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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?

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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.


Example:

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.

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  • $\begingroup$ 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. $\endgroup$ – Meni Rosenfeld Jun 19 '17 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. $\endgroup$ – Hritik Narayan Jun 19 '17 at 9:34
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    $\begingroup$ Ms Paint diagramms are better than no diagramms, kudos for making them $\endgroup$ – HopefullyHelpful Jun 19 '17 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? $\endgroup$ – Rumplestillskin Jun 22 '17 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. $\endgroup$ – Hritik Narayan Jun 22 '17 at 11:06
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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.

Drawing

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