Is Gravitational Attraction Negligible in Oscillating Systems?

[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.

 

[TSny]

The time period for the initial circular orbit comes out be 0.0129 seconds, much less than that for the elliptical orbit.

I'm getting a period of .0456 s for the initial circular orbit. But I could have made a mistake.

 

[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]

A pretty picture, to be sure, but there is a lot of physics going on here:

a) Each mass is attracted by the other three gravitationally;
b) each mass is repelled by the other three electrostatically; and
c) each mass is attracted to the center charge electrostatically.
d) Then, an orbit of any kind requires a specific initial tangential velocity at a specific distance to determine that orbit.

Has your simulation provided for all these forces and initial velocities?

I note that if Gm² = kq², each of the four masses always has zero force on it due to the other three. Then each mass has an inverse-square force law as provided by its own charge q and that of the center charge -q. Unfortunately, I don't see Gm² = kq² as given.

So, again, have you accounted for all the forces and initial velocities?

 

[mfb]

Gravitational force is completely negligible. If you care about the 5th decimal place, (EM) radiative effects would be way more important for any reasonable setup.

b) each mass is repelled by the other three electrostatically; and
c) each mass is attracted to the center charge electrostatically.

That gives the radial force.

d) Then, an orbit of any kind requires a specific initial tangential velocity at a specific distance to determine that orbit.

As you can see from the result, he took the right velocity to get the 4:1 distance variation.

I note that if Gm² = kq², each of the four masses always has zero force on it due to the other three.

That is wrong, both electrostatic and gravitational forces point inwards, they cannot add to zero. Anyway, gravity does not matter here.

 

[TSny]

As mfb points out, the gravitational force is assumed to be neglected. Otherwise, you could not get a numerical answer based on the information given. [You could imagine an idealized system where the gravity and electrical forces are similar. For example, bowling balls with charges on the order of a nC. ]

The important thing is that when you add up all of the electric forces on one of the moving particles, the net electric force is toward the central charge and varies as the inverse square of the distance r from the central charge: F = λ/r². See post #1 where Pranav gives the explicit expression for λ.

 

[ehild]

=Pranav-Arora;4446563]I should have asked this earlier. Why a and b are the arithmetic and geometric means of the initial and final radii respectively?[/QUOTE ]

Hi Pranav, see picture. The semi-major axis is 2a. 2a =r1+r2. For all points of the ellipse, the sum of distances from the foci has to be the same. The closest point to the left focus is A, at distance r1. The farthest point is at distance r2. A is at distance r1 from the left focus and at distance r2 from the right one. The sum of distances from the foci is r1+r2=2a. The sum of distances of point C from both foci is 2c=2a. So c=a. Form the yellow right triangle, b²=c²-(a-r1)²=a²-(a-r1)²=r1(2a-r1)=r1r2.

ehild

 

[rude man]

Gravitational force is completely negligible. If you care about the 5th decimal place, (EM) radiative effects would be way more important for any reasonable setup.

How do you know? m could be huge ... and the given constraint does not preclude a large m. Just make L0 small ...

That is wrong, both electrostatic and gravitational forces point inwards, they cannot add to zero. Anyway, gravity does not matter here.

The gravitational force between any two masses = Gm²/r². This is an attractive force. The electrostatic force between the same two masses = kq²/r². This is a repellent force. They cancel each other if Gm² = kq².

 

[mfb]

How do you know? m could be huge ... and the given constraint does not preclude a large m. Just make L0 small ...

I know because it is a homework question, the central mass is not given and gravity is not mentioned.

The gravitational force between any two masses = Gm²/r². This is an attractive force. The electrostatic force between the same two masses = kq²/r². This is a repellent force. They cancel each other if Gm² = kq².

The central charge is negative, the outer charges are positive. If you add the forces, you get an attractive force. Please read the problem statement, it is explained there.

 

[rude man]

I know because it is a homework question, the central mass is not given and gravity is not mentioned.

Unlike you, I and everyone else were not privy to that bit of information. I assumed the central mass = 0 but not the other four. I see no justification for that assumption. Once you mention "mass", gravity is kind of implied, is it not?

The central charge is negative, the outer charges are positive. If you add the forces, you get an attractive force. Please read the problem statement, it is explained there.

I acknowledge that if gravitational attraction is ignored then the remaining (electrostatic) forces are central. But I don't see ignoring gravity 'explained there'.

EDIT: in fact, I acknowledge that the forces are central even with the inclusion of gravity. But T cannot be determined without assuming zero gravitational attraction among the masses.

 

[mfb]

Unlike you, I and everyone else were not privy to that bit of information.

Actually, everyone else understood that gravity is supposed to be neglected.

I assumed the central mass = 0 but not the other four. I see no justification for that assumption.

Me neither.

Once you mention "mass", gravity is kind of implied, is it not?

No.

But T cannot be determined without assuming zero gravitational attraction among the masses.

Assuming zero, or neglecting it, right. Note that the problem statement said "particles". If those are elementary particles, gravity is weaker by ~40 orders of magnitude.

 

[ehild]

kq²/mLo=10⁴ was given. That means that gravitational force between two neighbouring particles is Fg=54q^2/Lo^6 and the electric force is 9˙10⁹ q^2/Lo^2. The ratio Fg/Fe=6˙10-9 q^2/Lo^4 . Fg and Fe are comparable if q/Lo^2 is of the order 10⁴, that is, the electric field of a particle at the position of its neighbour is about 10¹⁴ N/C.


ehild

 

[mfb]

Your formulas are not right. Try to add units and you will see the issue.

The gravitational force has to scale with $\frac{Gm^2}{L_0^2}$ and not with q or some 6th power of the length.

