<p>Such an object has a blackbody luminosity of 0.0014 watts, emitted mainly in the microwave region of the spectrum (and emits almost nothing at optical and UV wavelengths).</p>
<p>At a distance of <span class="math-container">$10^6$</span> km, the flux received at Earth would be <span class="math-container">$10^{-22}$</span> Watts per square metre and the object would subtend a solid angle of <span class="math-container">$6\times 10^{-18}$</span> steradians on the sky.</p>
<p>Your big problem in observing such an object is that is is almost exactly the same temperature as the cosmic microwave background and so you do need to (almost) resolve it to see it. Something like the <a href="https://en.wikipedia.org/wiki/Event_Horizon_Telescope" rel="nofollow noreferrer">Event Horizon Telescope</a> (a network of radio telescopes spread across the globe), that has enough angular resolution to resolve the object (it will have a diameter of 2 milli-arcseconds) and good sensitivity in the microwave region. This might just be able to pick out a small spot that is slightly warmer than the cosmic microwave background.</p>
<p>The alternative is to look for it occulting a background object, which is easily possible if the suspected position is known.</p>
<p>In that case, the telescope that you need for the job is a telescope that can observe the background object and therefore judge if it has been occulted or not—which would be judged by a dimming of the light received.
So this could be an optical or infrared telescope for example.</p>
<p>There isn't any need to be able to resolve the 10-m foreground object in this case—in fact, this technique can be used to estimate the size of the occulting object (see for example <a href="https://ui.adsabs.harvard.edu/abs/2015MNRAS.446..932L/abstract" rel="nofollow noreferrer">Liu et al. 2015</a>). Given the typical angular size of a star—the stellar images would have an equivalent size of less than a metre at <span class="math-container">$10^6$</span> km, so the occultations could be total.  The only reason you might need a large telescope is that you are far more likely to see occultations of the more numerous faint stars, and the occultations are likely to be very rapid, so that very short exposure times would be needed.</p>

CC BY-SA 4.0
2023-10-01T17:06:55.077Z
Based on https://www.hnsky.org/star_count the chance it occludes a known magnitude 16 star is only 1 in 2e-9. So we're looking at it occluding very dim stars. Extrapolating based on the chart (each magnitude has 2.2-2.7 times the stars per degree^2 as the previous magnitude), we could expect it to occlude some magnitude 32-37 stars. So how long would you have to point the Event Horizon Telescope at it before you could confirm the occlusions?
causative
CC BY-SA 4.0
2023-10-01T17:43:48.087Z
@causative the EHT would be used to directly image the object. I have no idea about how feasible that would be I'm afraid. The chance it will occult a star of a given magnitude increases with time, it is not a fixed number, since the object will move with respect to the background stars.
ProfRob
CC BY-SA 4.0
2023-10-01T17:49:50.690Z
As I understand it, the EHT has difficulty with objects that are rapidly changing.  That's why the Milky way's black hole was so much harder to image than M51.  I'd imagine that using such an interferometer to image a rapidly moving asteroid would be ... challenging.
James K
CC BY-SA 4.0
2023-10-01T19:03:37.920Z
@JamesK the object isn't rapidly changing. The telescopes have to slew to follow M87 and the GC. Usually not a problem to change the tracking rate for a predictable solar system object.
ProfRob
CC BY-SA 4.0
2023-10-01T14:00:00.733Z
2023-10-02T06:41:18.410Z
2023-10-02T06:41:18.410Z
ProfRob
<p>A radar telescope.</p>
<p>One million km is close (in space terms that's just over a <a href="https://www.youtube.com/watch?v=Gd6aLnPHqeE" rel="noreferrer">tad</a>) and well within the range of planetary radars.  As an example, the [asteroid 2021 PJ1 was detected by radar using the 70 m &quot;<a href="https://phys.org/news/2021-09-planetary-radar-1000th-near-earth-asteroid.html" rel="noreferrer">Deep Space Station 14&quot; in California</a>.  It is about 20 m across and was detected at a range of about 1.7 million km.  Since the radar signal scales inversely with the fourth power of the distance, but only with the square of the diameter, a 10 m asteroid will be detectable at 1 million km.</p>
<p>You won't get a nice picture (it's too small for that), but the radar will tell you the exact position and velocity of the asteroid, and so allow you to plot its future trajectory.</p>
<p>Other methods of detection are too haphazard.  You've said that it is occluded or reflects no sunlight. It is far too cold to emit enough radiation to pick out against the background, and the chance of a stellar occultation is too low. Of course, it will eventually cover a star, but how long have you got? There just aren't enough stars!  So radar is the way to go.</p>

