Particle in an Orbit: Central Force Motion and Radial Equation

[CAF123]

Homework Statement

A particle of mass m moves under the influence of a central force $$\underline{F(r)} = -mk\left[\frac{3}{r^2} - 2\frac{a}{r^3}\right] \hat{r}$$

The particle is projected from point A, at a distance a from the centre of force O, in a direction perpendicular to OA, its projection velocity is half of the velocity required for a circular orbit of radius a.

i) The radial equation of motion is $$\ddot{r} - \frac{9}{4} \frac{ka}{r^3} = -\frac{3k}{r^2}$$
Define $u(\theta) = 1/r(t)$. Show that the above equation reduces to the linear form $u''(\theta) + 9u(\theta) = 12/a.$

ii) Solve the equation in i) and obtain $$r = \frac{3a}{4 - cos(3 \theta)}$$ as the orbit.

The Attempt at a Solution

i)By the chain rule, $\dot{r} = -\frac{u'}{u^2} \dot{\theta}$ and $$\ddot{r} = -\left[\frac{u^2 u'' - 2u (u')^2}{u^4}\right] \dot{\theta}^2 + \frac{u'}{u^2} \ddot{\theta}$$ I seem to have theta dot and double dot terms that are not part of the expression I need to show. Did I do the differentiation right? If so, how do I reexpress the theta dot terms?

ii)Solving the second order linear ODE gives $$u(\theta) = A \cos(3 \theta) + B \sin (\theta) + 4/3a$$. For r, I then get $$\frac{3a}{3aA \cos(3\theta) + 3aB \sin (3 \theta) + 4}$$ I can then let aA = C, aB = D and I then thought about putting the two trig terms together into a single trig function. But I can't do this, because the final expression is solely in cos(3θ) with no phase angle. Also, they have a negative somewhere. The only way I see right now of getting a single trig term is to do as I said above but this will give a phase. (which is not what the question wants)

Many thanks.

 

[voko]

For i), recall that the central force field has a particular integral of motion.

For ii), consider the initial conditions given.

 

[CAF123]

Hi voko,

For i), recall that the central force field has a particular integral of motion.

So, I determined the angular momentum of the particle previously as $L = m \sqrt{ka}/2$. Then $$\dot{\theta} = \frac{\sqrt{ka}}{2} u^2 \Rightarrow \ddot{\theta} = \frac{ka}{2}u^3 u'$$
Inputting my expressions into the eqn give $$(-uu' - 2u'^2)ka u + u'^2 ka u /2 - 9kau^3/4 = -3ku^2$$ which nearly cancels the u' term but not quite.

EDIT: I got it: I missed a factor of 1/4 in the first term. Thank you.

For ii), consider the initial conditions given.

Sorry, I forgot to say $\theta$ is defined from the line OA. Can I say $r(0) = a, \dot{r}(0) = 1/2 \sqrt{k/a}$, where r = r(θ)?

EDIT: The first IC gives me the coefficient of cos -1 which is what I want, but the second IC I posted does not give zero for the coefficient of sin. $\dot{r}(0) = 0$ would give me what I want, but would the velocity at theta = 0 be zero? Since it doesn't actually say where in time the particle gets launched, can I arbritarily say not at theta = 0 and thus the IC is sensible?

 

[voko]

i)By the chain rule, $\dot{r} = -\frac{u'}{u^2} \dot{\theta}$ and $$\ddot{r} = -\left[\frac{u^2 u'' - 2u (u')^2}{u^4}\right] \dot{\theta}^2 + \frac{u'}{u^2} \ddot{\theta}$$

The last term has the wrong sign here.

 

[CAF123]

The last term has the wrong sign here.

Why? If it was negative then the u' term would not cancel?
EDIT: I see what you mean - it was just a typo because I have in the derivation in my book.

 

[voko]

its projection velocity is half of the velocity required for a circular orbit of radius a.

Find the condition on velocity for circular motion.

 

[voko]

Why? If it was negative then the u' term would not cancel?
EDIT: I see what you mean - it was just a typo because I have in the derivation in my book.

You then made another sign error :)

 

[CAF123]

Find the condition on velocity for circular motion.

That would be : $v = (\pm) \sqrt{\frac{F(r) a}{m}}$ as the tangential speed of the particle if it was to do circular motion. For the projection velocity, I just take half of this.
Is the IC $\dot{r}(0) = 0$suitable? Since there is no indication of the particle being projected at θ = 0, I think I may assume that at such an angle, nothing happens.

 

[voko]

That would be : $v = (\pm) \sqrt{\frac{F(r) a}{m}}$ as the tangential speed of the particle if it was to do circular motion. For the projection velocity, I just take half of this.

