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Homework Help: 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


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    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


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    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.
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