Bulb distance, d (cm)
13.5
Measurement #1
12.0
Measurement #2
11.0
Measurement #3
12.5
Measurement #4
13.0
Measurement #5
12.4
Average Value =
Table 1: Photometer measurements (typical values).
Measuring the Sun's Luminosity
1. Solving for the Sun's luminosity (or the solar luminosity) requires several steps, which are outlined below.
(a.) What is the average distance between the 300-watt bulb and your photometer (in cm)? (Get used to
writing down the parameter, in this case the distance, as equal to a numerical value, in this case your average
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measurement, along with its units, in this case cm.) How many meters is this? The Sun is about 1.5 X 10
km from the Earth. How many meters is this? (Again, make sure you write that the Sun's distance is equal to
a numerical value along with its units.)
(a) dBulb from Table 1 above is 12.4 cm = 0.124 m
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DSun = 1.5 x10 km This is equivalent to 1.5x1011 meters
(b.) Write down Equation (3) of the handout. Define each parameter in words; for instance “d” represents
what quantity? Identify each parameter in terms of your experimental quantities; for instance, “d =”' the
numerical value you got for your average measurement. (Don't forget units.) Be careful; “Lsun” is the
unknown quantity you are trying to solve.
Lsun = Lbulb D2sun/d2bulb
Plug the numbers into the right hand side of the equation you just wrote down. This will be
your derived value of the solar constant, Isc, because you have set the distance of the light bulb so its
illumination of the paraffin is the same as that of the sun.
= 300 watts (1.5x1011 m)/(0.124m)2
= 4.4x1026 Watts
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2. (a.) The accepted value of the Sun's luminosity is Lsun = 3.8 X 10 W. What is the ratio of your
derived value of the Sun's luminosity to the accepted value? (Solve for the ratio.)
4.4x1026/3.8x1026 = 1.16 so my value is 16% greater
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(b.) Why do you think the ratios you derived in parts (a.) and (b.) of this question are not equal to 1? That is,
why do you think your derived values for the Sun's luminosity and the solar constant in parts (c.) and (d.) of
the previous question are different than the average accepted values? (Hint #1: The light bulb filament is
cooler than the surface of the Sun, so it is only 1/3 as efficient in converting watts to visible light. Hint #2:
Think about the sources of error in measurement, etc.)
Need at least one of these:
-- errors in measuring distance between paraffin block and light bulb (had to estimate location of
center
of bulb)
-- light bulb is cooler and more redder (or more orange) in color which made comparison to Sun
difficult
and indicated that a different fraction of its output lies in the visible part of the spectrum
-- might have been some absorption in the earth’s atmosphere
-- light bulb might be getting old and no longer emits 300 Watts
3. Use your derived value of the Sun's luminosity Question #1 to predict how long the Sun will continue to
burn. Recall that the Sun is converting 4 hydrogen atoms to one helium atom in its core, and that a small
amount of matter is converted to energy in this process. Einstein’s famous equation, E = mc2, expresses how
much energy is produced when mass is converted to energy. A Watt is the same as 1 Joule /second and a
Joule is a unit of energy. The Sun’s luminosity can be equated to the energy produced from the rate at which
mass is being converted in the core of the Sun:
Lsun = ___ Energy/second = ____ kg c2/second
Lsun/c2 = ___ kg/second
The value of c, the speed of light, is 3x108 meters/second. Substitute for c and use your measured value for
LSun to compute the rate at which mass is being converted to energy in the Sun’s core:
LSun = 4.4x1026 Watts
4.4x1026 Watts = 4.9x109 kg / sec
(3x108 m/ s)2
Rate of mass conversion to energy = ___4.9x109___kg/second
When hydrogen is converted to helium, only a small fraction of the original mass of hydrogen is converted
to energy; most of the mass remains as helium. Only 0.7% of the hydrogen mass is converted to energy. At
what rate is hydrogen participating in the conversion of H to He:
4.9x109 kg/s / .007 = 7.0x1011 kg/sec
Rate of hydrogen conversion to helium = __7.0x1011 kg/sec
Only the hydrogen in the core of Sun where the temperature and pressure are sufficiently high can be
converted to helium. The core contains 10% of the mass of the Sun. The mass of the Sun is 2 x1030 kgs.
Compute the mass of the Sun’s core.
2
.1 x 2x1030 kgs = 2x1029 kgs
Mass of the core of the Sun = ___2x1029 __kg
If hydrogen is being converted to helium at the rate you calculated above, how long will it take to convert all
the mass in the core of the Sun to helium? (express your answer first in seconds and then convert to years).
2x1029 kgs / 7.0x1011 kgs/sec = 2.9x1017 seconds
Time to convert all the mass in the core = __2.9x1017__seconds
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1 year = 365 days x 24 hr/day x3600 seconds/hr = 3.15x10 seconds in a year
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2.9x10 seconds / 3.15x10 seconds/yr = 9.1x10 yrs
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= ___9.1x10 __years
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