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

$$T=\frac{2Am}{L}$$

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

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

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

$$T=\frac{2Am}{L}$$

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

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

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 $1/2 rdr sin\beta$ where $\beta$ is the angle that $d\vec r$ makes to $\vec r$. In other words $dS = 1/2 \vec r \times \vec{dr}$

Since:$L = m\frac{d\vec{r}}{dt}\times \vec{r}$:

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

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 $\theta$ is

$$S = \frac{1}{2}\theta r^2$$

then

$$\frac{dS}{dt}=\frac{1}{2}r^2 \frac{d\theta}{dt} + r\frac{dr}{dt}\theta$$

Why is this wrong?

 

[tim_lou]

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

$$\vec{L}=\frac{1}{2}mr^2\frac{d\theta}{dt}$$
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 $$v=\omega r$$, however, when r is not constant, omeaga should be:
$$\omega=\frac{\vec{r}\times{\vec{v}}}{r^2}$$