How to Convert Units in the Equation \(\alpha = a/d\)?

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Homework Help Overview

The discussion revolves around the equation \(\alpha = a/d\), where \(\alpha\) is initially expressed in radians, and both \(a\) and \(d\) are in meters. Participants explore whether this equation holds true when \(\alpha\) is expressed in arcseconds, \(a\) in astronomical units (AU), and \(d\) in parsecs.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conversion of units and the relationships between radians, arcseconds, AU, and parsecs. There are attempts to manipulate the equation and questions about why certain units do not cancel as expected.

Discussion Status

Some participants provide definitions and relationships between the units involved, suggesting that radians and arcseconds are both units of plane angles. There is an ongoing exploration of how to justify the equation under different unit systems, with no explicit consensus reached on the resolution of the unit cancellation issue.

Contextual Notes

The original poster notes that the problem requires starting with specific units (radians for \(\alpha\), meters for \(a\) and \(d\)), which adds constraints to the discussion. Participants are also considering the definitions of the units involved in the context of the problem.

tony873004
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Starting fromthe equation [tex]\alpha=a/d[/tex] with [tex]\alpha[/tex] in radians and a and d in meters, show that the equation is also valid if [tex]\alpha[/tex] is expressed in arcseconds, a is in AU and d is in parsecs.

Would this be the proper way to show this?

[tex]\alpha=a/d[/tex]
[tex]radians=meters/meters[/tex]
[tex]4.8481*10^{-6} radians / arcsecond = \frac{1.49598*10^{11}m/AU}{3.0857*10^{16}m/pc}[/tex]

Divide the numbers and cancel the m's



[tex]4.8481*10^{-6} radians / arcsecond = 4.8481*10^{-6}AU/pc[/tex]

Cancel the numbers
[tex]radians / arcsecond = AU/pc[/tex]
But radians is still there in the left part of the formula! What did I do wrong?
 
Last edited:
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U don't need any #-s.Just the definition of a parallaxis arcsecond:

[tex]1\mbox{parsec}=:\frac{1\mbox{AU}}{1\mbox{arcsecond}}[/tex]

Daniel.
 
dextercioby said:
U don't need any #-s.Just the definition of a parallaxis arcsecond:

[tex]1\mbox{parsec}=:\frac{1\mbox{AU}}{1\mbox{arcsecond}}[/tex]

Daniel.
Thanks, Dex. The problem says we have to start with alpha in radians, a and d in meters and justify it that way. That's why I did it the way I did. I just don't know why the radians won't drop off, like the intuitive answer says they should.
 
They do.Both radians & arcsecond are plane angle units...There's a connection between them

[tex]2\pi \ \mbox{radians}<--------------------->(180\cdot 3600) \ \mbox{arcseconds}[/tex]...

Daniel.
 
Just write down the identities:

[tex]4.8481 \cdot 10^{-6} \mbox{rad} = 1 \mbox{arcsecond}, \;<br /> 1.49598 \cdot 10^{11} \mbox{m} = 1 \mbox{AU}, \;<br /> 3.0857 \cdot 10^{16} \mbox{m} = 1 \mbox{parsec}, \; <br /> <br /> \Longrightarrow \frac{\mbox{arcsecond}}{4.8481 \cdot 10^{-6}} = 1 \mbox{rad} = \frac{1 \mbox{m}}{1 \mbox{m}} = <br /> \frac{\left(\frac{\mbox{AU}}{1.49598 \cdot 10^{11}}\right)}{\left(\frac{\mbox{parsec}}{3.0857 \cdot 10^{16}}\right)}[/tex]

[tex] \Longrightarrow 2.0627 \cdot 10^{5} \mbox{arcsecond} = 2.0627 \cdot 10^{5} \frac{\mbox{AU}}{\mbox{parsec}}[/tex]

[tex] \Longrightarrow 1\mbox{arcsecond} = \frac{1\mbox{AU}}{1\mbox{parsec}}[/tex]
 
Last edited:

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