Mhm, C Division has a blanket sig figs policy. Whether it's enforced is another story.Adi1008 wrote:Magikarpmaster629 wrote:The apparent position of star A from Earth is measured in March, and then again 6 months later in September. If the apparent position of star A changed by 0.036 arcseconds, what is the approximate distance from the sun to star A in lightyears?d (in parsecs) = 1/p (in arcseconds). p = 0.036 arcseconds. Substituting this in, we get ~27.77 parsecs. Multiply by 3.26 to get LY, giving us [b]90.55 LY[/b]. Do we need to worry about sig figs in astro?
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Re: Astronomy C
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Re: Astronomy C
Thanks Skink. Guess my answer should have been this:Skink wrote:Mhm, C Division has a blanket sig figs policy. Whether it's enforced is another story.Adi1008 wrote:Magikarpmaster629 wrote:The apparent position of star A from Earth is measured in March, and then again 6 months later in September. If the apparent position of star A changed by 0.036 arcseconds, what is the approximate distance from the sun to star A in lightyears?d (in parsecs) = 1/p (in arcseconds). p = 0.036 arcseconds. Substituting this in, we get ~27.77 parsecs. Multiply by 3.26 to get LY, giving us [b]90.55 LY[/b]. Do we need to worry about sig figs in astro?
d (in parsecs) = 1/p (in arcseconds). p = 0.036 arcseconds. Substituting this in, we get ~27.77 parsecs, [b]but 0.036 has 2 sig figs so 28[/b].Multiply by 3.26 to get LY and round to 2 sig figs, giving us [b]91 LY[/b].
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Re: Astronomy C
Adi1008 wrote: Suppose a person is looking at two stars, A and B, and perceives A to be 10 times brighter than B. The product of the magnitudes of the stars is 9, and both magnitudes are positive. What are the magnitudes of both stars?
A: 2, B: 4.5 From the equation for apparent magnitude [math]m_a - m_b = -2.5 \log_{10}\left(\frac{I_a}{I_b}\right)[/math] we see that [math]m_a - m_b = -2.5[/math] (because the common log of 10 is just 1) and since we're given [math]m_am_b=9[/math], algebra tells us that there are two solutions [math](m_a,m_b)=\pm(2,4.5)[/math] only one of which is positive.
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Re: Astronomy C
That's a good point. I forgot to specify in the question that I was talking about apparent magnitude. Sorry for any confusion. Also, how did you LaTeX the math in your answer?finagle29 wrote:Adi1008 wrote: Suppose a person is looking at two stars, A and B, and perceives A to be 10 times brighter than B. The product of the magnitudes of the stars is 9, and both magnitudes are positive. What are the magnitudes of both stars?A: 2, B: 4.5 From the equation for apparent magnitude [math]m_a - m_b = -2.5 \log_{10}\left(\frac{I_a}{I_b}\right)[/math] we see that [math]m_a - m_b = -2.5[/math] (because the common log of 10 is just 1) and since we're given [math]m_am_b=9[/math], algebra tells us that there are two solutions [math](m_a,m_b)=\pm(2,4.5)[/math] only one of which is positive.
The answer you got was right, so your turn now.
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Re: Astronomy C
Code: Select all
[math]LaTeX code here[/math]
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Re: Astronomy C
finagle29 wrote:Which DSO was the IAU petitioned to be renamed Gallifrey?Code: Select all
[math]LaTeX code here[/math]
HD 106906 b.
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Re: Astronomy C
Correct! Your turn.
It's no problem, I like to see pretty math as much as the next person.
It's no problem, I like to see pretty math as much as the next person.
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Re: Astronomy C
What is unique about the inclination of 55 Cancri e's orbit? What effect was used to determine the inclination?finagle29 wrote:Correct! Your turn.
It's no problem, I like to see pretty math as much as the next person.
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Re: Astronomy C
Adi1008 wrote:What is unique about the inclination of 55 Cancri e's orbit? What effect was used to determine the inclination?finagle29 wrote:Correct! Your turn.
It's no problem, I like to see pretty math as much as the next person.
Its inclination is 83 degrees, which nearly perpendicular to the plane of the debris disk. It was found using the observation of 55 cancri e's transit.
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Re: Astronomy C
The inclination is like that because of the gravitational pull of the red dwarf in the system about 1000 AU away, 55 Cancri B (just something I thought was cool). I was looking for something more specific for the effect - it's an actual name and stuff, not something that general.Magikarpmaster629 wrote:Adi1008 wrote:What is unique about the inclination of 55 Cancri e's orbit? What effect was used to determine the inclination?finagle29 wrote:Correct! Your turn.
It's no problem, I like to see pretty math as much as the next person.Its inclination is 83 degrees, which nearly perpendicular to the plane of the debris disk. It was found using the observation of 55 cancri e's transit.
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