# What is an arcsec and how do I convert it into AU or m?

• Stargazing
• marc873a
marc873a
I'm trying to calculate a few different things about sgr a* using the stars that orbit it. I have found this table which includes all the stars that orbit:
https://en.wikipedia.org/wiki/Sagittarius_A*_cluster

My problem is, that the semi-major axis is measured in arcsec, and I need it in AU or m. What is arcsec and how do I convert it into the other units?

An arcsec is a measure of an angle equal to 1/60 of an arcminute, which in turn is 1/60 of a degree.

Okay, thanks. How do I convert that into a distance unit like AU or m?

marc873a said:
How do I convert that into a distance unit like AU or m?
You need to know how far you are from the separated objects, in AU or m.
Then you can use trigonometry, or the approximation of the sin(small angle) in radians.

They have different physical dimension, so generally you don’t.

If you want to convert to an actual semimajor axis you need not only the distance, but also the angle the orbital plane makes with the direction from the Earth.

Okay so the distance from earth to sgr a* is 26670 ly or about 2.523 * 10^20 m away. According to the wiki site I linked, the star, S2, has a semi-major axis of 0.1251 arcsec.
What would the distance from S2 to sgr a* be?

Orodruin said:
you need not only the distance, but also the angle the orbital plane makes with the direction from the Earth.

Orodruin said:
If you want to convert to an actual semimajor axis you need not only the distance, but also the angle the orbital plane makes with the direction from the Earth.
Not in this case, from the look of it. The value provided for a appears to be the actual inferred orbital element, and not an observable.
I suspect it's given in arcseconds only because the scale of the orbits is sensitive to the uncertainty in the distance to the central black hole.

marc873a said:
What would the distance from S2 to sgr a* be?
What you can do is take the angular size for a from the table, treat it as if it were perpendicular to your line of sight, and use the small-angle approximation (look at the 'geometric' section): ##a=\theta d##, where a is the size of the semi-major axis in AUs, d is the distance to the black hole in AUs, and ##\theta## is the value for a in radians. You need to convert arcseconds to radians (multiply by ##\frac{\pi}{648000}##).

Or, you can just go to the column labelled q, and use eccentricity to get the semi-major axis: ##q=a(1-e)##
This column already assumes some particular value(s) for the distance to the black hole.

Thank you very much :)

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