Particle under central force with u(θ) = 1/r substitution

[voko]

Since r is periodic w.r.t. the angle, the orbit has to be closed. But it is not repeated every 120 degrees. It is repeated every 360 degrees.

 

[CAF123]

Since r is periodic w.r.t. the angle, the orbit has to be closed. But it is not repeated every 120 degrees. It is repeated every 360 degrees.

Maybe I misused the word 'repeated' and I see what you mean. This is the last part of this question:

The angle between successive maxima and minima is called the apsidal angle. Derive an expression for $dr/d\theta = \dot{r}/\dot{\theta}$ by extracting $\dot{r}$ from the expression for the energy and using $\dot{\theta} = h/r^2$. Use this expression to determine the apsidal angle and compare with your result previously.

The energy is conserved so I was given that $E = T + U,\,U = mk[\frac{9}{8} a/r^2 - 3/r]$. T = 0 initially (since $\dot{r}$= 0) and so E = U = -15mk/8a, with the condition r(0) = a.

My problem is that when I rearrange for $\dot{r}$, it appears I will have a negative under the square root. I believe what I am rearranging is the following: $$-\frac{15}{8} mk/a = \frac{m}{2}\dot{r}^2 + mk[\frac{9}{8}\frac{a}{r^2} - \frac{3}{r}]$$ Rearranging gives $$\dot{r} = \sqrt{-\frac{15}{8} mk/a - 3mk/r - 9mka/8r^2}$$ which is a negative under the square root.

EDIT: The potential U was given to us in the question as $mk [\frac{9}{8} a/r^2 - 3/r]$

 

[voko]

Without analyzing it further, there is clearly a sign error in the 1/r term after rearranging.

 

[CAF123]

Without analyzing it further, there is clearly a sign error in the 1/r term after rearranging.

I get the apsidal angle to be $\theta = \frac{2 \pi}{3}$. This agrees with the difference between the maxima and minima earlier. Just one more question: what exactly does the question mean by the geometrical significance of the max and min r? I would say that the particle can never go below 3a/5 and never beyond a, but this seems to be more of a physical significance rather than the geometry.

 

[voko]

Well, I cannot really say what the questions means. But at this stage, you might want to read this:

http://en.wikipedia.org/wiki/Newton's_theorem_of_revolving_orbits

Pay attention to Fig. 10, green orbit.

 

[CAF123]

Well, I cannot really say what the questions means. But at this stage, you might want to read this:

http://en.wikipedia.org/wiki/Newton's_theorem_of_revolving_orbits

Pay attention to Fig. 10, green orbit.

The green orbit is precisely the form of the orbit in this question.

 

[CAF123]

The only geometrical statement that I could make would involve the other two trajectories, but I am not sure if this is what is needed (since they don't mention the other two orbits in the question)

 

[voko]

Perhaps you could compare that to an elliptical orbit.

 

[CAF123]

Perhaps you could compare that to an elliptical orbit.

So maybe that the particle on the elliptical orbit has the same r(min) and r(max)

 

[CAF123]

The apsidal angle is defined as the angle between successive minima and maxima so the answer was pi/3. We just defined the apsidal angle properly today in lectures and the method used to obtain the apsidal angle was an integral. I understand this method. The way I found the angle (which I did before seeing the integral method) was as follows:

Since we have a functional form for r, I just subbed it into the resulting energy equation. I then solved for r'(θ) =0. This gave me the n2π/3. Since we know this corresponds to maxima as shown earlier, and since we know there is a change in concavity between two maxima, then this must correspond to a minima (as my sketch of the orbit showed). Hence to get the apsidal angle I would then divide by 2 to get nπ/3. Want the difference between successive max/min so I take the angle to be π/3.

Is this okay? (Very recently, I have become aware that some of the TAs at my university mark like computers and I think the method they will mark against will be via the integral). So if you think it is okay, I will argue will them. Obviously, I am aware that my method is not the most general, but I did what I did before going over the integral method.

Many thanks.

 

[voko]

There is nothing wrong with your method. You found an explicit expression for minima and maxima, and that is enough to find the apsidal angle. If there was not a specific requirement to use a particular method to solve the problem, you should be fine. After all, this is physics, and there is usually more than one way to solve a problem.

 

[CAF123]

Thanks voko.