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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. I suppose

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. 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 ago
  I 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 ago
    I 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 ago
    Well, 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 ago
      That'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 ago
  I 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 ago
    You 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 ago
    This 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 ago
      I don't think the visualisation shows any light source. The radar reflection from the asteroid just looks like it is "lit up"
    - ceejayoz 9y ago
      The 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 ago
Yes. The radar illumination comes from the Sun.
db48x 9y ago
No, 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 ago
Phil 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 ago
Here'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 ago
  Thank you very much! Awesome explanation.

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 ago
I 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 ago
  I 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 ago
  Well, 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 ago
    That'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 ago
I 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 ago
  You 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 ago
  This 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 ago
    I don't think the visualisation shows any light source. The radar reflection from the asteroid just looks like it is "lit up"
  - ceejayoz 9y ago
    The 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 ago

Yes. The radar illumination comes from the Sun.

db48x 9y ago

No, 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 ago

Phil 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 ago

Here'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 ago
Thank you very much! Awesome explanation.