CALCULATING THE POWER OUTPUT
OF THE SUN
ÓDecember 2000, Douglas A.
Fowler
INTRODUCTION
Work, energy, and power
What follows is a very brief review of the
most basic definitions of work and power.
This is not intended to be complete and you are encouraged to read up on
this in a good introductory physics text, for example, Giancoli (1998).
If you exert a constant force F over some distance d, the work that you do, w, is defined by the dot product of the
force vector and the displacement vector:
w = F×d
It
can be shown that this formula is equivalent to
w = Fdcosq
or
simply
w = Fd
when
the force vector and displacement are parallel and in the same direction, i.e.,
cosq = 1. Hence, in the SI system, work is given in units of
newton×meters, N×m; 1 N×m is called a joule, J,
i.e., 1 N×m = 1 J. Work on some object results in changes in the
energy of that object:
w = DKE +
DPE + Q
where
DKE is the change in the
object’s kinetic energy, DPE
is the change in its potential energy, and Q
is the energy lost to dissapative forces like friction. This is a version of what is often called the
work-energy theorem. From it follows the simple definition of
energy that one often hears: energy is the capacity to do work, and work
results in changes in energy.
Power is defined as the time rate of
change of energy production or consumption.
In the language of calculus this is written as
P = dE/dt
where
P is power, E is energy, and t is
time. Algebraically, this is
approximated as
The
symbols DE stands
for the ‘change in energy’, that is the amount of energy produced or consumed
by some process; Dt
stands for the time interval over which this occurs. The units of power are watts, symbolized as
W; 1 J/s = 1 W. We must be careful with
the terms ‘produced’ and ‘consumed’; we should really say transformed. For example, a
100 W light bulb transforms the energy supplied to the filament by an electric
current into radiant energy (with a spectrum corresponding to a blackbody at the same temperature as the
bulb’s filament). 100 W means that this
transformation happens at a rate of 100 J/s.
Now, you could lift a 1 kg mass through a
height of 1 m, and do this in one second.
1 kg weighs 9.8 N at the surface of the Earth. You then must lift with
an upward force equal to 9.8 N. (We, of
course, are completely aware of lifting with just a little more than this 9.8 N
to get the mass into motion, but this can be neglected when compared to the
work done throughout the rest of the lift.)
As a lifting machine you would be operating with a power output of 9.8 N×m/s, that is 9.8 J/s or 9.8 W. A kilogram weighs about 2.2 pounds in the
English system. So lifting this weight
to a height of 1 m (a distance that is just under 1 yard plus 3 and 3/8 inches)
once each second is equivalent to 9.8 W. If you wanted to lift weights to
generate the electricity needed to light the 100-watt bulb, you would have to
raise a weight of 22.4 pounds up 1 meter once each second for as many seconds
as you wanted the light to stay on!
The Solar constant
The amount of solar power intercepted by
the Earth has been determined to be 1370 watts per square meter. (Fix,
2001). Abell, Morrison, and Wolff (1991)
briefly discuss how measurements can be made from which to determine this
value, called the solar constant. Another way to describe the solar constant is
as follows: an area of 1 square meter placed at the distance the Earth is from
the Sun, 1 astronomical unit (1 AU), and facing normal to Sun’s rays, will
intercept 1370 W. Letting f stand for the solar constant, we can
write
f = 1370 W/m2
Observing the Sun
We will make observations of the Sun by
direct viewing through a filter attached to a telescope. As the saying goes, DO NOT TRY THIS AT HOME,
without the proper equipment. You are
encouraged to read Jeff Medkeff’s article (1999) “A Beginner’s Guide to Solar
Observing” in the June, 1999 issue of Sky and Telescope. We use a circular, full-apeture glass filter
mounted over the front corrector plate of an 8-inch Celestronâ Schmidt-Cassegrain
telescope. The filter itself consist of
a glass plate machined flat, with its front surface coated with evaporated
metals. This produces a highly
reflective surface, allowing only 0.00001 of the incoming sunlight to pass
through the filter and into the telescope.
The filter is placed over the front corrector plate so that no
unfiltered sunlight enters the telescope and cooks any of its optics. For a
description of this kind of filter, see Chou (1998) or MacRobert (1999).
Sunspots
The first thing you will notice are
sunspots. This will not be our focus here, and the interested reader is
encouraged to look up this topic in any of the astronomy texts listed in the
references.
Limb darkening
The visible layer of the Sun, what we
think of as its surface, is called the photosphere.
The
gaseous photosphere is less than 500 kilometers thick (Seeds, 2000) and may be
as little as only 200 kilometers in thickness (Fix, 1999). This is much less than 1/10 of 1 percent of
the Sun’s radius.
