The escape speed of a projectile on the earth | vedclass

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(B) The escape velocity of a projectile from the Earth is $v_{esc} = 11.2 \; km/s$.The projection velocity of the projectile is $v_p = 3 v_{esc}$.Let

Vedclass · Accepted Answer

$31.68$

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(B) The escape velocity of a projectile from the Earth is $v_{esc} = 11.2 \; km/s$.The projection velocity of the projectile is $v_p = 3 v_{esc}$.Let the mass of the projectile be $m$.According to the law of conservation of energy,the total energy at the Earth's surface equals the total energy far away from the Earth (where gravitational potential energy is zero).$\frac{1}{2} m v_p^2 - \frac{G M m}{R} = \frac{1}{2} m v_f^2 + 0$Since the escape velocity is defined as $v_{esc} = \sqrt{\frac{2GM}{R}}$,we have $\frac{GM}{R} = \frac{v_{esc}^2}{2}$.Substituting this into the energy equation: $\frac{1}{2} m v_p^2 - \frac{1}{2} m v_{esc}^2 = \frac{1}{2} m v_f^2$.$v_f = \sqrt{v_p^2 - v_{esc}^2} = \sqrt{(3 v_{esc})^2 - v_{esc}^2} = \sqrt{8 v_{esc}^2} = v_{esc} \sqrt{8}$.$v_f = 11.2 	imes 2.828 = 31.68 \; km/s$.

The escape speed of a projectile on the earth's surface is $11.2 \; km/s$. $A$ body is projected out with thrice this speed. What is the speed (in $km/s$) of the body far away from the earth? Ignore the presence of the sun and other planets.

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