# Circular orbit around a point that is not in the center

It is the first time I post here, but I used to post in the Mathematics forum (I finished my Math degree two years ago and decided to study Physics now).

The following question is from an exam I took two weeks ago and it gave me headaches:

A particle moves in a circular orbit of radius $$a$$ under a central potential which attracts it to a point $$O$$, which is not in the center of the circumference. Let $$v_1$$ and $$v_2$$ be the maximum and minimum speed. Compute the kinetic moment, areal velocity and orbit period. Could this central potential be a gravitational or electrostatic potential?

How could this situation happen? I don't understand how it can be a circular orbit while the attracting point $$O$$ is not in the center.

When I was in Mathematics, I studied a course on Celestial Mechanics and I am familiar with gravitational orbits, but I think maybe I am taking things for granted that do not happen in other kinds of orbital motion.

• Draw a picture. Use concepts and principles which you know to be true. Don't assume a specific force function beyond central. Crunch, crunch, crunch Feb 7 at 13:17
• I am interested in any answers, and it cannot be a gravitational or electrostatic potential, as the solutions of those potentials afaik are conic sections, and the problem gives fixed r (not eliptical orbit) Feb 7 at 13:45
• Maybe you can use this that I found searching arxiv.org/pdf/physics/0410149.pdf Feb 7 at 14:19
• @JonathanJeffrey: I'm pretty sure that their answer is wrong. All orbits in a harmonic potential are ellipses centered at the origin. This can be seen by writing the equations of motion in Cartesian components, where they are just three harmonic oscillator equations with the same frequency ($\ddot{x} = - \omega^2 x$, etc.) Feb 7 at 16:23
• It's not clear if point "O" is fixed or free. Circular two-body orbits of a variety of types exist where both mutually orbit their barycenter. If point "O" is fixed this is impossible. If point "O" is in motion, then a solution is possible.
– J...
Feb 8 at 14:15

We can actually find the force law which produces any curve $$r(\theta)$$ using the Binet equation, as outlined in my previous answer. In this case, we have $$r = r_0 \cos \theta + \sqrt{a^2 - r_0^2 \sin^2 \theta}$$ where $$r_0$$ is the distance from the origin to the center of the circle (displaced along the $$x$$-axis), and $$a > r_0$$ is the radius of the circle. After a significant amount of algebra (confession: I used Mathematica), we find that we must have $$F(r) = \frac{8 a^2 L^2 r}{\mu (a^2 - r_0^2 + r^2)^3},$$ which implies a potential of $$U(r) = - \frac{2 a^2 L^2}{\mu (a^2 - r_0^2 + r^2)^2}.$$

It may look a bit odd to have the properties of a particular particle's orbit in a potential that's supposed to apply to all particles. A better way to look at this is that such trajectories are possible if the potential is of the form $$U(r) = - \frac{k}{(b^2 + r^2)^2},$$ But there are constraints on the shapes of the eccentric-circle trajectories; for a fixed potential, only eccentric circles with $$a^2 - r_0^2 = b^2$$ are possible. Moreover (as was pointed out in the comments), most trajectories in this potential will not be eccentric circles, and in fact will not even be closed orbits. The eccentric-circle trajectories arise only for special initial conditions; in particular, a particle describing an eccentric-circle trajectory with parameters $$a$$ and $$r_0$$ must have $$L^2/\mu = k/2 a^2$$.

For those who are interested, I have provided Mathematica code to do numerical integration of this potential. It certainly looks like an eccentric circle.

F[r_, bb_, kk_] = -D[-kk/(bb^2 + r^2)^2, r]
a = 2; r0 = 1; b = Sqrt[a^2 - r0^2]; k = 1; l = Sqrt[k/(2 a^2)];
tmax = 100;
soln = NDSolve[{r''[t] == F[r[t], b, k] + l^2/r[t]^3,
ph'[t] == l/r[t]^2, r[0] == a - r0, r'[0] == 0, ph[0] == 0}, {r[t],
ph[t]}, {t, 0, tmax}]
ParametricPlot[{r[t] Cos[ph[t]], r[t] Sin[ph[t]]} /. First[soln], {t,
0, tmax}]

• This is cool! My interpretation is that if any particle started in an eccentric-circle trajectory in this potential will have the same distance of closest approach $b$, which means for a orbit of $Feb 7 at 17:28 • @ytlu: I've edited by last paragraph to (hopefully) make my point clearer. As far as the form for$r(\theta)$, this can be obtained simply by taking the Cartesian equation$(x - r_0)^2 + y^2 = a^2$, substituting$x \to r \cos \theta, y \to r \sin \theta$(with$\theta$implicitly measured from the$x$-axis), and then finding the roots of the resulting quadratic equation in$r$. Feb 7 at 20:11 • @Thomas: While the force law I've derived is a weird central force to consider, it is still conservative. Any central force law$\vec{F}(\vec{r}) = F(r) \hat{r}$is conservative, since any central vector field is curl-free ($\vec{\nabla} \times \vec{F} = 0$.) Also, can you elaborate on why it's obvious that you couldn't have a central force of this form? Feb 8 at 19:22 • I've made an animation showing the actual motion of the particle, using the solution given by the Mathematica code: link. It makes the speed difference between periapsis and apoapsis visible. Might be a useful addition to the answer. Feb 8 at 20:21 • The code to generate the animation (as an addition to the code in this answer) can be seen here. Feb 8 at 20:31 Let $$d=2a$$ be the diameter of the circle, and let $$d=f+c$$, where $$f,c>0$$. $$f$$ will be the distance from the center of the central potential to the point on the circle farthest from it, and $$c$$ the distance from the center of the central potential to the point on the circle closest to it. Note that the angular momentum $$L$$, and thus areal velocity (which is $$L/2m$$), is conserved for any central potential, and that $$v_f f= v_c c = L/m$$, where $$v_f$$ is the speed at the far point and $$v_c$$ is the speed at the close point. (We don't have to worry about cross-products here, because the orbit curve is perpendicular to the vector from the attractive point to the orbit curve at these two points.) If you do a little bit of thinking, you will see that $$v_f$$ is the minimum speed and $$v_c$$ is the maximum speed. Because the orbiting object is always increasing its distance from the attractive point as it moves away from the close point, the attractive potential will slow it down until it reaches the far point, at which point the attractive potential starts to reel it in, increasing the speed again. Now you can compute the area of the circle from $$\pi a^2$$, and divide by the areal velocity $$L/2m$$ to find the orbital period. (I'm not sure what the kinetic moment means in this case, I'll edit this answer if you can explain its definition to me.) We can tell this does not follow an inverse distance potential, because it is a bound orbit that does not form an ellipse with the orbit center at one of the focii, and thus is not from classical gravitation or from Coulomb's law. (There are several proofs of such an elliptical orbit on the web.) A circle is an ellipse with both focii at its center, but since the center of the central potential is not at the center of the circular orbit, the orbit fails the test to be due to an inverse distance potential. • Thank you for your explanation! I am not sure about what kinetic moment means too (maybe I mistranslated from Spanish and I had to write "kinetic momentum" instead). I planned to write linear and angular momentum if I had known how to solve this problem and let the professor decide if that was what he was asking for. Feb 7 at 14:54 • @MiguelMars: According to Wikipedia, "Momento cinético" = "momento angular" = "angular momentum". Feb 7 at 16:04 • It may be an electrostatic potential that is not due to some point charges, though, right? Fill space with circles that obey these types of orbits, and that gives you a value for the force field at any point, and then one can just compute$\nabla \cdot E\$, and find the charge density distribution to produce that motion Feb 8 at 18:23
• @JerrySchirmer I suppose, as long as the charge distribution is spherically symmetric and produces a force field that matches Michael's derivation. But speculation like that is out of scope of OP's question. Feb 8 at 18:39
• @JerrySchirmer Were dealing with a central potential, so by definition, the potential must be spherically symmetric. See Michaels answer for the explicit potential. Feb 8 at 19:24

This answer addresses the question of "how could this possibly happen?". The answer is that although it may seem physically contrived today, in fact for thousands of years this sort of phenomenon was actually invoked in an essential way in our best theories of astronomy! (And at the same time, even the astronomers using it sometimes worried that it seemed contrived.)

The thing that really needs to be said here is that this setup is called circular motion around an equant, and it was actually used, along with the more famous epicycles, in Western astronomical models from the time of Ptolemy (~200 BCE?) up until Kepler (~1500 CE). Remember that before Newton, astronomical models were (from our perspective today) basically just pure geometry, motivated by "aesthetic" considerations like "circles are perfect, and heavenly bodies are perfect, so naturally heavenly bodies always exhibit uniform circular motion" rather than what we'd recognize as principled physical theory today. (This is obviously an anachronism -- in those days there was really no such thing as "principled physical theory" as we know it today -- so what else were they supposed to do?!) Equants and epicycles were essentially geometric constructions which could be sorta, kinda motivated from these principles, used to overfit the observational data, which was precise enough to make its deviation from simple circular motion apparent (not to mention that geocentrically, epicycles were basically required to explain retrograde motion). My understanding is that for the thousands of years they were used, there was always a certain amount of unease with "aesthetic" acceptability of epicycles and equants. But nobody came up with anything better, so they stuck.

I'm sure I've been told that even Copernicus needed to use epicycles since he didn't have ellipses like Kepler. Maybe the fact that he still needed them was even a sticking point in widespread adoption of his heliocentric ideas? I'm not sure, but I wouldn't be surprised if Copernicus also used equants. And to hammer home the point -- even Copernicus and Kepler were pre-Newton, and their models were still what we'd think of as "pure geometry" -- no real physical mechanism or anything. They just managed to come up with new "aesthetic sensibilities" leading to geometric constructions which fit the data more naturally. In fact, I suspect it would be fair to say that Copernicus and Kepler's motivations for their models stemmed partly from precisely this sort of "aesthetic unease" that epicycles and equants tended to elicit from people.

Anyway, when you work through the geometry, just appreciate that for thousands of years, every apprentice astronomer in the Western world churned through the same geometric considerations that you're thinking about now, and then applied them constantly for the rest of their careers!

This focuses on finding the potential simply using some geometrical connection and conservation laws. The geometric arrangement is shown in the following figure:

Notations:

• From the center of force which is located $$x$$ to the right of the circle's center. $$(r, \theta)$$ is the polar coordinate from $$x$$. The hatted $$\hat{r}$$ and $$\hat{\theta}$$ are the corresponding unit vectors.
• Polar coordinate from the center of circle, the same point $$(r, \theta)$$ is denoted as $$(a, \phi)$$, $$a$$ is the radius. Their unit vectors are denoted as $$\hat{a}$$ and $$\hat{\phi}$$.

The first conserved quantity is angular momentum in the $$\hat{z}$$ direction (out of paper):

$$\tag{1} L = m r v_\theta = m r^2 \dot{\theta}.$$

The information at $$r=a + x$$ and $$r = a - x$$ defines the value for angular moment:

$$\tag{2} L = m r^2 \dot{\theta} = m (a-x) v_1 = m (a+x) v_2.$$

One of the three paramters $$v_1$$, $$v_2$$ and $$x$$ can be solved as the rest two's. The angular speed $$\dot{\theta}$$ can be written as a function of $$r$$ from Eq.(2):

$$\tag{3} \dot{\theta} = \frac{ (a-x) v_1}{r^2} = \frac{ 2 v_1 v_2}{v_1 + v_2} \frac{a}{r^2}$$

Then, note that it is a circular motion around the circular center, therefore the velocity is always in a tangential direction $$\hat\phi$$, $$v = v_\phi$$. The angular velocity $$v_\theta$$ is the projection of $$v$$ in the $$\hat\theta$$ direction. Thus:

$$\tag{4} v_\theta = v \cos \ \beta; \text{ therefore } v = \frac{ v_\theta}{\cos \ \beta} = \frac{ r \dot\theta}{\cos \ \beta} = \frac{(a-x) v_1}{ r \cos \ \beta}$$

