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(A) The escape velocity of a planet is given by $v = \sqrt{\frac{2GM}{R}}$.Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $M_p$ and $R_p$

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$2$

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(A) The escape velocity of a planet is given by $v = \sqrt{\frac{2GM}{R}}$.Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $M_p$ and $R_p$ be the mass and radius of planet $P$.Given: $M_e = 9M_p$ (so $M_p = \frac{M_e}{9}$) and $R_e = 2R_p$ (so $R_p = \frac{R_e}{2}$).The escape velocity on Earth is $v_e = \sqrt{\frac{2GM_e}{R_e}}$.The escape velocity on planet $P$ is $v_p = \sqrt{\frac{2GM_p}{R_p}} = \sqrt{\frac{2G(M_e/9)}{(R_e/2)}} = \sqrt{\frac{4GM_e}{9R_e}} = \frac{2}{3} \sqrt{\frac{GM_e}{R_e}}$.Since $v_e = \sqrt{\frac{2GM_e}{R_e}}$,we have $\sqrt{\frac{GM_e}{R_e}} = \frac{v_e}{\sqrt{2}}$.Substituting this into the expression for $v_p$: $v_p = \frac{2}{3} \left( \frac{v_e}{\sqrt{2}} \right) = \frac{\sqrt{2} v_e}{3} = \frac{v_e}{3} \sqrt{2}$.Comparing this with $\frac{v_e}{3} \sqrt{x}$,we get $x = 2$.

If the Earth has a mass nine times and a radius twice that of a planet $P$. Then $\frac{v_e}{3} \sqrt{x} \; ms^{-1}$ will be the minimum velocity required by a rocket to escape the gravitational force of $P$,where $v_e$ is the escape velocity on Earth. The value of $x$ is

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