## Astronomy C

Ashernoel
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### Re: Astronomy C

embokim wrote:Suppose that now after a series of observations astronomers have deduced some properties of this pair of
stars. The parallax angle of this system is 0.0555…”. The mean separation of these two stars appears to
be 1.11”.

What is the semimajor axis of this binary star system, in AU?

How would you solve this question?
Thank You so much!
Ok so its a few steps but nothing crazy.

First you want to take the parallax angle and find the distance in parsecs to the system. d =1/theta.
Once you have the distance to the system, you can use the Small Angle formula to find the D, diameter in parsecs (or AU ish, there is a trick here to avoid conversions but lets keep it simple). 1.11" = 206265D/d

Now that you have the diameter of the system, the semi major axis is half of that and in AU, so use parsec to AU conversion and divide by 2. This problem makes a lot of assumptions about the observers being on earth and that the system's major axis is moving along a "transverse" line, but this method should work for the information that is given.
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Unome
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### Re: Astronomy C

On the same test: can anyone explain how to do 2.a and 2.c?

(that was a great test btw, I ended up scoring ~55 or so).
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embokim
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### Re: Astronomy C

Thank you so much! it helped me so much
But may I ask, what is the small angle formula?
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Ashernoel
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### Re: Astronomy C

Unome wrote:On the same test: can anyone explain how to do 2.a and 2.c?

(that was a great test btw, I ended up scoring ~55 or so).
for 2a you look at the peak in B and see that it is slightly above 2446560 JD. Convert a value of around 2446562 JD to the Gregorian calendar and you should get around may 11th, 1986, which is probably pretty close.

As For 2c, either use the paper to reference the material, or get it wrong. The distance modulus used in the paper is used from Surface Brightness Fluctuations to the galaxy, which aren't given in the problem.

Hope this helps!
embokim wrote:Thank you so much! it helped me so much
Awesome
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embokim
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### Re: Astronomy C

For Newton's Law of Gravitation in what unit is R? (star separation distance)
Thanks
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Unome
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### Re: Astronomy C

embokim wrote:Thank you so much! it helped me so much
But may I ask, what is the small angle formula?
The small angle formula is an approximation of trigonometry. The basic concept is based upon something like this. $O$ (usually written as $d$) represents the linear diameter of the object being measured. This could be a single object, like the moon, or the separation between two stars in a binary system. $A$ (usually written as $D$) is the distance to the object, and $\theta$ is of course the angle. You can write $\tan{\theta} = \frac{O}{A}$ as an equation representing the system. This works if $d$ and $D$ are measured in the same units, and if $\theta$ is measured in radians.

The small angle formula is an approximation for very large values of $D$ (and hence very small values of $\theta$) where the formula is approximately $\theta=\frac{O}{A}$.

To get a more useful form of this equation, note that 1 radian = 206265 arcseconds, and 1 parsec = 206265 AU. This means that the function is also valid if $d$ is is AU, $D$ is in parsecs, and $\theta$ is in arcseconds. These are the units you're most likely to see in problems.
embokim wrote:For Newton's Law of Gravitation in what unit is R? (star separation distance)
Thanks
It can be in any unit, so long as G (the gravitational constant) is expressed in the same units; the units should cancel out to end up as units of force.
Ashernoel wrote:
Unome wrote:On the same test: can anyone explain how to do 2.a and 2.c?

(that was a great test btw, I ended up scoring ~55 or so).
for 2a you look at the peak in B and see that it is slightly above 2446560 JD. Convert a value of around 2446562 JD to the Gregorian calendar and you should get around may 11th, 1986, which is probably pretty close.
How do I convert? I can't find any reliable ways of doing it (and I'm not entirely sure how to derive the formula).
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CVMSAvalacheStudent
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### Re: Astronomy C

embokim wrote:Thank you so much! it helped me so much
But may I ask, what is the small angle formula?
I can't write it but you can search it up
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### Re: Astronomy C

embokim wrote:Thank you so much! it helped me so much
But may I ask, what is the small angle formula?
Short answer: if you have a small angle on a right triangle, the angle is approximately equal to the far side multiplied by the near side:

So in the above, x = A/B. Note here that x is in radians, and A and B must be in the same units. Often, in astronomy, you'll be given arcseconds instead of radians, so you'll have to convert arcseconds to radians by multiplying by 206265 (units of arcsec/rad).

Now the long answer. Note: this may be a little confusing at first depending on your math background... I recommend drawing a little triangle and following along on your own if this is the case. You have been warned.

The long answer involves explaining why the small angle formula works, which goes into a bit of trigonometry. In case you haven't seen trig yet, basically each trigonometric relation (sine, cosine, etc) is a ratio of two sides of a right triangle. So using the above angle x, the sine of x (sin x) is the length of the opposite side (A) divided by the length of the hypotenuse, the cosine of x (cos x) is the length of the adjacent side (B) divided by the length of the hypotenuse, and the tangent of x (tan x) is the length of A divided by the length of B.

Now, if x is really, really small, there are three approximations that arise: sin x = x, cos x = 1, and tan x = x (these are technically based on limits, which are part of calculus). Therefore, instead of saying tan x = A/B, you can say x = A/B, which is the small angle approximation that is used in the small angle formula.

The reason I went through this whole explanation is because the small angle approximation is used a lot in different areas of engineering, materials, and physics at the university level. You may not see it a ton in high school, depending on how advanced your school goes, but it's a good thing to be aware of by the time you encounter professors who are using it!

EDIT: Also see Unome's response
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### Re: Astronomy C

What is the difference among the 3 variants of Kepler's Law? They all produce different answers..

1. p^2 = a^3

2. p^2 = (4 pi^2 a^3) / (GM_total)

3. M_total = a^3 / p^2

EastStroudsburg13
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### Re: Astronomy C

y_utsumi wrote:What is the difference among the 3 variants of Kepler's Law? They all produce different answers..

1. p^2 = a^3

2. p^2 = (4 pi^2 a^3) / (GM_total)

3. M_total = a^3 / p^2
3 only works if M_total is in solar masses, p is in earth years, and a is in AU.

If the system has a solar mass of 1 (i.e. our solar system), then 3 simplifies to 1.

2 is the true full form of the law, and the units can be anything as long as they cancel correctly. So for example, if your G is in Nm^2/kg^2, then your a has to be in meters and your p has to be in seconds.
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