Download ASTROPHYSICS LAB: THE LUMINOSITY OF THE SUN

ASTROPHYSICS LAB: THE LUMINOSITY OF THE SUN
June 1, 1999
by John Kolena, NC School of Science & Mathematics
GOAL: To measure the luminosity of the sun with a wax photometer and a standard luminosity
source (for comparison). The diameter of the sun will be determined with a pinhole
camera.
Light from a 200-watt light bulb and the sun will shine on opposite sides of a wax photometer (a
flux-measuring device). The distance between the photometer and the 200-W bulb will be varied
until the flux of the bulb at the photometer matches the flux of the sun at the photometer. See
picture below.
the sun
200 W light
bulb
EQUIPMENT:
wax
photometer
200-watt unfrosted light bulb with socket, cord, and clamp
wax photometer ( = 2 slabs of wax separated by opaque aluminum foil)
meter stick
(pinhole camera)
PROCEDURE: (remember lab journal expectations)
(To determine the luminosity of the sun, we need to know the distance of the sun from the earth.
We will assume that this distance is known.)
1) Clamp the bulb onto something stable, such as the fence around the observing platform.
2) How should the bulb’s filament be oriented to the photometer face? Why? How/where
should the photometer be oriented relative to the bulb and sun? Why?
3) Note any observations that you can make about the colors of the light observed in the two
halves of the photometer.
4) One partner should be the flux judge and decide where the photometer should be held so that
the
sun and bulb have equal fluxes on the two photometer faces. This person should then hold
the
photometer steady at this distance while the other partners measure and record the
appropriate
data.
5) Repeat step 4 until each partner estimates the position of balanced flux (sun vs. light bulb) at
least
three times.
6) (pinhole camera option)
In advance of the lab, figure out how to measure the diameter of the sun with the pinhole
camera.
Assume that the distance to the sun is known. Use the pinhole camera to obtain an image
of the
sun. Carefully measure the appropriate data.
RESULTS and ANALYSIS
1) Explain your color observations using appropriate laws.
2) Using the average value of your group's measured distances (of balanced solar and bulb flux),
calculate the luminosity of the sun.
3) Give one good reason that would explain why your calculated value of the solar luminosity
could be higher than the accepted value. Explain clearly why your calculated value could be
higher. Be as quantitative as possible. (“Human errror” never counts as a good reason for
why an experimental value disagrees with an accepted value.)
4) Repeat question 3 with “higher” replaced by “lower.”
5) Mars is about 1.5 times farther away from the sun than earth. If you repeated this experiment
on Mars, how far would you have to hold the photometer away from the 200-W bulb so that
the two sides of the photometer were illuminated with equal flux?
(Use the accepted value of the solar luminosity in answering this question.)
6) (pinhole option) Determine the radius and the temperature of the sun from the additional data
collected with the pinhole camera.
CONCLUSION
Teacher notes:
EQUIPMENT:
150-watt bulbs can also be used if 200-W bulbs are unavailable. Special sockets are necessary
with 200-watt bulbs. And these bulbs get extremely hot.
details of a pinhole camera are below in extension section
PROCEDURE:
2) Students will have the most difficulty in keeping the bulb center, the wax photometer, and the
sun in a straight line. The large faces of the wax photometer should be kept perpendicular to the
bulb-photometer-sun axis.
RESULTS and ANALYSIS:
1) On a clear day, the photometer side facing the sun should be bluish, although answers such as
white, bluish-white, bluish-gray are often encountered. The photometer side facing the bulb,
however, should be a distinct orange, although responses such as yellow or yellow-orange are not
uncommon.
Wien’s law predicts the answer to this question:
 max flux
2.9  106 nm K
T

(1)
temperature
 max flux
spectral peak
sun
6000 K
480 nm
blue
tungsten filament bulb
3000 K
960 nm
infrared (visible
peak in red)
our eyes play an additional role in the color perception of wax’s reflected light... because our eyes
perceive yellow better than blue, the solar side of the wax may appear white.... because our eyes
perceive yellow better than red, the bulb side will appear yellow or orange
2) Because the flux of the bulb matches the flux of the sun at the photometer location,
Fsun

