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.

  • $\begingroup$ I was always taught that the Magnetic Field does NO work. If a charged object is increasing its speed, that's supposed to be due to only electric field effects. I've never seen what I would call a proof of this 'theorem' though. $\endgroup$ Commented Aug 18, 2015 at 19:18
  • $\begingroup$ Do a "current balance" experiment in which two wires repel each other due to their magnetic interaction. One wire can lift the other wire. So it does work. However, for a free (charged) particle moving in a magnetic field the magnetic force is always perpendicular to the velocity since $\vec{F}_B = q\vec{v} \times \vec{B}$. But work is a dot product between $\vec{F}$ and $\vec{v}$. So, for the charged particle in the magnetic field the magnetic field does no work since since $\vec{v}$ is perpendicular to $\vec{F}_B$. $\endgroup$ Commented Aug 19, 2015 at 2:57

3 Answers 3


I write this because it is too long for a comment


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!


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.


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)

  • $\begingroup$ You probably shouldn't use a lowercase $c$ for an arbitrary constant. It will almost universally be confused with the speed of light in vacuum, especially as we are talking about electrodynamics here. $\endgroup$ Commented Aug 18, 2015 at 19:20
  • $\begingroup$ @KevinDriscoll I forgot, physicist site.... :P $\endgroup$
    – Alec Teal
    Commented Aug 18, 2015 at 19:20

Weirdly I just wrote about this:

Magnetic force storage or amplification question


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)


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.


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.

  • $\begingroup$ I really appreciate the simplistic approach to your answer! As a follow up to one of your points... if there were no air resistance, the acceleration of gravitational/magnetic pull would continue at an exponential/linear rate? until infinity? (and that's why there's no terminal velocity of magnetic force acceleration?) $\endgroup$
    – MegaMark
    Commented Aug 18, 2015 at 18:48
  • 2
    $\begingroup$ The magnetic force between two objects is not proportional to $1/x^2$. Unless you've found a couple of magnetic monopoles, in which case you should alert the Nobel committee. $\endgroup$ Commented Aug 18, 2015 at 18:55
  • $\begingroup$ @MichaelSeifert fixed. $\endgroup$
    – Alec Teal
    Commented Aug 18, 2015 at 19:14
  • $\begingroup$ @AlecTeal Yes, if you have 2 magnetic dipoles it is $\sim x^{-3}$ but then your $k$ is a vector-valued quantity; the force is not spherically symmetric. $\endgroup$ Commented Aug 18, 2015 at 19:14
  • $\begingroup$ @KevinDriscoll I've made it clear that we now assume the magnets are parallel and (if extended) colinear. At an instant this is true. If only one is a magnet it is true. $\endgroup$
    – Alec Teal
    Commented Aug 18, 2015 at 19:19

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:

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

  2. 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.


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