I do not think this is useful as an expression. Use the equation for r to find the condition on circular velocity. But here is what puzzles me: how come you know the angular momentum if you do not know the initial velocity?

Is the IC $\dot{r}(0) = 0$suitable?

No. The ODE is 2nd order, so ICs must be zero and first order.

But here is something for you to think about. The initial velocity is perpendicular to the radius. Does this contain any useful information on $\dot{r}$?

 

[CAF123]

I do not think this is useful as an expression. Use the equation for r to find the condition on circular velocity. But here is what puzzles me: how come you know the angular momentum if you do not know the initial velocity?

The initial velocity is$v_{proj} = \frac{1}{2} \sqrt{\frac{k}{a}}.$ Then I computed L. Sorry for my confusion.

No. The ODE is 2nd order, so ICs must be zero and first order.

I initally had $\ddot{r}(0) = 0$, but that was a typo: I meant $\dot{r}(0) = 0$. The correction appeared on the quote.

But here is something for you to think about. The initial velocity is perpendicular to the radius. Does this contain any useful information on $\dot{r}$?

The angular momentum vector is orthogonal to both the radius and initial velocity. So $\dot{r}$ is parallel to the angular momentum vector.

 

[voko]

$\dot{r}$ cannot be parallel to anything, it is a scalar. Think about the components of the full vector velocity - can you obtain any useful condition on $\dot{r}$ if the velocity is perpendicular to the radius?

 

[CAF123]

$\dot{r}$ cannot be parallel to anything, it is a scalar. Think about the components of the full vector velocity - can you obtain any useful condition on $\dot{r}$ if the velocity is perpendicular to the radius?

I see, the velocity is perpendicular to the radius (to begin with), so then $\dot{r}(0)= 0$ ? $\dot{r}$ only becomes non zero when the particle starts to become displaced?

 

[voko]

$\dot{r}$ is the radial velocity. If the total velocity is perpendicular to the radius, then the radial component is zero, $\dot{r} = 0$. Another thing to note is that the latter condition means $r$ is at a minimum or a maximum.

 

[CAF123]

$\dot{r}$ is the radial velocity. If the total velocity is perpendicular to the radius, then the radial component is zero, $\dot{r} = 0$. Another thing to note is that the latter condition means $r$ is at a minimum or a maximum.

Would the particle ever subtend from it's initial direction (assuming ideal conditions)? (The force it experiences is not a function of it's height above the projection point, so I am unsure).

If this is the case, then why would be the particle trajectory even be called an orbit? Is it classed as an open orbit?

 

[voko]

I do not understand the question.

 

[CAF123]

I do not understand the question.

The initial velocity of the particle is perpendicular to the radius. Is this the case for all t? If so, why would the path be called an orbit? (Since if the particle stays in one direction, it wouldn't be orbitting anything).

EDIT: The particle will not stay in one direction for all t since there exists a minimum and maximum radius as you said. (which I computed below)

 

[CAF123]

I now have to sketch the orbit $$r = \frac{3a}{4 - \cos(3\theta)}$$ It is similar in style to a form in my notes which is a hyperbola although I am not exactly sure this form resembles a hyperbola.

The minimum value of r is at 3a/5 and the max at a. I also have to describe the geometrical significance of these numbers. I would assume that since the particle is under no repulsive force (it only experiences a force in the radial direction inwards), then the maximum r makes sense. However, it does not seem 'obvious' why the min is at 3a/5.

These max and min occur at $\theta = \frac{n 2 \pi}{3}$ and $\theta = \frac{\pi(2n + 1)}{3}$ respectively.

 

[voko]

The initial velocity of the particle is perpendicular to the radius. Is this the case for all t? If so, why would the path be called an orbit? (Since if the particle stays in one direction, it wouldn't be orbitting anything).

There is only one orbit with the velocity always perpendicular to the radius: the circle.

EDIT: The particle will not stay in one direction for all t since there exists a minimum and maximum radius as you said. (which I computed below)

The particle can move in one direction only when the angular momentum is zero. In all the other cases there will be some planar motion, bounded or unbounded depending on its energy.

 

[CAF123]

There is only one orbit with the velocity always perpendicular to the radius: the circle.

I have been assuming this whole time that the particle was being projected in the upwards direction. (i.e something like in z direction). No wonder you were confused - it must have sounded like complete nonsense. Sorry! (The question makes a whole lot more sense now..)

 

[voko]

A hyperbola is unbounded. In this case, however, r is clearly bounded.

At what values of the angle do you have minima, and at what angles do you have maxima? Where are the inflection points?

