Orbits and eccentricities
Published on Jan 20, 2007 at 10:58 pm.
13 Comments.
Filed under comets, physics.
“When is Comet McNaught coming back?” everyone is asking me. It has been mostly cloudy for a month in many places in the Northern Hemisphere. Most people that I know missed the comet, and I only saw it the one time from the airplane’s window. The farther south, the harder it was to see, so the more critical the timing. Now, it is a Southern Hemisphere object. So, people are asking how long until it returns. Well, first of all, while a few comets return every few years, these short period comets are typically not very bright, and any of consequence have long since been discovered. So, you can expect any comet that is newly discovered that turns out so bright to be one with a long period. In other words, it won’t likely be coming back in your lifetime. Sorry. But, that’s the way it is. But, C2006 P1, Comet McNaught, is one of a few comets that will never come back. Period. I don’t mean just that it won’t come back in your lifetime, or even within thousands or millions of years. Nor do I mean that it won’t be back before the Sun swells and makes Earth uninhabitable, or even until after the Sun dies. No, I meant what I said. It will never be back. It has picked up escape velocity, and is heading out of the Solar System, to drift through interstellar space. If it ever does make a swing through a stellar system again in the far distant future, it will be around another star, not the Sun. So, how do I know that this is a one time event?
To answer that, I’ll say a few words about orbits. Johannes Kepler discovered back in the Seventeeth Century that the orbits of comets and planets around the Sun are ellipses, with the Sun at one focus of the ellipse. The point on the orbit closest to the Sun is the perihelion of the orbit (P), and the point on the orbit farthest from the Sun is the aphelion of the orbit (A). As it turns out, this statment also applies to satellites and moons orbiting planets, and pretty much anything orbiting anything.
The semimajor axis of the orbit is defined as half of the long length of the ellipse. Since the long length of the ellipse runs from perihelion to aphelion, you can find the semimajor axis by adding these numbers and dividing by two. 
This is sort of like the average distance from the Sun (There are multiple ways of defining “average distance” and most of them give a value other than the semimajor axis, but thinking of it as the average distance is useful for a first look at orbits.) Johannes Kepler further worked out that there is a relationship between the period of the orbit, T, and the semimajor axis. The period (measured in years) squared is equal to the semimajor axis (measured in astronomical units) cubed. Note that this is a relationship that only works for planet, asteroid, and comet orbits around the Sun. 
A more correct form of this relationship was worked out years later by Isaac Newton. He showed that, indeed, the period squared is proportional to the semimajor axis cubed, but the proportionality involves the Universal Gravitational Constant the the total mass of the system. This more general form of this relationship can be used for any orbit around anything, but I’ll save that for another day. Right now, I am talking about orbits around the Sun.
Notice that elliptical orbits can come in many varieties. Some are nearly circular, and others are very elongated ellipses. A measure of how elliptical an orbit is its eccentricity, e. 
For a closed orbit, such and an ellipse or a circle, the eccentricty can be computed from the perihelion and aphelion, as in this equation. But, what do I mean by a closed orbit? A closed orbit is one that repeats and has a period. But not all orbits do that. Before I get to that, though, let’s look at the eccentricity. In a circle, all points of the circle are equidistant from the center of the circle. So, A and P would be equal. The eccentricity, thus would be equal to zero. But the more elliptical an orbit, the greater the difference between the aphelion and the perihelion. So, the numerator of the equation gets larger. The bigger the eccentricity, the more elliptical the orbit. Eccentricities of ellipses are in the range 0<1.
But, what makes orbits elliptical? Consider first a circular orbit. In a circular orbit, gravitational forces provide the centripetal force to keep an object in orbit. 
The gravitational force is exactly equal to the centripetal force, and in that case the object is moving at a speed determined solely by the Universal Gravitational Constant, the mass of the Sun (or whatever else is being orbited), and the radius of the orbit as in this equation. If an object is moving more slowly than this, then the gravitational forces are in excess of that needed to keep the body in orbit, so it is pulled closer to the Sun. In doing so, it speeds up. If it is moving faster than the value given by this equation, then the gravitational forces are not quite large enough to keep the object orbiting at that distance, so it moves outward, slowing down. In an elliptical orbit, objects will continually be speeding up as they move closer to the Sun and slowing down as they move farther from the Sun. But, this works only up to the point that a body is moving slower than the escape velocity. 
The escape velocity is given by an equation that looks very similar to the velocity of a circular orbit, only with an extra factor of the square root of 2. If an object is moving at the escape velocity, then it is going fast enough that it never slows down and stops. In this case the aphelion distance will be infininte. You can see what this does to the eccentricity. Look at the eccentricity equation. If A is infinite (or nearly so) the (A-P) is basically A, and (A+P) is also basically A. That means that the eccentricty is essentially A/A which equals one. So, e=1 defines such an orbit. This is a parabola. But, what if the velocity at perihelion is not just equal to the escape velocity, but is in fact larger? In that case, the object will shoot outward in a flatter curve than a parabola, and will leave the Solar System faster. Such and orbit is a hyperbola, and it has an eccentricity greater than 1. For any conic section other than a parabola, the following equation 
relates the perihelion distance to the eccentricty. The term phi is the angle of the object from the perihelion. The term a is the semimajor axis of an ellipse, and half the distance between the focii for a hyperbola. But what would a parabolic or hyperbolic orbit imply? After all, such orbits don’t actually have an aphelion. They are one time events, and they never repeat. C/2006 P1, Comet McNaught, is an example of such an orbit. How does this happen?
Well, lets think back to comets. Comets are believed to reside in two possible reservoirs: the Kuiper Belt and the Oort Cloud. The Kuiper Belt is likely populated by comets thrown outward during the formation of the Solar System. It is a toroidal regions surrounding the Solar System filled with a large number of small icy bodies. Hypothetical until just about a decade ago, we now know of many hundreds of Kuipter Belt objects. The Oort Cloud is a little less certain. The Oort Cloud is a roughly spherical shell of comets surrounding the Solar System. How the comets got there is a matter of debate. Did they form there or did they get tossed out during the formation of the Solar System? At any rate, the Oort Cloud is much farther out. When something disturbs the orbit of one of these bodies in the Kuipter Belt or the Oort Cloud, it dives in closer to the Sun, and becomes a comet. But, these bodies will have elliptical orbits. The Oort Cloud comets will have far more elliptical orbits, with eccentricities approaching 1. An example of such a comet is C/1927 X1 (Comet Skjellerup-Maristany). (Thanks to Darren Addy for mentioning this in a comment on my first Comet McNaught post.) Comet Skjellerup-Maristany has a very high inclination orbit, as expected of Oort Cloud comets, and an eccentricity of 0.99984. This comet comes closer than 0.18 AU of the Sun, much like C/2006 P1, the current Comet McNaught, which also has a very inclined orbit. But for such a close approach, an Oort cloud comet might be expected to have an eccentricity even closer to 1. Skjerllerup-Maristany only goes out 2200 AU, a bit close for the Oort cloud. But, that is easy to explain if something were to rob it of some momentum coming towards or heading out from the Sun. What could do that? Jupiter or Saturn can interact with comets passing by. As the comets pass near Jupiter or Saturn, these gas giants can either pull on the comet speeding it up, or pull on it slowing it down, depending upon the relative positions of the comet and the planet at the time. Comet McNaught had an eccentricity e=1.00003. This is just hyperbolic, so the comet may have just barely picked up a bit more momentum and energy on the way in. It has swung by the Sun, and is now on its way out of the Solar System.
So, that means the C/2006 P1 will simply keep going. It will slow down, moving slower and slower, but it will never stop and come back. It will barely escape the Sun’s gravitational bonds, and will soar through interstellar space, a frozen chunk of material, gradually cooling to within a few degrees of absolute zero. I don’t know how long it might go until it gets caught in another star’s gravitational well, if it ever does, and loops by as an interstellar interloping comet there.
