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Home , Kirkwood gap, Orbital resonance

An Analytic Model of Stable and Unstable Orbital Q. Xue, A. Dutta, J. Kim, K. Thaman, S. Jain, and V. Jayam (Dated: 10 February 2020) The purpose of this paper is to develop a model for , both when it leads to unstable chaotic and when it leads to stable configurations. The study of resonance is a great interest in as it can give clues to the formation of the early by analyzing where certain objects are, and where there is a characteristic lack of objects. This paper applies elementary techniques from Newtonian mechanics to accurately and reliably predict the relative strengths of Kirkwood gaps in the main belt. It also builds off of prior work to create a more updated and complete approximation of the liberation period in stable resonant systems such as ’s and .

I. INTRODUCTION

The solar system is a complex dynamic system. In short time scales, it appears to be a reliable and pre- dictable clockwork system. However, on the timescale of millions of , the pulls from even the tiniest of can lead to chaotic patterns. But through this planetary anarchy, stable patterns arise. Orbital resonance occurs when the orbital periods of two orbiting bodies are in the ratio of two integers. As a result, they will exhibit periodic motion relative to each other. Typically, gentle tugs from other planets can be mostly ignored because the direction of the force is ran- dom and after a very long time, the time average net force would be zero. However in resonant orbits, even FIG. 1. A histogram of Kirkwood Gaps and the locations of the tiniest tugs can be amplified to large degrees as they orbital , plotted with Mathematica[4] will always occur at the same location in each resonant period. One of the most well known examples are the Kirk- wood Gaps in the (a list of key terms is : , , and are locked located in the glossary at the back). At certain semi- in a 1:2:4 resonance. There are many other instances of major axes, their orbital frequencies will form an integer moons and even planets in stable resonance with each ratio with that of . For example, a resonant fre- other. See table I for a list of major stable resonant quency of 2:1 means that, for every two orbits the inner systems in our solar system. asteroid makes, Jupiter will make one . In figure 1, selected resonant frequencies are plotted against a his- System Resonance togram showing the abundance of in the 2 au Jupiter Io - Europa - Ganymede 4:2:1 to 2.5 au range. At these locations, there is a scarcity Cassini Division - 2:1 Mimas - 2:1 of asteroids, caused by the unstable of these reso- Saturn nances and will migrate away. These gaps, while clearly - 2:1 noticeable, are not all the same strength. This paper Titan - Hyperion 4:3 - 3:2 will attempt to build a model to explain these apparent relative strengths. TABLE I. Selected examples of resonant systems in the solar This unstable resonance is caused when objects can- system.[5] not maintain a continuous resonance with other planets. This is seen in the Kirkwood gaps as asteroids in reso- nance with Jupiter will receive so much energy by passing by it regularly that it will fly out of orbit. This causes II. AN ELEMENTARY MODEL the the gaps inside the asteroid belt. Only the Asteroids, (which are in a 1:1 ratio with Jupiter) found at the Lagrange points of Jupiter, are able to maintain a First, we shall present a simplistic model that can ex- continuous resonance. plain the relative strengths of the Kirkwood Gaps. Con- There are also cases in which orbital resonance leads to sider an asteroid of mass m in a circular orbit with a stable configurations. For example, Neptune and Pluto radius of r1. Moving into a frame co-rotating with the are locked in a 3:2 resonance. Even more striking, the asteroid, the net force is zero. Then, introduce Jupiter, with mass M  m, in another circular orbit with ra- 2 dius r2. By Kepler’s third law, the ratio of their orbital Using equation 1, we can eliminate r1: frequencies is given by ! GMm cos α − (n/m)2/3  3/2 F = f1 r2 m r 2 3/2 = = (1) r2 (n/m)4/3 + 1 − 2(n/m)2/3 cos α f r n 2 1 (5) The asteroid will experience resonance if m and n can The angle α can only take certain discrete values when be expressed as relatively prime integers. For every or- resonance is reached. We have: bit the asteroid makes, Jupiter will complete n/m of an 2πk orbit. Let us assume that Jupiter and the asteroid start α = (6) off at ; in other words, the three bodies are m co-linear. Every time the asteroid makes one full rev- for 0 < k < m. Substituting this in and getting rid of olution, Jupiter will cycle between m locations, equally the constant, we have: spaced around its orbit. We base our model on the as- sumption: cos 2πk − f 2/3 F ∝ m (7) r 3/2 If we superimpose the larger object at all the f 4/3 + 1 − 2f 2/3 cos 2πk  locations it can be in when the smaller ob- m ject makes one full revolution starting from n where f ≡ m . Taking the average of the radial forces conjunction, then the strength of orbital res- experienced at all m locations give us: onance directly depends on the average of all the superimposed forces. m 2πk 2/3 1 X cos m − f Fr,avg ∝ (8) m 4/3 2/3 2πk 3/2 k=1 f + 1 − 2f cos m

r2 d

α θ r1

FIG. 3. Average force at all possible resonant frequencies with m, n < 10. The size of the bubble represents the relative strengths. Frequencies that are not relatively prime, or exist outside of the main belt are not included.

FIG. 2. The inner mass, on average, experiences an outward The results are plotted in figure 3. The resonant fre- radial force due to the outer mass. quencies shown, listed from strongest to weakest are 2:1, 3:1, 5:2, 7:3, 9:4, 8:3, 4:1, 7:2. Qualitatively, they line up with observations, as shown in figure 1; however, there By symmetry, only the radial components of the force will are a few discrepancies when compared to the quantita- show up in the average. The radial force at an arbitrary tive data. location is: The number of asteroids within a 0.05 AU range of GMm each was recorded and listed and sorted in F = · cos θ (2) r d2 table II. While there is no direct parameter that gives the strength of each resonant frequency, we can assume the We have: number of asteroids within a certain resonant frequency q is inversely proportional to the strength. We see that the d = r2 + r2 − 2r r cos α (3) 1 2 1 2 elementary model built is extremely accurate in predict- r cos α − r cos θ = 2 1 (4) ing the relative strengths! Everything is in the correct d order, with the exception of the 4:1 resonant frequency. 3

Resonant Semi-major Number of Superimposed Strength Hyperion Frequency axis (AU) Asteroids (force) 4:1 2.06 29 0.39 2:1 3.27 67 3.46 3:1 2.49 459 0.94 5:2 2.82 614 0.68 Titan 7:3 2.95 752 0.56 9:4 3.02 1495 0.51 8:3 2.70 1705 0.40 7:2 2.25 2777 0.29 Saturn TABLE II. The resonant frequencies and semi-major axes of major kirkwood gaps and the number of asteroids (±0.05 AU) in the range, pulled from Mathematica. The average super- imposed forced is also shown for comparison and has units where G = M = rearth = 1

This discrepancy can be explained by considering the FIG. 4. Hyperion gains or loses angular depend- effects of other planets. One peculiar interest is . ing on the setup, which can rotate its orbit clockwise or coun- It is relatively large and is quite close to this Kirkwood terclockwise gap, which roughly only 1 AU away. However, the most important factor is that Earth is in a 12 : 1 resonant orbital with Jupiter. Therefore, not only are asteroids in this gap resonant with Jupiter, it is also in a 3 : 1 is right before the conjunction point, while Titan, which resonance with Earth, which we have already established is faster, trails right behind Hyperion. As a consequence, has a strong effect. Hyperion speeds up, which makes the conjunction move Another explanation of why observations show so little towards the apocenter of the orbit. Another case ex- asteroids in this region is simply because this is the lower ists if the conjunction exists when Hyperion moves from limit of the asteroid belt. Any asteroids that go past apocenter to pericenter. In this scenario, Titan’s pull in- this point can be strongly perturbed by the pull of . creases the gravitational energy and angular momentum, While other gaps have asteroids migrating across it from slowing the down. However, this effect still moves both sides, this gap only has asteroids migrating from one the conjunction to apocenter. side.

III. STABLE RESONANCES A. Liberation Period This periodic amplification can wreck havoc to objects in a perfectly circular orbit, but it can lead to stable The underlying dynamics of the restricted three body resonance as well in certain cases where the orbits have problem is complex, and it is difficult to find a closed a nonzero eccentricity. This is common between different form expression for the period of liberations discussed planets and moons. above. We shall follow the techniques outlined by Agol A famous example of stable resonances is found in the et al. 2005. However, they provided only the general pro- , or to be more specific, Hyperion and portionality between liberation period, eccentricity, and Titan. Hyperion gains an uncertainty of 0.123 in its the resonant frequency. We wish to take a step forward orbit due to the alluding orbital resonance caused from and investigate the general proportionality factor, and Titan. In turn, we can see a near perfect circular orbit how it depends on the semi-major axis and if and how it produced from Titan, and an elliptical orbit created by depends on the amplitude of oscillations.[1] Hyperion. To get deeper into our analysis, let us fig- Unlike the main asteroid belt example, Titan’s circular ure out what happens at the conjunction points of our orbit is inside Hyperion’s orbit, which has a high eccen- model. If the conjunction point is located when moving tricity. We shall choose units such that G = MSaturn = from pericenter to apocenter, Hyperion receives an out- rTitan = 1. Titan has a mass µ and it is in a 4 : 3 resonant ward velocity component while the radial component of orbital with Hyperion. For the sake of generalization, let Titans gravitational attraction pulls Hyperion inwards. their orbital resonances be j+1 : j. This is an example of Because of this, Hyperion’s energy and angular momen- a first-order resonance and as shown in figure 3, it is one tum decrease which, in turn, also decreases the semi- of the strongest types. One further modification is that major axis and period of Hyperion. This effect is also en- instead of having a constant angular frequency, we shall hanced because the point where the pull is the strongest denote ωT and ωH for the average angular frequencies of 4

Titan and Hyperion, respectively. We have: Using a first-order power expansion:

ω j + 1 a 3/2 ∆θ = 2πj + 2πj (16) T = = H (9) ωH j aT Note that the first term is irrelevant, since m is an inte- ger, the first-term tells us that Hyperion will return to We can rewrite the ratio as its original position a total of m times. However, we are interested in the net change in its angle between succes- aH /aT = 1 + r/aT ≡ 1 + x (10) sive conjunctions, and that’s given by the second term. From now on, we can say: where r is the difference between the two semi-major axes. Using a first-order expansion, we get: ∆θ = 2πj (17)

ωT j + 1 3/2 3x Let the total angle subtended in one liberation be φ. Note = = (1 + x) ≈ 1 + (11) ωH j 2 that this will be four times the angular amplitude. There- fore, we need This first order expansion works because we are only in- terested in an order of magnitude calculation. In most N = φ/∆θ (18) cases, rH and rT are both significantly far away from the central body, such as Saturn, so their percent difference is successive conjunctions in order to complete one full rev- negligible. In our case, it leads to a 1% error (x = 0.22)! olution. Therefore, the total time to complete one cycle Rearranging and solving for x, we get: is: φ 2 Tliberation = TconjunctionN = (19) x = (12) j(ωT − ωH ) 3j We can eliminate ωT by substituting in equation 11 and Next, we calculate the time between conjunctions. We equation 12. This gives us: can accomplish this by switching into a frame co-rotating with Hyperion. In this new frame, Titan’s mean angular 4π 2πj Tconjunction = = (20) velocity would be ωT − ωH , and the time it takes for 3xωH ωH Titan to make one full revolution (equivalent to the time and it takes to meet up with Hyperion again) is: 2πj φ φ 2π Tliberation = TconjunctionN = · = (21) T = (13) ωH ∆θ ωH  conjunction ω − ω T H Following Agol et al, in the high eccentricity limit of Tis- Let us assume that on average, the angle of Hyperion’s serand’s relation (this is satisfied for e > µ1/3, which is orbit changes by ∆θ between successive conjunctions. valid in our case), we have: Thus, ∆θ is given by: ∆(x2) = ∆(e2) 2πω (2x)∆x = (2e)∆e ∆θ = ω T = H (14) H conjunction ω − ω e T H ∆x = ∆e x However, note that equation 9 no longer holds after Hy- perion’s orbit changes by ∆θ. As discussed earlier, this is Determining ∆e is a difficult task. One method pro- because its period will slightly increase or decrease as an- posed by Agol et al is to use an average impulse-based gular momentum is removed or added from the system. approach.[1] They claim the change in eccentricity be- After dividing, we write tween successive conjunctions is given by: µ ω j + 1 ∆e/conjunction = t (22) H = (15) x2 encounter ωT j +  We have tencounter = 1 which is a reasonable assumption where   1. Plugging this in: due to the units we are working in, so the total change in eccentricity is: 2π ∆θ = µN ωT /ωH − 1 ∆e = 2 (23) 2π x = j+1 1 j+ − 1 From equation 12 we see that x ∼ j . Plugging this in gives: 2π(j + ) = 1 −  ∆x = eµj3N (24) 5

Using equation 18 we can substitute for N to get: 22. It is unlikely that the change in eccentricity is directly proportional to the impulse that is provided and the use eµj2φ ∆x = (25) of the force as: 2π F = µ/x2 From Kepler’s third law and equation 9, we have: is overly generous. For one, this gives the force only clos- −3/2 j ∼ rH (26) est approach of the two orbits. It is unlikely that con- junctions occur at these close approaches so the force is The differential change would be: an overestimation. Next, as the two moons move around Saturn, the angles they make relative to their respec- −3 −5/2 −3 −5/2 dj =  ∼ rH drH =⇒  ∼ a ∆x (27) tive paths constantly change. Thus, the force calculated 2 2 would give us only the magnitude, not the strength in Dividing the two equations, taking the absolute value any particular direction. Because of this, it is once again gives: another overestimation. By overestimating the effects a single conjunction can  3∆x ∼ (28) have, the final answer will show less conjunctions are j 2rH needed for the angle φ to be subtended and thus com- plete one whole liberation cycle. Isolating for ∆x and setting it equal to 25 gives:

2r eµj2φ 3eµj3φ H = =⇒ 2 = (29) IV. GLOSSARY 3j 2π 4πrH Plugging it into equation 21 gives: Conjunction Point - The point at which all three √ celestial objects are collinear. 2φ πrH Kirkwood Gaps - The gaps in the distribution of Tliberation = (30) p 3 ωH 3eµj φ asteroids in the Asteroid Belt with respect to their semi- major axis. We can simplify this even further if we substitute, ac- Liberation Period - The period of time for the orbits cording to Kepler’s third law: in a resonant system to return to its initial state.

−3/2 Orbital Frequency - The amount of orbits completed ωH = r (31) per unit of time. Apocenter - The point in an elliptical orbit that is to get: farthest from the inner body. √ 2 Pericenter - The point in an elliptical orbit that is 2 πφrH Tliberation = (32) closest to the inner body. p3eµj3

Hyperion has an eccentricity of e = 0.123 [6], Titan V. CONCLUSION has a mass µ = 0.000237mSaturn, semi-major axis of rH = 1.21rTitan.[3] We also have φ = 2.51 [2] and the 4:3 resonance means mj3. Plugging these values in, we In this paper, we have analyzed and created a math- get: ematical model for two cases of orbital resonance in the Solar System: Kirkwood Gaps in the Asteroid Belt, and Tliberation = 169.33 Saturn’s two moons Hyperion and Titan. The former was modeled as two concentric and circular orbits with Using the natural units we’ve chosen, the the outer mass, Jupiter, much larger than the inner mass, of Titan is: an asteroid. By taking the average radial component of force that Jupiter exerts on the asteroid at regular time TTitan = 6.283 → 15.95 days intervals, we found that, at certain resonant frequencies, Using this conversion factor, we get: the average force is much stronger than at others. To a degree of remarkable accuracy, these points corresponded

Tliberation = 1.18 years with the points at which the Kirkwood Gaps occur. Secondly, the Saturn-Titan-Hyperion system was mod- Although the actual liberation period is t = 1.75 years, eled as a large mass, Titan, in a circular orbit, and a this is surprisingly close![2] It is correct to an order of smaller mass, Hyperion, in a larger and elliptical orbit. magnitude, though it is an underestimation. During the subsequent 4:3 resonant motion, Hyperion Multiple crude estimates and approximations and they periodically gains and loses energy, causing its orbit to could have factored in varying amounts of error. How- change shape. The time for the system to completely re- ever, the main source of error is probably from equation turn to its initial state, also known as the liberation pe- 6 riod, was successfully approximated to an order of mag- REFERENCES nitude. During the process of finding an appropriate model, [1] Eric Agol et al. “On detecting terrestrial planets with timing one option was using the Lagrange Planetary Equations, of transits”. In: Monthly Notices of the Royal which give a much more accurate description of the mo- Astronomical Society 359 (2 May 2005). tion of the system. However, the equations were too long [2] Riccardo Bevilacqua. “Resonances and Close Approaches. I. and complicated, even when approximated to the first The Titan-Hyperion Case”. In: The Moon and the Planets 22 order. However, a complete analysis would require them (2 Apr. 1980). and the next step in our research is to implement the La- [3] R Jacobson. “The Field of the Saturnian System from Satellite Observations and Spacecraft Tracking Data”. In: The grange Equations in order to create a much more accurate Astronomical Journal 132 (6 Nov. 2006). model of both stable and unstable orbital resonance. [4] Wolfram—Alpha Knowledgebase. Data Source Information. [5] S Peale. “Orbital resonance in the solar system”. In: Annual review of astronomy and astrophysics 14 (1976). [6] P Thomas and Black Nicholson. “Hyperion: Rotation, Shape, and Geology from Voyager Images”. In: Icarus 117 (1 Sept. 1995).