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The masses and radii of the earth and moon are $\left({M}_{1}, {R}_{1}\right)$ and $\left({M}_{2}, {R}_{2}\right)$ respectively. Their centres are at a distance ' ${r}$ ' apart. Find the minimum escape velocity for a particle of mass ' ${m}$ ' to be projected from the middle of these two masses:
${V}=\frac{1}{2} \sqrt{\frac{4 {G}\left({M}_{1}+{M}_{2}\right)}{{r}}}$
${V}=\sqrt{\frac{4 {G}\left({M}_{1}+{M}_{2}\right)}{{r}}}$
${V}=\frac{1}{2} \sqrt{\frac{2 {G}\left({M}_{1}+{M}_{2}\right)}{{r}}}$
${V}=\frac{\sqrt{2 {G}}\left({M}_{1}+{M}_{2}\right)}{{r}}$
Solution
$\frac{1}{2} {mV}^{2}-\frac{{GM}_{1} {m}}{{r} / 2}-\frac{{GM}_{2} {m}}{{r} / 2}=0$
$\frac{1}{2} {mV}^{2}=\frac{2 {Gm}}{{r}}\left({M}_{1}+{M}_{2}\right)$
${V}=\sqrt{\frac{4 {G}\left({M}_{1}+{M}_{2}\right)}{{r}}}$
Similar Questions
Match the column $-I$ with column $-II$ For a satellite in circular orbit,
| Column $-I$ | Column $-II$ |
| $(A)$ Kinetic energy | $(p)$ $ – \frac{{G{M_E}m}}{{2r}}$ |
| $(B)$ Potential energy | $(q)$ $\sqrt {\frac{{G{M_E}}}{r}} $ |
| $(C)$ Total energy | $(r)$ $ – \frac{{G{M_E}m}}{{r}}$ |
| $(D)$ Orbital energy | $(s)$ $ \frac{{G{M_E}m}}{{2r}}$ |
(where $M_E$ is the mass of the earth, $m$ is mass of the satellite and $r$ is the radius of the orbit)
