I'm currently reading about orbits of near-Earth satellites and some terminology is getting thrown around that I'm not sure I understand what they actually mean:

The Earth's monopole moment and the Earth's quadrupole moment?

What are some easily understood explanations of the above terms?


2 Answers 2


A monopole (gravitational) of a system is basically the amount of mass-energy the system has.

A dipole is a measure of how the mass is distributed away from some center.

The quadrupole moment describes how stretched out the mass distribution is along an axis. Quadrupole would be zero for a sphere, but non-zero for a rod, for instance. It is also non-zero for the Earth, because the Earth is an oblate spheroid.

The gravitational contribution from a quadrupole falls of faster than that of a monopole. (which is why the Earth's quadrupole moment is important for studying satellites and not really for studying the moon, owing to the $r^{-3}$ dependency of the contribution to the potential)

Quadrupoles and other higher order moments are important in GR because the change in their distribution can produce gravitational waves.


Let's consider two cases, in both the cases, the large bodies are of mass $M$ and the small one of mass $m$, and the small one is on the line of symmetry at a distance $r$.

Case 1: No quadrupole moment. enter image description here

The force here is a simple: $$\frac{GMm}{r^2}$$.

Case 2: Non-zero quadrupole moment. (the larger spheres are separated by some distance $2R$.)enter image description here

The force in this case is: $$\frac{2GMmr}{(r^2+R^2)^{3/2}}$$

This, for large $r$, can be approximated to (two term series expansion): $$F \sim \frac{2GMm}{r^2}-\frac{3GMmR^2}{r^4}$$

The weird term here is because of the quadrupole moment of the system. As you go further away ($r>>R$), the force, $F$ is more or less: $$F \sim \frac{2GMm}{r^2}$$

This is why the "quadrupole moment effect" falls off with distance.

Apologies for the obnoxious MS Paint diagrams.

  • $\begingroup$ The potential due to quadrupole moment falls off as $r^{-3}$. But you've talked about force, which falls off as $r^{-4}$. In your example, the two-term expansion of the force is $\frac{2GMm}{r^2}-\frac{3GMmR^2}{r^4}$, where the second term is due to the quadrupole moment. $\endgroup$ Commented Jun 19, 2017 at 9:33
  • $\begingroup$ Oops, yes. I'll fix that. I talked about the force because I'm trying to show the effect of the asymmetry in general, not specifically the potential. $\endgroup$ Commented Jun 19, 2017 at 9:34
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    $\begingroup$ Ms Paint diagramms are better than no diagramms, kudos for making them $\endgroup$ Commented Jun 19, 2017 at 9:55
  • $\begingroup$ @HritikNarayan Just a quick follow up Q: Are these two sentences equivalent: 1. The forces on body A are due to the central force, the non-spherical geometry of the Earth. 2. The forces on body A are due to the Earth's monopole and quadropole moment? $\endgroup$ Commented Jun 22, 2017 at 11:03
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    $\begingroup$ The force, at the end of the day, is due to the mass distribution. These differences arise depending on the way we study it. So yes, the statements are equivalent. $\endgroup$ Commented Jun 22, 2017 at 11:06

Imagine having a mass distribution $\rho(x,y,z)$ around the origin O and we want to calculate the potential energy and force at a certain point P on the z-axis. The potential energy can easily expressed by the integral: $$U=-GM\int_{V}\frac{\rho(x,y,z)}{R}dv$$ However this intgral might be difficult to calculate and it's often easier to express the integrand by a series, this is called multipole expansion and can be done for both the gravitational force as the electrostatic force.

Because of the law of cosines, we express R in function of $\theta$, $r'$ and r: $R^2 = r^2 +r'^2 - 2rr'\cos(\theta)$, now we can simplify this integral by using this indentity and a Taylor series: $$\frac{1}{R} = \frac{1}{r}\frac{1}{\sqrt{1+\alpha}} = \frac{1}{r}\left(1-\frac{1}{2}\alpha+\frac{3}{8}\alpha^2-...\right)$$ where $\alpha=\left(\frac{r'}{r}\right)^2-\frac{2r'}{r}\cos(\theta)$

The potential energy now becomes: $$U=\frac{-GM}{r}\int_{V}\rho dv+\frac{-GM}{r^2}\int_{V}r'\cos(\theta)\rho dv+\frac{-GM}{r^3}\int_{V}r'^2\frac{3\cos^2(\theta)-1}{2}\rho dv +...$$ As you can see, in every term the the power of r becomes smaller and smaller. Often we rewrite this expression as: $$U=-GM\left(\frac{C_0}{r}+\frac{C_1}{r^2}+\frac{C^2}{r^3}...\right)$$ Where $C_0$ is the monopole moment, $C_1$ the dipole moment,$C_2$ the quadrupole moment and so on. These can be easily interpreted for wich I refer to @Hritik Narayan.



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