Is it possible for a charged, fast-moving object to slow down and enter geo-stationary orbit?

I've had a wild idea which I can not discuss at length in this forum, but it comes down to the following problem:

A sphere of radius R=~10μm and mass m=~10-16Kgr is travelling towards the earth at v = ~10^8m/sec. The sphere carries a charge Q and intersects the earth's magnetic field perpendicularly, at a distance r, such that it enters an elliptic orbit. Given an electron wave function of ~5eV, is there any possible configuration of Q and r that would allow the sphere to end up in a stable geo-stationary orbit?

Would releasing charges at particular points of each rotation make any difference?

Even though I did get a physics BS 13 years ago, I took a different direction and the calculations required are way beyond my current capabilities. If anyone finds the problem interesting, I'll be glad to hear from you.

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@ChristopherAkritidis: Welcome to SciComp! This question is cool, but has no direct computational science component. It's definitely more suitable for physics. Everyone else: if you think the question is more suitable for physics, flag the question or vote to close; up-voting a comment is nice, but flagging notifies mods, and five non-mod close votes will close a question. –  Geoff Oxberry Apr 29 '13 at 23:39
Why these particular numbers? Why is this idea one that you "can not discuss at length in this forum?" –  Ben Crowell Apr 29 '13 at 23:42
What you need is a good ratio of capacitance to mass. You need to avoid discharge at the same time, limiting the mass you have to move. Practically, this would be a large spherical wire mesh surrounding a micro satellite, large and light to the extent of what could withstand the acceleration. Even then, the movement due to the magnetic field is nothing like you've imagined because the field lines are not oriented in a helpful way and they are probably useless anywhere near GEO. –  AlanSE Apr 30 '13 at 0:05
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1 Answer

Your sphere is traveling at $10^8 \, \mathrm{m/s}$ (33% the speed of light) but Earth's escape velocity is $11200 \, \mathrm{m/s}$. There is absolutely no way the sphere could ever enter any sort of orbit around the Earth, regardless of the interaction between it and the Earth's magnetic field.

$$KE_{relativistic} = \left( \left( \frac{1}{\sqrt {1 - \frac{v^2}{c^2}}} \right) - 1 \right)mc^2 = 8.7 \cdot 10^{16} \, \mathrm{J}$$

That's 20 megatons of TNT. Even if the magnetic field could perform some sort of braking, the amount of heat generated to dissipate that much energy would induce nuclear fusion.

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What you said, but also note that magnetic fields can do no work. The force of a magnetic field on a charged particle is always perpendicular to the velocity so the energy doesn't change. What you can do is use the magnetic field to induce a current and dissipate that, but for a body of the OP's stated size this would be very ineffective and totally incapable of dissipating the required amount of energy. –  Michael Brown Apr 30 '13 at 3:01
@MichaelBrown exactly. The magnetic fields can either sap kinetic energy to ultimately heat the sphere or cause it to spin, transferring linear momentum to angular momentum. There is simply too much kinetic energy for either of these to have any major effect. –  Brandon Enright Apr 30 '13 at 3:03
So, there is something wrong with the following calculations that estimate the required charge density for a circular orbit? Is it that we're not talking about a point charge? The sphere intersects the earth's magnetic field perpendicularly, at a distance r = ~10^6m. At that radius, the magnetic field is B = ~10^-5T The charge of the sphere is such that it enters a circular orbit. Fc=m*v*v/r=qvB => q=mv/rB => q = ~(10^-16*10^8)/(10^6m*10^-5T) = ~10-9C The surface charge density is then D = Q / 4πR^2 = ~(10^-9)/(10*10^-10)m^2 = ~1C/m^2 –  Christopher Akritidis Apr 30 '13 at 10:13
I apologize for the second comment in a row, but I was not aware of inductive heating, until I read Brandon Enright's comment. I assume the excess thermal energy would slowly dissipate via infrared radiation, right? Would it just take too much time to lose all that energy or is the heating effect negligible in this case? Could you possibly point me to the equations I could use to calculate it? –  Christopher Akritidis Apr 30 '13 at 11:35
@Brandon What did you use for m in the energy calculation? $$KErelativistic = ((\frac{1 }{\sqrt{1 - v^2/c^2}})-1)mc^2 = ((\frac{1 }{\sqrt{1 - 0.09}})-1)*10^{-16}*9*10^{16}J = 0,43J$$ –  Christopher Akritidis Apr 30 '13 at 16:41
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