I'm trying to write the gravitational potential of two masses $m_1$ and $m_2$ using the multipolar expansion of the potential:

$$V = G\sum\frac{1}{r^{n+1}}\int r'^n P_n(\cos \theta) \rho(r') dV'$$

I was able to derive the expression for the expansion but my issue is when considering the terms alone (monopole, dipole, quadrupole term ...). I am aware that the terms come from the gravitational moments, but I don't understand how we achieved these relations. The monopole, dipole, quadrupole moments are respectively:

$$M = \int{\rho(r') dV'} = m_1 + m_2$$ $$P = \int r' \rho(r') dV' = \left(\frac{m_1m_2}{m_1 + m_2} - \frac{m_1m_2}{m_1 + m_2}\right)l = 0$$ $$Q = \int r'^2 \rho(r')dV' = \frac{m_1m_2}{m_1 + m_2}l^2$$

$l$ is the distance between the two masses. I get the left side of the equations, but it's the right side that I don't get.

Any explanation on this is much appreciated.


1 Answer 1


In layman terms

$$ V = V_{\rm monopole} + V_{\rm dipole} + V_{\rm quadrupole} + \cdots $$

Each term is then found by fixing $n$

  1. $n=0$

    $$ V_0 = \frac{G}{r}\int {\rm d}^3r'\; \rho(r') = \frac{GM}{r} $$

  2. $n=1$

    $$ V_1 = \frac{G}{r^2}\int {\rm d}^3r'\; \rho(r') = 0 $$

  3. $\ldots$

  • $\begingroup$ i'm okay with the monopole term, but how are we including the masses in the othr terms? $\endgroup$ Commented Nov 14, 2016 at 17:51
  • $\begingroup$ Through the density $\rho(r)$. The information about the mass distribution is contained in the density factor, each term in the expansion knows about the mass $\endgroup$
    – caverac
    Commented Nov 14, 2016 at 18:33

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