Photons are
emitted by the atoms of the hot gasses in the photosphere. A photon can travel only so far through these gasses before it is
absorbed by another atom. The average
length of this distance is called the mean
free path by physicists. Let’s
represent this mean free path as a distance d. When we look at the center of the Sun’s
image, we are seeing photons emitted from a depth d, that is, we can see to a depth d straight into the solar photosphere. But when we look farther to the edge of the
Sun (called the Sun’s limb), that
same distance d, now at a shallow
oblique angle, allows us to look only to a far lesser depth than when we looked toward the center of the Sun’s
disk. We see light from this shallower
depth where the gasses are cooler. The
cooler gasses emit photons at a longer wavelength causing the limb of the Sun
to look not nearly as bright as the central part of its image. This is called limb darkening. Abell,
Morrison, and Wolff (1991) give a good description, with diagrams, of limb
darkening.
For our purposes, we want to see whether
the limb darkening is uniform in all directions outward and across the Sun’s
image when viewed through a telescope. This will give us some clue as to
whether we can assume that the Sun radiates its energy evenly in all directions
out into space. Make these observations and see what you think.
THE PROBLEM
Using the solar constant, along with your
assumption about whether the Sun radiates uniformly in all directions,
calculate the total power output of the Sun. To set up this calculation, ask
yourself how much solar power passes through 1 square meter at the Earth’s
distance from the Sun. Then ask how much power would be intercepted by 1 square
meter placed at this same distance but in a direction directly outward from the
Sun’s north pole. Another clue will come from the idea of the Dyson sphere.
The Dyson Sphere
A clue as to how to approach this problem
can be found in the speculative concept
of
a Dyson sphere. In the late 1950’s
and throughout the 1960’s, physicist Freeman Dyson took an active part in
discussions concerning the possibility of detecting extraterrestrial
intelligence. Dyson (1959, 1966) argued that members of a technically advanced
civilization, faced with the Malthusian social and economic pressures resulting
from the overpopulation of their homeworld, might literally take apart the
other planets of their solar system and use these materials to construct a
sphere that would completely enclose their sun. Such a sphere, constructed with
a radius equal to the semi-major axis of the orbit of this civilization’s home
planet, would then be able to collect the entire energy output of the central
sun. Further discussion of this idea can be found in Dyson (1979), Niven
(1974), Sagan (1973), and Sullivan (1966). Lemarchand (1992) provides a good
more recent overview of the problems of detecting extraterrestrial
civilizations (if they even exist!).
Science fiction writer Larry Niven (1970,
1974) based his novel Ringworld in
part on Dyson’s ideas. Niven envisioned a somewhat less ambitious civilization
building a wide circular band centered on the sun, rather than a full sphere.
This band would again be constructed with a radius equal to the semi-major axis
of the orbit of the civilization’s home planet. This ‘ringworld’ would be
essentially a wide equatorial slice of a Dyson sphere.
Mass equivalence
Once we know the total power output of the Sun, we
can calculate the rate at which mass is converted to energy by the nuclear
reactions at Sun’s core. For this we use Einstein’s famous equation, perhaps
the only equation that most lay readers know from its status as a cultural
icon:
E = mc2
Here
we let E stand for the energy
produced by the Sun in 1 second. The amount of mass converted to this energy is
given by m. The speed of light in a
vacuum is represented by c. Einstein
showed in 1905 that lightspeed squared is essentially a conversion factor
between units of mass and equivalent units of energy: mass and energy can be converted
back and forth between one another with the restriction that the total mass/energy in any physical process
remains the same. Interested readers are encouraged to consult Einstein (1961)
or any good physics text covering modern physics such as those listed in the
references: Beiser (1987), Bernstein, Fishbane, and Gasiorowicz (2000),
Giancoli (1998), Tipler and Llewellyn (1999), and Weidner and Sells
(1980).
DISCUSSION and FURTHER
CALCULATIONS
The proton-proton chain
The Sun’s energy is produced deep in the
solar core where temperatures reach 15 million Kelvins (K) and the pressure is
great enough to compress hot ionized gas to a density of nearly 150 g/cm3
(Fraknoi, Morrison, and Wolff, 1997) - about 150 times that of water or 13 times
the density of lead. The ionized gas itself is essentially a gas of free
protons, electrons, and helium nuclei. The high density greatly increases the
odds that two protons collide; the high temperature means that these particles
are moving fast enough so that when they do collide, they can get close enough
for the short-range, attractive
strong-nuclear force to overcome the electrostatic repulsion between the
two positively-charged protons. We can represent the resulting process, called
the proton-proton chain, in three
steps, described below.
In the first step, two protons (1H)
collide and interact to form a nucleus of deuterium
or heavy hydrogen (2H),
a positron (e+), and a
neutrino (n). In the second step, a
deuterium nucleus collides and interacts with another proton to produce a
helium-3 nucleus (3He). In the process, one or more gamma-ray
photons (g) are released.
Lastly,
two helium-3 nuclei collide and
interact to produce a single stable helium-4 nucleus (4He) along
with a couple of leftover protons. Since the last step requires the input of
two helium-3 nuclei, we must run the second step twice. But to run this second
step twice, we need two deuteriums which requires that we also run the first
step twice. This is symbolized below by writing both the first and second steps
in parentheses with the notation 2 ´ in front.
