<p>I had a chat with a European PhD student who plans to make an attempt to find Oort cloud objects in data from the Gaia space telescope. This could be possible thanks to microlensing events when an Oort object transits (near) a background star and relativistically magnifies the star's light for a moment.</p>

<p>Best case is that in a few years we will have a map of a statistically useful number of Oort cloud objects. Enough to claim that we have "seen" it.</p>

CC BY-SA 4.0
2020-09-23T15:49:42.363Z
That an interesting idea, microlensing is a hot topic, I don't suppose there's any news of results yet?
uhoh
CC BY-SA 3.0
2014-05-14T09:19:54.323Z
2014-05-14T09:19:54.323Z
LocalFluff
<p>Put simply, it is because the objects that make up the Oort Cloud, remnants from the formation of our sun, are both too small and too faint for us to detect. They are faint because of their vast distance away. There is minimal absorption of the sun's light and even less is reflected back. So little light is reflected back that there is nothing for even our most advanced telescopes to see. </p>

CC BY-SA 3.0
2015-04-02T00:29:39.477Z
2015-04-02T00:29:39.477Z
PJS1987
<p>The angular resolution of the telescope <em>really has no direct bearing on our ability to detect Oort cloud objects</em> beyond how that angular resolution affects the depth to which one can detect the light from faint objects. Any telescope can detect stars, even though their actual discs are way beyond the angular resolution of the telescope.</p>

<p>The detection of Oort cloud objects is simply a question of detecting the (unresolved) reflected light in exactly the same way that one detects a faint (unresolved) star. Confirmation of the Oort cloud nature of the object would then come by observing at intervals over a year or so and obtaining a very large ($&gt;2$ arcseconds) parallax.</p>

<p>The question amounts to how deep do you need to go? We can do this in two ways (i) a back of the envelope calculation assuming the object reflects light from the Sun with some albedo. (ii) Scale the brightness of comets when they are distant from the Sun.</p>

<p>(i) The luminosity of the Sun is $L=3.83\times10^{26}\ W$. Let the distance to the Oort cloud be $D$ and the radius of the (assumed spherical) Oort object be $R$.
The light from the Sun incident on the object is $\pi R^2 L/4\pi D^2$.
If we now assume that a fraction $f$ of this is reflected uniformly into a $2\pi$ solid angle. This latter point is an approximation, the light will not be reflected isotropically, but it will represent some average over any viewing angle.</p>

<p>To a good approximation, as $D \gg 1$ au, we can assume that the distance from the Oort object to the Earth is also $D$. Hence the flux of light received at the Earth is
$$F_{E} = f \frac{\pi R^2 L}{4\pi D^2}\frac{1}{2\pi D^2} = f \frac{R^2 L}{8\pi D^4}$$</p>

<p>Putting some numbers in, let $R=10$ km and let $D= 10,000$ au. Cometary material has a very low albedo, but let's be generous and assume $f=0.1$.
$$ F_E = 3\times10^{-29}\left(\frac{f}{0.1}\right) \left(\frac{R}{10\ km}\right)^2 \left(\frac{D}{10^4 au}\right)^{-4}\ Wm^{-2}$$</p>

<p>To convert this to a magnitude, assume the reflected light has the same spectrum as sunlight. The Sun has an apparent visual magnitude of -26.74, corresponding to a flux at the Earth of $1.4\times10^{3}\ Wm^{-2}$. Converting the flux ratio to a magnitude difference, we find that the apparent magnitude of our fiducial Oort object is <strong>52.4</strong>.</p>

<p>(ii) Halley's comet is similar (10 km radius, low albedo) to the fiducial Oort object considered above. Halley's comet was observed by the <a href="http://www.eso.org/~ohainaut/nice/phot-27-03.html" rel="noreferrer">VLT in 2003</a> with a magnitude of 28.2 and at a distance of 28 au from the Sun. We can now just scale this magnitude, but it scales as distance to the power of <em>four</em>, because the light must be received and then we see it reflected.
Thus at 10,000 au, Halley would have a magnitude of $28.2 - 2.5 \log (28/10^{4})= 53.7$, in reasonable agreement with my other estimate. (Incidentally my crude formula in (i) above suggests a $f=0.1$, $R=10\ km$ comet at 28 au would have a magnitude of 26.9. Given that Halley probably has a smaller $f$ this is excellent consistency.)</p>

<p>The observation of Halley by the VLT represents the pinnacle of what is possible with today's telescopes. Even the Hubble deep ultra deep field only reached visual magnitudes of about 29. Thus a big Oort cloud object remains <strong>more than 20 magnitudes below</strong> this detection threshold!</p>

