# Finding the period of an orbit ##r=a(1+\cos\theta)##

Davidllerenav
Homework Statement:
A particle with angular momentum describes the orbit ##r = a (1 + \cos\theta)##. Find the central force that produces this orbit. Find the period of the orbit. Determine the minimum energy that the particle must have to escape from this orbit.
Relevant Equations:
##\frac{l^2}{m}\left(\frac{d^2u}{d\theta^2}+u\right)=-\frac{\partial V}{\partial u}##
##t=\sqrt{\frac{m}{2}}\int_{r_0}^{r}\frac{dr'}{\sqrt{E-V_{eff}}}##
I've already found the potential and force that produce the given orbit. my results were:
##V=-\frac{al^2}{mr^3}##
##\vec{F}=-\frac{-3al^2}{mr^4}\hat{r}##​
Now, I've been trying to find the period using the equation
##t=\sqrt{\frac{m}{2}}\int_{r_0}^{r}\frac{dr'}{\sqrt{E-V_{eff}}}##​
Using ##r_0=r_{min}=a## and ##r=r_{max}=2a##, and multiplying by 2 to make a full orbit, I end up with the integral
##T=2\sqrt{\frac{m}{2}}\int_{a}^{2a}\frac{dr}{\sqrt{E+\frac{al^2}{mr^3}-\frac{l^2}{2mr^2}}}##​
The problem is that I have no idea on how to integrate this. Is there a trick or substitution to integrate it, or am I wrong somewhere?

Staff Emeritus
Homework Helper
Homework Statement:: A particle with angular momentum describes the orbit ##r = a (1 + \cos\theta)##. Find the central force that produces this orbit. Find the period of the orbit. Determine the minimum energy that the particle must have to escape from this orbit.
Have you plotted this function? It doesn't seem correct to me. When ##\theta = \pi##, you get ##r=0##, so the potential and force would both be undefined if your expressions are correct.

• Davidllerenav and hutchphd
Davidllerenav
Have you plotted this function? It doesn't seem correct to me. When ##\theta = \pi##, you get ##r=0##, so the potential and force would both be undefined if your expressions are correct.
Yes, is a cardioid, I think. I hadn't thought about the problem at r=0, with both potential and force being undefined there. And at the same time, it means that the orbit goes through the nucleus. But that's the orbit in the problem statement.

Homework Helper
Gold Member
The integral will simplify if you find an explicit value for the energy E. Try evaluating E at ##\theta = 0##.

The origin does appear to be a singular point. Just close your eyes when the particle passes through the origin. [EDIT: Should the integration be from 0 to ##2a## instead of ##a## to ##2a##?]

Davidllerenav
The integral will simplify if you find an explicit value for the energy E. Try evaluating E at ##\theta = 0##.

The origin does appear to be a singular point. Just close your eyes when the particle passes through the origin. Well, I don't have any function of E or something. Can I just write it as the sum of the kinetic energy in one dimension plus the effective potential? The problem is still solvable even if there's a singularity at the origin?

Responding to your edit: Yes, I forgot that when ##\theta=\pi##, ##r=0##.

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Well, I don't have any function of E or something. Can I just write it as the sum of the kinetic energy in one dimension plus the effective potential?
Yes. Show that the term ##\frac{l^2}{2mr^2}## in the effective potential energy equals the part of the kinetic energy due to the velocity component perpendicular to the position vector.
The problem is still solvable even if there's a singularity at the origin?
Yes, I think so. I don't see anything particularly bad in having the speed approach infinity as the particle approaches the origin. This problem is for a Newtonian world! But I haven't thought much about it.

Davidllerenav
Yes. Show that the term ##\frac{l^2}{2mr^2}## in the effective potential energy equals the part of the kinetic energy due to the velocity component perpendicular to the position vector.
Ok, that can be done through the Lagrangian, I think.
Yes, I think so. I don't see anything particularly bad in having the speed approach infinity as the particle approaches the origin. This problem is for a Newtonian world! But I haven't thought much about it.
Well, that's strange. I guess since it is classical mechanics, we aren't considering the limit of speed as the speed of light?

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Ok, that can be done through the Lagrangian, I think.
Express the angular momentum in terms of ##r ## and ##\dot \theta## and then I think you can see why the term ##\frac{l^2}{2mr^2}## is equal to the part of the kinetic energy due to the angular motion.

Well, that's strange. I guess since it is classical mechanics, we aren't considering the limit of speed as the speed of light?
Yes, that's right.

