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Step-by-Step Solution
Step 1: Identify the System
We have four spheres of mass $m$ placed at the corners of a square of side $d$, and a fifth sphere of mass $M$ placed at the center of the square. We want to determine the total gravitational potential energy of this five-mass system.
Step 2: Recall the Formula for Gravitational Potential Energy Between Two Masses
The gravitational potential energy $U$ between two point masses $m_1$ and $m_2$ separated by a distance $r$ is given by:
$ U = - \dfrac{G m_1 m_2}{r} $
Step 3: Compute Potential Energy Between Each Corner Mass and the Center Mass
Each of the four corner masses $m$ is at a distance $\dfrac{d}{\sqrt{2}}$ from the center (since the diagonal of a square of side $d$ is $d \sqrt{2}$, and the distance from the center to any corner is half the diagonal, i.e., $\dfrac{d\sqrt{2}}{2} = \dfrac{d}{\sqrt{2}}$). Therefore, the potential energy between each corner mass $m$ and the center mass $M$ is:
$ U_{\text{corner-center}} = - \dfrac{G m M}{\bigl(\frac{d}{\sqrt{2}}\bigr)} = - \dfrac{\sqrt{2} \, G \, m \, M}{d} $
Since there are four such corner masses, the total contribution from all corner-center interactions is:
$ U_{\text{corner-center (total)}} = 4 \times \left(- \dfrac{\sqrt{2} \, G \, m \, M}{d}\right) = - \dfrac{4 \sqrt{2} \, G \, m \, M}{d} $
Step 4: Compute Potential Energy Among the Four Corner Masses
The four corner masses form a square of side $d$. We must account for the potential energy between each pair of corner masses:
Each corner mass has a nearest distance $d$ to two other masses on its adjacent corners.
Each corner mass also has a diagonal distance $\sqrt{2}\, d$ to the mass on the opposite corner.
Number of pairs with side $d$: There are 4 edges in the square, each edge has a pair of corner masses at distance $d$, giving 4 pairs.
Number of pairs with diagonal $\sqrt{2}\, d$: There are 2 diagonals, on each diagonal there is a pair of corner masses at distance $\sqrt{2}\, d$, giving 2 pairs.
Hence:
Potential energy from pairs at distance $d$:
$ U_{d} = 4 \times \left(- \dfrac{G m \, m}{d}\right) = - \dfrac{4 G m^2}{d} $
Potential energy from pairs at distance $\sqrt{2}\, d$:
$ U_{\sqrt{2} d} = 2 \times \left(- \dfrac{G m \, m}{\sqrt{2}\,d}\right) = - \dfrac{2 G m^2}{\sqrt{2}\, d} = - \dfrac{2 \sqrt{2} \, G m^2}{2 d} = - \dfrac{2 G m^2}{\sqrt{2} \, d} $
Summing these corner-corner interactions:
$ U_{\text{corner-corner}} = - \dfrac{4 G m^2}{d} - \dfrac{2 G m^2}{\sqrt{2} \, d} $
Step 5: Sum All Contributions
The total gravitational potential energy $U_{\text{total}}$ is the sum of:
Potential energy of each corner mass with the center mass ($4 \sqrt{2}$ term)
All pairwise potential energies among corner masses themselves
So,
$
U_{\text{total}}
= \underbrace{\left(- \dfrac{4 \sqrt{2} \, G m \, M}{d}\right)}_{\text{corner-center}}
\;+\; \underbrace{\left(- \dfrac{4 G m^2}{d} - \dfrac{2 G m^2}{\sqrt{2} \, d}\right)}_{\text{corner-corner}}
$
Combine the terms in a common factor $-\dfrac{G m}{d}$:
$
U_{\text{total}}
= - \dfrac{G m}{d}
\left\{ 4\sqrt{2}\,M + \left(4 + \sqrt{2}\right)\,m \right\}
$
Step 6: Present the Final Answer
The total gravitational potential energy of the system is thus:
$ \displaystyle \boxed{ - \dfrac{G m}{d} \left[ (4 + \sqrt{2})\, m + 4 \sqrt{2}\, M \right] }$
