User:Sławomir Biały/Gravitational potential

In classical mechanics, the gravitational potential at a location represents the work (energy) per unit mass as an object moves to that location from a reference location. It is analogous to the electric potential with mass playing the role of charge. By convention, the gravitational potential is defined as zero infinitely far away from any mass. As a result, it is negative elsewhere.

In mathematics the gravitational potential is also known as the Newtonian potential and is fundamental in the study of potential theory.

Potential energy
The gravitational potential (V) is the potential energy (U) per unit mass:


 * $$U = m V $$

where m is the mass of the object. The potential energy is the negative of the work done by the gravitational field moving the body to its given position in space from infinity. If the body has a mass of 1 unit, then the potential energy to be assigned to that body is equal to the gravitational potential. So the potential can be interpreted as the negative of the work done by the gravitational field moving a unit mass in from infinity.

In some situations the equations can be simplified by assuming a field which is nearly independent of position. For instance, in daily life, in the region close to the surface of the Earth, the gravitational acceleration can be considered constant. In that case the difference in potential energy from one height to another is to a good approximation linearly related to the difference in height:
 * $$\Delta U = mg \Delta h$$

Mathematical form
The potential V at a distance x from a point mass of mass M is
 * $$V = -\frac{GM}{x},$$

where G is the gravitational constant. The potential has units of energy per unit mass; e.g., J/kg in the MKS system. By convention, it is always negative where it is defined, and as r tends to infinity, it approaches zero.

The gravitational field, and thus the acceleration of a small body in the space around the massive object, is the negative gradient of the gravitational potential. Because the potential has no angular components, its gradient is:
 * $$\mathbf{a} = -\frac{GM}{x^3} \mathbf{x} = -\frac{GM}{x^2} \hat{\mathbf{x}},$$

where x is a vector of length x pointing from the point mass towards the small body and $$\hat{\mathbf{x}}$$ is a unit vector pointing from the point mass towards the small body. The magnitude of the acceleration therefore follows an inverse square law:
 * $$|\mathbf{a}| = \frac{GM}{x^2}.$$

The potential associated with a mass distribution is the superposition of the potentials of point masses. If the mass distribution is a finite collection of point masses, and if the point masses are located at the points x1, ..., xn and have masses m1, ..., mn, then the potential of the distribution at the point x is:
 * $$V(\mathbf{x}) = \sum_{i=1}^n -\frac{Gm_i}{|\mathbf{x} - \mathbf{x_i}|}.$$

In the vector diagram, cm denotes the point mass and px denotes the point at which the potential is being computed. If the mass distribution is given as a mass measure dm on three-dimensional Euclidean space R3, then the potential is the convolution of &minus;G/|y| with dm. In good cases this equals the volume integral
 * $$V(\mathbf{x}) = -\int_{\mathbf{R}^3} \frac{G}{|\mathbf{x} - \mathbf{r}|}\,dm(\mathbf{r}).$$

If there is a function &rho;(r) representing the density of the distribution at r, so that dm(r) = &rho;(r)dv, where dv is the Euclidean volume element, then the gravitational potential is
 * $$V(\mathbf{x}) = -\int_{\mathbf{R}^3} \frac{G}{|\mathbf{x}-\mathbf{r}|}\,\rho(\mathbf{r})dv \, .$$

If V is a potential function coming from a continuous mass distribution &rho;(r), then &rho; can be recovered using the Laplace operator &Delta; using the formula:
 * $$\rho(\mathbf{x}) = \frac{1}{4\pi G}\Delta V(\mathbf{x}),$$

This holds pointwise whenever &rho; is continuous and is zero outside of a bounded set. In general, the mass measure dm can be recovered in the same way if the Laplace operator is taken in the sense of distributions. Consequently, the gravitational potential satisfies Poisson's equation. See also Green's function for the three-variable Laplace equation and Newtonian potential.

Spherical symmetry
A spherically symmetric mass distribution behaves to an observer completely outside the distribution as though all of the mass were concentrated at the center, and thus effectively as a point mass, by the shell theorem. On the surface of the Earth, the acceleration is given by so-called standard gravity g, approximately 9.8 m/s2, although this value varies slightly with latitude and altitude: the magnitude of the acceleration is a little larger at the poles than at the equator because the Earth is an oblate spheroid.