Davidllerenav
Express the angular momentum in terms of ##r ## and ##\dot \theta## and then I think you can see why the term ##\frac{l^2}{2mr^2}## is equal to the part of the kinetic energy due to the angular motion.
Ok. So since the angular part of the kinetic energy is given by ##\frac{1}{2}mr^2\dot{\theta}## and ##\theta## is cyclic, then ##mr^2\dot{\theta}=l=\text{constant}##, then ##\dot{\theta}=\frac{l}{mr^2}\Rightarrow \dot{\theta}^2=\frac{l^2}{m^2r^4}##, then it implies that ##\frac{l^2}{2mr^2}=\frac{1}{2}mr^2\dot{\theta}^2##. That's what I got.

Yes, that's right.
Makes sense.

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Ok. So since the angular part of the kinetic energy is given by ##\frac{1}{2}mr^2\dot{\theta}## and ##\theta## is cyclic, then ##mr^2\dot{\theta}=l=\text{constant}##, then ##\dot{\theta}=\frac{l}{mr^2}\Rightarrow \dot{\theta}^2=\frac{l^2}{m^2r^4}##, then it implies that ##\frac{l^2}{2mr^2}=\frac{1}{2}mr^2\dot{\theta}^2##. That's what I got.
Looks good. Since## \frac{1}{2}mr^2\dot{\theta}^2## is the part of the KE due to the angular component of the velocity, this shows that the expression ##\frac{m}{2}\dot r^2 + V_{\rm eff}## equals the actual total energy of the particle.

Davidllerenav
Looks good. Since## \frac{1}{2}mr^2\dot{\theta}^2## is the part of the KE due to the angular component of the velocity, this shows that the expression ##\frac{m}{2}\dot r^2 + V_{\rm eff}## equals the actual total energy of the particle.
So, given that ##t=\sqrt{\frac{m}{2}}\int_{r_0}^{r}\frac{dr'}{\sqrt{E-V_{eff}}}## and ##E=\frac{1}{2}m\dot{r}^2+V_{\text{eff}}##, replacing in the integral, I would have
##t=\sqrt{\frac{m}{2}}\int_{r_0}^{r}\frac{dr'}{\sqrt{\frac{1}{2}m\dot{r}^2}}=\int_{r_0}^{r}\frac{dr'}{\dot{r}}##​
Then, writing ##\dot{r}=\frac{dr}{dt}##, it would become an integral with respect to ##t##, right?

Homework Helper
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So, given that ##t=\sqrt{\frac{m}{2}}\int_{r_0}^{r}\frac{dr'}{\sqrt{E-V_{eff}}}## and ##E=\frac{1}{2}m\dot{r}^2+V_{\text{eff}}##, replacing in the integral, I would have
##t=\sqrt{\frac{m}{2}}\int_{r_0}^{r}\frac{dr'}{\sqrt{\frac{1}{2}m\dot{r}^2}}=\int_{r_0}^{r}\frac{dr'}{\dot{r}}##​
Then, writing ##\dot{r}=\frac{dr}{dt}##, it would become an integral with respect to ##t##, right?
This will just lead to the identity ##t = t##. Instead, try to find the value of the constant ##E##. You can pick any point on the orbit. What would be a point that will make it easy to determine ##E##? Is there a point where the value of ##\dot r## is easy to determine? What is the value of ##V_{\rm eff}## at that point?

Davidllerenav
This will just lead to the identity ##t = t##. Instead, try to find the value of the constant ##E##. You can pick any point on the orbit. What would be a point that will make it easy to determine ##E##? Is there a point where the value of ##\dot r## is easy to determine? What is the value of ##V_{\rm eff}## at that point?
Well, since ##\dot{r}=-a\dot{\theta}\sin\theta##, then I guess that at ##\theta=\frac{\pi}{2}## would be a good choice.

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Well, since ##\dot{r}=-a\dot{\theta}\sin\theta##, then I guess that at ##\theta=\frac{\pi}{2}## would be a good choice.
You would need to know the value of ##\dot \theta## at this point in order to determine ##\dot r## at this point. I was thinking of a different point on the orbit where the value of ##\dot r## can be seen without any calculation. (Or, use your equation for ##\dot r## in your last post, but pick a different value of ##\theta##.)

Davidllerenav
You would need to know the value of ##\dot \theta## at this point in order to determine ##\dot r## at this point. I was thinking of a different point on the orbit where the value of ##\dot r## can be seen without any calculation. (Or, use your equation for ##\dot r## in your last post, but pick a different value of ##\theta##.)
##\theta=0## the, right? That would make ##\dot r=0## and ##r=a##, that is constant so it makes sense.

