Units for Semi-major Axis

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  • #1
ckirmser
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I'm trying to map out local space using Astrosynthesis and have been looking up stellar data online. When I come to companion stars, rather than use the RA and Dec for the star - only its position at the time of observation - I use the orbital data relative to the primary.

One of the measurements used is the Semi-major Axis, 'a.' Unfortunately, this measurement is not always given in a standard unit of length. In fact, most of the time it is given in " (seconds of arc) and, occasionally, in ' (minutes of arc).

My problem is - what's the arc?

I get that a minute is 1/60th and that a second is 1/3600th, but of what? What is the baseline that I use to determine the size of the semi-major axis?

Thanx in advance for any help!
 

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  • #2
Janus
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Degrees. One minute of arc is 1/60th of a degree and 1 sec of arc is 1/3600 of a degree. So for instance the apparent angular size of the Moon is ~1/2 of a degree or ~30 min of arc.
 
  • #3
ckirmser
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Well, yes. Thanks for the response, but, what is the arc? How long is it? When the data says the semi-major axis is 5", then, well, 5 seconds of arc of what?

Is it 5 seconds of arc of the circle made using the distance to the object as a radius? Is it 5 seconds of arc of the orbital circumference of the object? What?

I need to turn the semi-major axis data into a length. So, how would seconds of arc translate to a length without some sort of baseline?
 
  • #4
glappkaeft
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I need to turn the semi-major axis data into a length. So, how would seconds of arc translate to a length without some sort of baseline?

You can't. The baseline is necessary.
 
  • #5
Janus
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Well, yes. Thanks for the response, but, what is the arc? How long is it? When the data says the semi-major axis is 5", then, well, 5 seconds of arc of what?

Is it 5 seconds of arc of the circle made using the distance to the object as a radius? Is it 5 seconds of arc of the orbital circumference of the object? What?

I need to turn the semi-major axis data into a length. So, how would seconds of arc translate to a length without some sort of baseline?

It sounds like they might be using parallactic angle measurements, which is basically the inverse of parsecs. So 1 arc-sec = 1 parsec. 2 arc-sec = 1/2 parsec, 1 arc-min = 1/60 parsec, etc. You sometimes get distances to stars given like this (as milliarc-seconds).
 
  • #6
ckirmser
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You can't. The baseline is necessary.
Yes, it is the baseline I'm trying to find.
 
  • #7
ckirmser
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It sounds like they might be using parallactic angle measurements, which is basically the inverse of parsecs. So 1 arc-sec = 1 parsec. 2 arc-sec = 1/2 parsec, 1 arc-min = 1/60 parsec, etc. You sometimes get distances to stars given like this (as milliarc-seconds).
I suspected that it might be in terms of the parallax, but I wasn't sure. I figured this was some standard expression and an astronomer would be able to say, "Well, duh, it's a fraction of the arc made by such-and-so." But, sometimes the measurement was given in AU, so that gave me pause before going with my hunch.

So, if I'm understanding you correctly, when the semi-major axis is given in ' or ", then that it means the minutes or seconds of arc as seen from Earth? I.e., a measurement of X" would be X times 1/3600th of the circumference of the imaginary circle created using the distance to the star as a radius?
 
  • #8
Janus
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I suspected that it might be in terms of the parallax, but I wasn't sure. I figured this was some standard expression and an astronomer would be able to say, "Well, duh, it's a fraction of the arc made by such-and-so." But, sometimes the measurement was given in AU, so that gave me pause before going with my hunch.

So, if I'm understanding you correctly, when the semi-major axis is given in ' or ", then that it means the minutes or seconds of arc as seen from Earth? I.e., a measurement of X" would be X times 1/3600th of the circumference of the imaginary circle created using the distance to the star as a radius?

