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This is going to be kind of long so let me know if I don't explain something well. I need this process doublechecked by mathematicians better than myself before I proceed any further. I'll be offering a bounty once MSE allows if I get a nice answer

In earlier physics classes, we learn inverse Square Laws Like Newton's and Coulomb's Laws that calculate a force between two (typically spherical) objects. In this process, spheres are approximated as points. However, this approximation yields a small error term. What if we could calculate the actual force and bypass this use of point approximations for a more accurate answer? That's our objective

Objective: Find the actual force between a point particle and a sphere of radius $\rho$ with center at distance $d$ from our point particle (which we'll place at the origin).

Assumptions: 1. Uniform mass/charge distribution throughout the sphere (constant density function) 2. The term sphere here is understood to mean a solid ball, I'll use them interchangeably

Procedure

Recall that the average unweighted value of $f(x)$ is given by the following:

$$\overline{f(x)}=\frac{1}{a-b}\int\limits_b^a f(x)\;dx$$

Suppose we don't want each value of $x\in[a,b]$ weighted identically. Here we introduce the idea of a density function (which we'll denote $\delta$).

Claim: thus, the average weighted value of $f(x)$, which we'll denote $\overline{f_w(x)}$, is the following:

$$\overline{f_w(x)}=\frac{\frac{1}{a-b}\int\limits_b^a f(x)\cdot \delta(x)\;dx}{\frac{1}{a-b}\int\limits_b^a \delta(x)\;dx}=\frac{\int\limits_b^a f(x)\cdot \delta(x)\;dx}{\int\limits_b^a \delta(x)\;dx}$$

Applying this method to our problem:

We proceed by dividing our sphere into an infinite collection of spherical caps with radius $r\in[d-\rho,d+\rho]$.

Let $\phi\in[0,\alpha]$ be the zenith angle between the positive x-axis and any point within our spherical cap. Now we consider our spherical cap as an infinite collection of circles. Recall that the y and z components of two diametrically opposed points on a circle cancel, meaning only their x component is counted toward the force. Thus, we aim to fins the average value of $\cos\theta$ along some spherical cap of fixed radius $r$.

$$\implies f(\phi)=\cos(\phi)$$

At each value $\phi\in[0,\alpha]$, we multiply $f$ by the circumference of the circle formed by the intersection of the given zenith angle and our spherical cap. Naturally this is $2\pi$ times the minimum distance of a point from the x-axis.

$$\implies \delta_c(\phi)=2\pi r\sin(\phi)$$

Thus, the following gives the average value of $\cos\phi$ along a spherical cap of radius $r$ and with $\phi\in[0,\alpha]$

$$\overline{f_w(\alpha)}=\frac{\int\limits_0^{\alpha}2\pi\rho\sin(\phi)\cdot\cos(\phi)\;d\phi}{\int\limits_0^{\alpha}2\pi\rho\sin(\phi)\;d\phi}$$

After integrating:

$$\overline{f_w(\alpha)}=\frac{1+\cos(\alpha)}{2}=\cos^2\left(\frac{\alpha}{2}\right)$$

Thus, the average value of $\cos(\phi)$ within a spherical cap has been found. Now for the second part, we will consider the following functions:

$$g(\alpha)=\frac{1+\cos(\alpha)}{2}\implies g(x)=\frac{(x+d)^2-\rho^2}{4dx}$$

Claim: Given a spherical cap where radius is $R$ and altitude is $h$, the surface area $A$ of the cap is given by the following:

$$A=2\pi R h$$

This can be seen by using rotational solids. We are going to consider the following density function that multiplies $g$ by the area of the spherical cap with radius $x\in[d-\rho,d+\rho]$:

$$\delta_s(x)=2\pi x\Delta x$$

We then see the following:

$$\Delta x=x-\frac{x^2+d^2-\rho^2}{2d}$$

The following integral gives the average value of $\cos(\phi)$ for the ball:

$$\frac{\frac{2\pi}{2\rho}\int\limits_{d-\rho}^{d+\rho}g(x)\cdot\delta_s(x)\;dx}{\frac{2\pi}{2\rho}\int\limits_{d-\rho}^{d+\rho}\delta_s(x)\;dx}$$

Substituting our values and integrating, we arrive at the following result which I've confirmed with an integral calculator:

$$\overline{g_w(d,\rho)}=1-\frac{\rho^2}{5d^2}$$

Therefore, the average value of $\cos(\phi)$ for every point within a ball of radius $\rho$ with center some distance $d$ from the origin is $1-\rho^2/5d^2$

Lastly, we use a similar process to find the average value of $1/r^2$ within our sphere. Consider the following function $h$

$$h(x)=\frac{1}{x^2}$$

We'll use the same $\delta_s$

$$\implies \overline h(x)=\frac{\frac{2\pi}{2\rho}\int\limits_{d-\rho}^{d+\rho}x\Delta x\cdot h(x)\;dx}{\frac{2}{3}\pi\rho^2}$$

Note that the integral we would see in the denominator corresponds with the average surface area of a spherical cap. We can simply divide the volume of the sphere by its diameter to obtain this value.

After integrating and verifying my results with an integral calculator, I obtain the following for $\overline{h_w}$:

$$h_w(d,\rho)=\frac{3}{2\rho^2}+\frac{3(\rho^2-d^2)}{4d\rho^3}\ln\left(\frac{d+\rho}{d-\rho}\right)$$

Thus, the respective gravitational and electric forces between our point particle and sphere are the following:

$$F_g=GMm\cdot\overline{g_w}\cdot\overline{h_w}=GMm\cdot\left[\frac{30d^3\rho -12d\rho^3+(15d^2-3\rho^2)(\rho^2-d^2)\ln\left(\frac{d+\rho}{d-\rho}\right)}{20d^3\rho^3}\right]$$ $$F_e=kQq\cdot\overline{g_w}\cdot\overline{h_w}=kQq\cdot\left[\frac{30d^3\rho -12d\rho^3+(15d^2-3\rho^2)(\rho^2-d^2)\ln\left(\frac{d+\rho}{d-\rho}\right)}{20d^3\rho^3}\right]$$

I need this procedure verified as a valid one so that I may continue with the second half of the project, which involves making the particle another sphere.

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  • $\begingroup$ I was under the impression that the result was exact, not an approximation. See e.g. Shell Theorem. $\endgroup$ – user575517 Aug 21 '18 at 17:20
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For the problem as posed,

Find the actual force between a point particle and a sphere of radius ρ with center at distance d from our point particle (which we'll place at the origin). Assumptions: 1. Uniform mass/charge distribution throughout the sphere (constant density function)

and provided $d\geq \rho$, the force is exactly the same as that between two point particles with the total mass of the distribution concentrated at the centre of the sphere. The proof was known to Newton, and may be derived by integration as shown here, or more concisely using Gauss's Law.

In the event that $d<\rho$, the force is equal to that which would be generated by that part of the spherical mass distribution enclosed by a concentric sphere of radius $d$, concentrated at the centre of the sphere.

The radial distribution of the mass is not critical, it is just necessary that it be spherically symmetric to get this particular result.

So it isn't an approximation, and any calculation of an error term should yield zero.

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