What does the monopole/quadrupole moment of the Earth signify? 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?
 A: 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.

A: 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. 
The force here is a simple: $$\frac{GMm}{r^2}$$.
Case 2: Non-zero quadrupole moment. (the larger spheres are separated by some distance $2R$.)
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.
