# Can we measure comets in a similar way to exoplanets?

A 2km comet at 40AU's can dim a sun sized star at 100 light years by 3%, and a 10km comet can cause an eclipse (is that right?) So would 2,3,10 km objects be easy to measure via star dimming if we had enough telescopic data from space?

Can we measure lines of stars that are eclipsed/dimmed in sequence by transiting comets?

Given 1000 frames of a small patch of space, where all the images are recorded into a 3D image stack (a 3D array, a voxel) X,Y and Time as dimensions, then an Intel processor can compare 20 million voxels per second of that collection of images to search for vector information of aligned patterns of star dimming, i.e. searching for vectors of comets.

If that process is repeated for many patches of space, perhaps we can detect many comets, and I don't know the real maths involved, if a satellite of that design would realistically only find 1 comet, or 5000?

It's for a theoretical comet apprehension system which uses 2/3 identical satellites each comprised 100Tb of SSD and a 20Mflops processors to search for vector information in the images.

Some kind of fairy idea going around in my head, because of Oumuamua and because I work on voxels and I know that image collections from space can be stacked into 3D arrays and be searched for vector patterns at 2million voxels per second on a 200W PC, an idea that I am confusedly astrally-projecting as a basis to find comets.

edit- A quantitative answer would be cool, say if the voids between stars on the milky way arm are 8 times the width of the comet, the comet would on average hit 2 stars every 8x8 frames.

• – rob Dec 15 '17 at 15:13

The thing about detecting exoplanets by eclipses is that the eclipses are repeatable. You have the same star, whose light curve dims in the same way with each eclipse, which gives you information about the planet's orbital period and its apparent size relative to the star. The Kepler mission observed many fluctuations in stellar brightness that were not associated with exoplanets, but instead with changes in the instrument, fluctuations in the actual brightness of the star, probably some other explanations that I can't think of off the top of my head, and actual mysteries.

Basically, if you can only observe it once, then you can't claim to explain it.

Occultations of distant stars by planets or comets don't give you a light curve: the star goes from "on" to "off". So a single occultation doesn't give you any information about the size, speed, or direction of the object doing the occulting. And the occultations are sufficiently rare that I think you'd be extremely unlikely to find a "line" of them by chance --- though that would make a fun data-mining project for an astronomy student.

What you can do, if you have an object that you already know about, is to make many observations of the same occulatation. That's the mission of the International Occultation Timing Association. If you click through to their web page you'll see this image:

Each line here shows the observed brightness of a particular star observed from a particular location at the same time. You can see clearly that some observers saw the star disappear very briefly, some for longer, and some (who might have been predicted to be in the very center of the path) saw no occultation at all. Assembling all these individual observations, properly synchronized, shows clearly the shadows of the two parts of a binary asteroid, each of order 50 km in diameter.

Note that 90 Antiope is a main-belt asteroid. The angular diameter decreases linearly as the distance increases, so an object the same physical size orbiting out by Neptune (perhaps fifteen times farther) would need fifteen times the density of observing locations to create an image with the same detail.

• That's very fantastic. So it's a relative thing. There can be some kind of graph which balances the statistical probability of an occultation versus the number of CMOS frames and observation times that are necessary to find 2km objects at far as Neptune. Today's SSD's of 15TB weigh 140g and take 1W at idle, 100W processors do 40,Giga comparison instructions per seconds, i.e. 4giga star brightness comparisons every second, so the processing power perhaps sufficient today, The limit is the quantum efficiency of the optics- can we voxel 50bn images? – aliential Dec 15 '17 at 5:45
• Homework for you: for all the $10^6$ known main-belt asteroids, compute (or look up) the number of occultations in a year of stars brighter than magnitude eight (or whatever cutoff you like) visible from somewhere in Earth. I think it is dozens. The Kuiper belt has much less density than the asteroid belt, and the objects have much smaller apparent size. – rob Dec 15 '17 at 12:37
• That's a steep assignment, for magnitude 20 and including 10-20% occultations perhaps it would be millions? I dont know where to start.For comets arriving from other stars from above the sun, i have to admit I'm hoping to get xenoDNA from comets from less crowded zones. – aliential Dec 15 '17 at 13:35
• Well, here are some lists that don't go down to magnitude 20 --- but those lists assume that you have the freedom to move your observing location anywhere on Earth at any time. For example, it looks like 284 Amalia has five predicted occultations between 2017-12-01 and 2018-01-31, but none visible from any particular location. Your project would require not only many observations to catch unpredicted occultations, but also many observing locations. – rob Dec 15 '17 at 14:45
• I'm speechless. what a brilliant resource. by travelling to a fixed place, there are over 2500 occultation events every year, averaging about 20 seconds, so at least 25k if they listed 2 seconds occultations, 250k if they measured 10% eclipses. Conversely, I'd divide that number by 50 for a stationary telescope. So 5k events every year at worst. Perhaps multiplied by 1-10k to account for 15-20 magnitude occlusions, it may be 5-10mn? – aliential Dec 15 '17 at 20:29

You should consider how fast a comet at 40 AU would move past a star. The visible arc of a sun sized star at 100 light years is about the same as a 9 km object at 40 AU. Objects at 40 AU orbit at about 4.7 km/s, so that would be a 2 second eclipse, if perfectly lined up, and more often than not it would be partially lined up, for even less time. A 2 second shift in brightness might be too short to get a good measurement on.

Related to your question, rogue planets are being surveyed by shading and/or by movement of the background star by gravitational lensing, but those partial shading or lensing require much larger, more massive objects and from our perspective, moving much slower across the sky. See Optical Gravitational Lensing Experiment or OGLE survey.

The 2 seconds of shading that a 40 AU object would give you in your scenario (or, for example, a 20 seconds of shading from a 4,000 AU object) would be more difficult to accurately identify. Mercury, for example, takes 8 hours to travel a distance equivalent to the diameter of the sun. 8 hours of temporary eclipse offers much more easily track-able data, especially given that it repeats on regular intervals every orbit.

• What about the ratio of the voids between stars at the arm of the milky way, it's perhaps only 8 times the width of a comet at the distance of Neptune? Then a comet would on average have converged on 2 stars every 8x8 = 64 frames. I don't have real astronomy maths programs on my pc, i don't know. my program has a limit at 10^14 zeros. – aliential Dec 15 '17 at 10:55
• I'm guessing the telescope is closer to the Sun and therefore moving a lot faster than the comet. However, the telescope could choose to point tangent to the orbit to zero this velocity out, or even at a slightly different angle to zero out both its orbital velocity and some average orbital velocity of the mid-Kuipter belt. So the duration of the occultation could actually be managed better than you describe. – uhoh Dec 15 '17 at 14:16
• @uhoh That's a good point. I hadn't considered the movement of the Earth. When it's nearly perpendicular it won't matter much. But when it's moving sideways to the comet passing over the star, that's worth taking into account and that throws my entire answer off by a fair bit. It's still generally true that 40 AU too close and they move across the sky to fast but my calculation needs to be recalculated. Good catch. – userLTK Dec 15 '17 at 19:01
• @com.prehensible Maybe I'm slow, but I've read your comment twice and I'm not 100% sure what you're asking. Perhaps word it as a new question? If I understand your question then I'm not sure how to calculate how often a comet would pass in-front of a star. There's so many stars with a wide variety of apparent sizes. I wouldn't know where to begin. – userLTK Dec 15 '17 at 19:47