# How do they figure out where the asteroids are going?

May 3, 2010 5:02 PM Subscribe

How much information is needed to calculate the orbit of, say, an asteroid or a comet? Is it a simple, automated process?

Say I get out my telescope and camera one night, and snap a picture of the sky. Then I do the same thing the next night, and notice something has moved. I've found an asteroid, lucky me, I'll name it after my cat. Then I do the same the the third night. So I now have three sets of coordinates, and the times they were measured.

Is this enough to calculate the orbit and future position of the asteroid, and whether it will eventually crash into Jakarta? Is calculating the orbit simply a matter of putting these coordinates into some software (which software?) or is it something some experts have to sweat over? Is more information needed, for example an estimation of the mass of the object?

Say I get out my telescope and camera one night, and snap a picture of the sky. Then I do the same thing the next night, and notice something has moved. I've found an asteroid, lucky me, I'll name it after my cat. Then I do the same the the third night. So I now have three sets of coordinates, and the times they were measured.

Is this enough to calculate the orbit and future position of the asteroid, and whether it will eventually crash into Jakarta? Is calculating the orbit simply a matter of putting these coordinates into some software (which software?) or is it something some experts have to sweat over? Is more information needed, for example an estimation of the mass of the object?

(In case it's not clear in my comment: objects orbit the Sun in ellipses. http://en.wikipedia.org/wiki/Elliptical_orbit)

posted by dfriedman at 5:16 PM on May 3, 2010

posted by dfriedman at 5:16 PM on May 3, 2010

Response by poster: The question about collision was just a bit of hyperbole. I'm more interested in just being able to determine where the asteroid will be, in general - is converting a series observations of an object into a prediction of where in the sky to look for it in 5 years time something that can be done quickly by an amateur at home? What information would they need? I have, now, come across some software that seems to be used for this, but the input data seems to come from professional organizations, rather than simply "Enter your coordinates here and click go!".

posted by Jimbob at 5:18 PM on May 3, 2010

posted by Jimbob at 5:18 PM on May 3, 2010

Predicting future locations is probabalistic; in other words, you get a calculation that tells you with, say, 99% probability, that Asteroid X will be in location Y in 201Z.

A quick google search yields this paper by the Jet Propulsion Laboratory. http://neo.jpl.nasa.gov/apophis/

As to software, I have no idea.

posted by dfriedman at 5:21 PM on May 3, 2010

A quick google search yields this paper by the Jet Propulsion Laboratory. http://neo.jpl.nasa.gov/apophis/

As to software, I have no idea.

posted by dfriedman at 5:21 PM on May 3, 2010

Best answer: The one measurement that's hard to get is distance to said asteroid. This can be found out with parallax. This page is pretty much a how-to on doing this, using a software package called Hands-On Universe. I'm not sure if it can calculate orbits, though.

posted by zsazsa at 5:24 PM on May 3, 2010

posted by zsazsa at 5:24 PM on May 3, 2010

Best answer: In principle 3 positions measured by telescope are all you need to determine an orbit. In practice your measured positions have errors, so there is an 'uncertainty ellipse' around the orbit. If you really wanted to know whether an object was going to hit the Earth, you'd want observations over as long a period of time as possible. Another complication is that asteroids and comets are influenced to some degree by nongravitational forces - a thermal force called the Yarkovsky effect for asteroids, and recoil due to release of gas and dust by comets. These forces depend on the physical properties of the asteroid or comet under study so modeling them introduces more parameters into your orbit fit.

Orbit determination by radar observations is somewhat different, because you also get velocity information via the Doppler shift of the reflected signal.

posted by lukemeister at 5:33 PM on May 3, 2010

Orbit determination by radar observations is somewhat different, because you also get velocity information via the Doppler shift of the reflected signal.

posted by lukemeister at 5:33 PM on May 3, 2010

Don't you also have to factor in the gravitational effect of other large objects that it passes by?

posted by alms at 5:51 PM on May 3, 2010

posted by alms at 5:51 PM on May 3, 2010

Orbits of 'small' objects are very hard to calculate accurately.

I looked into this many years ago, and my dim memories are that finding the rising and setting times of the moon, accurate to a minute or so, even a few years in the past or future required accounting for the gravitation influence of the the earth and the sun (of course) but also of Venus, Mars, and Jupiter

I found some old code:

Jupiter is represented by N= in the above--you can kill 20 minutes figuring out what planets the other letters represent.

