View Full Version : How much energy is received by the planets

How much energy is received by the planets each year in total? Im mostly interested in Earth. I've come across quite a few numbers which change some from source to source. A lot of them don't make it clear if it's the amount of energy received on the surface or the atmosphere either. And most of them tell you per square meter or square centimeter instead of telling you how much in all.

eburacum45

2004-Mar-11, 08:49 PM

I am interested in this too;

the figures which I carry around in my head as a rough guide might not be much good to you, but if anyone knows any better/more accurate figures I would be very grateful.

The Earth has a cross-sectional area one billionth of the size of a dyson sphere 1 au in radius, so receives 1x 10^-9 of the total energy output of the sun.

The Sun puts out as much energy as contained in 4 million tonnes of mass per second- so

the Earth receives four kilograms worth of energy per second; the equivalent of 2 kilograms of antimatter fully annihilated.

approximately...

daver

2004-Mar-11, 08:58 PM

I am interested in this too;

the figures which I carry around in my head as a rough guide might not be much good to you, but if anyone knows any better/more accurate figures I would be very grateful.

Hmm, that's the right ballpark. If you figure 1 kW/m/m, and 4000 miles diameter, you get about 1.5 kg of energy per second. It shouldn't be too hard to work out the figures for the other planets (if you have their radius in terms of an earth radius, and their distances in AU, just calculate 1.5 * (Radius/Distance)**2).

JohnOwens

2004-Mar-11, 09:52 PM

Solar luminosity: 3.827e26 W

(Using R for orbital radius, r for planetary radius)

Solar flux (I'm not sure that's the right term): L_S/(4*pi*R^2) = L_S/(4*pi*R^2)

Area of planetary silhouette: A = pi*r^2

Total power: P = pi*r^2*L_S/(4*pi*R^2) = L_S*r^2/(4*R^2) = (L_S/4)*(r^2/R^2)

Let's use a figure for the (L_S/4) part, then we just have to plug in the R & r later:

3.827e26W/4 = 9.567e25W

Plugging in 6,288,150 m for r, and 1.49598e11 m for R, for Earth,

P = 9.567e25W*((6.288e6m)^2/(1.496e11m)^2) = 9.567e25W*1.767e-9 = 1.735e17W

or, if you prefer, using E=m*c^2 therefore P=(m/t)*c^2, 1.931 kg/s.

Easiest way to work it out for the solar system as a whole would be to add up all the (r^2/R^2) figures for the planets, then multiplying by the (L_S/4) figure.

Mercury: r_Merc^2/R_Merc^2 = (2,439,700m)^2/(5.791e10m)^2 = 1.775e-9

Venus: 3.128e-9

Earth: 1.813e-9

Moon (if you want to get fancy): 1.350e-10

Mars: 2.221e-10

Jupiter: 8.435e-9

Saturn: 1.778e-9

Uranus: 7.925e-11

Neptune: 3.018e-11

Pluto: 3.697e-14 (gonna get cold there! :wink: )

Add them all up, you get 1.7396e-8, which, multiplied by (L_S/4), gives you (tah-dah!) 1.664e18 W, or 18.52 kg/s.

For any other individual planet, you can just multiply the 9.567e25 W figure by the number in the list above. Admittedly, it won't help you finding out how much of that gets to the surface, though. That shouldn't be a significant difference anywhere besides Venus and Earth, though (depending how you define the surface & atmosphere distinction of the gas giants).

Added: Oops! Suface area of a sphere is 4*pi*r^2, no division by 3, that's only for volume. Big mistake. Number overhaul time....

daver

2004-Mar-12, 01:28 AM

Interesting. I ran through my other numbers, and I don't seem to have left off a pi anywhere, so my figure for solar insolation at earth radius must be off by about a factor of 4. Perhaps the figure I'm remembering was only for visible light.

JohnOwens

2004-Mar-12, 01:37 AM

Interesting. I ran through my other numbers, and I don't seem to have left off a pi anywhere, so my figure for solar insolation at earth radius must be off by about a factor of 4. Perhaps the figure I'm remembering was only for visible light.

I think that would be because you used 4000 miles as diameter, when it's pretty nearly the radius of Earth. Do the numbers click into place better now? :)

eburacum45

2004-Mar-12, 08:30 AM

The Earth reflects 30% of the light incident upon it (using the figure for bond albedo rather than geometric albedo here) so 70 percent of the energy is absorbed by the Earth; a lot of that energy is used to power the atmosphere, but quite a lot of it reaches the ground, even it it has been diffused by clouds.