CC BY-SA 4.0
2023-10-01T18:27:33.363Z
Detection by planetary radar only works if you know the asteroid's position to better than the beam width of a few arcminutes - it couldn't be used to survey for an unknown object at that distance.
astrosnapper
CC BY-SA 4.0
2023-10-01T18:47:16.387Z
The OP says that the exact position is known, "you suspect there's something there and you can point your telescope right at it and you only want to confirm"  I grant that that's not ultrarealistic, but then nor is an albedo=0 object.
James K
CC BY-SA 4.0
2023-10-01T23:18:41.967Z
Wouldn't an albedo of zero mean it wouldn't reflect the radar pulse either, though?
Hearth
CC BY-SA 4.0
2023-10-01T23:31:37.003Z
Good answer. What was the energy of the outgoing radar pulse? Did it need a really big transmitter too? Wikipedia article says that Goldstone has an 80kW transmitter, is that would they would have used for the radar pulse, or something else? @Hearth Yes but I wasn't really specific about that.
causative
CC BY-SA 4.0
2023-10-02T00:03:15.963Z
So, Arecibo could detect OP's suggested object, if it weren't damaged ?
Criggie
CC BY-SA 4.0
2023-10-02T02:24:48.610Z
@Hearth:  albedo refers to the reflectivity in the visible band.  The reflectivity can vary greatly with frequency.  If the albedo is zero all incoming visible light is absorbed and the energy will be radiated at longer wavelengths, infrared or microwave depending on its temperature.  My vague recollection is that at the earth's distance from the sun the temperature is somewhat above 200K but below 273K
Ross Millikan
CC BY-SA 4.0
2023-10-03T00:31:00.703Z
@Criggie and OP: Goldstone's DSS-14 planetary radar transmitter is ~450kW when all the klystrons are operating properly (there have been issues in the past). Arecibo's radar was 1MW transmitted power when it was operating. The 80 kW figure is for talking to spacecraft and for the smaller 34-m antenna. Arecibo very likely would have been able to detect it, assuming it was in the right part of the sky; Arecibo wasn't very steerable beyond a limited amount by moving the receiving cabin above the (sunk in the ground) dish
astrosnapper
CC BY-SA 4.0
2023-10-01T17:44:36.170Z
2023-10-02T17:07:56.110Z
2023-10-02T17:07:56.110Z
James K
<p>There's a 10m diameter object 1 million km away from you, stationary with respect to you. The object is dim: it is at 3K and is not reflecting any light towards you. So you have to pick it out by the way it occludes background stars and galaxies.</p>
<p>What kind, and what size, of telescope would be able to achieve this?</p>
<p>I'm guessing that you would want a large ultraviolet telescope. Picking a UV telescope at random for comparison, <a href="https://asd.gsfc.nasa.gov/archive/galex/Documents/instrument_summary.html" rel="noreferrer">GALEX</a> had 25cm^2 effective area and an angular resolution of 6 arc seconds. This resolution would need to be at least 2900 times better to spot the object. So you need a UV telescope with an effective area of at least 25cm^2 * 2900^2 = 21025 m^2, corresponding to a radius of about 82 meters.</p>
<p>Is that about right? Or could you make do with a smaller telescope by increasing the exposure time, decreasing the field of view, or something else?</p>

