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
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
$2\sqrt {\frac{G}{d}({M_1} + {M_2})} $
$2\sqrt {\frac{{2G}}{d}({M_1} + {M_2})} $
$2\sqrt {\frac{{Gm}}{d}({M_1} + {M_2})} $
$2\sqrt {\frac{{Gm({M_1} + {M_2})}}{{d({R_1} + {R_2})}}} $
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
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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