a little challenge....

Question

irishboy123 · Answer

A particle of unit mass moves under a central attractive force $\frac{\gamma}{r^{\alpha}}$, where r and $\theta$ are polar coordinates of particle in its plane of motion.  

Show that: 

$\frac{d^2 u}{d \theta^2} + u = \frac{\gamma}{h^2} u^{\alpha - 2}$


where $u = \frac{1}{r}$ 

and h is the angular momentum of the particle about the origin

and $\alpha, \gamma$ are constants


my solution looks OK but is very inelegant.....

astrophysics · Answer

Hehe, I actually remember some of this kind of stuff, this essentially leads you to find a force law, essentially you're using the equation $$\frac{ d^2 }{ d \theta^2 } \left( \frac{ 1 }{ r } \right)+\frac{ 1 }{ r }=\frac{ \mu r^2 }{ l^2 }F(r)$$ If I'm right haha, I'll have to come back and check this later as I don't have much time right now.

michele_laino · Answer

here is my answer:

irishboy123 · Answer

that looks really cool.  never occurred to me to do it that way.

i'll have a look and maybe stuff my own on here.....

once again, game set & match to the maestro!

irishboy123 · Answer

i did this using the polar EoM's straight of the bat; but the lagrangian is *the* coolest way i've seen yet of getting to those equations.....

michele_laino · Answer

yes! It is the standard method, which starts from the equation of Newton.
good job! :) @IrishBoy123 
oops... in my paper I have made a typo, here is the revised version of it: