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JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Question Bank Gravitation Potential

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    The masses and radii of the earth and moon are \[{{M}_{1}},\,{{R}_{1}}\] and \[{{M}_{2}},\,{{R}_{2}}\] respectively. Their centres are distance d apart. The minimum velocity with which a particle of mass m should be projected from a point midway between their centres so that it escapes to infinity is   [MP PET 1997]

    A)             \[2\sqrt{\frac{G}{d}({{M}_{1}}+{{M}_{2}})}\]  

    B)               \[2\sqrt{\frac{2G}{d}({{M}_{1}}+{{M}_{2}})}\]

    C)             \[2\sqrt{\frac{Gm}{d}({{M}_{1}}+{{M}_{2}})}\]

    D)               \[2\sqrt{\frac{Gm({{M}_{1}}+{{M}_{2}})}{d({{R}_{1}}+{{R}_{2}})}}\]

    Correct Answer: A

    Solution :

        Gravitational potential at mid point             \[V=\frac{-G{{M}_{1}}}{d/2}+\frac{-G{{M}_{2}}}{d/2}\] Now, \[PE=m\times V=\frac{-2Gm}{d}({{M}_{1}}+{{M}_{2}})\] [m = mass of particle] So, for projecting particle from mid point to infinity             \[KE\,=\,|\,PE\,|\]             \[\Rightarrow \,\frac{1}{2}m{{v}^{2}}=\frac{2\,Gm}{d}({{M}_{1}}+{{M}_{2}})\]   \[\Rightarrow \,v=2\sqrt{\frac{G\,({{M}_{1}}+{{M}_{2}})}{d}}\]       


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