A question about charged bodies/gravitation

[fluidistic]

I've learned that the electrostatic force is about $$10^{39}$$ times greater than the gravitational force.
I guess the Earth is electrically neutral, but what about the Sun? Or any other celestial body? For example what about a system in which there's 2 stars?
The motion of 2 non neutral bodies would be the consequence of the electrostatic force between them, much more than the gravitation. Is this the case?

 

[Doc Al]

I've learned that the electrostatic force is about $$10^{39}$$ times greater than the gravitational force.

That's true for elementary particles, perhaps, but not composite bodies. (Charges can cancel--mass cannot.)

The motion of 2 non neutral bodies would be the consequence of the electrostatic force between them, much more than the gravitation. Is this the case?

It depends on how much charge they have.

 

[fluidistic]

That's true for elementary particles, perhaps, but not composite bodies. (Charges can cancel--mass cannot.)

You're right.

It depends on how much charge they have.

Yes. I'm asking if the Sun is a ionized plasma, or at least if it is not neutral (and so even though it is a composite body, charges wouldn't all cancel each other).
This would almost imply that as many stars are similar to the Sun, they should repeal each other more than the gravity attract themselves.

I'm not asking only for the Sun, but any other celestial body.

 

[Doc Al]

I'm not aware of the sun, or any other stars, having a significant net charge. (Perhaps you can ask that question in the Astronomy forums.)

 

[fluidistic]

I'm not aware of the sun, or any other stars, having a significant net charge. (Perhaps you can ask that question in the Astronomy forums.)

Ok thanks. I think I'll wait some time and then ask for a moderator to move the thread if I get no other responses.
By the way, thanks for the response.

 

[Unit]

I can make no significant contribution except:

"a CHARGED STAR? is that what you're saying?! with oppositely-charged PLANETS? like a total electromagnetic solar system?! man, that would be EPIC!"

I don't think it's possible though because of what Doc Al said, though. BUT STILL theoretical situations are awesome. EXCEPT the moment this planet touched something else, it would lose its charge :O

 

[belliott4488]

The point made above is important - mass doesn't cancel, so with something as large as a star the gravitational force is significant. You could figure out how much excess charge two stars of known mass would need to exert the same electrostatic force on each other as the gravitational force. I'm guessing it would be a lot of charge ...

 

[fluidistic]

I can make no significant contribution except:

"a CHARGED STAR? is that what you're saying?! with oppositely-charged PLANETS? like a total electromagnetic solar system?! man, that would be EPIC!"

I don't think it's possible though because of what Doc Al said, though. BUT STILL theoretical situations are awesome. EXCEPT the moment this planet touched something else, it would lose its charge :O

Not exactly. I think it's quite improbable that a planet is charged, I was thinking of stars or maybe other celestial bodies (except neutron stars, planets and probably many others).

The point made above is important - mass doesn't cancel, so with something as large as a star the gravitational force is significant. You could figure out how much excess charge two stars of known mass would need to exert the same electrostatic force on each other as the gravitational force. I'm guessing it would be a lot of charge ...

Well, unless I'm wrong I'm tempted to say that 1 atom in $$10^{39}$$ is enough to equalize the gravitational force with the electrostatic one.

 

[belliott4488]

Well, unless I'm wrong I'm tempted to say that 1 atom in $$10^{39}$$ is enough to equalize the gravitational force with the electrostatic one.

I don't think so. Do you mean one atom has lost one electron? The charge of the electron is very small, so if you have one star with one excess electron and another missing one electron (so they are oppositely charged and attract each other), at astronomical distances that force will be negligible.

Think of this: if you rub a glass rod to put a static charge on it, you will have put many electrons on it, but it will still exert only enough force to lift a small piece of paper. Now separate the paper and the charged rod by 100,000,000 km - what's the force now (remember the inverse square law)? Negligible. On the other hand, gravity is strong enough to pull anything larger than a small piece of paper right off the rod.

Again: that ratio of $$10^{39}$$ applies only to elementary particles, which have tiny masses but relatively large charges (relative to their masses). A star has a lot more mass, but likely the same order of magnitude net charge, at most.

 

[fluidistic]

I don't think so. Do you mean one atom has lost one electron? The charge of the electron is very small, so if you have one star with one excess electron and another missing one electron (so they are oppositely charged and attract each other), at astronomical distances that force will be negligible.


