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&#39…

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)

Aicharya · 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 &rArr; v s = 2 G M e R e + 2 G M s r Given M s = 3 &times; 10 5 M e and r = 2.5 &times; 10 4 R e &rArr; v s = 2 G M e R e + 2 G 3 &times; 10 5 M e 2.5 &times; 10 4 R e = 2 G M e R e 1 + 3 &times; 10 5 2.5 &times; 10 4 = 2 G M e R e &times; 13 &rArr; v s = 42 k m / s

A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun…

Solution