# How could we observe the Oort cloud, if it exists?

The oort cloud has long been hypothesized as being around a light year away from the sun, but no observations have yet been made. Is it completely impractical with current science to make such an observation, or are there techniques we could try to use to observe it (and thus confirm or deny its existence?)

• An observation like, say, this? hubblesite.org/newscenter/archive/releases/2004/14/image/d
– Erik
Sep 17, 2013 at 20:02
• Dec 11, 2018 at 15:02
• "The oort cloud has long been hypothesized as being around a light year away from the sun..." just to clarify, it's thought to extend about this far but it's inner edge may only 0.05 light years away. Jan 23, 2019 at 8:37
• The correct answer to this question is here: astronomy.stackexchange.com/a/8685/6 Feb 28, 2020 at 16:34

Well, we can't observe it with the Hubble telescope. It's just too small.

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 $$3×10^{15}$$ 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:

$$φ = 2.06×10^{-5} = 0.00002~\text{arc-seconds}$$

The resolving power of Hubble is $$0.1~\text{arc-seconds}$$, 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:

$$w = 11.6 / 0.00001 = 1.16×10^6~\text{cm} = 11.6~\text{km}$$

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

• completely wrong... the resolution is about recognizing shapes, not about detecting things. the same calculation "proves" that the hubble telescope can't detect stars, even though we can see stars with our naked eyes. but yes, we still can't see the oort cloud objects. Jan 22, 2019 at 23:37
• Want to throw that in an answer @szulat?
– user12
Jan 22, 2019 at 23:38
• maybe later :-) Jan 23, 2019 at 0:44
• Just want to second @szulat's comment that this answer is wrong and based on a misunderstanding of the diffraction limit. The article on astroengine is incorrect. Clearly Hubble (and we with our naked eye) can see far away stars away well below the diffraction limit. Feb 28, 2020 at 12:37
• And also, the surface brightness of the tiny target would be very low too, no - at 3 000 000 Gm distance, the incident energy from the Sun is 1/400,000,000 that at Earth (150 Gm), and over 200,000 times lower than at Pluto and the Kuiper Belt (about 6500 Gm distant), simply by using the inverse-square law? And thus necessitate a telescope of even bigger size still? In effect, you're looking for an object illuminated by starlight. Mar 2, 2020 at 7:56