The situation wrt asteroids that might hit the earth is much worse, of course--all those numbers in the above formula are not known with any degree of accuracy (the moon has been observed for milennia; the typical asteroid for months).

But we all live with the possibility that some kind of rock, big or small, will smash us on the head at any old time. Or some asshole might leave an SUV full of firecrackers and stuff a block from where you work. It's all not worth thinking about.

posted by hexatron at 6:07 PM on May 3, 2010 [2 favorites]

I looked into this many years ago, and my dim memories are that finding the rising and setting times of the moon, accurate to a minute or so, even a few years in the past or future required accounting for the gravitation influence of the the earth and the sun (of course) but also of Venus, Mars, and Jupiter

I found some old code:

// Lunar azimuth,declination,radius: double L= fmod(0.606434+0.03660110129*T, 1)*2*PI; double M= fmod(0.374897+0.03629164709*T, 1)*2*PI; double F= fmod(0.259091+0.03674819520*T, 1)*2*PI; double D= fmod(0.827362+0.03386319198*T, 1)*2*PI; double N= fmod(0.347343-0.00014709391*T, 1)*2*PI; double G= fmod(0.993126+0.00273777850*T, 1)*2*PI; double V=0.39558*sin(F+N)+0.08200*sin(F)+0.03257*sin(M-F-N) +0.01092*sin(M+F+N)+0.00666*sin(M-F) -0.00644*sin(M+F-2*D+N)-0.00331*sin(F-2*D+N) -0.00304*sin(F-2*D)-0.00240*sin(M-F-2*D-N) +0.00226*sin(M+F)-0.00108*sin(M+F-2*D) -0.00079*sin(F-N)+0.00078*sin(F+2*D+N); double U= 1-0.10828*cos(M)-0.01880*cos(M-2*D)-0.01479*cos(2*D) +0.00181*cos(2*M-2*D)-0.00147*cos(2*M) -0.00105*cos(2*D-G)-0.00075*cos(M-2*D+G); double W=0.10478*sin(M)-0.04105*sin(2*F+2*N)-0.02130*sin(M-2*D) -0.01779*sin(2*F+N)+0.01774*sin(N) +0.00987*sin(2*D)-0.00338*sin(M-2*F-2*N) -0.00309*sin(G)-0.00190*sin(2*F)-0.00144*sin(M+N) -0.00144*sin(M-2*F-N)-0.00113*sin(M+2*F+2*N) -0.00094*sin(M-2*D+G)-0.00092*sin(2*M-2*D); double S; S=W/sqrt(U-V*V); azimuth= L+atan2(S,sqrt(1-S*S)); S=V/sqrt(U); declination=atan2(S,sqrt(1-S*S)); radius=60.40974*sqrt(U); where T seems to be (Julian Day) - 2451545 (T is measured in days)I'm not quite

*this*clever--it's all taken from the excellent Meeus' Astronomical AlgorithmsJupiter is represented by N= in the above--you can kill 20 minutes figuring out what planets the other letters represent.

The situation wrt asteroids that might hit the earth is much worse, of course--all those numbers in the above formula are not known with any degree of accuracy (the moon has been observed for milennia; the typical asteroid for months).

But we all live with the possibility that some kind of rock, big or small, will smash us on the head at any old time. Or some asshole might leave an SUV full of firecrackers and stuff a block from where you work. It's all not worth thinking about.

posted by hexatron at 6:07 PM on May 3, 2010 [2 favorites]

Best answer: The keywords you're missing here are ephemerides and orbital elements. An observer will go (literally to this website, which handles all new minor planet observations) and enter the derived ephemerides from their observations. This will then be pushed out to all the minor planet astronomers in the world, and some of them will repeat the observations. Over time, more and more accurate orbital elements are obtained.

As the Minor Planet Center is an official arm of the International Astronomical Union, it is considered authoritative. Its director, Brian Marsden^, is arguably the most experienced person on Earth at calculating the risk of asteroid impacts. Additionally, NASA separately operates the Near Earth Object Program, which is tasked with tracking and warning of near-earth asteroid impact risks. They operate a computer system which keeps located large objects in a database that is constantly updated. Note that the smaller the object, the more likely that an encounter with a larger object such as the Moon will introduce a deviation into its orbit. The impact risk to earth is calculated using a variety of methods including the Palermo and Torino scales. Note how far out they calculate the risk.