CC BY-SA 4.0
2023-10-01T10:14:53.163Z
Nice question. But why would it gjave to be UV, based on the temperature it would be 996 nm in IR wavelength considering it is an ideal blackbody according to wiens law?
Arjun
CC BY-SA 4.0
2023-10-01T13:08:45.723Z
Is the aim to spot this object, or resolve it?  The "not reflecting any light" is problematic since real objects are never truly black. Even comets have an albedo of 0.05, and they are about as black as coal.  Next, do you know where this object is?  It is a lot easier to spot something when you know where to look for it - or are you asking about chance discovery?  I don't think any asteroid has ever been discovered by occupation of a star. Such events are to rare, too transient and too local.
James K
CC BY-SA 4.0
2023-10-01T15:48:53.307Z
@Arjun I figured you would want a UV telescope because the object is too dim to pick up on IR, and because angular resolution scales based on wavelength, so to achieve a given resolution a UV telescope can be smaller than an IR one. (Also wouldn't it be 966 micrometer light as a blackbody?)
causative
CC BY-SA 4.0
2023-10-01T17:23:10.850Z
@JamesK Let's make it easy and say that you suspect there's something there and you can point your telescope right at it and you only want to confirm. About reflection, just for the sake of the scenario you could say that it's in the shadow of something.
causative
CC BY-SA 4.0
2023-10-02T13:50:38.040Z
Can you explain what you mean by "stationary with respect to you"?  Because I would have thought if it's stationary then it can't occlude anything. It needs to be moving against a relatively stationary background in order to pick up the fact that a light just went out.
Darth Pseudonym
CC BY-SA 4.0
2023-10-02T16:06:05.937Z
@DarthPseudonym You know that a light went out if you have a good enough picture of what that section of sky looked like before the object was there. I mean, it wasn't always there, and it might be moving slightly, but it's close enough to stationary to make it easier to image.
causative
CC BY-SA 4.0
2023-10-02T17:33:57.350Z
@causative Yes, that is correct. But one very essential factor influencing the frequency range of the telescope, is because the object is even dimmer at the UV range, for comparison (I computed the spectral density per wavelength of an ideal blackbody using Planck's law), while in IR range the spectral radiance is almost a 100 flicks, in the UV range it is just 0.06 flicks. Thank you!
Arjun
CC BY-SA 4.0
2023-10-02T18:10:33.990Z
@Arjun Well, yeah, I'm saying it's basically invisible on both IR and UV, but the stars it is occulting are visible on UV and a UV telescope to spot that occultation would be smaller than an IR telescope to spot the IR occultation.
causative
<telescope><asteroids><space-telescope>
What kind of telescope would be needed to image a 10m dim object 1 million km away?
causative
# What kind of telescope would be needed to image a 10m dim object 1 million km away?

(Asked by: username_0 on 2023-10-01)

There's a 10m diameter object 1 million km away from you, stationary with respect to you. The object is dim: it is at 3K and is not reflecting any light towards you. So you have to pick it out by the way it occludes background stars and galaxies.

What kind, and what size, of telescope would be able to achieve this?

I'm guessing that you would want a large ultraviolet telescope. Picking a UV telescope at random for comparison, [GALEX](https://asd.gsfc.nasa.gov/archive/galex/Documents/instrument_summary.html) had 25cm^2 effective area and an angular resolution of 6 arc seconds. This resolution would need to be at least 2900 times better to spot the object. So you need a UV telescope with an effective area of at least 25cm^2 \* 2900^2 = 21025 m^2, corresponding to a radius of about 82 meters.

Is that about right? Or could you make do with a smaller telescope by increasing the exposure time, decreasing the field of view, or something else?

