|
|
Here I am again in this mean old town And you're so far away from me And where are you when the sun goes down You're so far away from me
So far away from me Dire Straits, So Far Away |
Assignments:Problem Set #3 due Thursday 5pm
Check out the Special Web Pages on The Observer's Triangle
|
In Class:
-----------------
The Observer's Triangle:
relationship between angle, distance, and size
To use this with stars,
-- need to know two of the three quantities
-- distance -- we're trying to find that
-- height (aka size) -- we could guess that all stars
are the size of the Sun
-- angular size (can't really measure that
stars are just points)
what about turning the triangle around?
-- like ships do for determining their position (DIAGRAM)
-- loction of lighthouse at one time
-- location of lighthouse later on
-- know how far you've traveled <-- that's w
-- calculate distance to lighthouse, w/ObsTri
--> angle measurement from two vantage points
Example: ship, lighthouse
pos'n A: lighthouse 1 degree east of N
travel 500 yds
pos'n B: lighthouse 1 degree west of N
make triangle ABlighthouse
angles at A and B are 89 degrees
angle at lighthouse must be 2 degrees
OBS TRI: 2/57.3 = 500 yds/R
R = 500 yds x (57.3/2)
= 14000 yds
--> Result of angle measurement from two vantage points
-- it should work for stars
-- we just need to measure their position from two
vantage points
-- Earth moves around Sun
-- make two measurements six months apart
-- difference in perspective = 2 AU = w !!
-- backwards ObsTri
-- measure the angle and we're done
-- it's actually the CHANGE in the position
of the star, due to the Earth's orbital
motion
-- the star doesn't move, our view angle does
-- the angle's gonna be small
-- how to measure the angle?
-- relative to horizon or zenith?
-- tough measure
-- timing, star rises and sets
-- big angle to measure to
very high accuaracy
-- could be done, but HARD
-- relative to the Sun?
-- Sun doesn't move
-- in principle a good choice
-- hard to measure Sun-star angle
-- stars not visible in day
-- need a reference nearby in the sky
-- sailors could use far away land
-- if really far, change in
angle to lighthouse would be
seen by its projection onto
the background land
-- land won't appear to move
because it's really far away
(make second ObsTri)
-- if land is far enough away
its change in apparent
angle is really small and
we can neglect it
-- just like binocular vision
-- hopping finger
-- angle relative to background
-- how about using really far away stars?
-- same principle applies
-- if far enough away, the distant
stars will not change position much
-- if stars all at different distances
they should all change position
by different amounts
-- ones that change a lot are close
-- ones that change a little are far
--> DISTANCE DETERMINATION BY THE METHOD OF PARALLAX
-- in principle, it's hard to know which star has the
apparent motion
-- especially is you look at only two stars
-- in practice, look at a bunch
-- you can figure out which ones have the
biggest aparetnt motion by comparing
the ensemble
-- the application of this method to the stars was
technically challenging
-- angles really small
-- need very accurate angle measurements
-- had to do, even with good telescopes
-- known phenomenon since antiquity
-- understood that heliocentric model predicted it
-- Earth is moving
-- should see this effect
-- took until 1830's (Bessel, Struve, Henderson)
for convincing measurements
-- Rstar/Rsun = >10^5 (even for the nearest)
-- really big distances ( >1pc = 3.26 ly)
-- compare with planets
-- Rpluto/Rsun = 30 or so
-- surprising results
-- universe must be REALLY BIG
-- universe must be REALLY empty
-- stars must be really bright
--as bright as our Sun
-- they ARE SUNS!
-- took until 1830's (Bessel, Struve, Henderson)
for convincing measurements
-- Rstar/Rsun = >10^5 !!!
-- really big distances
-- compare with planets
-- Rpluto/Rsun = 30 or so
-- surprising results
-- universe must be REALLY BIG
-- universe must be REALLY empty
-- stars must be really bright
--as bright as our Sun
-- they ARE SUNS!
How do we conclude that just cuz they're far away, they're bright?
DISTANCE DIMMING LAW
naively, it makes sense
-- the further away from something you are, the fainter it
appears
-- but how to quantify this dimming?
consider a lonely star
-- gives off energy at a rate = L (luminosity) Watts
watts = energy/second <-- rate at which a star produces energy
-- energy (as photons) flows outward equally in every direction
no preferred direction
same amount everywhere
-- if we put a big balloon around the star, we'd catch every photon
-- all of the energy from the star would have to pass through
the balloon somewhere
-- furthermore, is the balloon is spherical
i.e, all parts equidistant from the star
each piece of the balloon will receive the same power
- ie., if 10 watts passes through this little piece,
then 10 watts passes through every peice of the
same size anywhere on the balloon.
-- if that small pice is one square meter, then we can
talk about how many watts passes through a square meter of
ballon, and it would be the same anywhere on the balloon
-- the energy passing through one square meter of the balloon
per second
would then be the total luminosity
(all of the energy per second)
divided by the total surface area of the balloon
(all of the square meters available)
flux = L/(4 pi r^2) units: W/m^2
we define this quantity as FLUX, and for us, it the quantity most closely
related to the brightness we see for any object
Our pupils have an area of 3 x 10^-6 m^2, and our retinas measure how
much energy/second (or power) impinges on them.
higher fluxes mean more power onto our retinas --> brighter
incidentally, this is why it's good that your pupils enlarge when it
gets dark. bigger opening --> sensitive to lower fluxes
also why telescopes are good. concentrate light from a really big
opening onto your eye. --> sensitive to much lower fluxes.
-----
For a source of constant L, FLUX drops as the
INVERSE SQUARE of the distance from the source
go twice as far away, source appears
1/2^2 = 1/4th as bright
Example: If the Sun were located at a distance of 10 pc,
what flux would we see?
L = 3.8 x 10^26 W
R = 10 pc = 10 x 3.09 x 10^16 m = 3.09 x 10^17 m
flux = L / (4 pi R^2) = 3.8 x 10^26/ (4 3.141 (3.09 x 10^17)^2)
= 3.18 x 10^-10 W/m^2
(compare with Solar value= 1300 W/m^2)
4 x 10^12 times fainter, or 31.6 magnitudes
learn to comprehend the MAGNITUDE SCALE
|