2 ´ ( 1H + 1H ® 2H + e+ + n)
2 ´ ( 1H + 2H ® 3He + g)
3He + 3He ® 4He + 1H + 1H
Running
the first and second steps twice requires the net input of six protons. But two
protons are returned with the completion of the third step. The net result of
this process can then be written as:
41H ® 4He
which
tells us that four hydrogen nuclei are fused to make one stable helium
nucleus.
In
stars with masses like that of the Sun or smaller, essentially all of their
energy is produced by this nuclear reaction, the proton-proton chain.
Mass defect
The conversion of mass into energy becomes
evident through what is called the mass
defect. If we take the at-rest mass of four neutral hydrogen-1 atoms
(Weidner and Sells, 1980), we have a total, in atomic mass units (u), given by:
4(1.007825u) = 4.031300u
A
single neutral helium-4 atom has a rest mass of 4.002603u (Weidner and Sells,
1980).
The
difference between the two is called the mass defect:
4.031300u - 4.002603u =
0.028697u
This
0.028697 atomic mass units represents the mass converted to energy each time
the proton-proton chain runs its course. We could use this fact to find the
number of helium-4 atoms produced each second by the nuclear reactions in the
core of the Sun.
The lifetime of the Sun
When
we have a reasonable estimate of the amount of mass converted by the Sun into
energy each second, it seems reasonable to ask how long the sun can last. Not
all of the Sun’s mass is available for nuclear fusion; only the portion of the
Sun’s material that is under enough pressure and at a high enough temperature
is subject to fusion. Seeds (2000) gives the minimum temperature for hydrogen
fusion at about 10 million K. Only about 34 % of the Sun’s mass, contained
within 20 % of its radius, is at a temperature of 10 million K or higher
(Seeds, 2000). We also know that the composition of the solar gasses is 71 %
hydrogen (Fix, 2001), but it has been estimated that at the core, only 34 % of
the core mass is hydrogen (Fix, 2001).
If we assume the Sun fuses hydrogen into helium at a constant rate over
the rest of its lifetime, how much longer will the sun last? Note that this
assumption is probably far too simple: about 5 billion years ago, when the Sun
began its main sequence life, the Sun’s power output was only about 70 % of
what it is now, so we know that the Sun’s power output is steadily
increasing.
REFERENCES
Abell,
G. O., Morrison, D., & Wolff, S. C. (1991). Exploration of the universe (6th ed.).
Beiser,
A. (1987). Concepts of modern physics (4th
ed.). New York
Bernstein,
J, Fishbane, P. M., & Gasiorowicz, S. (2000). Modern physics, Upper Saddle
River, NJ: Prentice Hall.
Chou,
B. W. (1998). Solar filter safety, Sky
and telescope, vol. 95, no. 2, p. 36-40,
June, 1998. Also On-Line at
http://www.skypub.com/sights/eclipses/solar/safety.html,
Contacted 12/20/00.
Dyson,
F. J. (1959). Search for artificial stellar sources of infrared radiation, Science,
vol. 131, p. 1667-1668.
Dyson,
F. J. (1966). The search for extraterrestrial technology, in Perspectives in modern
physics (essays in honor of Hans Bethe), R. E. Marshack (Ed.), New York
Wiley and Sons.
Dyson,
F. J. Disturbing the universe, New York
Einstein,
A. (1961). Relativity: the special and
general theories, New York
Publishers.
Fix,
J. D. (2001). Astronomy: journey to the
cosmic frontier (2nd ed., updated). Boston
McGraw Hill.
Fraknoi,
A., Morrison, D., & Wolff, S. (1997). Voyages
through the universe. Fort
Worth, TX: Saunders College
Giancoli,
D. C. (1998). Physics: principles with
applications (5th ed.). Upper Saddle
River, NJ: Prentice Hall.
Lemarchand,
G. A. (1992). “Detectability of extraterrestrial technological activities,” On-
Line at
http://www.coseti.org/lemarch1.htm, Contacted 12/21/00.
MacRobert,
A. M. (1999). “Solar filters: which is best?,” Sky and telescope, vol. 98, no.
1, p. 63-67, July, 1999. Also On-Line at
http://www.skypub.com/resources/
testreports/accessories/9907tp_filters.html, Contacted 12/20/00.
Medkeff,
J. (1999). “A beginner’s guide to solar observing,” Sky and telescope, vol. 97,
no. 6, June 1999.
Niven,
L. (1970). Ringworld. New York
Niven,
L. (1974). “Bigger than worlds” in A hole
in space. London
Niven,
L. (1990). Author’s introduction to “From Ringworld”
in N-space , New York
Tom Doherty Associates.
Sagan,
C. (1973). The cosmic connection. New York
Seeds,
M. A. (2000). Horizons: exploring the
universe (6th ed.). Pacific Grove ,
CA
Brooks/Cole.
Sullivan,
W. (1966). We are not alone, New York
Tipler,
P. A. & Llewellyn, R. A. (1999). Modern
physics (3rd ed.). New York
Freeman.
Weidner,
R. T. & Sells, R. L. (1980). Elementary
modern physics (3rd ed.). Boston
and Bacon.
