Is Gravitational Attraction Negligible in Oscillating Systems?

[TSny]

From Kepler's third law
P^2 \propto a^3 \Rightarrow P^2=ka^3
where $a=(L_0/\sqrt{2}+L_0/(4\sqrt{2}))/2=5L_0/(8\sqrt{2})$ and k is some constant.

How should I determine k here?

I discussed a similar question before where I had to find this k. Please look at this thread: https://www.physicsforums.com/showthread.php?t=700759 (post #5) Can I use the same way here?

Yes. Compare Newton's law of gravity to the force expression you got for this problem to find the correspondence between GM in the gravity problem and the constants k, q, and m in this problem. See https://www.physicsforums.com/showthread.php?t=700759 (post #1)

 

[rude man]

Here we have a periodic orbit with each charge having a maximum distance from the central chatge (apoapsis) and minimum distance from the central charge (periapsis) so we have an elliptical orbit.

I don't see the described motion (of each corner +q charge) as an ellipse. Rather, a spiral with decreasing and increasing radii.

And if it's an ellipse following Kepler, the central negative charge must be located at a focus, not at the center. I don't see that departure from symmetry here.

I would not assume that this configuration is self-sustaining. The motion is described, not justified in any way.

EDIT: since the period of the L contractions and expasions is not given, we are entitled to assume any value we want. I choose an arbitrarily low period such that there is no change in L for any interval of observation. Which brings me back to my early post.

 

[mfb]

I don't see the described motion (of each corner +q charge) as an ellipse. Rather, a spiral with decreasing and increasing radii.

Each point-charge has an ellipse as path.

And if it's an ellipse following Kepler, the central negative charge must be located at a focus, not at the center. I don't see that departure from symmetry here.

The 4-fold symmetry is conserved, but the 4 point-charges have different orbits (rotated by 90°)

I would not assume that this configuration is self-sustaining. The motion is described, not justified in any way.

It is possible (but unstable).

EDIT: since the period of the L contractions and expasions is not given, we are entitled to assume any value we want. I choose an arbitrarily low period such that there is no change in L for any interval of observation. Which brings me back to my early post.

We cannot assume values, we can calculate those values.

 

[rude man]

We cannot assume values, we can calculate those values.

OK, but the system with constant L is stable. So you're saying that a slight perturbation from the stable, constant-L configuration will result in variation of L from L0 to L0/4 exactly, all by itself? And at a determinable period? Really hard for me to swallow.

 

[mfb]

OK, but the system with constant L is stable.

Independent of the minimal and maximal separation, the system is labile - it can exist, but any deviation from the symmetry will break it.

So you're saying that a slight perturbation from the stable, constant-L configuration will result in variation of L from L0 to L0/4 exactly, all by itself?

No.

And at a determinable period?

Sure. As long as we have the symmetry, each particle experiences a central 1/r^2-force. We can just solve the Kepler problem for one particle, the other 3 particles have the same orbital dynamics.

 

[Saitama]

The area is correct. Show details of getting ω_f. The formula for vpf in #11 was still correct.

ehild

I had $$v_{pf}^2=\frac{kq^2(3\sqrt{2}-4)}{10mL_0} \Rightarrow \omega_f^2L_0^2/2=\frac{kq^2(3\sqrt{2}-4)}{10mL_0}$$
It is given in the question that $kq^2/(mL_0^3)=10^4 s^{-2}$, plugging this, I reached the value of $\omega_f$.

 

[ehild]

I had $$v_{pf}^2=\frac{kq^2(3\sqrt{2}-4)}{10mL_0} \Rightarrow \omega_f^2L_0^2/2=\frac{kq^2(3\sqrt{2}-4)}{10mL_0}$$
It is given in the question that $kq^2/(mL_0^3)=10^4 s^{-2}$, plugging this, I reached the value of $\omega_f$.

That is correct. What is then rf²ω_f?

ehild

 

[Saitama]

That is correct. What is then rf²ω_f?

ehild

I guess I made a mistake in my previous post. $r_f=L_0/\sqrt{2}$, correct?

This time I get,
\frac{L_0^2}{2}\sqrt{2\times 10^3 (3\sqrt{2}-4)}=\frac{2\pi}{P}\frac{5L_0^2}{32}
Solving for P, $P=0.0891 sec$. Can you please check my post #28 and see if I used the right values for $a$ and $b$?

 

[WannabeNewton]

What did you get using Kepler's 3rd law?

 

[Saitama]

What did you get using Kepler's 3rd law?

I couldn't apply that to this problem so I continued with the second law.

 

[ehild]

I guess I made a mistake in my previous post. $r_f=L_0/\sqrt{2}$, correct?

This time I get,
\frac{L_0^2}{2}\sqrt{2\times 10^3 (3\sqrt{2}-4)}=\frac{2\pi}{P}\frac{5L_0^2}{32}
Solving for P, $P=0.0891 sec$. Can you please check my post #28 and see if I used the right values for $a$ and $b$?

That looks correct.

ehild

 

[rude man]

Independent of the minimal and maximal separation, the system is labile - it can exist, but any deviation from the symmetry will break it.

