1. (2 pt) Betelgeuse (the red star in Orion's armpit) has a resolvable disk with a diameter
of 0.054 arc seconds, which is just barely discernable by Hubble Space Telescope.
Astronomers use both radians (a big unit of angle) and arc seconds (a small unit of angle).
You need to understand both of them.
A radian is a unit of angular measure, defined so that a circle = 2&Pi radians, = 360 degrees,
so that one radian = 180 / &Pi = 57.296... degrees (you can use 57.30).
An Arc second is 1/60th of an arc minute, which is 1/60th of a degree.
Given that, how many arc seconds are in a radian? _______________________
So now calculate the actual diameter in km of Betelgeuse if it is at a distance of 643 LY (light years).
Show your steps, and use the small-angle approximation d/D = angle in radians if d/D is small.
Here d is the diameter of Betelgeuse and D is the distance to it; both must be in the
SAME units, so you need to know the number of km in a Light Year - (the distance light travels in a year).
1 LY = (3.0 E5 km/s) * 365.25 days/Y * 24(hr/day) * 60(min/hr) * 60(sec/min) = 9.47 E 12 km.
_____________________________
2. (1 pt) The "apparent magnitude" (m) of a star is a logarithmic measure of how bright a star
appears. Logarithmic means that the step are not linear in the value of the brightness, but that each
step is a fixed multiple of the previous step. The brightest stars have magnitudes of -1 to 0,
and the dimmest naked-eye stars have magnitudes of 5 or 6. Each 5 steps in magnitude
corresponds to a factor of 100 in brightness. Thus each step is a ratio of the fifth root
of 100, or about 2.512 times the brightness. Thus, if one star is a magnitude of 12 and the other is a magnitude
of 4, then difference of magnitudes is 8, which means that the ratio of apparent brightness
is greater than 100 (which would be 5 steps) and closer to 10000 (which would be 10 steps).
Do to it exactly, the brightness of star 1 divided by the brightness of star 2 is given by
B1/B2 = (10**((m2 - m1)/2.5)). Or, log (B1/B2) = ((m2 - m1)/2.5).
What is the exact brightness ratio for these two stars? ____________________
(Note: some calculators allow 10**y or 10^y
where y is a non-integer. Other calculators may just use the "antilog" function.
Remember the definition of log (common log): if y = 10**x, then log y = x .)
3. (1 pt) By using a large aperture, we collect more light from a star, allowing us to see much
dimmer stars. If the Rice telescope is 16 inch diameter and your eye pupil is only .25 inch,
what is the ratio of area of light collected of the telescope compared to your eye? ___________________
(remember: you want the ratio of the AREAs, not the ratio of the diameters!)
This then can allow us to see much dimmer stars.
If with your naked eye you can see a 5th magnitude star, how dim a star can we see from
the Rice observatory - that is, what is the magnitude of the dimmest star you can see from
Rice? _____________________
4. (2 pts) We also can see dimmer stars by integrating the light over time, by having long-time
exposures for CCD's (previously, for camera film). If you can have a 4-hour time exposure
on an earth-based telescope, what is the ratio of time between that and your eye integrating
time (1/30 sec)? ______________________
How many magnitudes in brightness does that correspond to? _________________
5. (2 pts) The "Absolute Magnitude" (M) is the apparent magnitude of a star
if you move it to a standard
distance of 10 pc (1 pc = 3.262 LY = 3.09 E16 m).
The "Distance Modulus", (m - M),
is the apparent magnitude m minus the absolute magnitude M, and is a logarithmic measure
of distance. m - M = ( 5 log (r/1pc) - 5 ) where r is measured in pc. Check: for r = 10 pc,
log r = 1, so m - M = 0. For r = 100 pc, log r = 2, so m - M = 5, so its true brightness
M is smaller than its apparent brightness m by 5 magnitudes, or a difference of a factor of
100. If the distance to Betelgeuse is 643 LY, what is that in parsecs? _______ pc.
Then, what is its distance modulus? _________
6. (1 pt) If Betelgeuse has an apparent magnitude of 0.5, what would its absolute magnitude be? ___________
Does this agree with any textbooks you find? (list the source ___________________
And the value ____________).
7. (2 pts) Estimates of the value of the Hubble constant H, are converging to about 71 km/s per Mpc
(megaparsec) (for an interesting history see
http://cfa-www.harvard.edu/~huchra/hubble/. This means, for a galaxy which is 1 Mpc
away, its speed away from us is 71 km/s, and farther away galaxies are receding proportionately
faster.
For this value of H, what is the age of the universe in years if the expansion has been
uniform in time? (1 / H, but watch your units!). __________________________
At that age, what is the knowable size
of the universe (the Hubble time times the speed of light). (Give in MPc _______ as well as km________).
Is that the same as the distance at which the recession speed equals the speed of light? _____
Please note: you *can* get a Doppler shift z = ΔΛ/Λo (That is, the change
in wavelength divided by the rest wavelength) for very distant galaxies greater than one!
In that case, the simple Doppler formula we did before has to be modified for high speeds (v never gets bigger than c!).
In that case, z = ΔΛ/Λo = -1 + SQRT ((1 - v/c)/(1 + v/c)).
This formula reduces to z = v/c for small v's.
Note: New ways of calculating H have arrived at 67 km/s/Mpc via the microwave background, or 74 km/s/Mpc using Cepheids. https://www.nasa.gov/feature/goddard/2019/new-hubble-constant-measurement-adds-to-mystery-of-universe-s-expansion-rate Stay tuned. See also: https://news.uchicago.edu/explainer/hubble-constant-explained.
Last updated 2/9/2022