# Understanding Mercury's Precession

by OnceKnown
Tags: mercury, perihelion, precession
 P: 20 I was recently given a lab to work at home but I am having trouble understanding the formulas that they have used. The link to this lab Assignment: http://classroom.sdmesa.edu/ssiegel/...Precession.pdf My problem relies to Query #1 of the lab assignment, with the equation $\frac{V(r)}{m}$ = - $\frac{M}{r}$ + $\frac{(\frac{L}{m})^{2}}{r^{2}}$ I derived it to be equal to: $\frac{d}{dr}$ $\frac{V(r)}{m}$ = $\frac{M}{r^{2}}$ - $\frac{2(\frac{L}{m})^{2}}{r^{3}}$ So I was told that M = mass of the Sun = 1.99 x 10$^{30}$ kg m = mass of Mercury = 3.28 x 10$^{23}$ kg L = angular momentum of Mercury = 9.11 x 10 $^{38}$ $\frac{kg m^{2}}{s^{2}}$ I equated the derived equation to 0 and solved for " r " to get: r$_{o}$ = $\frac{2(\frac{L}{m})^{2}}{M}$ which r$_{o}$ = 7.729, which should be the radius at the effective potential minimum But that doesn't make sense at all since the number is so low and the units doesn't seem right. Can someone help with this?