Wednesday, July 20, 2011

Tackling Simple Astronomy Problems (4)

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. 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 "distance modulus" which makes use of the absolute magnitude M (see previous blog on this) 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

Sample problems:

(1) A star has an apparent magnitude m = +3.5. Find its absolute magnitude M if its distance = 10 pc.

Solution:

The absolute magnitude is defined as the magnitude the star would have at a distance of 10 pc. Since the star is already at that specified distance, then M = +3.5, or the same as its apparent magnitude.

(2) Barnard's star has an absolute magnitude of +13.2 and an apparent magnitude m = +9.5. Find its distance in LIGHT YEARS.

Solution:

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:

1/p = D = 1.81 pc

But 1 pc = 3.26 Ly, so D = (1.81 pc)(3.26 Ly/pc) = 5. 9 LY

Extra Problems for the motivated reader:

(1) A star with an apparent magnitude of +1.0 is located at a distance of 40 pc. What is its absolute magnitude M? What is its distance modulus?

(2) Find the apparent magnitude of the star Vega if its absolute magnitude is +0.3 and its parallax angle p = 0."123.

(3) The star Pollux in the constellation Gemini has a parallax angle p = 0."093. Find its distance in light years.

(4)Altair in the constellation Cygnus is 16.4 LY distant with an apparent magnitude m = +0.8 and absolute magnitude M = +2.3. Verify its absolute magnitude using the inverse square law for light.