Fbulb
Lsun
4  d s2 p

Lbulb
4  db2 p
Lsun

Lbulb
 ds p 


 db  p 
(3)
2
generally, the experimental result for Lsun is within a factor of 3-4 of the accepted value of
the
solar luminosity, Lsun = 3.9 x 1026 watts.
3) and 4)
a) reasons for the calculated answer being too high
(1) bulb luminosity is not primarily in the visible spectrum:
the fluxes at the photometer are matched by the observer visually, but the 200-W output (and
also the solar output; see (b) below) are not visual outputs.
% output in the
uv
visual
infrared
sun (T = 6000 K)
14
42
44
bulb (T = 3000 K)
0
11
89
because only 11% of the bulb’s luminosity is in the visual, the number substituted for L bulb
should have been only 11% of 200 W (or 22 W).... if 22 W had been substituted for L bulb
the result for Lsun would have come out 9 x smaller....
there is a similar, counterbalancing effect (see below) due to the fact that the sun does not
emit the majority of its luminosity in the visible either.....
this has the greatest effect on the calculation of L sun
b) reasons for the calculated answer being too low
(1) solar luminosity not primarily in the visible part of the spectrum
for the same reason described in 3a(1) above, taking into account the fact that only about
40%
of the solar luminosity is in the visible, this effect results in Lsun coming out about 2.5 x too
low...
the net effect of 3a(1) and 3b(1) is that the calculated L sun would come out 3.5x too high....
(2) presence of clouds
clouds are a bit tricky.... if the clouds are thick and dark (in which case, the experiment should
probably have been postponed), some sunlight is blocked from reaching the earth (and is
instead reflected back into space or absorbed by the cloud and then re-emitted in the invisible
infrared).... in this case the sun will not appear as bright as its accepted “L sun”
if clouds are high and bright (cirrus, for example), they may actually reflect sunlight (directed
toward them by the blue-scattering atmosphere) that would have normally escaped back into
space.... this reflection should have little effect on the calculated value of L sun (since this
scattered light was originally sunlight)
(3) sunlight reflected light by the earth (up toward the bulb side of the photometer)
this effectively increases the luminosity directed toward the bulb side of the photometer
beyond its “200 W”... therefore, Lbulb should have been considered to be higher than its rated
“200 W”
c) reasons for the calculated answer being too low or too high or have no effect
(1) the orbit of the earth around the sun is not perfectly circular; on or around January 3, the earth
is closest to the sun (1.7% closer than average)... on or around July 4, the earth is farthest from
the sun (1.7% farther than average).... roughly half way between these two dates (i.e., early April
and early October), the earth is at the average distance from the sun....
in any case, this is a very tiny effect overwhelmed by other major effects
(2) the blue sky scatters skylight
this effect should be considered irrelevant.... after all, the blue skylight is originally sunlight;
this blue sky light is “recorded” by the wax photometer, which is why the solar side of the
photometer will look bluish.... some blue scattered light also probably reaches the bulb side of
the photometer and has the same effect as discussed in 3b(3)
5) equations 2, 3, and 4 still apply....
db  p
 ds  p
Lb
Ls
because ds--p is 1.5x larger (and Lb and Ls remain the same), db-p is 1.5x larger
(5)
6) pinhole camera
if the class has also discussed the concept of pinhole cameras, the radius of the sun and the
temperature of the sun can also be determined
the length of the side of the box to be pointed toward the sun should be as large as possible
(preferably a meter in length); the pinhole size needs to be large enough to produce a visible
image of the sun on the opposite side of the box on the screen... however, the larger the pinhole
size, the fuzzier (and harder to measure) the sun’s image
to sun
pinhole in
aluminum foil
only open side of
the box
<
look this way
white cardboard
screen
from the pinhole geometry,
sun (diameter Dsun
)
ds-p
 sun   image
Dsun
ds  p

Dimage
pinhole
di  p
di-p
Dsun
 ds  p 

 Dimage 
 di  p 
image of sun (diameter Dimage
)
Rsun

Dsun
2
from the Stefan-Boltzmann law,
Lsun
2
4
 4  Rsun
 Tsun
and then
Rsun

Lsun
4
4   Tsun