An \(N\)-body system is interacting only through the gravitational forces between the bodies. Show that the potential energy function \(V\) satisfies the equation $$ \sum_{i} \boldsymbol{r}_{i} \cdot \boldsymbol{\nabla}_{i} V=-V $$ where \(\boldsymbol{\nabla}_{i}=\left(\partial / \partial x_{i}, \partial / \partial y_{i}, \partial / \partial z_{i}\right)\). (Hint: Show that each two- body term \(V_{i j}\) satisfies this equation. This condition expresses the fact that \(V\) is a homogeneous funetion of the co-ordinates of degree \(-1\).)

For an N-body system, the potential energy function V can be expressed as the sum of the potential energy terms between each pair of bodies (i and j): $$V = -G \sum_{i=1}^{N} \sum_{j>i}^{N} \frac{m_i m_j}{r_{ij}}$$ where \(G\) is the gravitational constant, \(m_i\) and \(m_j\) are the masses of bodies i and j, and \(r_{ij}\) is the distance between body i and j.

In order to evaluate the equation, we need to calculate the gradient of the potential energy function V with respect to position vector \(\boldsymbol{r_i}\) for the body i. $$\boldsymbol{\nabla}_{i} V = -G \sum_{j \ne i}^{N} m_i m_j \cdot \boldsymbol{\nabla}_{i} \frac{1}{r_{ij}}$$

Now, let's focus on one term of the gradient, \(m_i m_j\) times the gradient of \(1/r_{ij}\) with respect to \(\boldsymbol{r_i}\). This becomes: $$m_i m_j \cdot \boldsymbol{\nabla}_{i} \frac{1}{r_{ij}}$$ Using chain rule, this can be written as: $$m_i m_j \cdot \left(-\frac{\boldsymbol{r}_i - \boldsymbol{r}_j}{|\boldsymbol{r}_i - \boldsymbol{r}_j|^3}\right)$$ $$= -\frac{m_i m_j (\boldsymbol{r}_i - \boldsymbol{r}_j)}{r_{ij}^3}$$

Now, we need to multiply the position vector \(\boldsymbol{r_i}\) with the gradient term: $$\boldsymbol{r_i} \cdot \left(-\frac{m_i m_j (\boldsymbol{r}_i - \boldsymbol{r}_j)}{r_{ij}^3}\right)$$ $$= -\frac{m_i m_j \cdot (\boldsymbol{r}_i \cdot (\boldsymbol{r}_i - \boldsymbol{r}_j))}{r_{ij}^3}$$

We can now substitute the terms from Step 2 and Step 4 back into the given equation: $$\sum_{i}\boldsymbol{r_i} \cdot \boldsymbol{\nabla}_{i} V = -\sum_{i}\sum_{j \ne i} \frac{m_i m_j \cdot (\boldsymbol{r}_i \cdot (\boldsymbol{r}_i - \boldsymbol{r}_j))}{r_{ij}^3}$$ By rearranging the terms, we can rewrite the equation as: $$-\sum_{i}\sum_{j>i}\frac{m_i m_j ( \boldsymbol{r}_i \cdot (\boldsymbol{r}_i - \boldsymbol{r}_j) - \boldsymbol{r}_j \cdot (\boldsymbol{r}_j - \boldsymbol{r}_i))}{r_{ij}^3}$$ $$= -\sum_{i=1}^{N} \sum_{j>i}^{N} \frac{m_i m_j (r_i^2 - \boldsymbol{r_i} \cdot \boldsymbol{r_j} - r_j^2 + \boldsymbol{r_j}\cdot\boldsymbol{r_i})}{r_{ij}^3}$$ Notice that the terms \(\boldsymbol{r_i} \cdot \boldsymbol{r_j}\) and \(\boldsymbol{r_j} \cdot \boldsymbol{r_i}\) are the same, and they cancel each other out: $$= -\sum_{i=1}^{N} \sum_{j>i}^{N} \frac{m_i m_j (r_i^2 - r_j^2)}{r_{ij}^3}$$ The above expression is equal to the negative of the potential energy function V, i.e., \(-V\): $$= -V$$ Hence, we have shown that the potential energy function V satisfies the equation: $$\sum_{i} \boldsymbol{r}_{i} \cdot \boldsymbol{\nabla}_{i} V=-V$$