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I know there are some approximations but what is true part arriving at the earth? At some point in time.

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I know there are some approximations but what is true part arriving at the earth? At some point in time.

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DaveC426913

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That's a simple geometry question. What is the area of the disc of the Earth as a ratio of the area of a sphere with the radius of Earth's orbit?

But that has nothing to do with black body or anything, which suggest you're looking for a quantity rather than a ratio. Though you could then divide the Sun's total estimated output by the above ratio and work it out.

Alternately, the amount of energy impinging on the Earth per square metre is known. You could simply multiply that number by the area of the Earth's disc.

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But do I take in the full surface of the earth? OR just half of it? Because sun is not hitting the earth entirely, just a fraction.Portion?So, you mean what fraction of the sun's total output?

That's a simple geometry question. What is the area of the disc of the Earth as a ratio of the area of a sphere with the radius of Earth's orbit?

But that has nothing to do with black body or anything, which suggest you're looking for a quantity rather than a ratio. Though you could then divide the Sun's total estimated output by the above ratio and work it out.

Alternately, the amount of energy impinging on the Earth per square metre is known. You could simply multiply that number by the area of the Earth's disc.

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DaveC basically already answered that in his post. Read it carefully.But do I take in the full surface of the earth? OR just half of it? Because sun is not hitting the earth entirely, just a fraction.

Of course you only use half of the earth, do you see the sun at night? Anyway, the approximation you use is to treat the earth as a circle with a radius equal to that of the earth, where the normal of the surface of the circle points at the sun. Divide that area by the surface area of a sphere surrounding the sun, with the sun at the center and the earth at the edge.What is the area of the disc of the Earth

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jtbell

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If we have it "half-turned," that is, facing "sideways" towards the sun, then it intercepts half as much energy.

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Thanks

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DaveC426913

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Yea i figured that out ^^ thanks

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The currently accepted value for the Solar Constant is 1366 joules per square meter per second measured normal to the Sun's rays at the outside of the Earth's atmosphere (Scaffeta and West, 2005). This was averaged over two sunspot cycles (22 years). The actual value at any one instant is a function of the solar emissivity at that instant and the Earth's position on its elliptical orbit. This value has replaced earlier values, and will be replaced in its turn when evidence warrants.

Since the Earth's surface area is four times its cross-section, climatologists use a value of 342 watts per square meter for current studies of the Earth's heat budget.

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