Michael King
Sep29-10, 04:12 PM
Hello all! It's been a long summer, and I thought I'd warm up things by going over Kepler's Laws.
I've been following the mathematical derivation in Introduction to Modern Astrophysics, and to be honest I am little stumped on a part of it:
We have the derived definition of angular momentum as
\vec{L} = \mu r^{2}\hat{r}\times\frac{d}{dt}\hat{r}
Then what happens is out of the blue, it seems, the author takes the cross product of the acceleration vector and angular momentum:
\vec{a}\times\vec{L} = -\frac{GM}{r^{2}}\hat{r}\times \left(\mu r^{2}\hat{r}\times\frac{d}{dt}\hat{r} \right)
Ugh, to be honest I am stumped at the physical significance of that cross product. I can understand that the result is a \vec{v}\times\vec{L} expression, but it just seems to have come out of nowhere and I don't know (physically) why we go through that process
For reference it is on page 44 of the red paperback (second) edition. It is under Chapter 2: Celestial Mechanics.
I've been following the mathematical derivation in Introduction to Modern Astrophysics, and to be honest I am little stumped on a part of it:
We have the derived definition of angular momentum as
\vec{L} = \mu r^{2}\hat{r}\times\frac{d}{dt}\hat{r}
Then what happens is out of the blue, it seems, the author takes the cross product of the acceleration vector and angular momentum:
\vec{a}\times\vec{L} = -\frac{GM}{r^{2}}\hat{r}\times \left(\mu r^{2}\hat{r}\times\frac{d}{dt}\hat{r} \right)
Ugh, to be honest I am stumped at the physical significance of that cross product. I can understand that the result is a \vec{v}\times\vec{L} expression, but it just seems to have come out of nowhere and I don't know (physically) why we go through that process
For reference it is on page 44 of the red paperback (second) edition. It is under Chapter 2: Celestial Mechanics.