A body of mass $$(2M)$$ splits into four masses $$\{m, M-m, m, M-m\}$$, which are rearranged to form a square as shown in the figure. The ratio of $$\frac{M}{m}$$ for which, the gravitational potential energy of the system becomes maximum is $$x : 1$$. The value of $$x$$ is _________.

1. Understand the System Configuration

Let the four masses be placed at the corners of a square of side $$d$$ in a cyclic order:

• Corner 1: $$m$$
• Corner 2: $$M-m$$
• Corner 3: $$m$$
• Corner 4: $$M-m$$

The distance between adjacent corners is $$d$$, and the distance between diagonally opposite corners is $$\sqrt{2}d$$.

2. Write the Expression for Total Gravitational Potential Energy ($$U$$)

The total gravitational potential energy of a system of particles is the sum of the potential energies of all unique pairs:

$$U = -\sum \frac{G m_i m_j}{r_{ij}}$$

Let's calculate the interactions by grouping them by distance:

• Four Adjacent Pairs (at distance $$d$$):
  • Between Corner 1 and Corner 2: $$-\frac{G m(M-m)}{d}$$
  • Between Corner 2 and Corner 3: $$-\frac{G(M-m)m}{d}$$
  • Between Corner 3 and Corner 4: $$-\frac{G m(M-m)}{d}$$
  • Between Corner 4 and Corner 1: $$-\frac{G(M-m)m}{d}$$
  Sum of adjacent interactions = $$-4 \frac{G m(M-m)}{d}$$


• Two Diagonal Pairs (at distance $$\sqrt{2}d$$):
  • Between Corner 1 and Corner 3 (both mass $$m$$): $$-\frac{G m^2}{\sqrt{2}d}$$
  • Between Corner 2 and Corner 4 (both mass $$M-m$$): $$-\frac{G (M-m)^2}{\sqrt{2}d}$$
  Sum of diagonal interactions = $$-\frac{G}{\sqrt{2}d} \left[ m^2 + (M-m)^2 \right]$$

Combining all pairs, the total potential energy expression is:

$$U = -\frac{G}{d} \left[ 4m(M-m) + \frac{1}{\sqrt{2}}(m^2 + (M-m)^2) \right]$$

3. Maximize the Potential Energy

Since gravitational potential energy is a negative quantity, maximizing the potential energy means making its absolute magnitude minimum, or directly finding the critical point where $$\frac{dU}{dm} = 0$$.

Let's differentiate the expression inside the bracket with respect to $$m$$ and set it to zero:

$$\frac{d}{dm} \left[ 4(Mm - m^2) + \frac{1}{\sqrt{2}}(m^2 + M^2 - 2Mm + m^2) \right] = 0$$

$$\frac{d}{dm} \left[ 4Mm - 4m^2 + \frac{1}{\sqrt{2}}(2m^2 - 2Mm + M^2) \right] = 0$$

Performing the differentiation step-by-step:

$$\left[ 4M - 8m + \frac{1}{\sqrt{2}}(4m - 2M) \right] = 0$$

$$4M - 8m + \frac{4}{\sqrt{2}}m - \frac{2}{\sqrt{2}}M = 0$$

$$4M - \sqrt{2}M = 8m - 2\sqrt{2}m$$

4. Determine the Ratio $\frac{M}{m}$

Factor out $$M$$ on the left side and $$m$$ on the right side:

$$M(4 - \sqrt{2}) = m(8 - 2\sqrt{2})$$

Notice that the right side can be factored further by pulling out a $$2$$:

$$M(4 - \sqrt{2}) = 2m(4 - \sqrt{2})$$

Canceling out the common term $$(4 - \sqrt{2})$$ from both sides gives:

$$M = 2m \implies \frac{M}{m} = 2$$

Given that the ratio $$\frac{M}{m} = x : 1$$, we find that $$x = 2$$.

Final Answer: 2