4
$\begingroup$

I was trying to better understand the concept of angular diameter and was hoping for some clarification. Given some object's coordinates in RA and dec, is it possible to find that object's angular diameter in a different unit, such as degrees?

EDIT: Thank you so much for the clarification! More specifically, if I know that a point is traveling in a circular path, is it possible to find that path's angular diameter from the equatorial coordinates?

$\endgroup$
3
  • $\begingroup$ Does the path appear circular on the sky, or does it appear as an ellipse which is assumed to be circular in space? $\endgroup$ – Mike G Jun 22 at 16:54
  • $\begingroup$ @MikeG I believe it is a circumpolar star with a path that is circular on the sky. $\endgroup$ – starstarstars Jun 22 at 17:28
  • $\begingroup$ Then the path's extent in declination is its angular diameter. If you're thinking of a star's diurnal motion around the pole, that circle's angular diameter is 2(90 - dec). $\endgroup$ – Mike G Jun 22 at 20:37
3
$\begingroup$

I was trying to better understand the concept of angular diameter and was hoping for some clarification. Given some object's coordinates in RA and dec, is it possible to find that object's angular diameter in a different unit, such as degrees?

If you only have RA and Dec of a single point as @JamesK's answer supposes you won't have a size.

But if you have the celestial coordinates of several points around its perimeter like a nebula or even a constellation, then yes! you can get an angular size!

Let's say there's a "rectangular" patch that extends from 05h 33m to 05h 37m in RA and from -6. to -5 deg in dec.

Since angles in declination are real angles, it will be 1 degree tall.

4 minutes of RA is also 1 degree if it's on the equator, but we have to multiply by cos(dec) because the lines squeeze together at the top and bottom.

In this case cos(5.5 degrees) is almost 1 (0.995) so we can ignore it. But if your patch of sky is farther from the celestial equator, just don't forget to convert RA to degrees then multiply by cos(dec).

If you have a complicated shape defined by several vertices, like "How many square degrees does constellation X cover?" That will be an excellent new question!

$\endgroup$
4
  • $\begingroup$ Thank you so much! If I know that a point is moving in a circular path, is it possible to find that path's angular diameter? $\endgroup$ – starstarstars Jun 22 at 16:15
  • $\begingroup$ @starstarstars yes if you provide enough information. If you can add some example numbers that explain the path, then I can try to help. $\endgroup$ – uhoh Jun 22 at 16:22
  • $\begingroup$ @starstarstars oh, in that case, no. The other answers apply because you have only one coordinate. If you had three or more points, you can find the area. But with only one point, it's just a dot. There's no information about the size. $\endgroup$ – uhoh Jun 22 at 16:30
  • 1
    $\begingroup$ I see, thank you for your clarification! $\endgroup$ – starstarstars Jun 22 at 16:55
3
$\begingroup$

Given someone's address, can you find that person's height?

The RA and dec is the position in the sky of an object, relative to the other stars. The angular diameter is the apparent size of the object in the sky, in degrees (or fraction of a degree). So it is not possible to convert one to another.

Given the position in the sky, it might be possible to look up information about an object, including its angular diameter, in a catalogue. Whether that is possible depends on whether you particular object is in the catalogue, and whether the catalogue contains the information that you need.

$\endgroup$
0
2
$\begingroup$

Angular diameter is independent of sky coordinates. If you divide an object's linear diameter by its distance in the same units, you get its angular diameter in radians.

Suppose we observe Venus (radius 6052 km) when it is 0.69 au = 103 million km away. At that distance, its angular diameter is $$ \frac{2 \times 6052 ~\text{km}}{1.03 \times 10^8 ~\text{km}} = 1.17 \times 10^{-4} ~\text{radian} = 24.2 ~\text{arcsec}$$

$\endgroup$
1
  • $\begingroup$ with the caveat that there is only a single coordinate in play and not a boundary or perimeter definition. $\endgroup$ – uhoh Jun 22 at 15:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.