1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sphere of constant density has zero force of attraction

  1. Mar 28, 2009 #1
    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 [tex]0[/tex].

    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 [tex]x = \sin{\phi}\cos{\theta},y = \sin{\theta}\sin{\phi},z = \cos{\phi},0\leq \phi\leq \pi, 0\leq\theta \leq 2 \pi[/tex], and let [tex]P = (0,0,a)[/tex] with [tex]0\leq a \leq 1[/tex]. With [tex]\rho[/tex]=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 [tex]u^2 = 1 + a^2 - 2a\cos\theta.[/tex] One finds that [tex]F = 0.[/tex])

    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 [tex]n -[/tex] 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,

    Last edited: Mar 28, 2009
  2. jcsd
  3. Mar 28, 2009 #2


    User Avatar
    Homework Helper

    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
    [tex] \textbf{a} = (r, \theta, \phi) = (a,0,0) [/tex]

    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 [tex] \textbf{r}' = (1, \theta, \phi) [/tex] ?

    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
  4. Mar 28, 2009 #3
    Hi lanedance,

    Is there a general formula that is used to get [tex]F[/tex]?
    Last edited: Mar 28, 2009
  5. Mar 28, 2009 #4


    User Avatar
    Homework Helper

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

    [tex] \textbf{dg} = -\frac{Gdm}{|r|^2}\hat{\textbf{r}} [/tex]
  6. Mar 28, 2009 #5
    I have realized that the formula given for [tex] F [/tex] above in the original problem statement doesn't even make sense--[tex]F[/tex] 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 [tex]V[/tex] and then find the gravitational force [tex] F= \nabla V [/tex]. It can be shown (below) that [tex]V[/tex] is constant and so [tex]F=0[/tex], the zero vector. As a basic setup, we have

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

    where [tex]\delta[/tex] is =1. Then with spherical coordinates this is
    [tex] -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}} [/tex].

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

    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,

    Last edited: Mar 28, 2009
  7. Mar 31, 2009 #6
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Sphere of constant density has zero force of attraction