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##\theta=0## , right?
That's a good choice.

That would make ##\dot r=0##
Yes

and ##r=a##
No

Davidllerenav
No
Oh sorry, my bad, I thought it was a sine insted of cosine in ##r##, the correct value would be ##r=2a##.

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So, now you have a point where you know both ##r## and ##\dot r##. You can use these values to determine ##E##.

Davidllerenav
So, now you have a point where you know both ##r## and ##\dot r##. You can use these values to determine ##E##.
Ok, so since ##E=\frac{1}{2}m\dot{r}^2-\frac{al^2}{m}\frac{1}{r^3}+\frac{l^2}{2m}\frac{1}{r^2}##, then at ##\theta=0##, ##E=-\frac{al^2}{m}\frac{1}{8a^3}+\frac{l^2}{2m}\frac{1}{4a^2}=-\frac{l^2}{8ma^2}+\frac{l^2}{8ma^2}=0##?

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Ok, so since ##E=\frac{1}{2}m\dot{r}^2-\frac{al^2}{m}\frac{1}{r^3}+\frac{l^2}{2m}\frac{1}{r^2}##, then at ##\theta=0##, ##E=-\frac{al^2}{m}\frac{1}{8a^3}+\frac{l^2}{2m}\frac{1}{4a^2}=-\frac{l^2}{8ma^2}+\frac{l^2}{8ma^2}=0##?
Yes, that's what I got, too. Interesting! With ##E = 0##, the integral for the time is not too bad.

Davidllerenav
Yes, that's what I got, too. Interesting! With ##E = 0##, the integral for the time is not too bad.
It would be ##T=2\sqrt{\frac{m}{2}}\int_{a}^{2a}\frac{dr}{\sqrt{\frac{al^2}{mr^3}-\frac{l^2}{2mr^2}}}##. I'll try it and see if I can integrate it the. I have a question, isn't it problematic that we used the energy at only ##\theta=0##?

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It would be ##T=2\sqrt{\frac{m}{2}}\int_{a}^{2a}\frac{dr}{\sqrt{\frac{al^2}{mr^3}-\frac{l^2}{2mr^2}}}##. I'll try it and see if I can integrate it the. I have a question, isn't it problematic that we used the energy at only ##\theta=0##?
Would E be different at any other point of the orbit?

Davidllerenav
Would E be different at any other point of the orbit?
I guess not? If I recall correctly, the energy is constant.

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Yes. There is only one force acting on the particle and the force has an associated potential energy function. So, the system is conservative.

Davidllerenav
Yes. There is only one force acting on the particle and the force has an associated potential energy function. So, the system is conservative.
Ok. Thanks! I'll try to do the integral then.

• TSny
Fred Wright
I have a problem with your orbit ##r=a(1+\cos(\theta))##. I plot the orbit with ##a=1##: How could this be an orbit of a central force? Not likely.
If, however, the orbit is described by ##r=\frac{p}{1+e \cos(\theta)} ##, you could show by analytical geometry that the orbit is an ellipse with ##e## the eccentricity. So work out the orbit of the particle in a central field using center of mass coordinates. You will get an expression for the angle in terms of the radial distance, total energy, angular momentum, potential field strength and mass. You will then be able to associate these parameters with ##p## and ##e##.

Homework Helper
Gold Member
I have a problem with your orbit ##r=a(1+\cos(\theta))##. I plot the orbit with ##a=1##:
View attachment 281355
How could this be an orbit of a central force? Not likely.
You can get a cardioid orbit from an inverse 4th power central force, ##F = - \large \frac{A}{r^4} \hat r##, where ##A## is a positive constant.

If you want this force to cause a particle of mass ##m## to trace the upper half of the cardioid ##r = a(1+\cos \theta)## starting from the position ##(r, \theta) = (2a, 0)##, then you would give the particle an initial velocity ##\vec v_0 = \sqrt{\frac{A}{12ma^3}} \hat \theta##

When the particle reaches the singular point ##r = 0##, it would require some sort of conjuring to make the particle rebound in just the right direction such that it can continue to trace the bottom half of the cardioid. We could fix a very small circular disk at the origin and assume that the collision of the particle with the disk is elastic. The particle would not quite make it to the origin, but if the disk has a small enough radius we wouldn't notice.

• Fred Wright