surface temperature of a planet, 1st part
We will try to estimate what the temperature of the earth's surface if it had no atmosphere. This is generally known as orbital temperature. Tuesday, October 13, 2009
we think the planet Earth as a sphere , our beloved star is a star class With these data, and what we saw on radiation, we may be able to obtain a reasonable estimate. 
\\ [E_ {R_ {UA} } = \\ frac {L_ \\ odot t} {4 \\ pi R_ {UA} ^ 2} \\]
The fraction of this energy that reaches Earth only depends on the relationship between the surface of the sphere and the exposed surface, ie the cross section of the Earth: $ \\ pi R_ \\ Oplus} ^ 2 $. To better understand this result, imagine that far off lantern illuminate a ball, on a wall form a circular shadow is equal to the radius of the ball and the surface of that circle is equal to the area of \u200b\u200blight absorption for the ball.
Take some numbers: the solar luminosity $ L_ \\ odot = 3.85 \\ times 10 ^ {26} $ W, the Earth-Sun distance (also known as Astronomical Unit): $ R_ {AU} = 1.5 \\ times 10 ^ {11} $ m (150 million kilometers), the Earth radius $ R_ \\ Oplus = $ 6371 km and take $ t = $ 1 second. Then:
\\ [E_ {earth} = \\ frac {3.85 \\ times 10 ^ {26} \\ mathrm {W} \\ times 1 \\ mathrm {s}} {4} \\ left (\\ frac {6371 \\ × 10 ^ 6 \\ mathrm {m}} {1.5 \\ times 10 ^ {11} \\ mathrm {m}} \\ right) ^ 2 \\]
\\ [E_ {tierrras} = 1.74 \\ times 10 ^ {17} \\ mathrm {J} \\]
Every second the Earth receives from the Sun $ 1.74 \\ times 10 ^ {17} $ joules, that is: 174,000,000,000,000,000 joules!! and this is every second. To put things in perspective, the average world energy consumption during
mass $ M_ \\ Oplus $ and radius $ R_ \\ Oplus $ that orbits the Sun at a distance $ R_ {AU} $. 
On the other hand, the Sun Let's start by determining what fraction of the energy emitted by the Sun is received on Earth at a given time $ t $. Recalling that the luminosity $ L_ \\ odot $ is the energy emitted by a star per unit time, the energy radiated in time $ t $ is just $ L_ \\ odot \\ times t $.
Since the solar emission is isotropic (equal in all directions), at a distance $ R_ {AU} $ del Sol solar energy will distributed evenly over the surface of a sphere of radius $ r = R_ {AU} $ and therefore the energy density on the surface of the sphere is:
\\ [E_ {R_ {UA} } = \\ frac {L_ \\ odot t} {4 \\ pi R_ {UA} ^ 2} \\] The fraction of this energy that reaches Earth only depends on the relationship between the surface of the sphere and the exposed surface, ie the cross section of the Earth: $ \\ pi R_ \\ Oplus} ^ 2 $. To better understand this result, imagine that far off lantern illuminate a ball, on a wall form a circular shadow is equal to the radius of the ball and the surface of that circle is equal to the area of \u200b\u200blight absorption for the ball.
Therefore, the energy reaching the Earth at the same time $ t $ is:
\\ [E_ {tierrras} = E_ {R_ {UA}} \\ times \\ pi R_ \\ ^ 2 Oplus = \\ frac {L_ \\ odot t} {4} \\ left (\\ frac {R_ \\ Oplus} {R_ {UA} \\ right) ^ 2} \\]
Take some numbers: the solar luminosity $ L_ \\ odot = 3.85 \\ times 10 ^ {26} $ W, the Earth-Sun distance (also known as Astronomical Unit): $ R_ {AU} = 1.5 \\ times 10 ^ {11} $ m (150 million kilometers), the Earth radius $ R_ \\ Oplus = $ 6371 km and take $ t = $ 1 second. Then:
\\ [E_ {earth} = \\ frac {3.85 \\ times 10 ^ {26} \\ mathrm {W} \\ times 1 \\ mathrm {s}} {4} \\ left (\\ frac {6371 \\ × 10 ^ 6 \\ mathrm {m}} {1.5 \\ times 10 ^ {11} \\ mathrm {m}} \\ right) ^ 2 \\]
Every second the Earth receives from the Sun $ 1.74 \\ times 10 ^ {17} $ joules, that is: 174,000,000,000,000,000 joules!! and this is every second. To put things in perspective, the average
2005 was 16 TW (16 Terawatts = $ 1.6 \\ times 10 ^ {13} $ watts): the illuminated surface of the Earth receives from the Sun about 11,000 times the world energy requirement!
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