JEE Advanced2017PhysicsGravitationActual
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 × 10 5 times heavier than the Earth and is at a distance 2.5 × 10 4 times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is v e = 11.2 k m s - 1 . The minimum initial velocity v s required for the rocket to be able to leave the
Options
- Av s = 22 k m s - 1
- Bv s = 72 k m s - 1
- Cv s = 42 k m s - 1
- Dv s = 62 k m s - 1
Correct answer
C. v s = 42 k m s - 1
Step-by-step solution
Given v e = 11.2 k m / s e c = 2 G M e R e 1 2 m v s 2 - G M s m r - G M e m R e = 0 - 0 where r = distance of rocket from Sun ⇒ v s = 2 G M e R e + 2 G M s r Given M s = 3 × 10 5 M e and r = 2.5 × 10 4 R e ⇒ v s = 2 G M e R e + 2 G 3 × 10 5 M e 2.5 × 10 4 R e = 2 G M e R e 1 + 3 × 10 5 2.5 × 10 4 = 2 G M e R e × 13 ⇒ v s = 42 k m / s