Sphere of constant density has zero force of attraction

[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.

Homework 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
\[<br /> F = - \int_0^{2\pi}d\theta\int_0^{\pi}\frac {(\cos\theta - a)(\rho\sin\theta)}{(1 + a^2 - 2a\cos{\theta})^{3/2}} d\phi.<br /> \]
(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!
Thanks in advance,Helsinki

 

[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}&#039; = (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?

 

[Helsinki]

Hi lanedance,

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

 

[lanedance]

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

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

 

[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).



Thanks in advance,


Helsinki

 

[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.