central force problem

**James0502** · January 2nd 2009, 11:54 AM

A particle of unit mass moves in a plane under a central force lamda^2 / r^5 where lamda is a constant.

Explain why (r dot)^2 + r^2.(theta dot)^2 - lamda^2 / (2r^4) = constant
done this part

The particle is projected from a point P at which r = a with speed lamda / (root2 a^2). Find a differential equation for r as a function of the polar angle theta along the orbit, and show that the orbit is a circle through O.

I'm not entirely sure how to start this part

my initial thought was to put the IC's into the above equation. But this simply gives me a constant of zero and then I cant see how to progress from there..

many thanks

**mr fantastic** · January 3rd 2009, 11:12 PM

Originally Posted by James0502

A particle of unit mass moves in a plane under a central force lamda^2 / r^5 where lamda is a constant.

Explain why (r dot)^2 + r^2.(theta dot)^2 - lamda^2 / (2r^4) = constant
done this part

The particle is projected from a point P at which r = a with speed lamda / (root2 a^2). Find a differential equation for r as a function of the polar angle theta along the orbit, and show that the orbit is a circle through O.

I'm not entirely sure how to start this part

my initial thought was to put the IC's into the above equation. But this simply gives me a constant of zero and then I cant see how to progress from there..

many thanks

Yes the constant is zero. So you have:

$(\dot{r})^2 + r^2 (\dot{\theta})^2 - \frac{\lambda^2}{2 r^4} = 0$ .... (1)

Now you should also know that

$r^2 \dot{\theta} = C$ .... (2)

where $C$ is a constant (conservation of angular momentum).

I think you might need to know the angle that the initial velocity makes with $\theta = 0$ before you can calculate the value of $C$ .

Now note:

$\dot{r} = \frac{d r}{d \theta} \cdot \frac{d \theta}{d t}$

Substitute from equation (2):

$= \frac{d r}{d \theta} \cdot \frac{C}{r^2} = -C \frac{d}{d \theta} \left( \frac{1}{r} \right)$

Substitute $u = \frac{1}{r}$ :

$= -C \frac{du}{d \theta}$ .... (3)

Substitute $u = \frac{1}{r}$ , equation (2) and equation (3) into equation (1) and simplify to get the required DE (well, it's a DE for $u$ as a function of $\theta$ - good enough).

Now show that the solution to this DE is a circle (note: the solution is in polar form).

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