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Step-by-Step Solution
Step 1: Identify the position from where the particle is projected
The particle of mass $m$ is placed exactly at the midpoint between Earth (mass $M_1$) and Moon (mass $M_2$). The distance between the centers of Earth and Moon is $r$, so the particle is at a distance $\frac{r}{2}$ from each mass.
Step 2: Write the total gravitational potential at the midpoint
The gravitational potential $U$ at a distance $d$ from a mass $M$ is given by
$$
U = -\frac{GM}{d}.
$$
Since the particle is in the gravitational field of both Earth and Moon:
Potential due to Earth:
$$
U_1 = -\frac{G\,M_1}{\tfrac{r}{2}} = -\frac{2G\,M_1}{r}.
$$
Potential due to Moon:
$$
U_2 = -\frac{G\,M_2}{\tfrac{r}{2}} = -\frac{2G\,M_2}{r}.
$$
Therefore, the total potential at the midpoint is
$$
U_\text{total} = \left( -\frac{2G\,M_1}{r} \right) + \left( -\frac{2G\,M_2}{r} \right)
= -\frac{2G\,(M_1 + M_2)}{r}.
$$
Step 3: Express the kinetic energy required for escape
For the particle to escape from this midpoint region to infinity (away from the influence of both masses), its initial kinetic energy must equal the magnitude of the total gravitational potential energy (since total energy at infinity is taken as zero). Thus,
$$
\frac{1}{2}\,mV^2 = \left| U_\text{total} \right|
= \frac{2G\,(M_1 + M_2)\,m}{r}.
$$
Step 4: Solve for the escape velocity
Rearrange the above equation to find $V$:
$$
\frac{1}{2}\,mV^2 = \frac{2G\,(M_1 + M_2)\,m}{r}
\quad\Longrightarrow\quad
V^2 = \frac{4G\,(M_1 + M_2)}{r},
$$
$$
V = \sqrt{\frac{4G\,(M_1 + M_2)}{r}}.
$$
Step 5: Final Answer
The minimum escape velocity from the midpoint between Earth and Moon is
$$
V = \sqrt{\frac{4G\,(M_1 + M_2)}{r}}.
$$
