Lectures - Monday and Wednesday, 11:00 AM - 12:15 PM

Lab - Tuesday, 4:10 PM -7 PM

- Energy cycle of absorbed sunlight, emitted heat, and transport drives weather and climate.
- Intensity of sunlight decreases with distance according to inverse square law.
- Clouds and surface types regulate absorption of sunlight.
- Emitted heat is very sensitive to small changes in temperature.
- Hot objects (e.g., Sun) radiate more, and radiate at shorter wavelengths than cool objects (e.g., Earth).
- Balance between fluxes of absorbed sunlight and emitted heat determine temperature of the planet.
- Trace gases are good absorbers of long wavelength radiation emitted by Earth, but generally poor absorbers of short wavelength radiation emitted by Sun.
- Surface temperature is warmer than planetary temperature because surface needs to balance incoming sunlight plus downward radiation of atmosphere.

This summary is meant to explain the way the Earth mean temperature is maintained through a radiative balance with the incoming solar radiation. The explanation is done using a simple model, much less complex than the way the real climate system works, but the model contains the basic ingredients needed to explain the problem.

The model we are using is of a spherical planet that rotates on its axis
and revolves around the Sun at the same time. We are also studying the **average
state of the planet**, without addressing the details of the distribution
of the climate with latitude. What we are seeking is to find under these
conditions what is the **average temperature near the surface **of
the planet. We will consider also how this temperature depends on the presence
of an atmosphere. In our calculations will make use of the laws of radiative
heat transfer as summarized in the notes for Lecture 1.

In the lecture notes we have already reviewed the subject of effective temperature.
Here we repeat the derivation for the sake of completion. As stated, *the
effective temperature of a planet is the temperature that would exist near
its surface under radiative balance with the incoming
solar radiation if the planet had no atmosphere*.
The only transformation to the radiation from the Sun considered in this
case is

Consider the following case:

- A planet with a radius
**R**is radiated by a solar flux of S Wm^{-2} - The
**albedo**of the planet is A

With the help of **Figure 1** below, we can see that the planet's
area that collects the sunlight is effectively the area of the planet's shadow
on a plane perpendicular to the incoming sunlight. Considering that a fraction,
A, is being reflected back into space, the overall energy received by the
planet is:

**(1) E _{s}=
(1-A) SπR^{2} **

where **E _{s }**is the total solar (shortwave) energy
collected by the planet per unit time (in units of W).

**Figure 1**: The area collecting sunlight is the cross sectional
area of the spherical planet. The area emitting planetary radiation is the
entire surface of the sphere.

Over a period of one planetary rotation around its axis, the entire surface
of the planet is exposed to the incoming radiation and warms up to a absolute
temperature, **T**. Averaged over the length of a full number
of rotation period the **entire plant's surface** (equal to
4πR^{2}) emits energy, which according to the Stefan-Boltzman
law is proportional to the fourth power of its temperature:

**(2) E _{p}=
4πR^{2 } σT^{4}**

where **EP _{ }**is the planetary (longwave) radiation
(in W) and

**(3) EP=E _{s} **

or when:

**(4) 4πR ^{2 }σT^{4 }=
(1-A) S πR^{2} **

Eliminating identical terms from both sides of the equation we obtain:

**(5) G
= σT _{e}^{4 }= (1-A) S / 4 **

Where we used the letter **G** to express the **radiative
flux at the ground** and identified the effective temperature by
the subscript **e**.

**The radiative flux (energy per unit area) emitted by the planet
is thus 4 times smaller than the amount of solar flux it absorbs**.

As described in the notes to lecture 1, when we apply this equation to Earth the calculated temperature we obtain is much lower than the observed temperature. As explained in the notes this is because the Earth's atmosphere contains gases that absorb the longwave radiation emitted from the Earth's surface. Let us now examine what this atmospheric absorption implies.

**Figure 2**: A plant with an atmosphere containing a single
layer of IR absorbing matter.

Consider **Figure 2** above. It shows the planet's surface
still receiving a solar flux of **(1-A) S **(accept for reflection,
expressed in the albedo** A**, the solar flux is hardly affected
by the presence of the atmosphere because only small segments of the shortwave
spectrum are absorbed by its constituents). The planet warms up and emits
a flux of longwave (**IR**) radiation that we will denote
by the letter **G**. In the atmosphere there are gases that
absorb longwave radiation. We will consider for a moment that these
gases are all arranged in a layer of some thickness above the surface and
spread equally over the entire planet. We will also assume that all the
longwave radiation is absorbed in that layer. As it absorbs the longwave
radiation the layer also warms up and by radiative laws also emits longwave
radiation. We will denote this flux by the letter **H**. Because
the layer is elevated from the surface, it emits radiation through both
its upper and lower surface in equal amounts. Thus overall the layer emits
a total flux of **2H** W m^{-2}. Using these considerations
we can find if the addition of longwave radiation absorbing gases to the
atmosphere changes the equilibrium surface temperature.

In this new situation **the outer surface of the absorbing layer
becomes the outer surface of the planet**, and as described in
section 1 above, **it must come to radiative balance with the radiation
absorbed by the planet**. Thus, in radiative balance and following
equation (5) above, we have:

**(6) H
= (1-A) S / 4 **

where the division by 4 represents the ratio between the effective planetary area for absorption of solar radiation and the planetary area emitting longwave radiation (Figure 1 above).

At the surface, the energy balance is:

**(7) G
= (1-A) S / 4 + F **

This is because all the incoming shortwave radiation, not reflected back
to space, is absorbed by the Earth's surface. Substituting for **H** from
equation (6) we get:

**(8) G
= 2 (1-A) S / 4 **

**The addition of one IR absorbing layer has thus changed the balance
of energy at the ground. It now must balance twice as much radiation
than before! **We can now use the Stefan-Boltzman law to find
out the new equilibrium temperature (the one affected by the greenhouse
effect) and get:

**(9) sT _{g}^{4 }=
2 (1-A) S / 4 **

Which implies that in the presence of one effective layer of an atmospheric
absorbent the temperature ( **T _{g}**) is larger than
the equivalent temperature (

When this calculation is repeated adding another layer, the temperature at the equilibrium temperature at the ground increases even more. With N layers, equation 9 becomes:

**(10) σT _{g}^{4 }=
(1+N) (1-A) S / 4 **

and **TG** becomes **(1+N)**^{1/4 }times
larger than **T _{e}**.

*Text by Yochanan Kushnir, 2000.*