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

Period of simple orbit in central potential

  1. Jul 23, 2006 #1


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In Symon's book 'Mechanics', he writes that for a body of mass m and angular momentum L in an orbit that does not intersect itself (i.e. a simple curve), the period of revolution T is related to the area A of the orbit by


    Is this exact as he seems to be implying? It seems to me that the correct formula would be obtained by considering the area S of a section of circle, and saying that the area of the orbit is related to the period of revolution by

    [tex]A=\int_0^T \frac{dS}{dt}dt = \int_0^T \frac{1}{2}r^2 \frac{d\theta}{dt}dt + \int_0^T \theta r\frac{dr}{dt}dt = \frac{L}{2m}\int_0^T dt + \int_0^T \theta r\frac{dr}{dt}dt = \frac{LT}{2m} + \int_0^T \theta r\frac{dr}{dt}dt[/tex]

    Am I missing something?
    Last edited: Jul 23, 2006
  2. jcsd
  3. Jul 24, 2006 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    The element of area swept out in time dt, where the radius is changing is not a right triangle. So the area dS is not 1/2 rdr. The area is [itex]1/2 rdr sin\beta[/itex] where [itex]\beta[/itex] is the angle that [itex]d\vec r[/itex] makes to [itex]\vec r[/itex]. In other words [itex]dS = 1/2 \vec r \times \vec{dr}[/itex]

    Since:[itex]L = m\frac{d\vec{r}}{dt}\times \vec{r}[/itex]:

    [tex]Area = \int_0^T \frac{dS}{dt}dt = \int_0^T \frac{1}{2}\vec{r}\times \frac{d\vec{r}}{dt}dt = \int_0^T \frac{L}{2m}dt = \frac{LT}{2m}[/tex]

    Last edited: Jul 24, 2006
  4. Jul 24, 2006 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I see what you're doing. You're taking S as the area of a triangle, while I was taking S as the area of a section. I figure since the area of a section subtended by and angle [itex]\theta[/itex] is

    [tex]S = \frac{1}{2}\theta r^2[/tex]


    [tex]\frac{dS}{dt}=\frac{1}{2}r^2 \frac{d\theta}{dt} + r\frac{dr}{dt}\theta[/tex]

    Why is this wrong?
  5. Jul 25, 2006 #4
    i think the problem probably is, in your derivation you assumed that:

    which isn't valid when r is not constant. the moment of inertia should become a tensor quantity. Since in the above derivation, it is assumed that [tex]v=\omega r[/tex], however, when r is not constant, omeaga should be:
    Last edited: Jul 25, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook