Distance Between Two F0 Stars: How to Calculate?

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To calculate the distance of a more distant F0 star, the inverse square law for brightness is essential. The closer star has a parallax of 0.05 arcseconds, indicating a distance of 20 parsecs, while the more distant star is 11.3 times less bright. The relationship between brightness and distance can be expressed as a ratio, where the brightness difference leads to a calculation involving the square of distances. To find the unknown distance, it's crucial to set up the proportionality correctly and ensure all units balance in the equations. Using the inverse square law will ultimately guide the solution to the problem.
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Homework Statement


A distant F0 star is 11.3 times less bright than a nearer F0 star that has a
stellar parallax of 0.05 arcseconds. What is the distance in parsec of the more distant F0 star?

Homework Equations


$$d=\frac{1}{D} {pc}$$
$$B_0/B_1 = 10^{\frac{(m_1-m_0)}{2.5}}$$

The Attempt at a Solution


I calculated the distance to the closet planet to be 20pc and the difference in magnitude to be 2.63. But what I can't seem to do is find the distance between the two planets. Could anyone please give me a nudge in the right direction.
 
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Taylor_1989 said:
I calculated the distance to the closet planet to be 20pc and the difference in magnitude to be 2.63. But what I can't seem to do is find the distance between the two planets. Could anyone please give me a nudge in the right direction.

I think you mean stars, not planets.

Review the inverse square law for brightness versus distance. Your class notes or textbook should cover this. If not, a web search will turn up plenty of hits. For example:

http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit1/bright.html
 
gneill said:
I think you mean stars, not planets.

Review the inverse square law for brightness versus distance. Your class notes or textbook should cover this. If not, a web search will turn up plenty of hits. For example:

http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit1/bright.html

I just had a look, so dose this mean that if it is 11.3 times less bright the if I take the square root of that and times it by 20pc I would get the correct ans. But surely that not correct because I need a proportionality constant right? It the less bright that is confusing me. Less bright in my notes means it a ratio, but I don't understand how that relates to the inverse square law
 
If you have a proportionality law, then when you form a ratio using two instances the constant of proportionality will cancel.
 
my second attempt, was this: $$11.3=\frac{1}{x^2}$$ , $$ x=\sqrt{\frac{1}{11.3}}$$
 
What was the motivation? What does x represent?
 
gneill said:
What was the motivation? What does x represent?
Okay so I drew it out from star 1. I said that if it is athe 1m then the brightNess would be B, I then said at 2m the brightness would be 1/2^2=1/4 because it has to cover and area of 4 squares, so by extending that I said at distance x it would be 1/x^2 the orginals brightness so the as I know the brightness to be a ratione of 11.3 I the said 11.3=1/x^2 and went from there.
 
Okay. And what will you do with this value of x in order to answer the problem?
 
gneill said:
Okay. And what will you do with this value of x in order to answer the problem?
I need to covert to parsec and then add to the 20pc. correct?
 
  • #10
I just had a throught could I not workout the intensity of the star 20pc aways from me then work it out from there using invelse square law.
 
  • #11
Taylor_1989 said:
I need to covert to parsec and then add to the 20pc. correct?
Your x is just a function of the ratio of brightness, and by itself contains no distance units. You need to use the known distance of the first star to set the distance scale. You would do better to write out the proportionality law for both cases and then form the ratio, solving for your unknown distance. That way the math takes care of all the logical steps.
 
  • #12
Taylor_1989 said:
I just had a throught could I not workout the intensity of the star 20pc aways from me then work it out from there using invelse square law.
The brightness is a measure of the intensity, and using the inverse square law to find the distance is what the problem intends. It seems that you've just realized what it's all about :smile:
 
  • #13
gneill said:
The brightness is a measure of the intensity, and using the inverse square law to find the distance is what the problem intends. It seems that you've just realized what it's all about :smile:

Ah okay: so I did this: 1/20 = 11.3/d^2 if I rearrange I get the ans correct?
 
  • #14
Not quite. You've squared one distance and not the other, so the units won't balance. Fix that and you'll be okay.

It really would be to your advantage to write out the proportionality law symbolically first, forming the ratio and solving for your unknown variable. Building equations using numbers and partial guesswork is fraught with peril; you shouldn't have to ask if what you're doing is going to work, since a clean derivation using symbols will demonstrate the correctness.
 
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  • #15
Sorry I ment to put the square in my bad.
 
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