Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mercury's precession

  1. Apr 25, 2010 #1
    When calculation GR, the component due to other planets is removed. In that calculation outer planets are assumed to be rings of mass and the effect of that mass is seen to have an effect on mercury precession :


    How does this square with the idea of the shell theorem that says you do not feel a gravitational force from the inside of a hollow sphere of mass :


  2. jcsd
  3. Apr 25, 2010 #2
    The authors of the paper abstract the other planets as circular rings , not as spherical shells
  4. Apr 25, 2010 #3
    Thanks. For some reason it seemed to me that if there is no force inside a 3d sphere of mass that there would be no force inside a 2d ring of mass. I will have to think about that for a while...
  5. Apr 25, 2010 #4
    You are absolutely right, look at the calculations in paragraph II, in the opinion of the authors, the ring does not behave like a shell, the resultant force is non-null. I think that the paper is wrong, and I think I know where the authors introduced an error. AmJPhys is not a model of correctness, quite the opposite.
    Last edited: Apr 25, 2010
  6. Apr 26, 2010 #5
    Here is a link to how a ring's gravitational field can be derived.

    Here is an application to the planets.
  7. Apr 26, 2010 #6

    D H

    Staff: Mentor

    No error here. That a ring does not act like a shell is correct.
  8. Apr 26, 2010 #7

    D H

    Staff: Mentor

    The force is not null except at the center of the ring. Take the gradient of the potential and you will see that this is the case.

    Rings are used to prove the shell theorem, but the rings are carefully constructed so that the point in question lies on the axis of each ring.
  9. Apr 26, 2010 #8

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I have just run through the (analytical) calculations. To the level of approximation in


    the results in this reference and in


    are the same.

    Again, you have jumped too quickly to a conclusion.
  10. Apr 26, 2010 #9
    Yes, the analytic formula of the infinitesimal mass, "dm" used in calculating the resultant force for a spherical shell is different from the formula for the "dm" in a ring. This is why rings behave differently from shells. This is why a ring does not exhibit zero force, though it has the same radial symmetry as a shell.
    Last edited: Apr 26, 2010
  11. Apr 26, 2010 #10
    Here is a simple visual demonstration that the gravity inside a ring is not null as inside a hollow sphere (in Newtonian terms):


    In the above image the particle at point P inside the hollow sphere shell is at different distances from the shell but the shell mass contained in the circular intersection of the cone on the left is greater than the shell mass intersected on the right and this difference in intersected shell mass exactly compensates for the differences in distance. The regions outside the intersecting cones are identical and trivially cancel each other out, no matter what arbitary angle is chosen for the intersecting cones. See http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravField.htm for the source of the diagram.

    If a section of the shell above and below point P (the green sections in the attached diagram) are removed, leaving an equatorial ring, then some of the compensating shell mass is removed (the light blue regions in the attached diagram) and the forces no longer cancel out. Anything in the plane of the ring, but not exactly at the centre, moves outwards as if there is a repulsive force at the centre of the ring.

    Attached Files:

    Last edited: Apr 26, 2010
  12. Apr 26, 2010 #11

    D H

    Staff: Mentor

    In short, to re-echo the claims of MIT students at the 1971 World Science Fiction Convention, "The Ringworld is unstable!"

    It is also important to remember that Mercury's perihelion precession was known to be anomalously large well before Einstein. (The anomaly was first reported in 1859.) Those physicists were quite adept with calculation; they had to be because they did not have those electronic marvels that we rely on (possibly overly so) these days.
  13. Apr 26, 2010 #12
    Yes, it is :-)

    Sure, the advancement of the perihelion was never in debate, just the approach used by the paper cited in the OP for calculating the force inside the ring. I find their approach inferior to the approach used for the shell calculation. The wiki calculation for the shell is quite elegant, makes a much better use of radial symmetry.
  14. Apr 26, 2010 #13

    D H

    Staff: Mentor

    The proof of the shell theorem such as that in wikipedia constructs the rings so that the point in question is along each of the rings axes. That construction ensures radial symmetry.

    There is no radial symmetry in this problem to take advantage of. Mercury is not located along the axis of the outer planet's orbits. It is more or less located on the orbital planes of those orbits. The projection of Mercury's location onto each of those orbital planes is not at the center of those orbits.
  15. Apr 26, 2010 #14
    This is quite obvious. Nevertheless, it is easy to adapt the wiki proof for the shell to the ring. The only change required is a recalculation of the infinitesimal mass element "dm".
  16. Apr 26, 2010 #15

    D H

    Staff: Mentor

    Try it. You're going to get something rather ugly. In fact, the integral that results does not have a simple solution. You will get what are called elliptical integrals.
  17. Apr 26, 2010 #16

    D H

    Staff: Mentor

    This isn't homework, so if you want I can give you a link to a solution. (You probably wouldn't believe me if I the words came from my fingers. Fortunately, the solution is out there on the 'net. Besides, I don't particularly want to typeset the LaTeX with my fingers when the solution is out there on the 'net.)

    Or you can try it yourself. When/if you get stuck I'll be glad to help.
  18. Apr 26, 2010 #17
    Of course I believe you.

    Thank you, I appreciate the offer, what I was saying is that I have already worked it out. In the wiki solution you need to replace the spherical mass element with the ring one. Everything else in the solution stays same.
  19. Apr 26, 2010 #18

    D H

    Staff: Mentor

    No, it does not. Show your work.
  20. Apr 26, 2010 #19
    I don't like your tone. Bye.
  21. Apr 26, 2010 #20

    D H

    Staff: Mentor

    Here's the calculation for the potential.

    First some definitions. We want to calculate the gravitational potential due to a uniform ring of mass at some point. Define z as the distance between the point and the ring plane and x as the distance between the point and the ring axis. Without loss of generality, define
    • [itex]\hat x[/itex] is the direction from the center of the ring to the projection of the point in question onto the ring plane,
    • [itex]\hat z[/itex] is parallel to the ring axis, with positive z in the direction of the point in question, and
    • [itex]\hat y[/itex] completing a right-hand system ([itex]\hat y \equiv \hat z \times \hat x[/itex]).

    We also need some definitions regarding the ring itself.
    • [itex]a[/itex] is the radius of the ring,
    • [itex]m[/itex] is the mass of the ring,
    • [itex]\rho[/itex] is the linear density of the ring ([itex]\rho \equiv m/(2\pi a)[/itex]).

    Potential obeys the superposition principle. Thus

    [tex]\phi(\mathbf x) = \int \frac{G\,dm}{r} =
    \int_0^{2\pi} \frac {G \rho a}{\sqrt{z^2+x^2+a^2-2ax\cos\theta}} d\theta[/tex]

    This is an elliptical integral.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Mercury's precession
  1. Mercury's precession (Replies: 2)

  2. Mercury's Precession (Replies: 7)

  3. Precession of Mercury (Replies: 1)

  4. Precession of Mercury (Replies: 9)