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# UH - GEOL 1302 - Class Notes - GEOL 1302, Week 4 Notes

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UH - GEOL 1302 - Class Notes - GEOL 1302, Week 4 Notes

##### Description: In this chapter, we created a very simple climate model based on the fact that the solar energy received by a planet (Ein) must be balanced by the energy that is radiating to space (Eout). For a planet
This preview shows pages 1 - 2 of a 4 page document. to view the rest of the content Chapter 4: A “simple” climate model Power of the Sun P = sT^4a = sT^4(4pr^2) σ = 5.670373 * 10−8 (W/m2)/K4  T = 5778 K  Radius of the Sun (r) = 6.9634 * 108 m 3.8510 * 1026 W ≈ 3.8 trillion trillion Watts  Earth’s solar constant (S) Earth-Sun Distance =150,000,000 km = 150*109 m Surface area of a sphere = 4*pi*r2 SA = 4*pi*(150*109 m)2 =2.83 * 1023 m2 Earth’s Solar Constant (S) = Power of Sun / Surface area sphere = =
3.8*1026 W / 2.8*1023 m^2 = 1,360 W/m^2
Solar constant on Venus (Sv) Venus-Sun Distance = 108,000,000 km = 108*109 m Surface area of a sphere = 4*pi*r^2 SA = 4*pi*(108*109 m)^2 = 1.47 * 10^23 m^2  Solar Constant Venus (Sv) = Power of Sun / Surface area sphere =
3.8*1026 W / 1.47*1023 m^2
- Sv = 2,600 W/m^2
- What is S?  1,360 W/m^2
Total power of Solar energy shining on Earth Solar constant = 1,360 W/m^2 What is Sunlight area of Earth? Radius of Earth (R) = 6,400 km = 6.4*10^6 m  Area of the Earth’s shadow ( A Es ) = pi*R2 = 1.3*10^14 m^2  Total power of Solar energy shining on Earth =
- = Solar constant (S) * Area of Earth’s shadow ( A
Es ) - = (1360 W/m^2) * (1.3*10^14 m^2) = 1.8*10^17 W
- = 180,000 trillion watts = 180,000 terawatts (TW)
Each day humans consume ≈ 15 TW.  So if we could capture 0.01% of solar energy shining on Earth, we
could satisfy all of the world’s current energy needs.
Earth’s Albedo (α) Some of the sunlight that shines on Earth is absorbed and the rest is
reflected back into Space. The reflectivity of a planet, or the fraction of reflected photons, is
called its’ “albedo” or “α”.
The global average albedo of Earth is approximately 0.3, meaning that
30% of photons are reflected.
- What percent of the photons are absorbed?
Fraction of photons absorbed = 1 – α = 0.7 or 70%.  Thus the total rate of energy absorbed by Earth ( E ¿ ) = E ¿  = S(1 – a)pR^2 E ¿ ) = 180,000 TW * 0.7 = 126,000 TW Energy Input ( E ¿ ) aka Power Input E ¿ ) = 180,000 TW * 0.7 = 126,000 TW  More useful to describe  E ¿  in units of Watts per square meter. E ¿ area = S ( 1−a ) p R 2 p R 2 = S ( 1−a ) 4 E ¿ / area = (1,360 W/m^2 * 0.7) / 4 = 238 W/m^2  How much is reflected back into space?
-
E reflected  = (1,360 W/m^2 * 0.3) / 4 = 102 W/m^2 How much is average total E reaching top of our atmosphere?
- 238 W/m^2 + 102 W/m^2 = 340 W/m^2
Energy Input ( E ¿ ) varies spatially So the Earth absorbs an average of 238 W/m^2 from the Sun, but that
does not mean that every square meter absorbs this amount.
- Where is it absorbing less?
- Where is it absorbing more?
Energy loss to Space In the early 1800s, Joseph Fourier, on of history’s greatest
mathematicians asked the following simple question:
- “Because energy is always falling on the Earth from the Sun, why
doesn’t the Earth heat up until it is the same temperature as the
Sun?”
The answer he determined is that the Earth is losing energy at the
same rate as it is receiving it from the Sun.
- Otherwise it would heat up.
- Therefore, the global average  E
¿  = 238 W/m^2 =  E out .

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##### Description: In this chapter, we created a very simple climate model based on the fact that the solar energy received by a planet (Ein) must be balanced by the energy that is radiating to space (Eout). For a planet
4 Pages 19 Views 15 Unlocks
• Notes, Study Guides, Flashcards + More!
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