Think of this: if you rub a glass rod to put a static charge on it, you will have put many electrons on it, but it will still exert only enough force to lift a small piece of paper. Now separate the paper and the charged rod by 100,000,000 km - what's the force now (remember the inverse square law)? Negligible. On the other hand, gravity is strong enough to pull anything larger than a small piece of paper right off the rod.

I was thinking that if a star was non negligibly positively charged for example, others would have the same properties (and not negatively charged), so they would repeal each other.
It's true that the force decreases fastly with distance but so does the gravitational force. The electrostatic force would always remain greater, no matter the distance separating the stars.

Again: that ratio of $$10^{39}$$ applies only to elementary particles, which have tiny masses but relatively large charges (relative to their masses). A star has a lot more mass, but likely the same order of magnitude net charge, at most.

Thanks for the information. I guess this explains why the gravitational force can alone give a satisfying model for stars motion. (I hope I'm not wrong on the last sentence! I've heard of dark matter... hmm). Maybe in binary system at least, which would show that indeed charges in common stars does not affect their motion.

 

[DrZoidberg]

Let's calculate the voltage necessary to make the electric force equal to the gravitational force.

Let's say there are 2 stars, both have the same properties as the sun and are at a distance of 10^10 m.
The mass of the sun is 1.9891 * 10^30 kg
The gravitational force is
F=G*m1*m2/r^2
G=6.674*10^-11 m^3*kg^-1*s^-2
so F = 2.64 * 10^30 N

The electric force is
F=k*q1*q2/r^2
k=8.99*10^9 N*m^2/C^2

If q1 and q2 are the same that means the charge would be
q=square root (F*r^2/k)
If I put in 2.64 * 10^30 N for the force and 10^10m for r I get
q = 1.71 * 10^20 C

The voltage of a sphere is equal to k*q/r
The radius of the sun is 6.96 * 10^8 m
so the voltage would be 2.21 * 10^21 V

That's very high. I don't think that could ever happen. The solar wind that is produced by every star would immediately remove any charge a star might have.

 

[Vanadium 50]

I'm not aware of the sun, or any other stars, having a significant net charge. (Perhaps you can ask that question in the Astronomy forums.)

It's known that most stars are close to electrically neutral - well below the level needed to have measurable orbital consequences - because of Stark effect measurements.

Of course, a star of one charge would preferentially repel ions of that charge and attract ions of the opposite charge and so would quickly neutralize this charge.

 

[Doc Al]

It's known that most stars are close to electrically neutral - well below the level needed to have measurable orbital consequences - because of Stark effect measurements.

Of course, a star of one charge would preferentially repel ions of that charge and attract ions of the opposite charge and so would quickly neutralize this charge.

Thanks, Vanadium. Makes sense.

 

[fluidistic]

It's known that most stars are close to electrically neutral - well below the level needed to have measurable orbital consequences - because of Stark effect measurements.

Of course, a star of one charge would preferentially repel ions of that charge and attract ions of the opposite charge and so would quickly neutralize this charge.

Thanks a lot, indeed. I've read yesterday night before to sleep that stars are ionized gas (plasma) but they are neutral.

Thank you all.

Just a little question : why taking out one electron in a sheet of paper won't have any effect at our scale if the electrostatic force is so high? Say you have a little sheet of paper with a $$e^{-}$$ charge which is in a vacuum. Now put a toy balloon in the vacuum room. Will the sheet of paper induce a charge in the balloon, if they are at about 20 m from each other? If so, will they move to meet in a time relatively short so that we can them moving without waiting for more than one day?

I know that if there is air between the 2 objects the attraction effect due to the electrostatic charge would almost disappear. But what if there is no air nor any matter?

I guess I should do a calculation similar to the one of DrZoidberg.

 

[dE_logics]

You would like to know apart from the standard that, there's 'dark energy' which is actually generated my 'dark matter' which make most of the universe.

However the thing is sill very much experimental...dark energy is the actual reason (according to NASA, conclusions governed from observations off hubble) why the universe is expanding.

 

[Naty1]

How about the charge of black holes?. The huge gravitational effects are always mentioned in descriptions but not charge, so I suspect it's modest but don't recall everseeing any quantative perspective.