-Astroprof
A Ler…-- Rastos de Luz on January 22, 2007 at 8:12 am: 1
[…] “All of McNaught“, no Astronomy Blog. Ainda sobre este tema “Photos of comet C/2006 P1 McNaught by its discoverer“, no The Planetary Society Blog; “Orbits and Eccentricities“, no Astroprof’s Page. […]
Muonraker on January 24, 2007 at 10:48 am: 2
Dear Astroprof
Is there any way of knowing, approximately, how fast C/2006 P1 has been travelling through space during the last two weeks or so as it has passed perihelion?
Astroprof on January 24, 2007 at 10:50 am: 3
It is currently moving just a shade over 63 km/s.
harold dougan on January 25, 2007 at 11:25 pm: 4
if the comet is still viewable from dallas fort worth area where should i look and at what time as of jan 26 2007
Astroprof on January 25, 2007 at 11:50 pm: 5
Comet McNaught is gone from the DFW area. It is a southern hemisphere object, and is now fading and getting farther away. We pretty much got clouded out during the best viewing.
Astroprof’s Page » Comet McNaught Update on February 28, 2007 at 3:28 pm: 6
[…] It’s been a while since my last postings on Comet McNaught (here and here). And, a reader asked a question on another post about where the comet is heading. I thought that others might be interested, so I looked up some information to share with everyone. […]
hajara on January 20, 2008 at 4:05 pm: 7
gravitational force keep the comet in orbit around the sun, when are these forces the largest?
Astroprof’s Page » The Kirkwood Gaps on March 17, 2008 at 8:13 pm: 8
[…] The gaps are not really gaps in the asteroid belt, but they appear to be when a histogram of the asteroids is plotted in a certain manner. Asteroids orbit the Sun in elliptical orbits. Some of the orbits are far more elliptical than others. If you simply plot the positions of asteroids, then the asteroid belt looks pretty much continuous. However, if you plot the distribution of asteroids as a function of their semimajor axes, then something interesting shows up. (The semimajor axis of an orbit is sort of like the average distance from the Sun. I wrote about orbits a little over a year ago. You can read that posting for more information about orbits and the semimajor axis of an orbit.) Plotted in this way, you see that certain semimajor axes don’t have very many asteroids in them. Here is such a plot from the IAU’s Minor Planet Center: […]
bobbys on March 22, 2008 at 9:59 am: 9
Do you actually understand this strange language
yawn yawn
Astroprof on March 22, 2008 at 11:35 am: 10
Nothing strange here. It is just English.
Astroprof’s Page » 2008 BT18 Passing Earth on July 13, 2008 at 1:15 pm: 11
[…] The asteroid 2008 BT18 was discovered January 31, 2008, but the LINEAR program. 2008 BT18 is one of those potentially hazardous objects. Tomorrow it passes Earth, but at a very safe distance. In fact, for as far as we can reliably compute its orbit, it will continue to miss Earth. In fact, it may never run into Earth. But, it comes close enough that it needs to be watched. Above is an orbital diagram, courtesy JPL, showing the current position of the body. As with many Earth crossing asteroids, its orbit is not just near Earth. In fact, the semimajor axis of its orbit is 2.22 AU, out in the asteroid belt. But, it has a very elliptical orbit, ranging from about 0.89 AU out to almost 3.55 AU. Interestingly, the orbit of 2008 BT18 has a semimajor axis that is very close to one of the asteroid belt’s Kirkwood gaps. That may have something to do with its large eccentricity. […]
Mike on February 8, 2009 at 5:34 pm: 12
I have a question pretaining to High Eccentricity orbits. What is the challenges of a High Eccentricity orbit on say satelites or plants etc. Is a plantary body in a high Eccentricity orbit likley to face additional challenges vs a geostationary orbit per say. do you know of any website/pages that may speak to this question. thank you
Astroprof on February 8, 2009 at 6:30 pm: 13
Mike, I am not sure of what you are asking. Planets can not be in geostationary orbits, as that would imply that they orbited Earth. As for satellites, the challenge is to place them into circular orbits. It is easier to place them into eccentric orbits. If the orbit is too eccentric, of course, then they might dip a bit too close to the atmosphere.