No.

Sure. As long as we have the symmetry, each particle experiences a central 1/r^2-force. We can just solve the Kepler problem for one particle, the other 3 particles have the same orbital dynamics.

OK. I think I finally see what you're getting at. Thanks.

 

[TSny]

Here's a video of the motion:

[I will delete this post if animations are not allowed to be uploaded or if it doesn't work for some reason. File size is 191.3 KB]

 

[WannabeNewton]

Oh my god TSny I've been spending the last hour trying to make an animation of the system motion; you are absolutely brilliant I love you :)! You're going to have to tell me what program you used to make that.

 

[TSny]

You're going to have to tell me what program you used to make that.

I used Mathematica. I searched the net today to find out how to export a Mathematica animation. I'm not an experienced programmer. Glad that it seems to be working.

 

[WannabeNewton]

Yeah it works great! Thanks. I'm horrible with Mathematica :p

 

[rude man]

Here's a video of the motion:

[I will delete this post if animations are not allowed to be uploaded or if it doesn't work for some reason. File size is 191.3 KB]

My thanks also!

 

[ehild]

Great animation! One understands the problem at once when looking at it. You are a genius, TSny!

ehild

 

[consciousness]

Here's a video of the motion:

[I will delete this post if animations are not allowed to be uploaded or if it doesn't work for some reason. File size is 191.3 KB]

Great animation! I have one doubt regarding this, is it necessary that the time period of the revolutionary motion(the time taken to complete one revolution about the center) when viewed separately from the to and fro motion of the particle will be same as the time period of the to and fro motion?

In other words is it necessary that when the particle crosses say the x-axis it will always be at the same distance from the center? Or is that just a coincidence?

 

[TSny]

In other words is it necessary that when the particle crosses say the x-axis it will always be at the same distance from the center? Or is that just a coincidence?

Since the net force on an orbiting particle is an attractive inverse-square force, the motion of the particle will be an ellipse that does not precess. So, it will cross the x and y-axis at the same points each revolution.

 

[consciousness]

Since the net force on an orbiting particle is an attractive inverse-square force, the motion of the particle will be an ellipse that does not precess. So, it will cross the x and y-axis at the same points each revolution.

Thanks!. This fact prompts me to search for an alternate solution to this problem.

If the four charges were revolving in a circular orbit initially and an impulse towards the central charge is provided to each simultaneously, would it be possible to get this exact situation?

If so would the time period of the motion change?(I tried to solve this problem by assuming that the time period doesn't change but got a wrong answer for my efforts)

 

[Saitama]

Thank you everyone for the help!

And nice animation TSny! :)

 

[TSny]

If the four charges were revolving in a circular orbit initially and an impulse towards the central charge is provided to each simultaneously, would it be possible to get this exact situation?

Yes. Since radial impulses will not change the angular momentum of the system, you would need to start in a circular orbit with the same angular momentum as for the final elliptical orbits.

If so would the time period of the motion change?(I tried to solve this problem by assuming that the time period doesn't change but got a wrong answer for my efforts)

Yes, the period will change. The radial impulses will increase the energy of the particles which will increase the period.

 

[Saitama]

I should have asked this earlier. Why a and b are the arithmetic and geometric means of the initial and final radii respectively?

 

[consciousness]

Yes, the period will change. The radial impulses will increase the energy of the particles which will increase the period.

I can confirm that. The time period for the initial circular orbit comes out be 0.0129 seconds, much less than that for the elliptical orbit. Why does increasing energy increase the time period?

 

[WannabeNewton]

I should have asked this earlier. Why a and b are the arithmetic and geometric means of the initial and final radii respectively?

http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/EllipticOrbits.pdf

 

[WannabeNewton]

The radial impulses will increase the energy of the particles which will increase the period.

In what sense is the phrase energy being used here? Doesn't an elliptical orbit have less total energy than a circular orbit (zero for circular and negative for elliptical)?

 

[voko]

In what sense is the phrase energy being used here? Doesn't an elliptical orbit have less total energy than a circular orbit (zero for circular and negative for elliptical)?

Zero energy is in a parabolic orbit. For a circular orbit, zero pertains to eccentricity, not to energy. For any given angular momentum, energy is minimal for a circular orbit, which is easily seen from the equation relating eccentricity, energy, and angular momentum.

 

[WannabeNewton]

Zero energy is in a parabolic orbit. For a circular orbit, zero pertains to eccentricity, not to energy. For any given angular momentum, energy is minimal for a circular orbit, which is easily seen from the equation relating eccentricity, energy, and angular momentum.

Oops yeah I was mixing up eccentricity and energy. Cheers.

 

[voko]

I can confirm that. The time period for the initial circular orbit comes out be 0.0129 seconds, much less than that for the elliptical orbit. Why does increasing energy increase the time period?

Total energy is directly related to the semi-major axis. This follows from considering conservation of energy and angular momentum at apoapsis and periapsis, try figuring that out.

The semi-major axis and period are related via Kepler's third law.