JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Gravitation

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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 1 0^{5}

times heavier than the Earth and is at a distance

2.5 \times 1 0^{4}

times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is

v_{e} = 11.2 km 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)

(A) $v_{S} = 22 km s^{- 1}$
(B) $v_{S} = 42 km s^{- 1}$ ✓ Correct
(C) $v_{S} = 62 km s^{- 1}$
(D) $v_{S} = 72 km s^{- 1}$

Answer: (B)

Solution

The escape velocity from the Sun-Earth system requires overcoming both gravitational potentials. Energy equation:

\frac{1}{2} m v_{es}^{2} = \frac{GM m}{R} + \frac{G ( 3 \times 1 0 ^{5} M ) m}{2.5 \times 1 0 ^{4} R}

. Since

v_{e}^{2} = \frac{2 GM}{R}

, we get

v_{es}^{2} = v_{e}^{2} (1 + \frac{3 \times 1 0 ^{5}}{2.5 \times 1 0 ^{4}}) = v_{e}^{2} \cdot 13

. Therefore

v_{es} = 13 \cdot 11.2 \approx 40.3 km/s

, closest to 42 km/s.

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Single correct option

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JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Gravitation

Solution

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