First understand the method of image charges. The idea behind the method is to bypass actually solving the differential equation with boundary conditions, and instead "cheat" by guessing the correct solution. To this end, we find a configuration of imaginary charges that together with the real ones will make the potential on all surfaces be what is given.
In your case, you have two surface, each with constant potential zero. For a plane, if you have a charge $q$ at $(x,y,z)$ and another one at $(x,-y,z)$, clearly the potential at $y=0$ will be zero. For the sphere, if a charge is at radius $R$ from the sphere with radius $a$, an image charge with charge $-qa/R$ should be placed at radius $a^2/R$, but the same idea remains.
Now, first add the image charge for the sphere. Now the potential on the sphere is zero, but on the plane, we still have a gradient. However, if you mirror both charges off the plane, you should see that the potential on both the sphere and the plane is zero.
Writing this all down, we have:
$$\begin{eqnarray}
\phi({\bf x}) &=& \frac{q}{\sqrt{(x-R\cos\alpha)^2+(y-R\sin\alpha)^2+z^2}} \\
&-& \frac{q}{\sqrt{(x-R\cos\alpha)^2+(y+R\sin\alpha)^2+z^2}} \\
&-&\frac{qa}{R\sqrt{(x-(a^2/R)\cos\alpha)^2+(y-(a^2/R)\sin\alpha)^2+z^2}} \\
&+&\frac{qa}{R\sqrt{(x-(a^2/R)\cos\alpha)^2+(y+(a^2/R)\sin\alpha)^2+z^2}}
\end{eqnarray}$$
Clearly, at $y=0$ we have $\phi=0$. Now, what about on the hemisphere? points on the hemisphere satisfy $z^2 = a^2 - x^2-y^2$, so that substituting leads us to the potential on the hemisphere:
$$\begin{eqnarray}
\phi({\bf x}_{sphere}) &=& \frac{q}{\sqrt{-2xR\cos\alpha -2yR\sin\alpha + R^2 +a^2}} \\
&-& \frac{q}{\sqrt{-2xR\cos\alpha +2yR\sin\alpha + R^2 +a^2}} \\
&-&\frac{qa}{R\sqrt{-2x(a^2/R)\cos\alpha -2y(a^2/R)\sin\alpha + (a^2/R)^2 + a^2}} \\
&+&\frac{qa}{R\sqrt{-2x(a^2/R)\cos\alpha +2y(a^2/R)\sin\alpha + (a^2/R)^2 + a^2}} \\
&=&0
\end{eqnarray}$$
By putting the $a/R$ into the square root.
To write down the multipole expansion, you just need to write down the Taylor expansion of the potential around $1/r$, with $r = \sqrt{x^2+y^2+z^2}$. This gives that any expression of the form:
$$\lim_{r\to\infty}\frac{1}{\sqrt{r^2+b}}= \frac{1}{r} - \frac{b}{2r^3} + \frac{3b^2}{8r^5} + \cdots$$
You can then use this expansion on the components of the potential to get your result.
The satellite will always be falling towards the Earth. The trick to achieving orbit to have enough tangential (horizontal) velocity to constantly 'miss' the Earth. To be in a state of free fall means that the only force acting on you is gravity. This is true in this case, since there is no friction, drag, etc in space.
So if you are constantly falling towards the Earth, you also need to move very fast tangentially, to always avoid crashing. This is true for satellites, the space station, and even the Moon!
Best Answer
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.