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If i know the radius r of a star in cm, how can I convert it to arcseconds?

For example, if I have a star with r = 3.18e13 cm, and distance to the star d = 220 parsecs, what is the relation to convert the radius from cm to arcseconds.

Thank you.

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  • $\begingroup$ Use trigonometry. $\theta \sim R/D$ $\endgroup$
    – ProfRob
    Feb 28, 2018 at 17:31
  • $\begingroup$ The definitions of parsec and astronomical unit may help. $\endgroup$
    – Mike G
    Feb 28, 2018 at 20:05
  • $\begingroup$ A radius in centimeters?? Ok, you can calculate with that too, but I wonder where that came from... $\endgroup$
    – user1569
    Mar 1, 2018 at 8:40

2 Answers 2

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Using basic circular maths: where $d$ is the distance of the star from the observer in AU, and $r$ is the star's radius in AU, and $a$ is the angle encompassed by the radius of the star in degrees:

$ r = \frac{a}{360}2\pi{d} $

Now rearrange it to make $a$ the subject:

$ a = \frac{180r}{\pi{d}} $

To get $a$ in arcseconds, you need to multiply the result by 3600 (because there are 3600 arcseconds in a degree):

$ a = \frac{648000r}{\pi{d}} $

Now, convert your numbers from cm and parsecs to astronomical units:

$r$ = 3.18e13cm = 2.126 AU

$d$ = 220pc = 4.538e7 AU

Put them into the equation:

$ a = \frac{648000\times2.126}{{4.538\times10^7}\pi} $

To get:

$ a = 0.00966327′′ $

Hope that helps!

Edit:

As Mike has pointed out, the final equation I've ended with can be further simplified if you use different units, to:

$ a = \frac{r}{d} $

where $r$ is in AU and $d$ is in parsecs, giving an answer $a$ in arcseconds.

This is because the equation $a = \frac{648000r}{\pi{d}}$ has $\frac{648000}{\pi}$ in it - which is the definition of a parsec in AU. Therefore, multiplying $d$ by this number to convert it from AU to parsecs, we get $a = \frac{648000r}{648000/\pi\times\pi{d}}$. Cancel out the pi on the bottom, and cancel the ${648000}$ on the top and bottom, and you're left with $a = \frac{r}{d}$ where $r$ is in au and $d$ is in pc.

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    $\begingroup$ I got the same result by using $a_{as} = r_{AU} / d_{pc}$. $\endgroup$
    – Mike G
    Mar 1, 2018 at 15:58
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    $\begingroup$ @MikeG Yes, I wasn't aware there was an actual equation for it; I just used my (limited) knowledge of how circles work to find an equation. I guess your one's more succinct! It makes sense, because the equation $a = \frac{648000r}{\pi{d}}$ has $\frac{648000}{\pi}$ in it - which is the definition of a parsec in AU. Therefore, multiplying $d$ by this number to convert it to parsecs, we get $a = \frac{648000r}{648000/\pi\times\pi{d}}$. Cancel out the pi on the bottom, and cancel the ${648000}$ on the top and bottom, and you're left with $a = \frac{r}{d}$ where $r$ is in au and $d$ is in pc. $\endgroup$
    – JThistle
    Mar 1, 2018 at 16:54
  • $\begingroup$ +1 Working things though step-by-step and explaining along the way is a great way to answer, and can be very helpful for the OP or for future readers. $\endgroup$
    – uhoh
    Mar 5, 2018 at 10:43
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2 * arctan(r / d)

Use Wolfram Alpha, it takes free-form units like cm, parsecs, etc and usually does the right thing. And you can specify the unit for the answer you're looking for (e.g. arcseconds).

http://www.wolframalpha.com/input/?i=2+*+arctan(3.18e13+cm+%2F+220+parsecs)+in+arcseconds

In this case, the answer is 0.019 arc seconds

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    $\begingroup$ Just to be pedantic, wouldn't it be 2 arctan(r/d) since arctan(r/d) is half the angle? Of course, in this case, the angles would be virtually identical. $\endgroup$
    – user21
    Mar 4, 2018 at 23:07
  • $\begingroup$ @barrycarter You're right. At such small angles it doesn't matter, but still the argument is valid in a strict sense. I'll fix my reply. Thanks! $\endgroup$ Mar 4, 2018 at 23:29
  • $\begingroup$ @uhoh Except Florin got 0.019, which about twice 0.009, which is what you got. His original formula was arctan(2r/d) $\endgroup$
    – user21
    Mar 5, 2018 at 15:14
  • $\begingroup$ @barrycarter oh, my eyes are playing tricks on me; better them than my brain at least. Never mind, thanks, etc.. ;-) $\endgroup$
    – uhoh
    Mar 5, 2018 at 15:22

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