Question

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

Short answer

The potential \( V = \frac{1}{r} \) satisfies Laplace's equation, as its Laplacian is zero.

Step-by-step solution

Define the Gravitational Potential

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

Express Laplace's Equation

Laplace's equation is given by \[abla^2 V = 0\]. To satisfy this equation, we need to find the Laplacian of our potential and show it equals zero.

Understand the Laplacian in Spherical Coordinates

Since the potential depends only on \(r\), we use spherical coordinates. In spherical coordinates, the Laplacian of a radially symmetric function \(f(r)\) is \[abla^2 f(r) = \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{d f}{dr} \right)\].

Apply the Laplacian to the Given Potential

Substitute \(V = \frac{1}{r}\) into the spherical coordinates Laplacian formula: \[abla^2 \left( \frac{1}{r} \right) = \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{d}{dr} \left( \frac{1}{r} \right) \right)\].

Compute the Derivatives

First, compute the first derivative: \[ \frac{d}{dr} \left( \frac{1}{r} \right) = -\frac{1}{r^2} \].Then compute the second derivative: \[ \frac{d}{dr} \left( -\frac{1}{r^2} \right) = \frac{2}{r^3} \].

Simplify the Expression

Substitute the second derivative back into the Laplacian formula: \[ abla^2 \left( \frac{1}{r} \right) = \frac{1}{r^2} \frac{d}{dr} \left( r^2 \left( -\frac{1}{r^2} \right) \right) = \frac{1}{r^2} \frac{d}{dr} (-1) = \frac{1}{r^2} \cdot 0 = 0\].

Conclude the Solution

Since the Laplacian of \( \frac{1}{r} \) is zero, \[abla^2 \left( \frac{1}{r} \right) = 0 \], it is proven that the gravitational potential satisfies Laplace's equation.

Key concepts

Laplace's Equation

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

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.

Laplacian

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.