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

Laplace's equation is a second-order partial differential equation. It is widely used in physics and engineering. The equation is: \[ abla^2 V = 0 \]. Here, \[ abla^2 \] (called the Laplacian) is the divergence of the gradient of a function. In simple terms, it measures how a function \(V\) deviates from its average value at a given point. Solutions to this equation, also known as harmonic functions, are important in fields like electrostatics, fluid dynamics, and gravitational potential theory.

In the context of gravitational potential, satisfying Laplace's equation implies that the potential \(V\) behaves in a manner consistent with the inverse-square law of gravitation.

Spherical coordinates are a system of coordinates that extends the concept of polar coordinates to three dimensions. They are especially useful for problems involving spherical symmetry. In spherical coordinates, any point in space is represented by three parameters: \[ (r, \theta, \phi) \], where \[ r \] is the radius or distance from the origin, \[ \theta \] is the angle in the \(xy\)-plane (similar to the polar angle), and \[ \phi \] is the angle from the \(z\)-axis (often called the azimuthal angle).

The notation uses:

• \textbf{r}: distance from the origin
• \textbf{θ}: angle in the \(xy\)-plane
• \textbf{ \phi}: angle from the \(z\)-axis

For functions that only depend on \(r\), spherical coordinates simplify many calculations. This is especially true for problems with radial symmetry, like the gravitational potential.

The Laplacian, denoted \[ abla^2 \], is a differential operator that appears frequently in physics. It is defined as the sum of the second partial derivatives with respect to each spatial coordinate. In Cartesian coordinates, the Laplacian of a function \(V(x, y, z)\) is given by: \[ abla^2 V = \frac{abla^2 V}{abla x^2} + \frac{abla^2 V}{abla y^2} + \frac{abla^2 V}{abla z^2} \]

In spherical coordinates, if a function only depends on \(r\), the Laplacian is simplified to: \[ abla^2 V = \frac{1}{r^2} \frac{d}{dr}(r^2 \frac{dV}{dr}) \].

When you apply this to \(V = -Gm/r\), you first find the first and second derivatives of \(V\) with respect to \(r\). After that, substitute these derivatives back into the spherical coordinates form of the Laplacian. When these steps are done correctly, you'll see that \[ abla^2 V = 0 \], proving that Laplace's equation is satisfied.