Parallax of the stars

[sunshinesunshine]

How do you measure the distance of a 4.22 ly star using parallax using the earth's orbital diameter and Hubble (.1 arcsec)?

[Strange]

https://en.wikipedia.org/wiki/Cosmic_distance_ladder#Parallax

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The most important fundamental distance measurements come from trigonometric parallax. As the Earth orbits the Sun, the position of nearby stars will appear to shift slightly against the more distant background. 

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The Hubble telescope WFC3 now has the potential to provide a precision of 20 to 40 microarcseconds, enabling reliable distance measurements up to 5,000 parsecs (20,000 ly) for small numbers of stars.^[4][5] By the early 2020s, the Gaia space mission will provide similarly accurate distances to all moderately bright stars.

 

[sunshinesunshine]

reliable distance measurements up to 5,000 parsecs (20,000 ly)

 

Could you show your calculations? I get a value of less than one ly.

[Strange]

Show us how you work it out and maybe we can see where you went wrong.

[Sensei]

I would start from drawing two right triangles on piece of paper.

In small scale e.g. a=10 cm, b= 1.5 cm (and c=sqrt(a^2+b^2) obviously). You can check angle using simple protractor.

If you will understand idea behind parallax in small scale, you should be able to solve it also with millions or billions of kilometers.

 

[sunshinesunshine]

Ladies first

[Strange]

2 minutes ago, sunshinesunshine said:

Ladies first

There is a page on how parallax is calculated here: https://en.wikipedia.org/wiki/Parallax

The relevant detail her: https://en.wikipedia.org/wiki/Parallax#Derivation

If you still get 1 light-year, show us what you are doing and maybe we can see where you go wrong.

But, if you don't actually want to do the math, you can get a computer to do it for you. Using the 20 microarcsecond resolution of Hubble:

1 au / tangent(20 microarcseconds) in light-years = 163,078

http://www.wolframalpha.com/input/?i=1+au+%2F+tangent(20+microarcseconds)+in+light-years

[sunshinesunshine]

A----earth orbital diameter in meters

 

θ----resolution of the Hubble

 

A/θ = B ------------------>1.4 x 10^10 (3600) / .02 = 2.52 x 10^15 m = .266 ly

 

 

 

[Strange]

Your value for A is out by a factor of 10. And I don't know where the (3600) comes from.

You are using 0.02 radians instead of 20 microarcseconds.

So:

A = 1.496 x 10¹¹ metres (note: it is the radius not the diameter)

θ = 20 microarcseconds = 20 x 10^-6 * pi / 648,000 = 9.7 x 10^-11 radians

A/θ = 1.496 x 10¹¹ metres / 9.7 x 10^-11 radians = 163,018 light-years

 

Edited April 18, 2018 by Strange

[sunshinesunshine]

θ = 20 microarcseconds = .02 arcsec = 9.7 x 10^-8 radians

When you make the approximation A/θ = B you must use the degrees that give you a lower number. So for .1 arcsec the maximum parallax distance is .1 ly.

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No, you need to use radians not degrees.

 

You need to use the value that give you the lower number when using the approximation.

 

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1 AU / 0.1 arcsec = 1 AU /  4.85x10^-7 radians = 32.6 ly (or calculated using degrees .1 ly).

 

Does that mean the maximum distance to a star is 33 ly? 

Edited April 18, 2018 by sunshinesunshine
Strange girl with a fuzzy

[Strange]

10 minutes ago, sunshinesunshine said:

θ = 20 microarcseconds = .02 arcsec

Do you know what "micro" means?

20 microarcseconds = 0.000020 arcseconds

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When you make the approximation A/θ = B you must use the degrees that give you a lower number. 

No, you need to use radians not degrees.

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So for .1 arcsec the maximum parallax distance is .1 ly.

1 AU / 0.1 arcsec = 1 AU /  4.85x10^-7 radians = 32.6 ly

I have no idea where you get 0.1 ly from.

Edited April 18, 2018 by Strange