From the triangle in the figure, we find the $$\cos\beta$$:

$$\cos\beta = \frac{a^2 + r^2 - x^2}{2 a r}$$

Substitute $$\cos\beta$$ into Eq.(4):

$$\tag{5} v = \frac{2 a (a-x) v_1}{a^2 + r^2 - x^2}.$$

The kinetic energy for the speed at position $$(r, \theta)$$:

$$E_K = \frac{1}{2}m v2 = m \frac{2 a^2 (a-x)^2 v_1^2}{(a^2 + r^2 - x^2)^2}$$

Now, using the conservation of energy, assuming the potential energy at position of $$r= a-x$$ is $$V_1$$:

$$\tag{6} V_1 + \frac{1}{2} m v_1^2 = V(r) + m \frac{2 a^2 (a-x)^2 v_1^2}{(a^2 + r^2 - x^2)^2}$$

Finally,

$$\tag{7} V(r) - V_1 = \frac{1}{2} m v_1^2 \{ 1 - \frac{4 a^2 (a-x)^2}{(a^2 - x^2 + r^2 )^2} \}.$$

It helps to build confidence by checking at $$r=r_1=a - x$$, both sides of Eq.(7) should be zero:

$$V(a-x) - V_1 = \frac{1}{2} m v_1^2 \{ 1 - \frac{4 a^2 (a-x)^2}{(a^2 - x^2 + (a-x)^2)^2} \} \\ 0 =\frac{1}{2} m v_1^2 \{ 1 - \frac{4 a^2 (a-x)^2}{(2 a (a-x))^2} \} =0.$$

Another harder exercise is to prove $$V(a+x) + \frac{1}{2}mv_2^2 = V_1 + \frac{1}{2}mv_1^2$$.

In a first order approximation, an ellipse with a small eccentricity can be represented by a circle with the focus displaced from the center.

This is the only scenario I can think of here.

• The displacement could be large, and we are not trying to approximate anything. The orbit, by definition, is already a circle. Feb 7 at 17:00
• @JohnathanJeffrey Then describe a general scenario where the conditions mentioned by the OP apply strictly. Feb 7 at 18:54
• Please note that Michael Siefert already calculated the required potential in Mathematica, so you can see his answer, and critique it if you don't agree with it. But you don't need to calculate the potential to answer the OP's question, as I showed. Feb 7 at 19:13

Angular moment conservation is based on central potential

$$\tau = \vec{r} \times \vec{F} = 0$$

Therefore, this problem follows law of conservaion of angular momentum, which leads to constant area swept per time.

Also, the conservation of total energy is based on the conservative force (derived from a potential). Thus, we have both laws of conservation.

Assume the force center is away from the geometric center a distance $$x$$ closed to the fast speed side, showing as the figure:

The constant area sweeping rate gives:

$$(a+x) v_2 = (a-x) v_1. \text {Solve x to be } x = a \frac{v_1 - v_2}{v_1 + v_2}.$$

Lets try a potential form of $$V(r) = C r^n$$. Then from the conservation of energy at hese two points:

$$\frac{1}{2} m v_2^2 + C \left( \frac{2 a v_1}{v_1+v_2}\right)^n = \frac{1}{2} m v_1^2 + C \left( \frac{2 a v_2}{v_1+v_2} \right)^n$$

One possible solution is $$n = 2$$. It is harmonic potentail, and the constant $$C = m\frac{(v_1+ v_2)^2}{8 a^2}$$:

$$V(r) = \frac{m(v_1 + v_2)^2}{8 a^2} r^2$$

This is a guess for curiosity purpose from only two special points on the orbital, based on the assumption of a power law of potential form. (I have a feeling that I examiner may be make this problem out of nothing.) But the following answers about the area rate and the period are not dependent on this guess.

The area sweeping rate is:

$$\frac{dA}{dt} = \frac{1}{2} \vec{r}\times \vec{v}= v_1 (a-x) = \frac{ a v_1 v_2 }{v_1 + v_2}$$

The period is then:

$${\pi a^2} / {\frac{ a v_1 v_2 }{v_1 + v_2}} =\frac{ a \pi (v_1 + v_2) }{v_1 v_2}$$

• All orbits in a harmonic potential should be ellipses centered at the origin (not off-center), so I think you've done something wrong here. Feb 7 at 16:21