Tuesday, June 18, 2013

Angular Measurement

Angular Measurement

Consider the following convention which has been with us since the
rise of Babylonian mathematics:

There are 360 degrees per circle.
Each degree can be further divided into 60 minutes (60'), each called
an arcminute.
Each arcminute can be divided into 60 seconds (60"), each called an arcsecond.
Therefore, there are 3600 arcseconds in one degree.

Some rough approximations:
A fist extended at arm's length subtends an angle of approx. 10º.
A thumb extended at arm's length subtends an angle of approx. 2º.
The Moon (and Sun) subtend an angle of approx. 0.5º.

Human eye resolution (the ability to distinguish between 2 adjacent
objects) is limited to about 1 arcminute – roughly the diameter of a
dime at 60-m.  Actually, given the size of our retina, we're limited
to a resolution of roughly 3'

So, to achieve better resolution, we need more aperture (ie., telescopes).

The Earth's atmosphere limits detail resolution to objects bigger than
0.5", the diameter of a dime at 7-km, or a human hair 2 football
fields away.  This is usually reduced to 1" due to atmospheric
turbulence.

The parsec (pc)

The distance at which 1 AU subtends an angle of one arcsec (1") is
definite as one parsec – that is, it has a parallax of one arcsec.

For example, if a star has a parallax angle (d) of 0.5 arcsec, it is
1/0.5 parsecs (or 2 parsecs) away.

The parsec (pc) is roughly 3.26 light years.

Distance (in pc) = 1 / d

where d is in seconds of arc.

Measuring star distances can be done by measuring their angle of
parallax – typically done over a 6-month period, seeing how the star's
position changes with respect to background stars in 6 months, during
which time the Earth has moved across its ellipse.

Unfortunately, this is limited to nearby stars, some 10,000.  Consider
this:  Proxima Centauri (nearest star) has a parallax angle of 0.75" –
a dime at 5-km.  So, you need to repeat measurements over several
years for accuracy.

This works for stars up to about 300 LY away, less than 1% the
diameter of our galaxy!
[If the MW galaxy were reduced to 130 km (80 mi) in diameter, the
Solar System would be a mere 2 mm (0.08 inches) in width.]

Apparent magnitude (m) scale

This dates back to the time of Hipparchus who classified things as
bright or small.
Ptolemy classified things into numbers:  1-6, with 1 being brightest.
The brightest (1st magnitude) stars were 100 times brighter than the
faintest (6th magnitude).  This convention remains standard to this
day.  Still, this was very qualitative.

In the 19th century, with the advent of photographic means of
recording stars onto plates, a more sophisticated system was adopted.
It held to the original ideas of Ptolemy

A difference of 5 magnitudes (ie., from 1 to 6) is equivalent to a
factor of exactly 100 times.  IN other words, 1st magnitude is 100x
brighter than 6th magnitude.  Or, 6th magnitude is 1/100th as bright
as 1st mag.

This works well, except several bodies are brighter than (the
traditional) 1st mag.

So….. we have 0th magnitude and negative magnitudes for really bright objects.
Examples:
Sirius (brightest star):  -1.5
Sun:  -26.8
Moon:  -12.6
Venus:  -4.4
Canopus (2nd brightest star):  -0.7
Faintest stars visible with eye:  +6
Faintest stars visible from Earth:  +24
Faintest stars visible from Hubble:  +28

The magnitude factor is the 5th root of 100, which equals roughly
2.512 (about 2.5).

Keep in mind that this is APPARENT magnitude, which depends on
distance, actual star luminosity and interstellar matter.
Here's a problem:  What is the brightness difference between two
objects of magnitudes -1 and 6?

Since they are 7 magnitudes apart, the distance is 2.5 to the 7th power, or 600.
For the math buffs:  the formula for apparent magnitude comparison:
m1 – m2 = 2.5 log (I2 / I1)

The m's are magnitudes and the I's are intensities – the ratio of the
intensities gives a comparison factor.  A reference point is m = 100,
corresponding to an intensity of 2.65 x 10^-6 lumens.

Absolute Magnitude, M

Consider how bright the star would be if it were 10 pc away.  This is
how we define absolute magnitude (M).

It depends on the star's luminosity, which is a measure of its brightness:

L = 4 pi R^2 s T^4

R is the radius of the body emitting light, s is the Stefan-Boltzmann
constant (5.67 x 10-8 W/m^2K^4) and T is the effective temperature (in
K) of the body.

So, a star's luminosity depends on its size (radius, R) and absolute temperature (T).

If the star is 10 pm away, its M = m (by definition).
m – M = 5 log (d/10)

We let d = the distance (in pc), log is base 10, m is apparent
magnitude and M is absolute magnitude.

A problem:  If d = 20 pc and m = +4, what is M?  (2.5)
And another (more challenging):
If M = 5 and m = 10, how far away is the star?  (100 pc)

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