Period of simple orbit in central potential

[quasar987]

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

[tex]T=\frac{2Am}{L}[/tex]

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?

 

[Andrew Mason]

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

[tex]T=\frac{2Am}{L}[/tex]

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?

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]

AM

 

[quasar987]

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]

then

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

Why is this wrong?

 

[tim_lou]

i think the problem probably is, in your derivation you assumed that:

[tex]\vec{L}=\frac{1}{2}mr^2\frac{d\theta}{dt}[/tex]
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:
[tex]\omega=\frac{\vec{r}\times{\vec{v}}}{r^2}[/tex]