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A rocket is launched normal to the surface of the Earth, away

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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 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)

v_{s} = 72 km s^{- 1}

v_{s} = 22 km s^{- 1}

v_{s} = 42 km s^{- 1}

v_{s} = 62 km s^{- 1}

Views: 6,106 students

Updated on: Apr 3, 2023

Book:

Mechanics Volume 2 (DC Pandey)

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Olivia Discussed

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 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)

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

Given,

v_{e} = 11.2 km / s = \frac{2 G M _{e}}{R _{e}}

From energy conservation,

K_{i} + U_{i} = K_{f} + U_{f} \frac{1}{2} m v_{s}^{2} - \frac{G M _{s} m}{r} - \frac{G M _{e} m}{R _{e}} = 0 + 0

Here,

r =

distance of rocket from sun

\Rightarrow v_{s} = \frac{2 G M _{e}}{R _{e}} + \frac{2 G M _{s}}{r}

Given,

M_{s} = 3 \times 1 0^{5} M_{e}

and

r = 2.5 \times 1 0^{4} R_{e}

\Rightarrow v_{s} = \frac{2 G M _{e}}{R _{e}} + (2 G) (\frac{3 \times 1 0 ^{5} M _{e}}{2.5 \times 1 0 ^{4} R _{e}})

= \frac{2 G M _{e}}{R _{e}} (1 + \frac{3 \times 1 0 ^{5}}{2.5 \times 1 0 ^{4}})

= \frac{2 G M _{e}}{R _{e}} \times 13

\Rightarrow v_{s} ≃ 42 km / s

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Question Text	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 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)
Updated On	Apr 3, 2023
Topic	Gravitation
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 3
Upvotes	329
Avg. Video Duration	6 min

Text solutionVerified

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