 

[CAF123]

A hyperbola is unbounded. In this case, however, r is clearly bounded.

After looking back through my old Calculus notes, I see that the eqn resembles that of an ellipse (eccentricity = 1/4, directrix x = -3a). Plotting this on an xy frame, I have intersections on the y-axis at $\pm 0.75 a$. On the x axis, intersections at a and 0.6 a, which match the min and max given below:

At what values of the angle do you have minima, and at what angles do you have maxima? Where are the inflection points?

Minima occur at $\theta = \frac{\pi (2n+1)}{3}$ and max at $\frac{n2 \pi}{3}$. I have not yet calculated the inflection points, but just by looking at the graph, I don't think there are any (I don't see any change in concavity)

 

[voko]

After looking back through my old Calculus notes, I see that the eqn resembles that of an ellipse (eccentricity = 1/4, directrix x = -3a).

An ellipse would have $\cos \theta$, you have $\cos 3 \theta$.

Minima occur at $\theta = \frac{\pi (2n+1)}{3}$ and max at $\frac{n2 \pi}{3}$.

How many of these do you have when $0 \le \theta \le 2 \pi$? Does that look an ellipse?

I have not yet calculated the inflection points, but just by looking at the graph, I don't think there are any (I don't see any change in concavity)

Since you have minima and maxima, the second derivative has to change its sign, so it has to have zeros.

 

[CAF123]

How many of these do you have when $0 \le \theta \le 2 \pi$? Does that look an ellipse?

For max (at r=a), this occurs at $\theta = 0, 2\pi$, so two points. For min (r = 3a/5), this occurs at only one point, $\theta = \pi$.

Does this not look like an ellipse? Starting at $\theta = 0$ and going anticlockwise, I get to the min at $\theta = \pi$ and then continuing, I get to the max again at $\theta = 2\pi$. So I have come back to my original point, with what looks like an elongated circle.

Since you have minima and maxima, the second derivative has to change its sign, so it has to have zeros.

$r''(\theta) = 0,$when $4 \cos (3 \theta) - \cos^2 \theta = 2 \sin^2 \theta$

 

[voko]

For max (at r=a), this occurs at $\theta = 0, 2\pi$, so two points. For min (r = 3a/5), this occurs at only one point, $\theta = \pi$.

I do not understand how you arrive at these. Here is the complete list of minima and maxima in the $2 \pi$ interval: $0, \ \frac 1 3 \pi, \ \frac 2 3 \pi, \ \frac 3 3 \pi, \ \frac 4 3 \pi, \ \frac 5 3 \pi$

$r''(\theta) = 0,$when $4 \cos (3 \theta) - \cos^2 \theta = 2 \sin^2 \theta$

How come you have $3 \theta$ in only one term?

 

[CAF123]

I do not understand how you arrive at these. Here is the complete list of minima and maxima in the $2 \pi$ interval: $0, \ \frac 1 3 \pi, \ \frac 2 3 \pi, \ \frac 3 3 \pi, \ \frac 4 3 \pi, \ \frac 5 3 \pi$

Yes, for some reason I was considering cos(θ). It is clear now why it is not an ellipse.

How come you have $3 \theta$ in only one term?

#

Rather, $4 \cos (3 \theta) - \cos^2 (3 \theta) = 2\sin^2 (3 \theta)$

 

[voko]

I am sure you can solve the equation.

 

[CAF123]

I am sure you can solve the equation.

I get $\theta = arccos(-2 + \sqrt{6})/3$ as the solution.

 

[CAF123]

After plotting some points (max, min points and some other theta), I get the orbit looking a bit like a fish with the tail at the left hand side and the tip of the head coinciding with r=a.

 

[voko]

It cannot have just this one solution. Between those 6 extrema, you must have 6 inflections.

But perhaps you could take an easier approach. From the form of the solution for r, it is obvious that the period of r is $\frac 2 3 \pi$. So you can just investigate what the graph looks like in $(-\frac 1 3 \pi, \frac 1 3 \pi)$; then you just rotate the figure by $\frac 2 3 \pi$ two times to obtain the full picture.

 

[CAF123]

So the trajectory is like a blob, which is is repeating every 120 degrees. And yes, what I gave was the principal solution - I should have said $\theta = arcccos(-2 + \sqrt{6}) + n \pi/3$

I presume that a closed orbit is an orbit where at two different theta, you arrive back at same r? If so, then can I say by inspection that this orbit is closed? I also have to mention the geometrical significance of these min and max r. Is it simply that r=a is the 'upper' bound for r and r=3a/5 is the 'lower' bound?

 

[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.