The gravitationalpotentialV at a distancex from apoint mass of massM can be defined as the workW that needs to be done by an external agent to bring a unit mass in from infinity to that point:^[1][2][3][4]

$${\displaystyle V(\mathbf{x})=\frac{W}{m}=\frac{1}{m}{\int }_{\mathrm{\infty }}^x\mathbf{F}\left({\mathbf{x}}^{\prime }\right)\cdot d{\mathbf{x}}^{\prime }=\frac{1}{m}{\int }_{\mathrm{\infty }}^x\frac{GmM}{x^{\prime 2}}d{x}^{\prime }=-\frac{GM}{x},}$$whereG is thegravitational constant, and F is the gravitational force. The productGM is thestandard gravitational parameter and is often known to higher precision thanG orM separately. The potential hasunits of energy per mass, e.g., J/kg in theMKS system. By convention, it is always negative where it is defined, and asx tends to infinity, it approaches zero.

Thegravitational field, and thus the acceleration of a small body in the space around the massive object, is the negativegradient of the gravitational potential. Thus the negative of a negative gradient yields positive acceleration toward a massive object. Because the potential has no angular components, its gradient is$${\displaystyle \mathbf{a}=-\frac{GM}{x^3}\mathbf{x}=-\frac{GM}{x^2}\hat{\mathbf{x}},}$$where x is a vector of lengthx pointing from the point mass toward the small body and${\displaystyle \hat{\mathbf{x}}}$ is aunit vector pointing from the point mass toward the small body. The magnitude of the acceleration therefore follows aninverse square law:$${\displaystyle \Vert \mathbf{a}\Vert =\frac{GM}{x^2}.}$$

The potential associated with amass 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, ..., x_n and have massesm₁, ...,m_n, then the potential of the distribution at the point x is$${\displaystyle V(\mathbf{x})=\sum _{i=1}^n-\frac{G{m}_i}{\Vert \mathbf{x}-{\mathbf{x}}_i\Vert }.}$$

If the mass distribution is given as a massmeasuredm on three-dimensionalEuclidean space R³, then the potential is theconvolution of−G/|r| withdm.^{[citation needed]} In good cases^{[clarification needed]} this equals the integral$${\displaystyle V(\mathbf{x})=-{\int }_{{\mathbb{R}}^3}\frac{G}{\Vert \mathbf{x}-\mathbf{r}\Vert }dm(\mathbf{r}),}$$where|x − r| is thedistance between the points x and r. If there is a functionρ(r) representing the density of the distribution at r, so thatdm(r) =ρ(r)dv(r), wheredv(r) is the Euclideanvolume element, then the gravitational potential is thevolume integral$${\displaystyle V(\mathbf{x})=-{\int }_{{\mathbb{R}}^3}\frac{G}{\Vert \mathbf{x}-\mathbf{r}\Vert }\rho (\mathbf{r})dv(\mathbf{r}).}$$

IfV is a potential function coming from a continuous mass distributionρ(r), thenρ can be recovered using theLaplace operator,Δ:$${\displaystyle \rho (\mathbf{x})=\frac{1}{4\pi G}\mathrm{\Delta }V(\mathbf{x}).}$$This holds pointwise wheneverρ is continuous and is zero outside of a bounded set. In general, the mass measuredm can be recovered in the same way if the Laplace operator is taken in the sense ofdistributions. As a consequence, the gravitational potential satisfiesPoisson's equation. See alsoGreen's function for the three-variable Laplace equation andNewtonian potential.

1. ^Marion, J.B.; Thornton, S.T. (1995).Classical Dynamics of particles and systems (4th ed.). Harcourt Brace & Company. p. 192.ISBN 0-03-097302-3.
2. ^Arfken, George B.; Weber, Hans J. (2005).Mathematical Methods For Physicists International Student Edition (6th ed.).Academic Press. p. 72.ISBN 978-0-08-047069-6.
3. ^Sang, David; Jones, Graham; Chadha, Gurinder; Woodside, Richard; Stark, Will; Gill, Aidan (2014).Cambridge International AS and A Level Physics Coursebook (illustrated ed.).Cambridge University Press. p. 276.ISBN 978-1-107-69769-0.
4. ^Muncaster, Roger (1993).A-level Physics (illustrated ed.).Nelson Thornes. p. 106.ISBN 978-0-7487-1584-8.
5. ^MacMillan, W.D. (1958).The Theory of the Potential. Dover Press.

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