The length has to drop out, as both forces are inverse square laws.

 

[rude man]

kq²/mLo=10⁴ was given. That means that gravitational force between two neighbouring particles is Fg=54q^2/Lo^6 and the electric force is 9˙10⁹ q^2/Lo^2. The ratio Fg/Fe=6˙10-9 q^2/Lo^4 . Fg and Fe are comparable if q/Lo^2 is of the order 10⁴, that is, the electric field of a particle at the position of its neighbour is about 10¹⁴ N/C.


ehild

Given was kq²/mL₀³ = 1e4 s^-2.
Gravitational and electrostatic forces are equal and opposite for the four masses (ignoring the center charge) iff Gm² = kq².

Looks like you are somebody other than myself who didn't "know" that gravity is to be ignored. Tsk tsk.

Sorry, Sir Isaac.

 

[ehild]

Your formulas are not right. Try to add units and you will see the issue.

You are right, I copied hastily. The correct formula is (hopefully) F_g≈54q⁴ /Lo⁸ (N)

The gravitational force has to scale with $\frac{Gm^2}{L_0^2}$ and not with q or some 6th power of the length.

The length has to drop out, as both forces are inverse square laws.

Thank you for teaching me the formula for the gravitational force. As the product kq²/mLo³ was given (10⁴ s^-2), m=kq²10^-4(s²)/Lo³. I substituted for m into the formula of the gravitational force. The two kinds of forces between neighbours are equal if q/Lo³≈1.3e4 C/m³. That can happen if 0.01 C charges are 1 cm apart, for example. (Hoping no mistakes now).

ehild

 

[mfb]

Without G, I don't see how the formula could get right.

$F_g=\frac{cGm^2}{L^2}$ as radial force towards the center, with some numerical prefactor c I am too lazy to evaluate. L=L₀
$F_c=\frac{c' k q^2}{L^2}$
I'll drop the numerical prefactors now.

Clearly, $\frac{F_g}{F_c}=\frac{Gm^2}{kq^2}$. Using $m=\frac{k q^2 \cdot 1s^2}{10^4 L^3}$, I get equality for
$$10^8 L^6=Gk q^2 1s^4$$
That leaves some non-trivial relation between the length and the charge.

Edit: I can confirm the pair (1cm, 0.01C). That is a really large charge, however. A smaller charge leads to a smaller influence of gravity (interesting relation...), the same is true for a larger separation.

 

[ehild]

Clearly, $\frac{F_g}{F_c}=\frac{Gm^2}{kq^2}$. Using $m=\frac{k q^2 \cdot 1s^2}{10^4 L^3}$, I get equality for
$$10^8 L^6=Gk q^2 1s^4$$
That leaves some non-trivial relation between the length and the charge.

G=6.67 e-11 Nkg^-2m²
k=9 e9 NC^-2 m².
Gk≈0.6 m⁶s^-4C^-2
10⁴Lo³≈0.775 q, q/Lo³≈1.3 10⁴ C/m³. That means m=11.7 e9 q. If we have 1 nC charges, the masses have to be almost 12 kg ...

ehild

 

[rude man]

Well, at least gravity has been allowed to insidiously enter the picture, PF's Walhalla's gods notwithstanding.

 

[TSny]

Another observation for fun:

We have $10^4 = \frac{kq^2}{mL_0^3} = \frac{kq^2}{Gm^2}\frac{Gm}{L_0^3}=(1) \frac{Gm}{L_0^3}$

Suppose the particles are spheres of density $\rho$ and radius $\small R$. Substituting $m = \frac{4}{3}\pi R^3 \rho$ and solving for $\rho$:

$\rho = \frac{3\cdot10^4}{4\pi G} (\frac{L_0}{R})^3$

We require $R< \frac{L_0}{8\sqrt{2} }$ in order for the spheres not to collide at closest approach distance of $\frac{L_0}{4\sqrt{2} }$.

Thus $\rho > \frac{3\cdot10^4(8 \sqrt{2})^3}{4\pi G}≈5\times 10^{16}$kg/m³ (This is approaching neutron star densities).

From $kq^2 = Gm^2$, $q = \sqrt{\frac{G}{k}} m$. If the spheres are uniformly charged with volume charge density $\rho_c$,

$\rho_c = \sqrt{\frac{G}{k}} \rho ≈ 4 \times 10^6$ C/m³ = 4 C/cm³. (Large!)

 

[rude man]

All this comedy could have been avoided if the problem had simply stated to ignore gravitational attraction.

kq^2 = Gm^2 merely equates the two forces among the four masses. It could be Gm^2 = 0.1kq^2 and still make an impact on T. Or even < 0.1.

 

[TSny]

kq^2 = Gm^2 merely equates the two forces among the four masses. It could be Gm^2 = 0.1kq^2 and still make an impact on T. Or even < 0.1.

Yes, we're just looking at the numbers roughly.

 

[mfb]

Thus $\rho > \frac{3\cdot10^4(8 \sqrt{2})^3}{4\pi G}≈5\times 10^{16}$kg/m³ (This is approaching neutron star densities).

I think that is a very good reason to ignore gravity.

All this comedy could have been avoided if the problem had simply stated to ignore gravitational attraction.

I still don't think it is necessary. For a mechanics problem on Earth (like a pendulum), do you need an explicit statement that you can neglect the gravitational influence of the moon?
For a double pendulum, do we have to add "neglect the gravitational attraction between the pendulum masses"?

As TSny showed, the gravitational force is more than 12 orders of magnitude below the electrostatic force for every material available on Earth (as density < 50000kg/m^3). Even without knowing this number, it should be obvious that gravity is negligible.