×105Me and r=2.5×104Re

vs=2GMeRe+2G3×105Me2.5×104Re

=2GMeRe×13vs42km/s



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