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Gravitation proof

  1. May 17, 2006 #1
    Please help me prove - A uniform spherical shell mass is cut into two pieces parallel to one of Symmetry axis. Prove that the gravitational field intensity due to the mass at the centre of the cross sectional plane parrallel to the symmetry axis is same for both the parts.
  2. jcsd
  3. May 18, 2006 #2
    Thats fairly simple. What have you done?
  4. May 22, 2006 #3
    My idea is from the fact that field is 0 internally at that point. So there must be two opposite vectors due to these two parts which must be equal in magnitude. But proving them separately using integration is becoming a bit difficult.
  5. May 23, 2006 #4
    Well I'm not sure if I understand you correctly. You said uniform spherical shell so that means a sphere with zero mass inside it. By symmetry and the shell theorem, the field at any point inside is zero. :smile: If you cut it along a diametrical plane, then the fields due to the two hemispherical shells thus formed must be zero at their (former) common center (by symmetry they must be along the axis passing through the center and perpendicular to the plane face).

    If you are talking of a solid sphere with uniform mass distribution (or even a mass distribution that is radially symmetrical--mass function of radius only and not any angular coordinate) then the field is zero only at its center. At points in the interior it varies directly with radial distance and outside it falls off as inverse square of radial distance. Now, if you cut a sphere along a diametrical plane then again by symmetry of mass distribution the field due to BOTH is zero at their common center. Is this what you're saying?

    For a hemisphere, the easiest way is to consider it to be made up of differential discs of varying radius and distance from the center located at the plane face of the hemisphere. (And for a hemispherical ring its even easier with the differential mass element being a ring). You need to know the gravitational field of a disc (or ring) though. If you know spherical coordinates, then this is easier: you just write the field due to some [itex]dm=\rho dV[/itex] where [itex]dV = r^{2}\sin\theta dr d\theta d\phi[/itex] (for more info google it up). Of course that brings in a lot of accounting to start with (3 integrals) but its neat and you're less likely to make mistakes with boundary terms.
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