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Name: William B.
Status: other
Age: 40s
Location: N/A
Country: N/A
Date: 3/7/2003

Does sunlight have weight? If so, how many ounces-pounds-tons of sunlight are falling on the surface of the Earth at any given moment?


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 piezo-electrical 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 hyper-atom 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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