gravitational potential of an extended body

From Physics wiki

Jump to: navigation, search

Let us determine the gravitational potential due to an extended body of mass M\, and density \rho(\mathbf{r}')\,. Evaluated at the point \mathbf{r}\,, it is equal to

\Phi(\mathbf{r})\, = -G\int\!d^3x'\,\frac{ \rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\,
= -\frac{G}{r}\int\!d^3x'\,\frac{\rho(\mathbf{r}')}{\left(1 - \frac{2 \mathbf{r}\cdot\mathbf{r}'}{r^2} + \frac{r'^2}{r^2} \right)^\frac{1}{2} }\,.

Denoting \cos\gamma = \hat{\mathbf{r}}\cdot\hat{\mathbf{r}}'\,, suppose that r' \ll r\, at all points where \rho(\mathbf{r}')\, has support. We may perform a spherical multipole expansion in powers of \frac{r'}{r}\,, to obtain

\Phi(\mathbf{r})\, = -\frac{GM}{r} - \frac{G}{r^2} \int\!d^3x'\, \rho(\mathbf{r}') r' P_1(\cos\gamma) - \frac{GM}{r^3} \int\!d^3x'\, \rho(\mathbf{r}') r'^2 P_2(\cos\gamma) + ...\,.

For the purpose of expressing the potential in terms of known quantities, we shall re-derive this result in Cartesian coordinates.

\Phi(\mathbf{r})\, = -G\int\!d^3x'\,\frac{\rho(\mathbf{r}')}{r\left(1 - \frac{2 \mathbf{r}\cdot\mathbf{r}'}{r^2} + \frac{r'^2}{r^2} \right)^\frac{1}{2} }\,
= -G\int\!d^3x'\,\frac{\rho(\mathbf{r}')}{r}\left(1 - \frac{2 \hat\mathbf{r}\cdot\mathbf{r}'}{r} + \frac{r'^2}{r^2} \right)^{-\frac{1}{2}} \,
\approx -G\int\!d^3x'\,\frac{\rho(\mathbf{r}')}{r} \left(1 + \frac{\hat\mathbf{r}\cdot\mathbf{r}'}{r} - \frac{r'^2}{2r^2} + \frac{3}{8} \frac{(2\hat{r}\cdot\mathbf{r}')^2}{r^2}\right)\,

Without loss of generality, place the origin at the center of mass. Then

\Phi(\mathbf{r})\, = -\frac{GM}{r} - \frac{GM}{2 r^3} \int\!d^3x'\, \rho(\mathbf{r}') \left( 3 (\hat\mathbf{r}\cdot \mathbf{r}')^2 - r'^2 \right)\,.
= -\frac{GM}{r} - \frac{GM}{r^3} \int\!d^3x'\, \rho(\mathbf{r}') r'^2 P_2(\cos\gamma)\,, where \cos\gamma = \hat{\mathbf{r}}\cdot\hat{\mathbf{r}}'\,.

Align \mathbf{\hat{z}}\, with the line joining the two masses. I.e., \mathbf{\hat{z}} = \mathbf{\hat{r}}\,. Then,

\Phi(\mathbf{r})\, = -\frac{GM}{r} - \frac{G}{2 r^3} \int\!d^3x'\, \rho(\mathbf{r}') \left( 3 (\hat\mathbf{r}\cdot \mathbf{r}')^2 - r'^2 \right)\,,
= -\frac{GM}{r} -\frac{G}{2 r^3} \int\!d^3x'\, \rho(\mathbf{r}') \left( 3 r'_i \hat{z}_i \hat{z}_j r'_j - r'_i \delta_{ij} r'_j \right)\,,
= -\frac{GM}{r} - \frac{G}{2 r^3} \int\!d^3x'\, \rho(\mathbf{r}') \left( 3 r'_i [\delta_{ij} - \hat{x}_i\hat{x}_j - \hat{y}_i \hat{y}_j ] r'_j - r'_i \delta_{ij} r'_j \right)\,, (completeness)
= -\frac{GM}{r} - \frac{G}{2 r^3} \int\!d^3x'\, \rho(\mathbf{r}') \left( 2 r'^2 - 3 [{r'^2}_x + {r'^2}_y] \right)\,,
= -\frac{GM}{r} - \frac{G}{2 r^3} (I_{xx} + I_{yy} + I_{zz} - 3 I_r)\,,

where I_r\, is the moment of inertia about the line joining the bodies. Our choice of \mathbf{\hat{z}} = \mathbf{\hat{r}}\, was arbitrary, and doesn't necessarily reflect the symmetry of the body (i.e., I_{xx}\, need not be principal moments of inertia). However, I_{xx} + I_{yy} + I_{zz} = \operatorname{tr}\, \mathbf{I}\, is invariant.

\Phi(\mathbf{r}) = -\frac{GM}{r} - \frac{G}{2 r^3} (\operatorname{tr}\, \mathbf{I} - 3 I_r)\,.

This is known as MacCullagh's Formula (1855).

on to Precession of the equinoxes

See also


References

Further reading:[1] [2] [3]

  1. ↑ Landau, L.D.; Lifshitz, E.M. (1997). Mechanics (3rd ed). Butterworth-Heinemann. ISBN 0-750-62896-0. 
  2. ↑ Herbert Goldstein, Charles P. Poole, John L. Safko (1980). Classical Mechanics (3rd ed). Addison Wesley. ISBN 978-0201657029. 
  3. ↑ Richard Fitzpatrick, Classical Dynamics: An intermediate level course, retrieved October 16, 2007
Retrieved from "http://www.physics.thetangentbundle.net/wiki/Classical_mechanics/gravitational_potential_of_an_extended_body"
Personal tools