Animated Solution for Physics - Gravitation: 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……… .

Imagine you are given a lump of clay of mass $2M$ and asked to break it into four pieces. You decide to make two pieces of mass $m$ and two pieces of mass $M-m$. Then, you place them at the corners of a square of side $d$. The question asks: how should you choose $m$ so that the gravitational potential energy of this system is maximized?

Visualizing the System First, let's draw the free body diagram. We have a square. Let's place the masses $m$ at opposite corners, and the masses $M-m$ at the other two opposite corners. This alternating arrangement is crucial because it dictates the distances between different pairs of masses.

Counting the Pairs To find the total gravitational potential energy of a system, we must account for every single pair of masses. The formula for the potential energy between two masses is $U=-\frac{G{m}_1{m}_2}{r}$.

For four masses, the number of pairs is given by $\frac{n(n-1)}{2}$, which equals 6 pairs. Let's break them down:

1. The Four Sides: There are four edges of the square. Each edge connects a mass $m$ to a mass $M-m$ at a distance $d$. Their total energy is:

2. The Two Diagonals: The diagonals have a length of $\sqrt{2}d$. One diagonal connects the two $m$ masses, and the other connects the two $M-m$ masses. Their energy is:

Adding these together gives us the total potential energy $U_T$:

It is simply a function of one variable: $m$. To find the maximum potential energy, we turn to our trusty tool: calculus. We need to differentiate $U_T$ with respect to $m$ and set it to zero.

We can factor out the constant $-\frac{G}{d}$. This means the derivative of the terms inside the bracket must be zero:

Don't make a silly mistake here!

Since $\frac{2}{\sqrt{2}}=\sqrt{2}$, we can simplify this to:

Now, let's group all the $M$ terms on one side and the $m$ terms on the other:

Notice how beautifully this simplifies! We can factor out a 2 on the right side:

The term $(4-\sqrt{2})$ cancels out perfectly from both sides, leaving us with:

The question states this ratio is $x:1$, which means $x=2$.

This problem is a beautiful blend of geometry, physics, and calculus. It teaches us the importance of systematically counting pairs and trusting the math to simplify elegant physical symmetries.