Most of the energy that reaches the surface is absorbed by the oceans.

JohnOwens

2004-Mar-12, 08:50 AM

Oh yeah, that's right: I've got a link to a lovely diagram if you want details of that kind of stuff.

Earth's Annual Global Mean Energy Budget (Kiehl and Trenberth, 1997) (http://www.cgd.ucar.edu/cas/abstracts/files/kevin1997_1.html)

Added: Alert! Alert! I made an error which throws a lot of the numbers I had up there off by a factor of 3. Corrected numbers are now in place. Don't know what this does to daver's numbers. Perhaps he really used 4000 miles for radius, but just typed it as diameter?

daver

2004-Mar-12, 05:31 PM

Interesting. I ran through my other numbers, and I don't seem to have left off a pi anywhere, so my figure for solar insolation at earth radius must be off by about a factor of 4. Perhaps the figure I'm remembering was only for visible light.

I think that would be because you used 4000 miles as diameter, when it's pretty nearly the radius of Earth. Do the numbers click into place better now? :)

No, I miswrote--i used 4000 miles as radius, not diameter (6400 km radius), which gives 1.3e14 m**2 area which gives 1.3e17 watts if you figure 1 kW/m**2 which still gives about 1.5 kg/sec.

[edit]

Oops, I should have read further before posting. Anyway, your conjecture is correct, i used the correct figure for the radius but incorrectly wrote diameter. Solar insolation is actually around 1.37 kW/m**2 at 1au, which raises the energy intercepted to about 2 kg/sec. That number might be round enough for me to remember.

JohnOwens

2004-Mar-13, 08:58 PM

Oh yeah, that's right: I've got a link to a lovely diagram if you want details of that kind of stuff.

Earth's Annual Global Mean Energy Budget (Kiehl and Trenberth, 1997) (http://www.cgd.ucar.edu/cas/abstracts/files/kevin1997_1.html)

One thing to add about that: It lists 342 W/m^2 for the insolation, which had me wondering about my math again. Then I figured that they probably averaged it over the entire surface of the Earth, including the night side. Did the calculations, and it came out just right. Aside from the night side, which would cut the 1360 W/m^2 (or 1367) in half, it's also due to the spherical shape; 1360 W/m^2 only actually happens at the point between the tropics of Capricorn and Cancer where the Sun happens to be directly overhead at the time. Everywhere else on the day side will only get, umm, I guess the easiest way to express it would be 1360 times the cosine of the angle between the point directly under the Sun and the location in question. Or the sine of the altitude of the Sun, if you prefer.

(I'm trying to work out a formula for the daily insolation for a given latitude and date myself, so I may go on at some length here. :-? )

tracer

2004-Mar-14, 12:48 AM

1360 W/m^2 only actually happens at the point between the tropics of Capricorn and Cancer where the Sun happens to be directly overhead at the time.

My Astro prof. at CSUN (ol' Adrian "Pizza" Herzog) called this point the "subsolar point." Is that the correct term for it?

JohnOwens

2004-Mar-14, 08:00 PM

1360 W/m^2 only actually happens at the point between the tropics of Capricorn and Cancer where the Sun happens to be directly overhead at the time.

My Astro prof. at CSUN (ol' Adrian "Pizza" Herzog) called this point the "subsolar point." Is that the correct term for it?

Beats me, but it sounds like it makes sense. If so, it'll be nice to know the proper term for it, instead of having to explain it every time I want to discuss it. :)

This website has some usefull information Following the Energy Trail (http://curriculum.calstatela.edu/courses/builders/lessons/less/biomes/SunEnergy_good.html). It gives numbers for different environments and latitudes. Below is an example of some of the information, I changed the column names so that they can fit, its best to look at the website.

Ecosystem Type K-cal. per square meter per year & day

Tropical Rain Forest 9000 25

Estuary 9000 25

Swamps and Marshes 9000 25

Savanna 3000 8

Decidous Temperate Forest 6000 16

Boreal Forest 3500 10

Temperate Grassland 2000 6

Polar Tundra 600 2

Desert <200 1

Latitude Biome K-cal. per square meter per year & day

0 Equator Rain Forest, desert 973090 6000 average

30 Degrees Deciduous Forest 843515 5200 average

60 Degrees Grasslands 70,000 4800 (midsummer)

3000 (average)

731 (midwinter)

70 Degrees Coniferous Forest 4100 (midsummer)

0 (midwinter)

80 Degrees Tundra 42,000- 3300 (midsummer)

0 (midwinter)

90 Degrees

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