This is Part 1 -- Part 2, on the "greenhouse effect," is HERE.
Let's start with Sun, and the quantity of energy it emits . . . .
Sun's rate of emission from the photosphere
Stefan-Boltzman Law (assuming emissivity = 1.0):
I = σ . T4
where: I = energy flux, Watts/m2; σ, constant = 5.67 x 10-8 W/m2/K4; T = Kelvin temperature of Sun's photosphere = 6000 K
I = (5.67 x 10-8) . (60004)
= 73.5 x 106 W/m2
Total energy emitted by Sun's photosphere
W = (energy per square meter) x (area of photosphere)
= (73.5 x 106) x (4 . π . r2); where r = 647 x 106 m
= 3.865 x 1026 W
Energy received by Earth
As the Sun's energy radiates in all directions, we can
think of it being spread out over the surface of sphere of ever increasing
volume and surface area.
At the distance of Earth, the sphere will have a radius
equal to Earth's average distance from the Sun:
i.e. 150 x 106 km, or 150 x 109 m
The surface area of this sphere will be:
4 . π . r2
4 . π . (150 x 109) 2
= 2.83 x 1023 m2
So, we are going to be spreading our 3.865 x 1026 W over this area.
Dividing the total energy by this area will give us the average flux of solar energy as it approaches Earth:
= 1367 W/m2
This is called "Earth's Solar Constant" (So)
Average energy falling upon Earth
The Solar Constant is not the energy that falls on a typical square meter of Earth -- the Solar Constant is measured at right angles to the Sun's beam, but this is not how the majority of Earth lies in relation to that beam -- most of the surface is set back at an angle, resulting in a lower intensity:
So, how do we calculate the average input of energy?
First -- note that the total energy intercepted by Earth
must be equal to:
W = (Solar Constant) x (area of Earth's disk)
By "Earth's disk" we mean the cross-sectional area of Earth's sphere, as shown below:
Think of placing a ball in front of a light -- the shadow it casts will be a circle (or disk) -- so the light being intercepted by the ball is a function of its circular cross-section, not its spherical shape.
Therefore, the total energy intercepted by Earth's disk is:
W = (So) . (π . r2)
where: r, radius of Earth
However, we need not actually compute this, because if we want to go on and calculate the average flux of energy input, we now have to spread the total energy intercepted over the actual spherical surface area of Earth:
Surface area of Earth: 4 . π . r2
Therefore, the energy falling on an average square meter:
Average temperature (given this rate of energy input)
Before we begin, we have to be careful and note that the energy available to heat Earth is likely not going to be (So / 4), because a certain fraction of this energy is going to be reflected back to space. This reflected energy , called "albedo," does not participate in heating Earth -- it is the "absorbed energy" that is available to do this.
Earth's "Planetary albedo" is estimated to be 30%, or 0.3. Therefore, the absorbed energy is 70%, or 0.7, of the incoming energy:
This is the energy available to heat Earth's surface.
For Earth's temperature to remain steady, this rate of input must be balanced by an equal rate of emission from Earth.
The Stefan-Boltzman Law allows us to find the temperature that corresponds to this rate of emission:
This is Earth's "equilibrium radiating temperature."
We derived it from the average rate of energy input, taking into consideration the
Even although we set the planetary albedo equal to its present value, we assumed there was no atmosphere to interfere with the incoming and outgoing energy in any way -- i.e. that the energy was incident directly upon Earth's surface, and that Earth's surface, in turn, could send its energy directly out to space, both without any atmospheric interference.
HERE is an exercise in calculating radiation equilibrium temperatures.
The diagrams below show the sensitivity of this temperature to changing planetary albedo and to changing Solar Constant. Note how sensitive the system is to changing albedo in the absence of an atmosphere.
Note how much the actual solar constant varies (see below) -- from the graph above, if equilibrium were attained at the top and bottom of each cycle, the resultant temperatures would vary between -18.14 to -18.18 oC.
Finally, we know that the Earth-Sun distance varies throughout the course of a year -- from 147m km at Perihelion (Jan 03) to 152 m km at Aphelion (July 05). If Earth attained temperature equilibrium at either of these distances (which it does not), then the resultant temperatures would be -19.8 and -15.5 oC, respectively.