

Sunlight and Weight
Name: William B.
Status: other
Age: 40s
Location: N/A
Country: N/A
Date: 3/7/2003
Question:
Does sunlight have weight? If so, how many
ouncespoundstons of sunlight are falling on the surface of the Earth
at any given moment?
Replies:
Hello,
Sunlight, composed of photons reaching the earth, has no weight. It has
energy, but from the equation E=mc2, an equivalent mass to this energy can
be found. This, however, does not mean that the mass of the earth increases.
To roughly estimate the incident solar power (energy per second), the
average radiation flux incident on the earth surface of about 1000
Watts/m2 is multiplied by one half of the area of the earth (5 E14 m2) to
yield 2.5E17 Watts. From E=mC2, the mass equivalent to this energy is
about 2.8 kg/seconds.
Ali Khounsary, Ph.D.
Argonne National Laboratory
This is one of the strange aspects of quantum mechanics. Sunlight,
composed of photons of various energies, has MOMENTUM but no MASS. That
is, it does not "weigh" anything in the usual sense that we use the term
"weigh" because the earth's gravitational field has almost no effect upon
photons. They are not attracted by the force of gravity. When we "weigh"
some object
that has a certain mass, we measure the force that the earth exerts upon
that mass. When our scale, or balance, comes to equilibrium the force of
gravity is just balanced by the extension/compression of a spring or by
the "weights" we place in the other pan of the balance. Of course
electronic balances operate on a somewhat different principle of
piezoelectrical effects, but the principle is the same. The mass comes to
rest, and hence has a "rest mass". Since photons, at least from the Sun,
always travel at the speed of light they can have no "rest mass". However,
they can and do exert a force, e.g. solar "winds". Quantum mechanics is weird.
Things get even weirder. Photons have a spin of "1" and things with
integral spin (as opposed to electrons and protons that have spin "1/2")
can all occupy the same energy level. Identical spin "1/2" cannot. That is
why we have atoms filling various shells as electrons are added. No two
electrons can have the same set of quantum numbers. At temperatures only
about 10^6 Kelvins quantum effects, not seen at higher temperatures,
become manifest. By some careful quantum mechanical "tricks" it is
possible to slow the photons down to essentially zero velocity. Certain
atoms can be strung together and behave as a single hyperatom composed of
hundreds of individual atoms. The atoms lose their individual identity! It
is not like they condense out from vapor to liquid. The individual atoms
no longer have an individual identity.
Under these conditions, our intuition has to be abandoned. It just does
not work. We have to hold on and let the mathematics tell us what to
"expect" because our intuition is not going to give us any insight, and in
fact will lead to all sorts of paradoxes and apparent contradictions. It
all started with the double slit experiment by Young. And physicists still
ponder and argue what Young's results "mean".
Vince Calder
Since sunlight has energy, it also has a mass associated with it as
indicated by Einstein's famous equation E = mc^2 or m = E/(c^2).
An elementary textbook says the sun converts 4.2 x 10^9 kg of mass to energy
every second. Using this number, we can estimate the amount of energy
(mass) from the sun hitting the earth by calculating the fraction of the
entire solid angle the earth intercepts as seen by the sun. Since the
diameter of the earth is about 1.3 x 10^7 m and it is 1.5 x 10^11 m from the
sun, it subtends an angle of about 8.7 x 10^5 radians. If we square this
angle and divide by 4 pi = 12.6, we get the solid angle fraction subtended
by the earth, which I calculate to be about 6 x 10^10 of the entire solid
angle. Multiplying this by the 4.2 x 10^9 kg burned by the sun every second
and we obtain 2.5 kg/s as the mass of the photons (light) from the sun
striking the earth every second.
There is another way to guesstimate this number using the rough estimate of
1 kW/m^2 as the energy of the sunlight striking the earth. Multiplying this
by the area of a disk with the area of the cross section of the earth (pi x
R^2 = 1.3 x 10^14 m^2) gives 1.3 x 10^17 J/s. Dividing this by the velocity
of light squared (c^2 = 9 x 10^16) gives 1.4 kg/s. I consider this to be in
excellent agreement with the figure of 2.5 kg/s obtained above.
Please let me know if you find this less than clear or would like more
information or explanation.
Best, Dick Plano, Professor of Physics emeritus, Rutgers University
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Update: June 2012