<p>The most feasible way of detecting Oort objects is when they occult background stars. The possibilities for this are discussed by <a href="http://arxiv.org/abs/0912.0948" rel="noreferrer">Ofek &amp; Naker 2010</a> in the context of the photometric precision provided by Kepler. The rate of occultations (which are of course single events and unrepeatable) was calculated to be between zero and 100 in the whole Kepler mission, dependent on the size and distance distribution of the Oort objects. As far as I am aware, nothing has come of this (yet).</p>

CC BY-SA 4.0
2022-07-22T20:20:09.083Z
Would it be worth asking a new question on how JWST might change this? Or does it not really have much of an impact here?
called2voyage
CC BY-SA 4.0
2022-07-22T20:56:03.060Z
@called2voyage JWST might beat HST by a magnitude or two, but not 20 magnitudes.
ProfRob
CC BY-SA 3.0
2015-01-26T12:53:07.383Z
2015-02-21T18:39:56.010Z
2015-02-21T18:39:56.010Z
ProfRob
<p>The Oort cloud is a hypothetical structure based on our observation of long-period comets. There are currently proposals to design probes to confirm the existence of the Oort cloud.</p>

<p><img src="https://upload.wikimedia.org/wikipedia/commons/0/03/Kuiper_oort.jpg" alt="Oort cloud"></p>

<p>Now, sending a probe would have other benefits, but why can't we observe the Oort cloud with a telescope?</p>

CC BY-SA 3.0
2018-01-21T15:22:52.250Z
@Marc This is not true if we use a form of propulsion that gets there within 50 years.
gerrit
CC BY-SA 3.0
2018-01-21T15:49:58.317Z
Where will this form of propulsion come if we never take the first step of building the best we can today and using it? If we never make a serious effort, we'll always be waiting for the perfect solution.
Marc
CC BY-SA 3.0
2014-05-14T09:30:52.073Z
I think an Oort probe in our lifetime is unrealistic and actually irrational! The Oort cloud starts about 2000 AU out. It would take generations to arrive there with foreseable propulsion technology. Even if launched today, it might well be surpassed by a far superior probe 50 years later. And besides, where to go if no target has been observed by telescope before? The Oort cloud is a very empty space. I'd like to see one of those proposals for an Oort probe, because I don't understand how the concept could work.
LocalFluff
CC BY-SA 3.0
2014-05-23T01:29:50.777Z
*"Even if launched today, it might well be surpassed by a far superior probe 50 years later."*  This will always be true, and is an argument for doing nothing forever.
Marc
<observational-astronomy><oort-cloud>
Why can't we observe the Oort cloud with a telescope?
called2voyage
# Why can't we observe the Oort cloud with a telescope?

(Asked by: username_0 on 2013-09-26)

The Oort cloud is a hypothetical structure based on our observation of long-period comets. There are currently proposals to design probes to confirm the existence of the Oort cloud.

![Oort cloud](https://upload.wikimedia.org/wikipedia/commons/0/03/Kuiper_oort.jpg)

Now, sending a probe would have other benefits, but why can't we observe the Oort cloud with a telescope?

**Tags:** observational-astronomy, oort-cloud

> **Comments:**
> **username_3**: I think an Oort probe in our lifetime is unrealistic and actually irrational! The Oort cloud starts about 2000 AU out. It would take generations to arrive there with foreseable propulsion technology. Even if launched today, it might well be surpassed by a far superior probe 50 years later. And besides, where to go if no target has been observed by telescope before? The Oort cloud is a very empty space. I'd like to see one of those proposals for an Oort probe, because I don't understand how the concept could work.  
> **username_2**: *"Even if launched today, it might well be surpassed by a far superior probe 50 years later."*  This will always be true, and is an argument for doing nothing forever.  
> **username_1**: @username_2 This is not true if we use a form of propulsion that gets there within 50 years.  
> **username_2**: Where will this form of propulsion come if we never take the first step of building the best we can today and using it? If we never make a serious effort, we'll always be waiting for the perfect solution.  

---

## 3 answers

### Answer by username_6 on 2015-01-26. Score: 18 | Accepted answer

The angular resolution of the telescope *really has no direct bearing on our ability to detect Oort cloud objects* beyond how that angular resolution affects the depth to which one can detect the light from faint objects. Any telescope can detect stars, even though their actual discs are way beyond the angular resolution of the telescope.

