# Mercury's precession

by zincshow
Tags: mercury, precession
 P: 94 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 : http://www.physics.princeton.edu/~mc..._47_531_79.pdf 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 : http://en.wikipedia.org/wiki/Shell_theorem Thanks
P: 1,568
 Quote by zincshow 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 : http://www.physics.princeton.edu/~mc..._47_531_79.pdf 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 : http://en.wikipedia.org/wiki/Shell_theorem Thanks
The authors of the paper abstract the other planets as circular rings , not as spherical shells
P: 94
 Quote by starthaus The authors of the paper abstract the other planets as circular rings , not as spherical shells
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...

P: 1,568
Mercury's precession

 Quote by zincshow 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...
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.
P: 100
 Quote by zincshow 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...
Here is a link to how a ring's gravitational field can be derived.
http://farside.ph.utexas.edu/teaching/336k/Newton/node127.html

Here is an application to the planets.
http://farside.ph.utexas.edu/teachin...n/node128.html
Mentor
P: 15,201
 Quote by starthaus 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.
No error here. That a ring does not act like a shell is correct.
 Mentor P: 15,201 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.
Mentor
P: 6,248
 Quote by starthaus Yes, this appears to be the correct solution to the problem, by using the non-null potential rather than the force. The paper is still wrong, the link you cite is correct.
I have just run through the (analytical) calculations. To the level of approximation in

http://www.physics.princeton.edu/~mc..._47_531_79.pdf

the results in this reference and in

http://farside.ph.utexas.edu/teachin...n/node127.html

are the same.

Again, you have jumped too quickly to a conclusion.
P: 1,568
 Quote by D H 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.
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.
 P: 3,967 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/cla.../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 Thumbnails
 Mentor P: 15,201 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.
P: 1,568
 Quote by D H In short, to re-echo the claims of MIT students at the 1971 World Science Fiction Convention, "The Ringworld is unstable!"
Yes, it is :-)

 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.
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.
 Mentor P: 15,201 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.
P: 1,568
 Quote by D H The projection of Mercury's location onto each of those orbital planes is not at the center of those orbits.
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".
 Mentor P: 15,201 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.
 Mentor P: 15,201 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.
P: 1,568
 Quote by D H 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.
Of course I believe you.

 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.
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.
 Mentor P: 15,201 No, it does not. Show your work.

 Related Discussions Classical Physics 2 Astronomy & Astrophysics 1 Special & General Relativity 1 Quantum Physics 0 General Physics 0