The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.
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The escape speed of a projectile on the earth’s surface is 11.2 km s-1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Updated On: Nov 3, 2023
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Solution and Explanation

Escape velocity of a projectile from the Earth, vescv_{esc} = 11.2  km/s11.2 \; km/s
Projection velocity of the projectile, vpv_p = 3vesc3v_{esc}
Mass of the projectile = mm
Velocity of the projectile far away from the Earth = vfv_f
Total energy of the projectile on the Earth = 12mvp212mvesc2\frac{1}{2}mv^2_p - \frac{1}{2}mv^2_{esc}
Gravitational potential energy of the projectile far away from the Earth is zero.
Total energy of the projectile far away from the Earth= 12mvf2\frac{1}{2}mv^2_f

From the law of conservation of energy, we have
12mvp212mvesc2\frac{1}{2}mv^2_p - \frac{1}{2}mv^2_{esc} = 12mvf2\frac{1}{2}mv_f^2 

vrv_r = vp2vesc2\sqrt{v_p ^2 - v_{esc}^2}

=(3vesc)2(vesc)2\sqrt{(3v_{esc}) ^2 - (v_{esc})^2}

=8vesc\sqrt 8 v_{esc}

=8×11.2\sqrt 8 \times 11.2 = 31.68  km/s31.68 \; km/s

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Concepts Used:

Escape Speed

Escape speed is the minimum speed, which is required by the object to escape from the gravitational influence of a plannet. Escape speed for Earth’s surface is 11,186 m/sec. 

The formula for escape speed is given below:

ve = (2GM / r)1/2 

where ,

ve = Escape Velocity 

G = Universal Gravitational Constant 

M = Mass of the body to be escaped from 

r = Distance from the centre of the mass