# Sphere of constant density has zero force of attraction

1. Mar 28, 2009

### Helsinki

Hi all-

I have the following problem that I am trying to understand:

1. Problem statement. Show that the force of attraction within a spherical shell of constant density is everywhere $$0$$.

2. Relevant equations
My understanding of the statement is that, for example, in a gravitational field, the sphere would not 'cave in' on itself because the patches of the sphere are of constant density. I have the solution (below). The problem is presented in the context of advanced calculus (after talking about the implicit function theorem, surfaces and surface area). The integration is easy but I don't understand how the integral for the force is derived.

Solution. Describe the shell by $$x = \sin{\phi}\cos{\theta},y = \sin{\theta}\sin{\phi},z = \cos{\phi},0\leq \phi\leq \pi, 0\leq\theta \leq 2 \pi$$, and let $$P = (0,0,a)$$ with $$0\leq a \leq 1$$. With $$\rho$$=density (mass per unit area), the component of the force at $P$ in the vertical direction is
$$$F = - \int_0^{2\pi}d\theta\int_0^{\pi}\frac {(\cos\theta - a)(\rho\sin\theta)}{(1 + a^2 - 2a\cos{\theta})^{3/2}} d\phi.$$$
(This may be integrated easily; for example, put $$u^2 = 1 + a^2 - 2a\cos\theta.$$ One finds that $$F = 0.$$)

Also, I'm interested as to why this is an advanced calculus problem. My guess is that it mathematically interesting insofar as it can be generalized to $$n -$$ spheres and one must be careful with calculus. I would appreciate any help on this, since I have to give a presentation, and I have no idea what is going on!

Helsinki

Last edited: Mar 28, 2009
2. Mar 28, 2009

### lanedance

hi Helsinki

it looks like your integral assumes a sphere of density ro, radius 1 and centred at (x,y,z) =(0,0,0).

then, without loss of generality, consider a point at (x,y,z) =(0,0,a). due to the symmetry of the sphere, this effectvely represents any point in the sphere, at distance |a| from the centre. If the offest was not aligned along the z axis, you could rotate the probalem so it is.

In spherical coordinates, this is the point
$$\textbf{a} = (r, \theta, \phi) = (a,0,0)$$

First due to the symmetry of the problem, it is clear the horizontal field is zero, so you only need to check the vertical field as the point a = (a,0,0).

can you define the field contribution at a for an infintesimal surface area element at point on the spherical shell $$\textbf{r}' = (1, \theta, \phi)$$ ?

this should lead you to your integral

as for generalisation to higher dimensions, to be honest, i don't know whether it does or not. However for some insight you could try setting up the same integral for a ring of even mass distribution. Does it still vanish?

Last edited: Mar 28, 2009
3. Mar 28, 2009

### Helsinki

Hi lanedance,

Is there a general formula that is used to get $$F$$?

Last edited: Mar 28, 2009
4. Mar 28, 2009

### lanedance

start with the gravitational field g, at r from a mass dm

$$\textbf{dg} = -\frac{Gdm}{|r|^2}\hat{\textbf{r}}$$

5. Mar 28, 2009

### Helsinki

I have realized that the formula given for $$F$$ above in the original problem statement doesn't even make sense--$$F$$ ought to be a vector field, and F as given above is certainly not! This is a formula that's given in the back of the book, and I think it's purposefully wrong since it seems way off base.

I think the correct approach is to write the gravitational potential $$V$$ and then find the gravitational force $$F= \nabla V$$. It can be shown (below) that $$V$$ is constant and so $$F=0$$, the zero vector. As a basic setup, we have

$$-V(0,0,a)=-\int\int\int_W \frac{\delta(x,y,z) dx dy dz}{\sqrt{x^2+y^2+(z-a)^2}}$$

where $$\delta$$ is =1. Then with spherical coordinates this is
$$-V(0,0,a)=-\int\int\int_W \frac{(\rho^2\cdot \sin{\phi}) d\rho d\theta d\phi}{\sqrt{1-2a \cos{\phi}+a^2}}$$.

At the end we should have something like
$$-V=(Gm)2\pi \rho_{(0,0,a)}$$ and so $$F=\nabla{V}=0$$.

I would appreciate if someone could validate what I have said-naturally I must fill in some work in the integration above. I believe the text is completely wrong--on the other hand I am unwilling to reject its answer since its a textbook, and textbooks are always right (or at least, never this wrong).

Helsinki

Last edited: Mar 28, 2009
6. Mar 31, 2009

### sundried

The given formula is just the component of the force in the z direction, which is all you need.

This problem is perhaps easier in cylindrical coords.