What is the furthest star or celestial object whose distance has been calculated with parallax and how does it compare to the theoretical limit using today's telescopes? And how exactly does telescope aperture relate to the maximum distance measurable (other than the bigger the aperture the bigger the distance)?
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asked May 1, 2019 at 10:31
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$\begingroup$ Your first question is a dupe of What's the farthest object as determined only by parallax?. $\endgroup$– pelaCommented May 1, 2019 at 10:50
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1$\begingroup$ Unclear what you mean. The best parallax precisions are discussed in answers to that question. There is no obvious "theoretical limit" to how accurately you can measure the position of an object, only technical and engineering limits that are continually being improved. We are already beyond limits where the bending of light by GR by the Sun and solar system objects must be taken account of. $\endgroup$– ProfRobCommented May 1, 2019 at 16:43
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1$\begingroup$ FWIW, we don't even have a very accurate distance measurement for the well-known and relatively nearby star Betelgeuse, which has parallax of around 4.51 ± 0.80 milliarc-seconds. $\endgroup$– PM 2RingCommented May 26, 2020 at 15:06
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1$\begingroup$ @PM2Ring That is because it is too bright. $\endgroup$– ProfRobCommented May 26, 2020 at 16:10
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1$\begingroup$ @Rob True, and its variable brightness doesn't help either. $\endgroup$– PM 2RingCommented May 26, 2020 at 16:22
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2 Answers
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Quick Google reveals a couple simple analyses. For example,
The Andromeda Galaxy, M31, is the nearest major galaxy to the Milky Way. The distance to M31 has been measured using other techniques to be 2.5⋅10^6 light years , or 7.6⋅10^5 parsecs. Using the slightly modified parallax formula, we can find the necessary parallax angle to measure the distance to Andromeda. $ p = \frac{1}{d} = > \frac{1}{7.6*10^5} parsec = 1.3 *10^{-6} arc-seconds $
This is an incredibly small angle. For comparison, the resolution of the Hubble Space Telescope is 0.05 arc-seconds, so even Hubble would not be able to detect the necessary angular shift of the nearest galaxy to effectively use parallax as a measure of its distance.
answered May 1, 2019 at 13:47
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1$\begingroup$ I think, measuring parallax does not need to measure the absolute angular position of the object on the sky. If there is an enough strong another source behind it, but enough far away, only their angle is also enough. This would enable for more better measurements. $\endgroup$– peterhCommented Sep 29, 2019 at 15:34
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Per Wikipedia's Gaia (spacecraft); Objectives which I linked to in the question What actually determines the angular uncertainty of the source of a detected gravitational wave?
- Determine the position, parallax, and annual proper motion of 1 billion stars with an accuracy of about 20 microarcseconds (µas) at 15 mag, and 200 µas at 20 mag.
20 (µas) is about $1 \times 10^{-10}$ radians. If the Earth's amplitude is 2 AU, then the farthest distance that could be detected is $2 \times 10^{10}$ AU.
If you want to measure to about 10% accuracy, then that distance is $2 \times 10^{9}$ AU or about 3,000 30,000 light years.
That sounds surprisingly far away!
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1$\begingroup$ But 20 microarcsec is not the most accurate parallax available. $\endgroup$– ProfRobCommented May 1, 2019 at 16:39
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1$\begingroup$ Yes VLBI is the record holder I think - as discussed in the closely linked question. In your answer, I note that a 200 $\mu$as parallax would be measured with 10% precsion. This corresponds to a distance of 5000 pc or $\sim$ 15,000 light years. $\endgroup$– ProfRobCommented May 2, 2019 at 10:02
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1$\begingroup$ Parallax is based on a triangle with a base of 1 au. $\endgroup$– ProfRobCommented May 2, 2019 at 10:15
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1$\begingroup$ @CarlWitthoft Stellar parallax is the angle in a triangle with a base of 1 au. That is what the parallax (and parallax precision) reported by Gaia is defined as. $\endgroup$– ProfRobCommented May 2, 2019 at 17:21
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1$\begingroup$ @RobJeffries But this question could certainly have a better answer, based on a better source than Wikipedia. $\endgroup$– uhohCommented May 2, 2019 at 23:21