**Tags:** telescope, asteroids, space-telescope
> **Comments:**
> **username_1**: Nice question. But why would it gjave to be UV, based on the temperature it would be 996 nm in IR wavelength considering it is an ideal blackbody according to wiens law?  
> **username_2**: Is the aim to spot this object, or resolve it?  The "not reflecting any light" is problematic since real objects are never truly black. Even comets have an albedo of 0.05, and they are about as black as coal.  Next, do you know where this object is?  It is a lot easier to spot something when you know where to look for it - or are you asking about chance discovery?  I don't think any asteroid has ever been discovered by occupation of a star. Such events are to rare, too transient and too local.  
> **username_0**: @Arjun I figured you would want a UV telescope because the object is too dim to pick up on IR, and because angular resolution scales based on wavelength, so to achieve a given resolution a UV telescope can be smaller than an IR one. (Also wouldn't it be 966 micrometer light as a blackbody?)  
> **username_0**: @JamesK Let's make it easy and say that you suspect there's something there and you can point your telescope right at it and you only want to confirm. About reflection, just for the sake of the scenario you could say that it's in the shadow of something.  
> **username_3**: Can you explain what you mean by "stationary with respect to you"?  Because I would have thought if it's stationary then it can't occlude anything. It needs to be moving against a relatively stationary background in order to pick up the fact that a light just went out.  
> **username_0**: @DarthPseudonym You know that a light went out if you have a good enough picture of what that section of sky looked like before the object was there. I mean, it wasn't always there, and it might be moving slightly, but it's close enough to stationary to make it easier to image.  
> **username_1**: @causative Yes, that is correct. But one very essential factor influencing the frequency range of the telescope, is because the object is even dimmer at the UV range, for comparison (I computed the spectral density per wavelength of an ideal blackbody using Planck's law), while in IR range the spectral radiance is almost a 100 flicks, in the UV range it is just 0.06 flicks. Thank you!  
> **username_0**: @Arjun Well, yeah, I'm saying it's basically invisible on both IR and UV, but the stars it is occulting are visible on UV and a UV telescope to spot that occultation would be smaller than an IR telescope to spot the IR occultation.  

---

## 2 answers

### Answer by username_2 on 2023-10-01. Score: 15 | Accepted answer

A radar telescope.

One million km is close (in space terms that's just over a [tad](https://www.youtube.com/watch?v=Gd6aLnPHqeE)) and well within the range of planetary radars. As an example, the [asteroid 2021 PJ1 was detected by radar using the 70 m "[Deep Space Station 14" in California](https://phys.org/news/2021-09-planetary-radar-1000th-near-earth-asteroid.html). It is about 20 m across and was detected at a range of about 1.7 million km. Since the radar signal scales inversely with the fourth power of the distance, but only with the square of the diameter, a 10 m asteroid will be detectable at 1 million km.

You won't get a nice picture (it's too small for that), but the radar will tell you the exact position and velocity of the asteroid, and so allow you to plot its future trajectory.

Other methods of detection are too haphazard. You've said that it is occluded or reflects no sunlight. It is far too cold to emit enough radiation to pick out against the background, and the chance of a stellar occultation is too low. Of course, it will eventually cover a star, but how long have you got? There just aren't enough stars! So radar is the way to go.

