Show that the gravitational potential \(V=-G m / r\) satisfies Iaplace's equation, that is, show that \(\nabla^{2}(1 / r)=0\) where \(r^{2}=x^{2}+y^{2}+z^{2}, r \neq 0\). (Also see Chapter 15, Section 8.)

The gravitational potential, denoted as \(V\), is a measure of the potential energy per unit mass at a point in a gravitational field. It's crucial because it helps in calculating the force experienced by a mass in the field.

The gravitational potential due to a point mass \(m\) at a distance \(r\) is given by:
\[ V = - \frac{Gm}{r} \]
Here, \(G\) is the gravitational constant, \(m\) is the mass, and \(r\) is the distance from the mass.

The negative sign indicates that the potential is lower (or more negative) closer to the mass. This is due to the nature of gravitational forces being always attractive. The potential approaches zero as \(r\) goes to infinity.

Cartesian coordinates, represented as \((x, y, z)\), are a system for defining the positions of points in three-dimensional space. This system utilizes three mutually perpendicular axes: x (horizontal), y (vertical), and z (depth).

In the context of gravitational potential, the distance \(r\) from a point to the origin in Cartesian coordinates is:
\[ r = \sqrt{x^2 + y^2 + z^2} \]
This transformation is key when analyzing problems using Laplace's equation in three dimensions. By expressing \(r\) in terms of \(x\), \(y\), and \(z\), we can perform calculus operations, like partial derivatives, more conveniently.

Partial derivatives measure how a function changes as one of its variables varies while keeping the others fixed. They are essential tools in multivariable calculus.

For a function \(f(x, y, z)\), the partial derivative with respect to \(x\) is denoted as \(\frac{\partial f}{\partial x}\). This represents the rate of change of \(f\) as \(x\) changes, holding \(y\) and \(z\) constant.

In the problem, we use partial derivatives to find how \(\frac{1}{r}\) changes with respect to \(x\), \(y\), and \(z\). The gradients involve compounds like:
\[ \frac{\partial}{\partial x} \left( \frac{1}{r} \right) = -\frac{x}{r^3}, \frac{\partial}{\partial y} \left( \frac{1}{r} \right) = -\frac{y}{r^3}, \frac{\partial}{\partial z} \left( \frac{1}{r} \right) = -\frac{z}{r^3} \]
These derivatives are critical to formulating the Laplacian of the potential.

The Laplacian, denoted as \(abla^2\), is a differential operator that combines the second partial derivatives of a function.

For a scalar function \(V\), the Laplacian in Cartesian coordinates is:
\[ abla^2 V = \frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} + \frac{\partial^2 V}{\partial z^2} \]
In our exercise, we apply the Laplacian to \(\frac{1}{r}\) and show that the result is zero, satisfying Laplace's equation:
\[abla^2 \left( \frac{1}{r} \right) = 0 \]
The steps to calculate this involve finding each second partial derivative, summing them up, and simplifying. Because \(x^2 + y^2 + z^2 = r^2\), simplifications lead to the desired result. This shows that \(\frac{1}{r}\) is a harmonic function, particularly significant in potential theory and physics.