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Question

(C) The energy required to escape the earth's gravitational field is $E_{esc} = \frac{1}{2} m V_{e}^2$.The initial kinetic energy given to the body is

Vedclass · Accepted Answer

$2 \sqrt{2} \,V_{e}$

Vedclass · Answer

(C) The energy required to escape the earth's gravitational field is $E_{esc} = \frac{1}{2} m V_{e}^2$.The initial kinetic energy given to the body is $K_i = \frac{1}{2} m (3 V_{e})^2 = \frac{9}{2} m V_{e}^2$.According to the law of conservation of energy, the final kinetic energy $K_f$ after escaping the gravitational pull is the difference between the initial energy and the energy required to escape:$K_f = K_i - E_{esc}$$\frac{1}{2} m V^2 = \frac{9}{2} m V_{e}^2 - \frac{1}{2} m V_{e}^2$$\frac{1}{2} m V^2 = 4 m V_{e}^2$$V^2 = 8 V_{e}^2$$V = \sqrt{8} V_{e} = 2 \sqrt{2} V_{e}$.

$A$ body is projected from the earth's surface with thrice the escape velocity. What will be its velocity when it escapes the gravitational pull?

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