No. The distance to the star is not involved with this "unit" of measurement. What you thinking about is apparent angular size. For instance, both the Sun and moon have an apparent angular size of ~ 30'. The actual diameter of the Sun is ~400 times that the Moon because it is ~400 times further away. The parallactic measurement is based on the idea of how much of a parallax would be measured in an object that distance away. The closer the object, the smaller the distance, and the larger the parallax. Apparent angular size is dependent on the distance to the star while with parallactic angle,a particular parallactic measurement always is equal to the same distance: 1"= 1 parsec, 1' = 1/60 parsec, 60' = 1/3600 parsec, etc. Converting parallactic angle to standard distance measurement uses a straightforward process:
1. convert the angle to seconds (I.E. 2'30'' = 150")
2. Take the inverse to get parsecs (1/150"= 0.00667 parsec)
3. Convert parsecs to unit of choice (0.00667 parsec = 0.21733 ly = 1375.4 AU)

Now, I don't know if the resource you are using uses this particular type of measurement. One way to determine this is if you have a multiple star system and the orbital periods are given. A shorter period linked with a larger angular measurement would be a strong indicator. Another would be no tendency for the angular measurements to be tied to star distance. (further star pairs tend to list smaller angular measurements than closer stars.) If no such pattern is apparent, then it is likely they are using a fixed unit.
 
  • #9
ckirmser
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No. The distance to the star is not involved with this "unit" of measurement. What you thinking about is apparent angular size. For instance, both the Sun and moon have an apparent angular size of ~ 30'. The actual diameter of the Sun is ~400 times that the Moon because it is ~400 times further away. The parallactic measurement is based on the idea of how much of a parallax would be measured in an object that distance away. The closer the object, the smaller the distance, and the larger the parallax. Apparent angular size is dependent on the distance to the star while with parallactic angle,a particular parallactic measurement always is equal to the same distance: 1"= 1 parsec, 1' = 1/60 parsec, 60' = 1/3600 parsec, etc. Converting parallactic angle to standard distance measurement uses a straightforward process:
1. convert the angle to seconds (I.E. 2'30'' = 150")
2. Take the inverse to get parsecs (1/150"= 0.00667 parsec)
3. Convert parsecs to unit of choice (0.00667 parsec = 0.21733 ly = 1375.4 AU)

Now, I don't know if the resource you are using uses this particular type of measurement. One way to determine this is if you have a multiple star system and the orbital periods are given. A shorter period linked with a larger angular measurement would be a strong indicator. Another would be no tendency for the angular measurements to be tied to star distance. (further star pairs tend to list smaller angular measurements than closer stars.) If no such pattern is apparent, then it is likely they are using a fixed unit.

My resource is simply Wiki.

For example, https://en.wikipedia.org/wiki/Luyten_726-8 is the link to Luyten 726-8, a binary containing UV Ceti and BL Ceti. UV Ceti is the companion and the Orbit data on the right hand side gives a semi-major axis of 1.95"

So, if I'm reading you right,1.95" would convert to 0.51 PC. But, that's over 1-1/2 light years and that just seems a very long ways to me. Is this a reasonable measurement for a binary system?
 
  • #10
Janus
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From the other numbers given in this reference it is clear that they are using the apparent angular size for the semi-major axis. (the method that does depend on the distance to the star.) Since they give the masses of the stars and the orbital period, you can work out the semi-major axis from that and get 5.22 AU. This matches the distance you get if if work it out by using that 1.95" as the apparent angular separation as seen from the Earth.
 
  • #11
snorkack
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I´d simply say - yes, it´s the angular distance, and baseline is the distance to Earth.
Note that it´s the angular distance that´s easier to measure. Because that´s what we see in the skies of Earth. The separation of, for example, Alpha Centauri components goes to about 20 arc seconds - easy to measure by telescope. And also importantly, both are visible which is why their separation and position angles can easily be measured.
Whereas the parallax is just 0,75 arc seconds. To be measured against nothing, of empty black sky background.
No wonder the parallax took a long time to measure.

Basically, angular sizes, with the distance from Earth as the baseline, are used whenever the true distance, and true linear size, are unknown - or unimportant. Often they are unimportant. For another example, the linear size of Venus is always the same. But it´s often useful to know the angular size of Venus. Of vourse, it changes only with distance. But the angular size of Venus is the more useful measurement.
 
  • #12
ckirmser
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From the other numbers given in this reference it is clear that they are using the apparent angular size for the semi-major axis. (the method that does depend on the distance to the star.) Since they give the masses of the stars and the orbital period, you can work out the semi-major axis from that and get 5.22 AU. This matches the distance you get if if work it out by using that 1.95" as the apparent angular separation as seen from the Earth.

Do you think, then, that the value given is merely the separation at the time of observation or the true extent of the semi-major axis?

In other words...
Question on semi-major axis.png

I'd presume the answer is B, because Wiki does define the value as a, the semi-major axis, but I wondered if that would be a safe assumption.
 

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