> **Comments:**
> **username_5**: Detection by planetary radar only works if you know the asteroid's position to better than the beam width of a few arcminutes - it couldn't be used to survey for an unknown object at that distance.  
> **username_2**: The OP says that the exact position is known, "you suspect there's something there and you can point your telescope right at it and you only want to confirm"  I grant that that's not ultrarealistic, but then nor is an albedo=0 object.  
> **username_6**: Wouldn't an albedo of zero mean it wouldn't reflect the radar pulse either, though?  
> **username_0**: Good answer. What was the energy of the outgoing radar pulse? Did it need a really big transmitter too? Wikipedia article says that Goldstone has an 80kW transmitter, is that would they would have used for the radar pulse, or something else? @Hearth Yes but I wasn't really specific about that.  
> **username_7**: So, Arecibo could detect OP's suggested object, if it weren't damaged ?  
> **username_8**: @Hearth:  albedo refers to the reflectivity in the visible band.  The reflectivity can vary greatly with frequency.  If the albedo is zero all incoming visible light is absorbed and the energy will be radiated at longer wavelengths, infrared or microwave depending on its temperature.  My vague recollection is that at the earth's distance from the sun the temperature is somewhat above 200K but below 273K  
> **username_5**: @Criggie and OP: Goldstone's DSS-14 planetary radar transmitter is ~450kW when all the klystrons are operating properly (there have been issues in the past). Arecibo's radar was 1MW transmitted power when it was operating. The 80 kW figure is for talking to spacecraft and for the smaller 34-m antenna. Arecibo very likely would have been able to detect it, assuming it was in the right part of the sky; Arecibo wasn't very steerable beyond a limited amount by moving the receiving cabin above the (sunk in the ground) dish  

---

### Answer by username_4 on 2023-10-01. Score: 9

Such an object has a blackbody luminosity of 0.0014 watts, emitted mainly in the microwave region of the spectrum (and emits almost nothing at optical and UV wavelengths).

At a distance of $10^6$ km, the flux received at Earth would be $10^{-22}$ Watts per square metre and the object would subtend a solid angle of $6\times 10^{-18}$ steradians on the sky.

Your big problem in observing such an object is that is is almost exactly the same temperature as the cosmic microwave background and so you do need to (almost) resolve it to see it. Something like the [Event Horizon Telescope](https://en.wikipedia.org/wiki/Event_Horizon_Telescope) (a network of radio telescopes spread across the globe), that has enough angular resolution to resolve the object (it will have a diameter of 2 milli-arcseconds) and good sensitivity in the microwave region. This might just be able to pick out a small spot that is slightly warmer than the cosmic microwave background.

The alternative is to look for it occulting a background object, which is easily possible if the suspected position is known.

In that case, the telescope that you need for the job is a telescope that can observe the background object and therefore judge if it has been occulted or not—which would be judged by a dimming of the light received.
So this could be an optical or infrared telescope for example.

There isn't any need to be able to resolve the 10-m foreground object in this case—in fact, this technique can be used to estimate the size of the occulting object (see for example [Liu et al. 2015](https://ui.adsabs.harvard.edu/abs/2015MNRAS.446..932L/abstract)). Given the typical angular size of a star—the stellar images would have an equivalent size of less than a metre at $10^6$ km, so the occultations could be total. The only reason you might need a large telescope is that you are far more likely to see occultations of the more numerous faint stars, and the occultations are likely to be very rapid, so that very short exposure times would be needed.

> **Comments:**
> **username_0**: Based on https://www.hnsky.org/star_count the chance it occludes a known magnitude 16 star is only 1 in 2e-9. So we're looking at it occluding very dim stars. Extrapolating based on the chart (each magnitude has 2.2-2.7 times the stars per degree^2 as the previous magnitude), we could expect it to occlude some magnitude 32-37 stars. So how long would you have to point the Event Horizon Telescope at it before you could confirm the occlusions?  
> **username_4**: @causative the EHT would be used to directly image the object. I have no idea about how feasible that would be I'm afraid. The chance it will occult a star of a given magnitude increases with time, it is not a fixed number, since the object will move with respect to the background stars.  
> **username_2**: As I understand it, the EHT has difficulty with objects that are rapidly changing.  That's why the Milky way's black hole was so much harder to image than M51.  I'd imagine that using such an interferometer to image a rapidly moving asteroid would be ... challenging.  
> **username_4**: @JamesK the object isn't rapidly changing. The telescopes have to slew to follow M87 and the GC. Usually not a problem to change the tracking rate for a predictable solar system object.  

---

