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Radar reveals two moons orbiting asteroid Florence
- jaquers 9y agoIs the 'light' reflecting off the asteroid emitted from the sun? I had never seen or even knew radar could distinguish value like what is shown here. I suppose terrestrial radar looks at much smaller objects and what it sees is emitted from the tower and bounced back in a straight line so that sort of makes sense. More info on the observatory: https://en.wikipedia.org/wiki/Goldstone_Solar_System_Radar https://en.wikipedia.org/wiki/Goldstone_Solar_System_Radar
- benjoffe 9y ago> Is the 'light' reflecting off the asteroid emitted from the sun? I had never seen or even knew radar could distinguish value like what is shown here. Yes it's emitted from the sun, it's clear given that the time-lapse shows "Inner moon eclipsed" when passing behind the site of the asteroid that is not lit.
- Teever 9y agoI don't understand how the phrase 'inner moon eclipsed' tells the reader that the source of radiation for this radar measurement is the sun.
- jaquers 9y agoI don't either :) but sort of a silly question I guess. I presumed it was the sun due to the angle of illumination as well as common sense that the sun is probably the brightest object in the solar system along multiple bands of light; I was really just amazed at the relatively large amount of detail in the image.
- devrandomguy 9y agoWell, if the illumination was from the radio telescope, then the moon would be hidden behind the parent asteroid when it is eclipsed. We would not see it go dark, it would just be obscured.
- azernik 9y agoThat's just an artifact of the visualization; the point of view is above the north pole of Florence, which points decidedly away from Earth (probably +/- a few degrees from normal to the plane of the ecliptic).
- Lordsandwhich 9y agoI was always under the assumption that with Radar images that it didn't rely on a light source as technically the radar is the source. So that the impression that there is a light an dark side was based purely on the motion towards and away from the receiver. But, you have pointed out something interesting and that would appear to indicate other wise. The clear rotation independent of source already shows that I am wrong on that. Apparently I don't understand this very well.
- RachelF 9y agoYou are correct, the "light" is the transmitted radar pulse from Goldstone, not from the sun. They do all sorts of fancy signal processing to get this sort of resolution. Because radar power received goes down with 1/(distance^4), the inverse square law, squared, this is hard to do at astronomical distances. The power level differential between transmitted and received power can be in the order of ~10^15. Goldstone transmits a 500kW radio pulse but will probably get nanowatts back, which is amplified millions of times by the big dish, amplifiers and signal processing.
- radicalbyte 9y agoThis thread is very confused. There are two things: the radar images are created with radio waves emitted by the radar installation. The second thing is that the visualisation shows a light source. They are unrelated to each other. I assume that the light source was added during rendering of the film and that it approximates the sun.
- TorKlingberg 9y agoI don't think the visualisation shows any light source. The radar reflection from the asteroid just looks like it is "lit up"
- ceejayoz 9y agoThe radar is the light source, just like shining a flashlight during the night at something. The actual solar illumination of the asteroid at the time was likely significantly different from what we see in these images.
- InclinedPlane 9y agoYes. The radar illumination comes from the Sun.
- db48x 9y agoNo, that's not showing light reflected from the sun. In the still image, the radar source (aka Earth) is to the left; in the animation it's at the top. In the still image, the vertical dimension is the bearing of the transmitted radar signal, while the horizontal dimension is distance away from Earth, as computed by the time of flight of the radar signals. In the animation those are swapped.
- verytrivial 9y agoPhil Plait to the rescue! http://www.syfy.com/syfywire/close-encounter-by-the-big-asteroid-florence http://www.syfy.com/syfywire/close-encounter-by-the-big-aste... Edit: I think it might be helpful to view this more like ultrasound image than a photograph. There is no "side" view. It has been reconstructed based upon the time it takes for the bounce to return.
- wiredfool 9y agoHere's a good explainer on what you're actually seeing in the radar image: http://www.planetary.org/blogs/emily-lakdawalla/2011/3248.html http://www.planetary.org/blogs/emily-lakdawalla/2011/3248.ht... * The signal is emitted from a radar installation on the earth, and reflected back to another dish. * Brightness is reflected signal strength. So surfaces that are perpendicular to us will reflect better. The less the surface normal points to us, the less reflection we see. * Up/Down is Time axis. The points closer to us will return sooner than points farther away. * Left/right is Doppler shift. Parts of the asteroid that are rotating away from us will show up on one side, parts rotating towards us will show up on the other side. (this is the vertically oriented image).
- jaquers 9y agoThank you very much! Awesome explanation.
- SCAQTony 9y agoImagine hijacking a 100-meter moon made of rare earth metals like gold, platinum, rhodium, ruthenium, iridium, etc. etc. More than enough to under write the mission and pay for a moon base.
- Terr_ 9y agoUnless you did it very secretly and slowly you'd might crash the market though.
- devrandomguy 9y agoImagine heating and spinning it into a thin foil disk, and then curving it into a slight parabola by gently accelerating the center point away from the sun. Solar-cooker death-ray for anyone who disputes your claim!
- DrScump 9y agoThe trick is getting it through the atmosphere intact.
- SCAQTony 9y agoThe Apollo module was 20,000-kg. We could do it in 50,000-lbs chunks but we would have to process it on the moon first so it wouldn't crater when it landed. If the commodities market tanks, well, we'll value add it by selling it as "all weather platinum or gold siding" at Home Depot. ;-) Apollo weight info: https://www.google.com/search?q=apollo+commad+module+weight&oq=apollo+commad+module+weight&aqs=chrome..69i57j0l2.6655j0j4&sourceid=chrome&ie=UTF-8 https://www.google.com/search?q=apollo+commad+module+weight&...
- Gravityloss 9y agoYou can just make a flattish shape out of it and most likely it will enter just fine.
- DrScump 9y agoYour link says 5,500 kg for the Command module, which is the only part that re-entered. And of that 5,500kg, maybe 300-400kg was actual payload (astronauts and samples).
- nomoon 9y agoThat's no moon.
- sgt 9y agoSo you created an account 'nomoon' simply to make this joke. I hope it was worth it.
- mukundmr 9y agoHow about we have satellites orbiting asteroids for a piggy back ride and then getting off the orbit when they are near a point of interest? My guess is the gravity would be too feeble.
- krastanov 9y agoYou would need to first match the asteroid's velocity, and by then you do not really need the asteroid. But something very similar to your idea is used constantly for interplanetary travel - gravity assist which uses other bodies' gravity to accelerate and change direction.
- abecedarius 9y agoKrastanov's answer is good, but it's worth adding that 'cyclers' are a similar but practical idea: https://en.m.wikipedia.org/wiki/Mars_cycler https://en.m.wikipedia.org/wiki/Mars_cycler
- hyperpallium 9y agoWhat's the surface gravity? https://en.wikipedia.org/wiki/3122_Florence https://en.wikipedia.org/wiki/3122_Florence Mass isn't given, but could estimate from volume, and typical asteroid density?
- bediger4000 9y agoHow do the orbits of the moons, and the asteroid's axis of rotation match up to the ecliptic? Do both moons orbit in the same direction? How eccentric are the orbits of the moons? What's the going theory on how such a small body can capture moons? It must not be too difficult, otherwise 3122 Florence wouldn't be the 3rd triplet spotted.