The detection of Oort cloud objects is simply a question of detecting the (unresolved) reflected light in exactly the same way that one detects a faint (unresolved) star. Confirmation of the Oort cloud nature of the object would then come by observing at intervals over a year or so and obtaining a very large ($>2$ arcseconds) parallax.

The question amounts to how deep do you need to go? We can do this in two ways (i) a back of the envelope calculation assuming the object reflects light from the Sun with some albedo. (ii) Scale the brightness of comets when they are distant from the Sun.

(i) The luminosity of the Sun is $L=3.83\times10^{26}\ W$. Let the distance to the Oort cloud be $D$ and the radius of the (assumed spherical) Oort object be $R$.
The light from the Sun incident on the object is $\pi R^2 L/4\pi D^2$.
If we now assume that a fraction $f$ of this is reflected uniformly into a $2\pi$ solid angle. This latter point is an approximation, the light will not be reflected isotropically, but it will represent some average over any viewing angle.

To a good approximation, as $D \gg 1$ au, we can assume that the distance from the Oort object to the Earth is also $D$. Hence the flux of light received at the Earth is
$$F\_{E} = f \frac{\pi R^2 L}{4\pi D^2}\frac{1}{2\pi D^2} = f \frac{R^2 L}{8\pi D^4}$$

Putting some numbers in, let $R=10$ km and let $D= 10,000$ au. Cometary material has a very low albedo, but let's be generous and assume $f=0.1$.
$$ F\_E = 3\times10^{-29}\left(\frac{f}{0.1}\right) \left(\frac{R}{10\ km}\right)^2 \left(\frac{D}{10^4 au}\right)^{-4}\ Wm^{-2}$$

To convert this to a magnitude, assume the reflected light has the same spectrum as sunlight. The Sun has an apparent visual magnitude of -26.74, corresponding to a flux at the Earth of $1.4\times10^{3}\ Wm^{-2}$. Converting the flux ratio to a magnitude difference, we find that the apparent magnitude of our fiducial Oort object is **52.4**.

(ii) Halley's comet is similar (10 km radius, low albedo) to the fiducial Oort object considered above. Halley's comet was observed by the [VLT in 2003](http://www.eso.org/~ohainaut/nice/phot-27-03.html) with a magnitude of 28.2 and at a distance of 28 au from the Sun. We can now just scale this magnitude, but it scales as distance to the power of *four*, because the light must be received and then we see it reflected.
Thus at 10,000 au, Halley would have a magnitude of $28.2 - 2.5 \log (28/10^{4})= 53.7$, in reasonable agreement with my other estimate. (Incidentally my crude formula in (i) above suggests a $f=0.1$, $R=10\ km$ comet at 28 au would have a magnitude of 26.9. Given that Halley probably has a smaller $f$ this is excellent consistency.)

The observation of Halley by the VLT represents the pinnacle of what is possible with today's telescopes. Even the Hubble deep ultra deep field only reached visual magnitudes of about 29. Thus a big Oort cloud object remains **more than 20 magnitudes below** this detection threshold!

The most feasible way of detecting Oort objects is when they occult background stars. The possibilities for this are discussed by [Ofek & Naker 2010](http://arxiv.org/abs/0912.0948) in the context of the photometric precision provided by Kepler. The rate of occultations (which are of course single events and unrepeatable) was calculated to be between zero and 100 in the whole Kepler mission, dependent on the size and distance distribution of the Oort objects. As far as I am aware, nothing has come of this (yet).

> **Comments:**
> **username_0**: Would it be worth asking a new question on how JWST might change this? Or does it not really have much of an impact here?  
> **username_6**: @username_0 JWST might beat HST by a magnitude or two, but not 20 magnitudes.  

---

### Answer by username_3 on 2014-05-14. Score: 15

I had a chat with a European PhD student who plans to make an attempt to find Oort cloud objects in data from the Gaia space telescope. This could be possible thanks to microlensing events when an Oort object transits (near) a background star and relativistically magnifies the star's light for a moment.

Best case is that in a few years we will have a map of a statistically useful number of Oort cloud objects. Enough to claim that we have "seen" it.

> **Comments:**
> **username_4**: That an interesting idea, microlensing is a hot topic, I don't suppose there's any news of results yet?  

---

### Answer by username_5 on 2015-04-02. Score: 0

Put simply, it is because the objects that make up the Oort Cloud, remnants from the formation of our sun, are both too small and too faint for us to detect. They are faint because of their vast distance away. There is minimal absorption of the sun's light and even less is reflected back. So little light is reflected back that there is nothing for even our most advanced telescopes to see.


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