The masses and radii of the earth and moon are $$(M_1, R_1)$$ and $$(M_2, R_2)$$ 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:

We begin by recalling the expression for the gravitational potential energy of interaction between two point masses. The formula is stated first:

$$U \;=\; -\dfrac{G\,M\,m}{d}$$

Here $$G$$ is the universal gravitational constant, $$M$$ is the source mass, $$m$$ is the test mass and $$d$$ is the separation of their centres. The negative sign tells us that gravity is an attractive interaction whose potential energy decreases (becomes more negative) as the distance decreases.

In the present problem two large bodies are involved. Their masses and radii are $$M_1,\,R_1$$ (Earth) and $$M_2,\,R_2$$ (Moon). The distance between their centres is given to be $$r$$. A particle of mass $$m$$ is released from the exact mid-point of the line joining the two centres, so its distance from each of the two bodies is clearly $$\dfrac{r}{2}$$.

We write the total initial gravitational potential energy of the particle due to both masses by adding the individual contributions:

$$U_{\text{initial}} \;=\; -\dfrac{G\,M_1\,m}{\dfrac{r}{2}} \;-\; \dfrac{G\,M_2\,m}{\dfrac{r}{2}}$$

Because the denominator $$\dfrac{r}{2}$$ appears in both terms we simplify each fraction by inverting and multiplying:

$$ U_{\text{initial}} = -\,G\,M_1\,m\left(\dfrac{2}{r}\right) -\,G\,M_2\,m\left(\dfrac{2}{r}\right) = -\,\dfrac{2Gm}{r}\left(M_1+M_2\right) $$

Thus the combined potential energy at the mid-point is

$$U_{\text{initial}} = -\dfrac{2Gm\,(M_1+M_2)}{r}. $$

Next we invoke the principle of conservation of mechanical energy. Let the particle be projected from the mid-point with a speed $$v_e$$ just sufficient to escape to infinity. “Just sufficient” (minimum escape condition) means that when it finally reaches infinity its speed falls to zero. We therefore have for the final state at infinity:

$$K_{\text{final}} \;=\; 0, \qquad U_{\text{final}} \;=\; 0.$$

The mechanical energy at the starting point equals that at infinity:

$$ K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}}. $$

Substituting the known expressions, we write

$$ \dfrac{1}{2}m\,v_e^{\,2} + \left(-\dfrac{2Gm\,(M_1+M_2)}{r}\right) = 0 + 0. $$

Now we isolate the kinetic term and solve for $$v_e^{\,2}$$ step by step:

$$ \dfrac{1}{2}m\,v_e^{\,2} = \dfrac{2Gm\,(M_1+M_2)}{r}. $$

We cancel the common factor $$m$$ from both sides:

$$ \dfrac{1}{2}\,v_e^{\,2} = \dfrac{2G\,(M_1+M_2)}{r}. $$

Multiplying both sides by $$2$$ gives

$$ v_e^{\,2} = \dfrac{4G\,(M_1+M_2)}{r}. $$

Finally, taking the square root, we obtain the minimum escape velocity:

$$ v_e = \sqrt{\dfrac{4G\,(M_1+M_2)}{r}}. $$

This expression matches Option A exactly.

Hence, the correct answer is Option A.