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Astr 209, FALL 2010
Lecture 3 (Sat Oct 2)continued
previously: the Sun’s luminosity (the number of energy-units the Sun radiates each
To calculate the Sun’s luminosity, we needed to know the number of solar energy-units
hitting a one-square-meter solar cell at the Earth’s distance from the Sun. This number
(1.366×10 J/s ) is called the solar constant.
Practical interlude: how much ground must you cover with solar cells to replace coal-
fired power plants? One such plant produces energy at the rate of about 1 GW
(1 gigawatt, or 10 watts or 10 joules/second).
About 1400 J hits a one-square-meter solar cell above the atmosphere each second. The
atmosphere reflects about 1/3 of the incident energy, so only about 1000 W reaches each
square meter of solar cell on Earth’s surface. Not all of this energy does useful things;
the solar cell has an efficiency, which can account for cloudy days, night-time, as well as
the efficiency inherent in the solar cell, (An efficient solar cell is at the heart of a digital
camera; the solar cell is sensitive to about 90 % of the incident light.)
A solar cell for producing electricity is not 100 % efficient (10 %), and 1% (10 %) might
intuitively seem too low. A representative ef1iciency is between no (zero) factors of 10m
and 2 factors of 10; use an efficiency of 10 , or just 10%.. In other words, each solar cell
produces “useful” energy at the rate of about 100 W (the same as the energy required to
operate a light bulb, or the same as the average rate at which a “normal” adult male
Each one-square-meter produces about 100 J/s; how many such cells do you need to
replace a coal-fired plant?
Think of the rate at which many solar cells cell must supply energy to replace a coal
plant as a long line; think of the rate at which ONE cell produces energy as a small line.
The number of cells required is also the number of small lines that fit into the big line.
10 W per cell Page 2 of 8
#cells = 2
= 10 cells, each with an area of 1 square meter
To replace one coal-fired plant, you cover an area of about 10 or 10×10 6square
meters. Think of this area as a square; how long is one side? About 3× 10 m
or about 3 km.
3× 3 10×1`0 square meters
You would need to cover an area about the size of Claresholm.
If you try to supply energy to our entire civilization (at the rate of about 10 W), you
would need a square 1000 km by 1000 km, roughly the size of Alberta, Saskatchewan,
and Manitoba combined. Page 3 of 8
Sarnia Photovoltaic Power Plant, The Biggest Ever
Posted 227 days ago around Biggest Power Plant on Admin
Enbridge Inc. declared that he has accomplished the enlargement of its solar energy
center in Sarnia, Ontario Canada, to produce the planet’s biggest photovoltaic solar farm.
Using 80 megawatts of result, the site currently produces sufficient energy to power a lot
more than 12, 000 houses. Initially a 20-megawatt facility constructed by former owner
First Solar Inc., the procedure was improved by Enbridge to 1. 3 million solar modules
throughout 943, 000 SQUARE METRES included in its push to get further into eco-
How (why) does the Sun shine? (the bottom line? nuclear fusion. But WHY do we think
We have an operational definition of an element, using spectra. In particular, we
can identify a flask of hydrogen.
(CLAIM) It turns out that matter is composed of atoms, as opposed to some sort
of paste. Page 4 of 8
(CLAIM) Hydrogen is the simplest element, composed of one proton, and one
electron “somewhere near” the proton. (J. J. Thomson discovered the electron in
1897, and in 1906 showed that each hydrogen atom has only one electron.)
In the Sun’s core, 4 protons (hydrogen nuclei) combine to form a nucleus of
helium (plus some other sub-atomic particles that have very little mass).
It turns out the mass of one helium nucleus is less than the original mass of
the 4 protons. The difference is 0.7% of the 4-proton mass. It further turns out that
when nature does this sort of thing, we are NOT converting mass into energy.
Instead, kg and joules are units for the same quantity (call it energy).
Analogy: cm and inches are different units for the same quantity –
length. If you have measured a length in cm, but want to express it
in cm, you must know a unit conversion factor (2.54 cm measures
the same length as one inch).
Similarly, each kilogram measures the same energy as 9×10 J.
(energy measured in J = energy measured in kg × 9×10)
This is one way to interpret Albert Einstein’s result of 1905:
E = mc .
It turns out that the unit-conversion factor is the square of the speed of light (c).
WHY this should be is a story for another course.
With this one piece of nuclear physics (mass of He is 0.7% less than the mass of 4
H nuclei) we can estimate how long the Sun will fuse hydrogen into helium, starting now.
1. Fusion happens only in a star’s core. Why? It turns out fusion requires
very high temperature, density, and pressure. We will see why later, when
we look at some of the details of fusion. So we have to estimate what
“core “ means. The core is not all of the Sun (100%, or 10 % of the Sun’s
mass); 1% (or 10 % just SOUNDS1too small; a first guess for the mass of
the Sun’s core is 10% (or 10 %) of the Sun’s mass. The Sun’s mass is
2×10 30kg, so
the core’s mass (10% of the total) is 2×10 kg
2. Hydrogen makes up 70 % (or a relative fraction of 0.7) of the Sun’s mass.
The mass of H in the core 0.7×2×10 kg, or1.4×10 kg. Page 5 of 8
3. Each time 4 hydrogen nuclei fuse to become one helium nucleus, the
Sun’s mass (kg-energy) decreases by 0.7% of the mass of hydrogen. The
total decrease in the Sun’s mass over its hydrogen-fusion lifetime is
0.7 ×1.4×10 29=10 2kg.
The Sun’s mass is decreasing! Does this affect the orbits of the planets?
No. Notice that the total decrease in the Sun’s mass over its lifespan is
only 0.1% of its mass today. The planets’ orbits are not noticeable
4. The Sun’s kg-energy (measured in kilograms) decreases as the Sun
radiates energy (usually measured in joules). To estimate the total energy
(in joules) radiated from now until the Sun runs out o