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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

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

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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 May 14 '14 at 9:30
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"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 May 23 '14 at 1:29

4 Answers 4

up vote 8 down vote accepted

Here's an answer I wrote for a question on Space.SE, but which applies equally well here. Let's talk about the Hubble space telescope, which would be much better at observing these comets than any ground telescope:


From astroengine.com

Using the equation: (d / D) × c = φ

where d is the diameter of the Oort Cloud comet (some estimates put this number at an upper limit of 300 km for the diameter of a cometary nucleus), $D$ is the distance from the Oort Cloud to Hubble (0.3 light years, or 3×1015 metres – distance at which it is theorized there is the highest density of Oort Cloud objects), c is a constant (c = 206265) and φ is the telescope resolution.

So what resolution do we need to image an Oort Cloud object, 300 km in diameter, from 0.3 light years away? If we plug in the numbers we get:

φ = 2.06×10-5 = 0.00002 arc-seconds

The resolving power of Hubble is 0.1 arc-seconds, and is therefore useless at detecting anything below this angular size; Oort Cloud comets (although pretty big at an upper limit of 300 km) simply cannot be observed by the world’s most advanced space-based optical observatory.

How big would a telescope have to be? Well, from the same article:

But how big would an Oort Cloud observing telescope have to be to resolve a cometary nucleus 300 km wide at a distance of 0.3 light years away? Using the simple relationship R = 11.6 / w, where R is the resolving power (R = 0.00002/2; the reason for halving our resolving power is given by Phil), and w is the width of the telescope mirror, we rearrange to get:

w = 11.6 / 0.00001 = 1.16×106 cm = 11.6 km

As you can see, such a telescope would be huge.

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It seems that this new telescope might be able to observe the Oort Cloud. –  called2voyage Dec 12 '13 at 17:33
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Why do we need to resolve Oort cloud objects to prove their existence? The statement about the HST is wrong. All stars have angular sizes smaller than this but I think we can safely say there are detected. The entire premise of this answer is incorrect. –  Rob Jeffries Jan 26 at 2:26
    
Lifted verbatim from an unreliable, but copyrighted, source –  Rob Jeffries Feb 15 at 19:46

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.

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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 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 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).

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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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