Is Gravitational Attraction Negligible in Oscillating Systems?

[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.