- #1
LAHLH
- 409
- 2
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
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