Perihelion precession of Mercury

  • Thread starter LAHLH
  • Start date
  • #1
409
1
Hi,

It's found it Carroll that [tex] r=\frac{L^2}{GM(1+e\cos{((1-\alpha)\phi)}} [/tex]. He states that this means we've found that during each orbit the perihelion advances by [tex]\Delta\phi=2\pi\alpha[/tex], I'm not sure that I follow this. Perihelion is the min value of r, and thus should correspond to whenever [tex]\cos{((1-\alpha)\phi)} [/tex] is at a max, namely [tex](1-\alpha)\phi_n =2n\pi[/tex] therefore [tex] \Delta\phi= \phi_{n+1}-\phi_{n}=\frac{2(n+1)\pi}{1-\alpha}-\frac{2n\pi}{1-\alpha}=\frac{2\pi}{1-\alpha}=2\pi+2\pi\alpha+O(\alpha^2) =2\pi(1+\alpha)[/tex]. Yet this is not what he states.....

Also more generally, the figure shown of the precession (5.6) seems to show the orbital plane itself changing, I was expecting things to remain in a plane, but the ellipse to move around (like flat flower petals), just wondering how I should visualise this precession and if anyone has any good links/vids to see it
 

Answers and Replies

  • #2
97
3
It's found it Carroll that [tex] r=\frac{L^2}{GM(1+e\cos{((1-\alpha)\phi)}} [/tex]. He states that this means we've found that during each orbit the perihelion advances by [tex]\Delta\phi=2\pi\alpha[/tex], I'm not sure that I follow this. Perihelion is the min value of r, and thus should correspond to whenever [tex]\cos{((1-\alpha)\phi)} [/tex] is at a max, namely [tex](1-\alpha)\phi_n =2n\pi[/tex] therefore [tex] \Delta\phi= \phi_{n+1}-\phi_{n}=\frac{2(n+1)\pi}{1-\alpha}-\frac{2n\pi}{1-\alpha}=\frac{2\pi}{1-\alpha}=2\pi+2\pi\alpha+O(\alpha^2) =2\pi(1+\alpha)[/tex]. Yet this is not what he states.....
What you've computed there is the total angle the planet has traversed from one perihelion to the next. If there was no precession at all, this would be 2pi, so the precession is the total angle minus 2pi. If you subtract 2pi from your expression, you get (2pi alpha)/(1-alpha), which is essentially 2pi alpha for sufficiently small alpha.

Also more generally, the figure shown of the precession (5.6) seems to show the orbital plane itself changing, I was expecting things to remain in a plane, but the ellipse to move around (like flat flower petals), just wondering how I should visualise this precession and if anyone has any good links/vids to see it
The orbital plane doesn't change. I haven't checked Carroll's figure, but it shouldn't show any change in the orbital plane, if it's just talking about the precession of an orbit.
 
  • #3
409
1
Thanks alot,makes sense..

Yeah I didn't think the plane should change, maybe I'm just looking at the fig weirdly, looks like a slinky to me, whereas I guess it's intended to look like flat flower petals
 

Related Threads on Perihelion precession of Mercury

  • Last Post
Replies
20
Views
537
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
3
Views
4K
  • Last Post
Replies
2
Views
3K
  • Last Post
2
Replies
38
Views
6K
  • Last Post
Replies
7
Views
3K
Replies
4
Views
749
  • Last Post
2
Replies
37
Views
9K
Top