Hot answers tagged

14

A parsec (abbreviated pc) is a unit of distance used by astronomers, cosmologists, and astrophysicists. 1 parsec is equal to $3.08567758 \times10^{16}$ meters, or $3.26163344$ light years (ly). A few typical scales to keep in mind: 1) The disks of galaxies like the Milky Way are a few 10's of kpc (that's pronounced kiloparsecs, which are 1000's of ...


6

Because a separation of 1 au subtends an angle of 1 arcsecond at 1 parsec. That is the definition of a parsec. As you move further away, the angle gets smaller by a factor of the increase in distance. Just draw a right-angled triangle where the base length is the distance and the height is the star-planet separation. Now increase the base length and ask ...


3

Parsec or Parallax-arcsecond is defined as the 'distance' of something that has a parallax angle of one arc second. The International Astronomical Union (IAU) define 1 parsec(pc) as: $$1 \text{pc} = 1/\tan(1'') \ \text{au}$$ $$\Rightarrow 1 \text{pc} = 1/\tan({1\over60}') \ \text{au}$$ $$\Rightarrow 1 \text{pc} = 1/\tan({1\over60\times60}^{\circ}) \ \text{au}...


3

For small angles the Laurent series of $1/ \tan(x)$ about zero begins: $$\frac{1}{\tan(x)} = \frac{1}{x} - \frac{x}{3} - ... = \frac{1}{x} \left( 1 - \frac{x^2}{3} -...\right)$$ so for small $\alpha$ the approximation $1/ \alpha$ will be high by about the fraction $\alpha^2/3$. This means that the error in the distance determined by parallax will be only ...


3

For angles of magnitude 1/50 arcsecond (of the order $10^{-7}$, the difference between $\tan\alpha$ and $\alpha$ is truly negligible (if alpha is in radians) and only a matter of a choice of units if you don't. It is quite correct to say: distance (in AU) = 1/angle (in radians) to get the distance from the angle in radians. Using parsecs a means that we ...


3

You can't, not very well anyway. That's simply because the velocity of an object isn't calculable from the Hubble constant until the so-called Hubble flow starts to dominate (which in turn only happens at cosmological distances, i.e. very far away). That said, if you are in the regime where Hubble flow dominates, you can do this easily. The Hubble constant ...


2

Basic angle calculations: $\mathrm{arc\ length} = r\theta$ (angle measured in radians) and for small angles the arc length approximates the chord length. The angle in radians is therefore $0.131501\pi/180=0.00223$, so the separation is $0.00223\times 879= 2.02 \mathrm{kpc}$


2

The pedagogical answer is this. You're confusing the width of something, with it's parallax! Here's the fixed background Here's something that happens to have width ... Here's something that has parallax ... That's the deal! Note that width can just be measured, using one printout of the photo and a wooden ruler. But parallax means two photos taken 180 ...


2

Ken G's answer is essentially correct, but there is one important thing to keep in mind: The distance $r$ between the galaxies that is interesting for you is the distance they had at the time they emitted the light you see (you don't care about what has happened to them since that time and how far apart they are today; you care about the physical properties ...


1

You asked for a dumb, tedious way of doing it. So here it is, in all its glory One parsec is defined as $$\dfrac{1}{\tan(1'')} \text{ AU}=\dfrac{648000}{\pi} \text{ AU}$$ As $1 \text{ AU} = 149597870700 \text{ m}$, one parsec is equal to $$\dfrac{648000}{\pi} \cdot 149\,597\,870\,700 = \dfrac{96\,939\,420\,213\,600\,000}{\pi} \text{ m}$$ One light year can ...


1

Remember that an arcsecond is an angle, so imagine a pie wedge that extends out from your observatory, having that angle inside the point of the pie wedge. That's all you can tell by looking at the sky, the angle in that pie wedge. Also, notice that a field on the sky is not a line, it's more like a circle, so the pie wedge is really a cone with that ...


1

Luminosities cannot be calculated from magnitudes without knowing the distance to the source (and perhaps something about extinction). I assume therefore that you are talking about absolute magnitudes in those filter band passes. Luminosity can also not be arrived at from a single absolute magnitude measurement in the way you suggest in your post, since ...


Only top voted, non community-wiki answers of a minimum length are eligible