__Question__: Can you please explain the idea of parallax and how it is used to measure the distances to nearby stars? Could you also give a specific example in how it might be used? Thanks!

__Answer__:

One of the most basic types of astronomy problem entails finding distances, say to nearby stars. The most intuitively obvious method is known as "trigonometric parallax" which is illustrated in the diagram below;

It is generally agreed this can apply to all stars within a distance of maybe 50 parsecs (pc) or those for which a measurable parallax angle p exists. The angle p is obtained by taking photographs of the same star six months apart (i.e. from opposite sides of Earth's orbit) and comparing the two positions. One can thereby obtain the distance, D from:

D = r/ tan (p)

The relationship is such that for p = 1 arcsec the distance of the star would be 1 parsec (e.g.

**par**-allax

**sec**-ond). an angle of 1 arcsec = 1" = 1/3600 degree. So we see it is an extremely tiny angle. similarly, if the angle p = 1/10" then D = 10 parsecs, so we perceive a reciprocal relationship such that D = 1/p", though we must ensure the units are consistent.

In many applications, the parallax angle p is merged with the equation for the "

If D is the distance, the usual expression for distance modulus is:

(m - M) = 5 log (D /10) = 5 log D - 5 log 10 = 5 log D - 5

**distance modulus**" which is a simple logarithmic equation that makes use of the absolute magnitude M and apparent magnitude m.If D is the distance, the usual expression for distance modulus is:

(m - M) = 5 log (D /10) = 5 log D - 5 log 10 = 5 log D - 5

But: D = 1/p

so:

(m - M) = 5 log (1/p) - 5

Or: (m - M) + 5 = 5 log p

so:

(m - M) = 5 log (1/p) - 5

Or: (m - M) + 5 = 5 log p

Consider the case of finding the distance (in light years) to Barnard's star, which has an absolute magnitude of M= +13.2 and an apparent magnitude m = +9.5.

We may use the parsec form of the distance modulus:

(m - M) = 5 log (1/p) - 5

(9.5 - 13.2) = 5 log(1/p) - 5

-3.7 = 5 log (1/p) - 5

5 log (1/p) = (5 - 3.7) = 1.3

log (1/p) = (1.3)/5 = 0.26

Taking anti-logs:

(m - M) = 5 log (1/p) - 5

(9.5 - 13.2) = 5 log(1/p) - 5

-3.7 = 5 log (1/p) - 5

5 log (1/p) = (5 - 3.7) = 1.3

log (1/p) = (1.3)/5 = 0.26

Taking anti-logs:

1/p = D = 1.81 pc

But 1 pc = 3.26 Ly,

But 1 pc = 3.26 Ly,

So D = (1.81 pc)(3.26 Ly/pc) = 5. 9 LY

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