Despite all this watching, and almost none of this existed in such rigorous form just ten years ago, the biggest risk is actually from the objects we don't yet know about -- the comets and space rocks that will fall in from the Kuiper Belt or farther out and come at us fast. Most asteroids are

posted by dhartung at 10:32 PM on May 3, 2010

As the Minor Planet Center is an official arm of the International Astronomical Union, it is considered authoritative. Its director, Brian Marsden^, is arguably the most experienced person on Earth at calculating the risk of asteroid impacts. Additionally, NASA separately operates the Near Earth Object Program, which is tasked with tracking and warning of near-earth asteroid impact risks. They operate a computer system which keeps located large objects in a database that is constantly updated. Note that the smaller the object, the more likely that an encounter with a larger object such as the Moon will introduce a deviation into its orbit. The impact risk to earth is calculated using a variety of methods including the Palermo and Torino scales. Note how far out they calculate the risk.

Despite all this watching, and almost none of this existed in such rigorous form just ten years ago, the biggest risk is actually from the objects we don't yet know about -- the comets and space rocks that will fall in from the Kuiper Belt or farther out and come at us fast. Most asteroids are

*generally*spinning around the Sun in the same direction as Earth and within a margin of Earth's orbital speed. Those coming in from the outer solar system, though, will come at us from oblique angles with a lot of kinetic energy.posted by dhartung at 10:32 PM on May 3, 2010

well...(as an aside, here), you're not going to be naming any astonomical bodies after your cat. if you discover an asteroid, you'll be invited to

also, if it's a comet you discover, you won't even get that privelege...it will be named after

posted by sexyrobot at 12:27 AM on May 4, 2010

*suggest*a name to the IAU, maybe they'll approve it, maybe they won't...but 'names of pet animals are discouraged'.also, if it's a comet you discover, you won't even get that privelege...it will be named after

*you*, though, and the next one or two folks that confirm the discovery (in general...there are a few exceptions). for example: Comet Shoemaker-Levy 9 (9 b/c it was the 9th comet discovered by that team...) (this comet collided with jupiter in 1994, as predicted by brian marsden...spectacular images in link)posted by sexyrobot at 12:27 AM on May 4, 2010

Response by poster: Okay, so I would take my observations, type them out in the format listed here, submit them here to generate a file I could put in software like this to calculate orbital elements for myself. Gotcha.

posted by Jimbob at 1:03 AM on May 4, 2010

posted by Jimbob at 1:03 AM on May 4, 2010

I did this by hand many years ago - in a high school program called SSP. The minimum number of observations is three but, yes, in practice, a few more are required in order to dampen any errors in observation.

Once you have repeated observations you can calculate the orbital elements which are the six numbers which uniquely define an orbit. My object, by the way, was the comet Giacobini-Zinner.

Any observations, even of a known object, are valued since they help add to the data of what is known as well as help track any changes that the orbit may have undergone.

posted by vacapinta at 1:55 AM on May 4, 2010

Once you have repeated observations you can calculate the orbital elements which are the six numbers which uniquely define an orbit. My object, by the way, was the comet Giacobini-Zinner.

Any observations, even of a known object, are valued since they help add to the data of what is known as well as help track any changes that the orbit may have undergone.

posted by vacapinta at 1:55 AM on May 4, 2010

*Is this enough to calculate the orbit and future position of the asteroid, and whether it will eventually crash into Jakarta? Is calculating the orbit simply a matter of putting these coordinates into some software (which software?) or is it something some experts have to sweat over? Is more information needed, for example an estimation of the mass of the object?*

I actually do something similar to this as a day job.

Yes, you just need 3 obs (of the kind you describe) to make an orbit. But you can't predict much from that and not just because of errors. For instance, a really bright, low mass object might experience more solar radiation pressure and so the orbit isn't predictable from gravity alone. For this, you'd need not just more obs but also to track it better over time. Or if there's outgassing that's basically like a rocket thrust.

For something big, heavy and relatively dark like an asteroid, a few well-spaced obs is probably fine for a really long prediction. You won't have to give an estimation of the mass, the software can figure it out for you (or the area/mass ratio or something similar).

posted by DU at 4:50 AM on May 4, 2010

This thread is closed to new comments.

Or are you asking how to determine if the plane of one ellipse (an asteroid) will intersect the plane of another ellipse (the Earth's orbit) and hit that second object? That's a much tougher mathematical calculation.

The Wikipedia entry on asteroid mitigation has some information about this second question: http://en.wikipedia.org/wiki/Asteroid_mitigation_strategies

posted by dfriedman at 5:13 PM on May 3, 2010