# 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 pu

Sara Fleming 2022-09-17 Answered
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.
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Simon Zhang
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.
$E=\frac{1}{2}m{v}^{2}$
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 $v+C=c\mathrm{ln}|t|$ where $C$ is an abitrary constant, note $t\ge 0$ so we can write $\mathrm{ln}\left(t\right)$
We are left with $v=c\mathrm{ln}\left(t\right)-C$
It is well known that
$\underset{t\to +\mathrm{\infty }}{lim}\left(\mathrm{ln}\left(t\right)\right)=+\mathrm{\infty }$
So there really is no "terminal velocity" (ignoring relativity of course)

We have step-by-step solutions for your answer!