Rate of Acceleration of an Object Pulled by Magnetic Force Gravity on earth pulls objects toward it with an acceleration of 9.8 m/s/s on Earth until the object reaches it's max potential free fall speed. (I call this terminal velocity though I think that this is wrong by definition)
a) Is there a similar equation for the acceleration of an object being pulled by magnetism?
b) Is there a "terminal velocity" of an object being accelerated by magnetism alone? (This would have to mean that the origin of the magnetic field has zero mass, so I'm thinking this is a theoretical question)
c) Is it possible (however unlikely) that one object can be held in orbit around another by magnetism with minimal gravitational influence? That is to say, can an equilibrium be found between the magnetic force pulling against an object and the velocity of an object. (best example, moon orbiting Earth because of an equilibrium of gravitational force and the velocity of the moon revolving around it)
Thanks in advance for any and all help!
P.S. It would also be interesting to know if there is a difference in the equations for magnetic repulsion, or if the forces are equal.
 A: Weirdly I just wrote about this:
Magnetic force storage or amplification question
Anyway.
For the first one! YES the equations for force due to magnetism and force due to gravity are both very similar, they take the form:
$$F=\frac{k}{x^3}$$ where $x$ is the distance from the source of gravity/magnetism.
I am a mathematician, so I am completely happy with that $k$ - in reality it'll be some negative (because $x$'s positive direction is away from the magnet, and the magnet pulls in $-$ this direction) number, further reasoning for this is discussed on the linked page. Physicists usually use more letters.
We assume infinitely thin magnet(s) with a centre of mass at the origin, and the other (which is a magnet, or item influenced by it) to be at a distance of $x$ away. Then this works, otherwise it involves nasty integrals.
(forgive my earlier mistakes, I was thinking of something different)
NEXT QUESTION
There is no "terminal velocity" due to gravity. What you have is a case like this:
$$\frac{dv}{dt}=-9.81+kv^2$$
Note k>0  
When $\frac{dv}{dt}=0$ we have $kv^2=9.81$ - the force up due to air resistance (which gets 4 times as hard if we go twice as fast in this case) negates the downward pull of gravity, thus we do not accelerate but maintain a constant velocity - this is terminal velocity. 
Air resistance isn't simply proportional to $v^2$, sometimes it's $v$ instead, generally it is neither (especially if the body is rotating).
Notice that, for $v=0$ we have $\frac{dv}{dt}=-9.81$, but if $v>0$ then we have $\frac{dv}{dt}>-9.81$ - it accelerates more slowly the faster it goes. 
This address your terminal velocity problem. 
NEXT QUESTION
Okay, go outside and put a hammer on some string and start swinging it around. Remember without any force the hammer will want to go off in a straight line. The fact it is spinning around means something is constantly pulling it towards you.
The string is providing a force on the hammer (towards you) that is (pretty much, because of air resistance) exactly $$F=\frac{mv^2}{r}$$
If the force inwards due to magnetism matches this then the hammer will spin on forever. 
I hope this helps. I sense however the differential equations are a little above you, which is why I have avoided (as much as I can) their discussion.
A: I write this because it is too long for a comment
Situation
You are putting $k$ watts (=joules/second) of kinetic energy into a body. We will show the acceleration decreases with time. 
$$E=\frac{1}{2}mv^2$$ so $$v=\sqrt{\frac{2E}{m}}$$ so $$\frac{dv}{dt}=\frac{m}{\sqrt{\frac{2E}m}}\cdot\frac{de}{dt}$$
(I can't be bothered to tidy this up - remember I'm from maths!
Now
Assume $m$ is constant, and $\frac{de}{dt}$ is also constant (call it $k$) (so you are putting in so many watts (=joules/second) of energy). Then $E=tk$ (because it's going up at a constant rate), thus we have:
$$\frac{dv}{dt}=\frac{mk}{\sqrt{\frac{2kt}m}}$$
You can see that as time goes on, $\frac{dv}{dt}$ becomes smaller and smaller. However it still tends towards infinity.
Proof
As you know, I do maths, so I cannot be bothered to simplify this, however I am sure we can both agree it's of the form:
$$\frac{dv}{dt}=\frac{c}{t}$$ for some $c$ that I am too lazy to come up with.
This is easy to solve:
We have: $dv=c\frac{1}{t}dt$, so $\int dv=c\frac{1}{t}dt$
evaluating we see $v+C=c\ln|t|$ where $C$ is an abitrary constant, note $t\ge 0$ so we can write $\ln(t)$
We are left with $v=c\ln(t)-C$
It is well known that $$\lim_{t\rightarrow+\infty}(\ln(t))=+\infty$$
So there really is no "terminal velocity" (ignoring relativity of course)
A: I'll just answer the part of your question about terminal velocity (better to call it terminal speed...).
The terminal speed of a falling object is not caused by anything special about gravity.  It is caused by the fact that for an object falling straight down there are two main forces acting on it:


*

*gravity (which points down as is constant to a good approximation)

*a drag force exerted by the air which points up and depends on speed
Under most circumstances the drag force is approximately proportional to the speed of the object squared.  So Newton's 2nd law for the falling object is (using up as positive)
$ma = -mg + C v^2$
where $C$ depends on the shape of the object, the cross-sectional area of the object and the density of the air.  Notice that at low speed we get
$ma = -mg$
so $a$ points down and $|a| = g$ independent of the mass.  But as $v$ gets large the magnitude of acceleration decreases.  The acceleration goes asymptotically to zero as
$v \rightarrow \sqrt{\frac{mg}{C}}$
So this limiting speed is called the terminal speed.
Now there is nothing special about gravity here.  Any time we have some combination of constant forces and a speed dependent force that act in opposite directions we can get a terminal speed.  So for an object falling through water there is gravity (constant, down) buoyancy (constant, up) and a drag (up, varies approximately linearly with speed).  So we can again get a terminal speed, but it comes out differently from what I did above.  For a rocket car we have thrust (approximately constant, forward), rolling friction (approximately constant, back) and drag (back, proportional to $v^2$) and again there will be a terminal speed.
So if you could arrange for a constant magnetic force pulling an object through  fluid this would result in a terminal speed because of the "competition" between the magnetic force and the fluid drag. 
