8.18 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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Vishal Baghel | Contributor-Level 10

4 months ago

Escape velocity of the projectile on the Earth's surface, $v_{e s c}$ = 11.2 km/s = 11.2 ${* 10}^{3}$ m/s

Projection velocity of the projectile = 3 $v_{e s c}$

Mass of the projectile = m

Velocity of the projectile far away from the Earth = $v_{f}$

Total energy of the projectile on the Earth = $\frac{1}{2}$ m ${v_{p}}^{2}$ $-$ $\frac{1}{2}$ m ${v_{e s c}}^{2}$

Gravitational potential energy of the projectile far away from the Earth is zero.

Total energy of the projectile far away from the Earth = $\frac{1}{2}$ m ${v_{f}}^{2}$

From the law of conservation of energy

$\frac{1}{2}$ m ${v_{p}}^{2}$ $-$ $\frac{1}{2}$ m ${v_{e s c}}^{2}$ = $\frac{1}{2}$ m ${v_{f}}^{2}$

${v_{p}}^{2} - {v_{e s c}}^{2} = {v_{f}}^{2}$

$v_{f}$ = $\sqrt{{v_{p}}^{2} - {v_{e s c}}^{2}}$ = $\sqrt{{{(3 v}_{e s c})}^{2} - {v_{e s c}}^{2}}$ = $\sqrt{8 *}$ 11.2= 31.68 km/s

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