Calculating Apparent Semimajor Axis: A Unit Conversion Challenge

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Homework Help Overview

The discussion revolves around the calculation of the apparent semimajor axis using unit conversions, particularly focusing on the parallax method of distance determination. Participants are exploring the conversion of arcseconds to parsecs and astronomical units (AU).

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the definition of arcseconds and their conversion to other units, questioning the calculation process and the origin of specific values. There are attempts to clarify the relationship between parallax measurements and the distances involved.

Discussion Status

Some participants have shared their calculation attempts and expressed confusion about unit conversions. Others have provided hints and suggestions for approaching the problem, noting that more context may be needed for complete understanding. There is acknowledgment of mistakes made in earlier calculations, and some participants report arriving at correct answers after revisiting their methods.

Contextual Notes

There is mention of a specific value (4.1 arcseconds) that seems to be crucial for the calculations, but its source is unclear. Participants also note the importance of understanding the definitions of units like parsecs and their relation to astronomical units.

arcTomato
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Homework Statement
Lead linear size of the semi-major axis by using precision parallax
Relevant Equations
precision parallax
Hello

I think this image's calculation is wrong.
I cannot convert units(What is arcsec??How convert to pc??)

How do you think ??I would like to see your calculation process.

I'm so grad if you show.

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What do you know about the parallax method of distance determination?

An arcsec (arcsecond) is one sixtieth of one sixtieth of one degree. That is:
$$arcsec = \frac{1°}{60 \times 60} = \frac{1°}{3600}$$
 
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gneill said:
What do you know about the parallax method of distance determination?

An arcsec (arcsecond) is one sixtieth of one sixtieth of one degree. That is:
$$arcsec = \frac{1°}{60 \times 60} = \frac{1°}{3600}$$

Thank you !
I think I know,But I cannot lead 14AU again.( I lead 10455AU😭)
Could you show me a process?
 
arcTomato said:
Thank you !
I think I know,But I cannot lead 14AU again.( I lead 10455AU😭)
Could you show me a process?
Helpers can only give hints and suggestions, we cannot do the work for you. If you show your calculation attempts in detail then we can point out where you might have gone wrong.

I suspect that there was more to the question that you haven't shown, since there's no apparent place that the 4.1 arcsec in question B came from. I imagine that the original problem statement gave the parallax reading for Procyon and the apparent angular measure of some body orbiting it.

The calculation shown in your image looks simple enough: ##4.1 \times 3.50 = 14.3 \text{(in AU)}##
 
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arcTomato said:
I think this image's calculation is wrong.
To me, both calculations in the picture looks correct. Although, there is no complete explanation provided, where the parallax in the second questions comes from, it apparently points to the fact that Procyon is binary star system and it refers to the semi-major axis of this binary (Although wiki mentions value ##4.3''## instead, but this is not important - use the value you are given).

If you are not sure why the sheer multiplication of the values in ##arcsec## and ##pc## works, you can convert the values to ##rad## and ##AU## respectively, before the multiplication.
 
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gneill said:
Helpers can only give hints and suggestions, we cannot do the work for you. If you show your calculation attempts in detail then we can point out where you might have gone wrong.

I suspect that there was more to the question that you haven't shown, since there's no apparent place that the 4.1 arcsec in question B came from. I imagine that the original problem statement gave the parallax reading for Procyon and the apparent angular measure of some body orbiting it.

The calculation shown in your image looks simple enough: ##4.1 \times 3.50 = 14.3 \text{(in AU)}##

Wow,this is so simple!
Finally I lead 14AU.
I realize my mistake about convertion.

I'm sorry for my bad English, and I appreciate for your help!
 
Whan I convert all units to ##rad## and ##AU##, I can lead right answer! thank you!

But,,,,
lomidrevo said:
If you are not sure why the sheer multiplication of the values in ##arcsec## and ##pc## works,

Yes, exactly!
My pic shows that we should use ##degree## and ##maters##?? It also works??
 
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arcTomato said:
Yes, exactly!
My pic shows that we should use degreedegreedegree and matersmatersmaters?? It also works??
Yes, because of the way that parsecs are defined in terms of radius of the Earth's orbit (1 AU by definition : look up the definition).
 
gneill said:
Yes, because of the way that parsecs are defined in terms of radius of the Earth's orbit (1 AU by definition : look up the definition).
I got it.
I have lead right answer again (in degrees) !
Thanks for your kindness!
 
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Happy to help!
 
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