Within a spherically symmetric mass distribution, it is possible to solve Poisson's equation in spherical coordinates. Within a uniform spherical body of radius R and density &sigma;, the gravitational force g inside the sphere varies linearly with distance r from the center, giving the gravitational potential inside the sphere, which is


 * $$V(r) = \frac {2}{3} \pi G \sigma (r^2-3R^2),\qquad r\leq R,$$

which smoothly connects to the potential function for the outside of the sphere (see the figure at the top).

General relativity
In general relativity, the gravitational potential is replaced by the metric tensor.

Multipole expansion
The potential at a point x is given by


 * $$V(\mathbf{x}) = - \int_{\mathbb{R}^3} \frac{G}{|\mathbf{x}-\mathbf{r}|}\ dm(\mathbf{r}).$$

The potential can be expanded in a series of Legendre polynomials. x is a vector from the center of mass to the point (i.e. "point x") at which the potential is being computed. r is a vector from the center of mass to the differential element of mass. The vector difference, x - r, thus emanates from the differential element of mass to the "point x" as can be seen in the vector diagram to the right where the "point x" is denoted as px. The denominator in the integral is expressed as the square root of the square to give
 * $$\begin{align}

V(\mathbf{x}) &= - \int_{\mathbb{R}^3} \frac{G}{ \sqrt{|\mathbf{x}|^2 -2 \mathbf{x} \cdot \mathbf{r} + |\mathbf{r}|^2}}\,dm(\mathbf{r})\\ {}&=- \frac{1}{|\mathbf{x}|}\int_{\mathbb{R}^3} G \, \left/ \, \sqrt{1 -2 \frac{r}{|\mathbf{x}|} \cos \theta + \left( \frac{r}{|\mathbf{x}|} \right)^2}\right.\,dm(\mathbf{r}) \end{align}$$ where in the last integral, r = |r| and &theta; is the angle between x and r.

The integrand can be expanded in a Taylor series in $$Z=\frac{r}{|\mathbf{x}|}$$. The calculation of the coefficients is straightforward but cumbersome. The calculations can be simplified by observing that the integrand is the generating function for the Legendre polynomials. Also without knowing anything about generating functions, Theorem 2 in section 10.8 on page 454 of can be used which proves that


 * $$\left(1- 2 X Z + Z^2 \right) ^{- \frac{1}{2}} \ = \sum_{n=0}^\infty Z^n P_n(X)$$

where the coefficients Pn are the Legendre polynomials of degree n. This shows that the Taylor coefficients are given by the Legendre polynomials in $$\cos \theta$$. So the potential can be expanded in a series which is convergent for at least positions x such that r < |x| for all mass elements of the system (i.e., outside a sphere, centered at the center of mass, that encloses the system):
 * $$ \begin{align}

V(\mathbf{x}) &= - \frac{G}{|\mathbf{x}|} \int \sum_{n=0}^\infty \left(\frac{r}{|\mathbf{x}|} \right)^n P_n(\cos \theta) \, dm(\mathbf{r})\\ {}&= - \frac{G}{|\mathbf{x}|} \int \left(1 + \left(\frac{r}{|\mathbf{x}|}\right) \cos \theta + \left(\frac{r}{|\mathbf{x}|}\right)^2\frac {3 \cos^2 \theta - 1}{2} + \cdots\right)\,dm(\mathbf{r}) \end{align}$$ The integral, $$\int r\cos \theta dm$$, is the component of the center of mass in the x direction; this vanishes because the vector x emanates from the center of mass. So, bringing the integral under the sign of the summation gives
 * $$ V(\mathbf{x}) = - \frac{GM}{|\mathbf{x}|} - \frac{G}{|\mathbf{x}|} \int \left(\frac{r}{|\mathbf{x}|}\right)^2 \frac {3 \cos^2 \theta - 1}{2} dm(\mathbf{r}) + \cdots$$

This shows that elongation of the body causes a lower potential in the direction of elongation, and a higher potential in perpendicular directions, compared to the potential due to a spherical mass.