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

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 $3 \times 10^{5}$ times heavier than the Earth and is at a distance $2.5 \times 10^{4}$ 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 Sun-Earth system is closest to
(Ignore the rotation and revolution of the Earth and the presence of any other planet)

TestHub · Accepted Answer

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

Physics - Gravitation Question with Solution | TestHub

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