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(D) Given: Escape velocity $= u$.The initial velocity of the projectile is $200 \%$ more than the escape velocity.$V_{	ext{initial}} = u + (200/100)u

Vedclass · Accepted Answer

$2 \sqrt{2} u$

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(D) Given: Escape velocity $= u$.The initial velocity of the projectile is $200 \%$ more than the escape velocity.$V_{	ext{initial}} = u + (200/100)u = u + 2u = 3u$.According to the Law of Conservation of Energy:$\frac{1}{2} m V_{	ext{initial}}^2 + (P.E.)_{	ext{initial}} = \frac{1}{2} m V_{	ext{final}}^2 + (P.E.)_{	ext{final}}$.At the surface of the planet,the potential energy is $P.E. = -\frac{GMm}{R} = -m u^2$ (since $u = \sqrt{\frac{2GM}{R}}$,so $u^2 = \frac{2GM}{R}$,which implies $\frac{GM}{R} = \frac{u^2}{2}$,thus $P.E. = -\frac{m u^2}{2}$).In interstellar space,the potential energy is $0$.$\frac{1}{2} m (3u)^2 - \frac{1}{2} m u^2 = \frac{1}{2} m V_{	ext{final}}^2$.$\frac{9}{2} u^2 - \frac{1}{2} u^2 = \frac{1}{2} V_{	ext{final}}^2$.$8 u^2 = V_{	ext{final}}^2$.$V_{	ext{final}} = \sqrt{8} u = 2\sqrt{2} u$.

Let the escape velocity of a body kept at the surface of a planet be $u$. If it is projected at a speed of $200 \%$ more than the escape speed,then its speed in interstellar space will be ...........

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