Distribution of incident solar radiation

The geographical variations of temperature are driven by variations in the amount of sunlight falling on each square meter of surface, and also by variations in albedo. Seasonal variations are driven by changes in the geographical distribution of absorbed sunlight as the planet proceeds through its orbit. Therefore, the starting point for any treatment of seasonal and geographical variation must be the study of how the light of a planet's sun is distributed over the spherical surface of the planet. This section dealso only with the distribution of incident sunlight, or insolation. The geographical distribution of the amount of sunlight absorbed is affected also by the distribution of the albedo. The albedo variations can also affect the seasonal distribution of solar forcing through seasonal variations in ice,snow, cloud and vegetation cover.

It will help to first consider an airless planet, so that we don't at once have to deal with the possible effects of scattering of the solar beam by the atmosphere. If our planet is far from its Sun, as compared to the radius of the Sun, the sunlight encountering the planet comes in as a beam of parallel rays with flux L. Even if the surface of the planet is perfectly absorbing, the sunlight the planet intercepts is not spread uniformly over its surface; per unit area, parts of the planet where the sun is directly overhead receive a great deal of energy, whereas parts where the Sun grazes the surface at a shallow angle receive little, because the small amount of sunlight intercepted is spread over a comparatively large area, as shown in Figure 7.3. The night side of the planet, of course, receives no solar energy at all.

To obtain a general expression for the distribution of incident solar radiation per unit of surface area, we may divide up the surface of the planet into a great many small triangles, and consider each one individually. The solar energy intercepted by a triangle is determined by the area of the shadow that would be cast by the triangle on a screen oriented perpendicular to the solar beam. To compute this area, suppose that one of the vertices of the triangle is located at Figure 7.2: Map of July-January surface air temperature difference.

Figure 7.2: Map of July-January surface air temperature difference. the origin, and that the two sides coming from this vertex are given by the vectors r\ and r2 By the definition of the cross product, the area of the triangle is given by 2An = (rl x r2) where n is the unit normal to the plane containing the triangle. To obtain the area of the shadow cast by the triangle, we apply the cross product to the projection of the vectors r{ and r2 onto the plane. These projections are given by f\ — zr\ • z and r2 — zr2 • z, where z is the unit vector pointing in the direction of the Sun. The cross product of these two vectors is

(rl — zr{ • z) x (r2 — zr^ • z) = r[ x r2 — (z x r2)(r[ • z) — (rl x z)(r2 • z) (7.1)

Now, the cross product of two vectors in the xy plane must point in the direction of the z axis. Hence, we can obtain the magnitude of the above vector by taking its dot product with z. This is very convenient, since the dot product of zz with the second two terms vanishes, leaving us with

2A± = z • (rl x r2) = 2Az • n = 2A cos(Z) (7.2)

where A± is the area of the shadow and Z is the angle between the normal to the patch of surface and the direction of the sun. This is known as the zenith angle. When the the zenith angle is zero, the Sun is directly overhead, and when it is 90o the sunlight comes in parallel to the surface and leaves no energy behind. Zenith angles greater than 90o are unphysical, since they represent light that would have to pass through the solid body of the planet in order to illuminate the underside of the surface; these are on the night side of the planet. If one draws a line from the center of the planet to the center of the Sun, the zenith angle will be zero where the line intersects the surface of the planet; this is the subsolar point. At any given instant, the curves of constant zenith angle make a set of concentric circles centered on the subsolar point, with a zenith angle of 90o along the great circle which at the given instant separates the dayside of the planet from the nightside. If the surface were in equilibrium with the instantaneous incident solar flux, the subsolar point would be the hottest spot on the planet, with temperature falling to zero with distance away from the hot spot. As the planet rotates through its day/night cycle, a given point of the surface is swept through a range of distances from the hot spot, leading to a diurnal temperature variation. As the planet proceeds through its orbit in the course of the year, the diurnal cycle will change as the orientation of the planet's rotation axis changes relative to the Sun. Insofar as the surface actually takes a finite amount of time to heat up or cool down, the diurnal cycle will be attenuated to one extent or another.

As the next step toward realism, let's now consider a rapidly rotating planet whose axis of rotation is perpendicular to the line connecting the center of the planet to the center of its Sun. If the axis of rotation is in fact perpendicular to the plane of the orbit, this situation prevails all year round; otherwise, the condition is met only at the equinoxes, and indeed the condition defines the equinoxes. We assume that the planet is rotating rapidly enough that the day-night difference in solar radiation is averaged out and the corresponding temperature fluctuations are small. In other words, the length of the day is assumed to be short compared to the characteristic thermal response time of the planet's surface, a concept which will be explored quantitatively in Section 7.4. Consider a small strip of the planet's surface near a latitude of angular width If a is the planet's radius, then the area of this strip is 2na2 cos(^)d^, if angles are measured in radians. The cross section area of the strip seen edge-on looking from the Sun determines the amount of solar flux intercepted by the strip. This area is 2a2 cos2(^)d^ when d^ is small. In consequence, the incident solar radiation per unit area at latitude ^ is L cos(^)/n. At the Equator, the solar radiation per unit area is L/n, which is somewhat greater than the value L/4 which we obtained in Chapter 3 by averaging solar radiation over the entire surface of the planet. If the planet has no atmosphere to transport heat or create a greenhouse effect, the equilibrium temperature as a function of latitude is

The temperature has its maximum at the Equator, and falls to zero at the poles.

Exercise 7.3.1 For the geometric situation described above, derive an expression for the cosine of the zenith angle as a function of latitude and longitude. Re-derive the expression for the daily-average distribution of solar absorption by averaging the cosine of the zenith angle along latitude circles.

Now we turn to the general case, in which the axis of rotation of the planet is not perpendicular to the plane containing the orbit. The angle between the perpendicular to the orbital plane, and the planet's axis of rotation, is known as the obliquity, and we shall call it 7. It can be regarded as constant over the course of a planet's year, though there are longer term variations which will be of interest to us later. The task now is to determine the solar zenith angle as a function of latitude, position along the latitude circle, and time of year.

Let the point P be the center of the planet, and S be the center of the sun. If we draw a line from P to S, it will intersect the surface of the planet at a latitude 5, which is called the latitude of the sun, or sometimes the subsolar latitude. It is a function of the orientation of the planet's axis alone, and serves as a characterization of where we are in the march of the seasons. If the obliquity of the planet is 7, then 5 ranges from 7 at the Northern Hemisphere summer solstice to —7 at the Southern Hemisphere summer solstice. Let Q be a point on the planet's surface, characterized by its latitude \$ and its "hour angle" h, which is the longitude relative to the longitude at which local noon (the highest sun position) is occurring throughout the globe. For radiative purposes, we just need to compute the zenith angle Z, defined previously. To get the zenith angle, we only need to take the vector dot product of the vector QS and the vector PQ. To do this, it is convenient to introduce a local Cartesian coordinate system centered at P, with the z-axis coincident with the axis of rotation, the x-axis lying in the plane containing the rotation axis and PS, and the y-axis orthogonal to the other two, chosen to complete a right-handed coordinate system.

First, note that by the definition of the dot product, 