Find force lawa from orbit equation

[KBriggs]

Homework Statement

The orbit of a particle moving on a central field is a circle passing through the origin, namely, $$r = r_0cos(\theta)$$. Show that the force law is inverse fifth power.

Homework Equations

$$\frac{d^2u}{d\theta^2} + u = \frac{-mF(u^{-1})}{L^2u^2}$$
$$u=r^{-1}$$

The Attempt at a Solution

I keep getting that it is inverse third power...

$$u = \frac{1}{r_0cos(\theta)}$$
then
$$\frac{d^2u}{d\theta^2} = u + \frac{tan^2(\theta)}{u}$$

so

$$F(u^{-1}) = \frac{-1}{m}\left(L^2u^2\left(u + \frac{tan^2(\theta)}{u}\right) + L^2u^2\right)$$

$$=\frac{-2L^2}{m}\left(u^3+utan(\theta)\right)$$

so $$f(r) = \frac{-2L^2}{m}\left(\frac{1}{r^3}+\frac{tan(\theta)}{r}\right)$$

Where am I going wrong?

 

[anathema]

Thank you for your post. In order to show that the force law is inverse fifth power, we need to find a relation between the force and the distance r. From your attempt, it seems that you have found a relation between the force and the angle \theta, which is not what we are looking for.

To show that the force is inverse fifth power, we can start from the equation of motion for a particle moving on a central field:

\frac{d^2r}{d\theta^2} + r = -\frac{mF(r)}{L^2}

where r is the distance from the origin, \theta is the angle, m is the mass of the particle, and L is the angular momentum. We can then substitute u = r^{-1} into this equation to get:

\frac{d^2u}{d\theta^2} + u = -\frac{mF(u^{-1})}{L^2u^2}

Now, since we know that the particle moves in a circular orbit, we can substitute the given relation r = r_0cos(\theta) into the equation of motion. This gives us:

\frac{d^2u}{d\theta^2} + u = -\frac{mF(r_0^{-1}cos(\theta)^{-1})}{L^2u^2}

Using the trigonometric identity cos(\theta)^{-1} = sec(\theta), we can simplify this to:

\frac{d^2u}{d\theta^2} + u = -\frac{mF(r_0^{-1}sec(\theta))}{L^2u^2}

Now, we know that the particle is moving in a circular orbit, so the acceleration is given by a = v^2/r, where v is the tangential velocity. We can express the tangential velocity in terms of u and \theta as:

v^2 = \frac{d}{d\theta}\left(\frac{1}{u}\right)^2 = \frac{1}{u^3}\left(\frac{du}{d\theta}\right)^2

Substituting this into the equation of motion, we get:

\frac{d^2u}{d\theta^2} + u = -\frac{mF(r_0^{-1}sec(\theta))}{

 

[tomcue]

Your mistake is in your substitution of u. u = r^-1 is not the correct substitution for this problem. The correct substitution is u = r_0cos(\theta). This will change your final result to be inverse fifth power.