# Solar energy absorbed by earth per unit of time

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I'm looking for numbers here. Amount of solar energy that is absorbed by the earth per second, hour, year or whatever. More numbers would be best, any would be useful. Reliable sources please.

And dont look this up on wikipedia. Their page on orders of magnitude of energy http://en.wikipedia.org/wiki/Orders_of_magnitude_%28energy%29 is obviously incorrect, and in fact im looking these up so I can fix the page.

Well I keep on finding 1370 W/m^2. However this seems to be solar energy incident on the earth, not energy that is actually absorbed. Comments, anyone?

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The earth's average albedo is about 0.30, meaning 30% of that energy is reflected, and 70% is absorbed.

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Yet actual solar energy absorbed also this varies by latitude, and possibly other factors. While I'm sure I could calculte an estimate by myeslf, I'm looking for an authoritative source for the number.

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']Yet actual solar energy absorbed also this varies by latitude, and possibly other factors[/b']. While I'm sure I could calculte an estimate by myeslf, I'm looking for an authoritative source for the number.

Isn't that considered in the average albedo, given that you use frontal area and not surface area?

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In one of the reseacrh paper I read, it reports average albedo to be ~30% with maximum change of 2%. At low altitude, it reports absorbed maximum solar energy to be 300 W/m2. If you wish, you can read more on pp 380-384 of this article.

Sen z,Progress in energy and combustion science,30(2004),367-416.

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Let the solar constant be s (1.37 kWm-2), the albedo be a (0.3) and the radius of the Earth R (6400km)

Energy input per second $= \pi R^2 (1-a) s = 1.23 \times 10^{17} W$ (roughly)

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• 2 weeks later...

The Sun's luminosity is 3.85e26 W

At a distance of 150,000,000,000 meters (~1AU), the solar flux is 1362 W / m^2

3.85e26 W / (4 * pi * 150,000,000,000^2) = 1362 W / m^2

The Earth presents a 2-d disk r=6371km to the Sun. So it will intercept pi * r^2 = pi * 150,000,000,000^2 = 127516117977447 square meters of sunlight.

127516117977447 m^2 * 1362 W/m^2 = 1.73676952685283E+17 W

So the Earth receives 1.73676952685283E+17 Joules of energy per second from the Sun. If it absorbs 70% of this energy then it absorbs 1.21573866879698E+17 Joules of energy per second.

I got the same answer as Severian.

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The Sun's luminosity is 3.85e26 W

At a distance of 150' date='000,000,000 meters (~1AU), the solar flux is 1362 W / m^2

3.85e26 W / (4 * pi * 150,000,000,000^2) = 1362 W / m^2

The Earth presents a 2-d disk r=6371km to the Sun. So it will intercept pi * r^2 = pi * 150,000,000,000^2 = 127516117977447 square meters of sunlight.

127516117977447 m^2 * 1362 W/m^2 = 1.73676952685283E+17 W

So the Earth receives 1.73676952685283E+17 Joules of energy per second from the Sun. If it absorbs 70% of this energy then it absorbs 1.21573866879698E+17 Joules of energy per second.

I got the same answer as Severian.[/quote']

Except the referee threw a flag for excessive significant digits.

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Except the referee threw a flag for excessive significant digits.

Hey, now, if the caluclator displays the digits, then they are significant.

I have calibrated instruments in accordance with an engineer who did just that, used all of the digits displayed to show upper and lower limits for a reading. I reminded him that the plant (and all the instrumentation and controls) was designed using a slide rule, and also about the 3rd digit is enough for the accuracy required of the instruments involved, and he was actually insulted that a mere technician would challenge his work.

He didn't even understand the difference between percent of reading and percent of range for accuracy determination. He was young, tho, and eventually decided to learn from us lowly techs.

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Hey' date=' now, if the caluclator displays the digits, then they are significant.

I have calibrated instruments in accordance with an engineer who did just that, used all of the digits displayed to show upper and lower limits for a reading. I reminded him that the plant (and all the instrumentation and controls) was designed using a slide rule, and also about the 3rd digit is enough for the accuracy required of the instruments involved, and he was actually insulted that a mere technician would challenge his work.

He didn't even understand the difference between percent of reading and percent of range for accuracy determination. He was young, tho, and eventually decided to learn from us lowly techs.[/quote']

I was a stickler for significant digits even back when I taught at nuke school (when it was in Orlando, '84 - '89)

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my lecturers at uni teach us to do the calculations with as many digits as are displayed on the calculator and only to round at the end. better accuracy that way.

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my lecturers at uni teach us to do the calculations with as many digits as are displayed on the calculator and only to round at the end. better accuracy that way.

Better, yes, but usually a waste of time. Once you get past 0.1%, it is usually overkill. As one of my technology professors put it, build it using 0.1%, then fine tune it if needed. He was the one who scared some in the class the first day by putting up a big formula for designing transistor amplifier cirucuits, then showed us that REAL engineers disregard all the factors of the formula that are not major contributors to the outcome. They use the simplified formula to design and build, and then go back and fine tune if it seems necessary.

Your lecturers are like a lot of engineers I have worked with, they get caught up in absolute perfection and maximum efficiency and overlook practicality.

But, if they are getting paid by the hour......

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There are plenty of measured quantities that need precision better than the number of digits on your calculator....

... this one springs to mind.

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There are plenty of measured quantities that need precision better than the number of digits on your calculator....

... this one springs to mind.

Your link doesn't work for me. I have spent more than a few years in a metrology lab, so I know where high accuracy is required. I suppose navigating in space would need extreme accuracy, tho.

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'']Well I keep on finding 1370 W/m^2. However this seems to be solar energy incident on the earth, not energy that is actually absorbed. Comments, anyone?

Well, do a quick check. Earth is 1.50e11 m from the Sun. The Sun's luminosity is 4e26 W. The solar irradiance at Earth's orbit is just the luminosity divided by the area of a sphere at Earth's radius. You get roughly 1370 W/m^2. So yes, that figure is solar output at the top of atmosphere.

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Your link doesn't work for me. I have spent more than a few years in a metrology lab, so I know where high accuracy is required. I suppose navigating in space would need extreme accuracy, tho.

Actually it's precision, not accuracy, that is being referred to in Severian's example. Accurate means the answer is right. Precise means you have a lot of digits past the decimal. One can be precise but not accurate, and accurate but not precise.

Space navigation probably requires both, as well as knowing whether you are using English or metric units.

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I still don't see how english units have not been completely abolished yet

I do not need any english units besides fahrenheit for the oven lol

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