where we drop the first term based on the assumption that the radius of the planet is a small fraction of its distance from the Sun. For the same reason, \QS\ in the denominator can with good approximation be replaced by \PS\, leaving the expression in the form of a dot product between two unit vectors. Letting n1 = PQ/\PQ\ and n2 = PS/\PS\, the unit vectors have the following components in the local Cartesian coordinate system.

n1 = (cos($) cos(h), cos($) sin(h), sin($)),n2 = (cos(5), 0, sin(5)) (7.6)

whence cos(Z) = cos($) cos(5) cos(h) + sin($) sin(5) (7.7) When cos Z < 0 the sun is below the horizon.

The cosine of the zenith angle attains a maximum value cos($ — 5) when h = 0, and a minimum value — cos($ + 5) when h = ±n. Both values are above the horizon when \$\ > \n/2 — 5\, corresponding to the perpetual polar summer day. Both values are below the horizon when \$\ > \n/2 + 5\, corresponding to the perpetual polar winter night. At the solstices, 5 takes on its extreme values of ±7. Therefore, perpetual day or night are experienced at some time of year for latitudes poleward of n/2 — 7. These circles are known as the Arctic and Antarctic circles on Earth. Apart from the case of perpetual day or night, there is a terminator which separates the illuminated from the dark side of the planet. The position of the terminator is given by cos ht = — tan($) tan(5) (7.8)

If ^ is the angular velocity of rotation of the planet, so that the planet's day is Tday = 2n/il time units, then the number of time units of daylight is 2ht/Q = (ht/n)Tday. We shall adopt the convention ht = 0 in the case of perpetual night, and ht = ±n for perpetual day.

Exercise 7.3.2 For a given latitude 0, what 5 yields the least hours of daylight? What 5 yields the most hours of daylight? The Earth's present obliquity is 23.5 degrees, and it's length of day is 23.94hours. Sketch a plot of the maximum and minimum hours of daylight vs. latitude for the Earth.

The diurnal variations of the zenith angle lead to hot days and cold nights. Where the thermal response time is long enough to average out an appreciable portion of the diurnal temperature variation, the daily mean incident solar flux is an informative statistic. Since the incident solar flux per square meter of surface is L cos Z, where L is the solar constant in W/m2, one can obtain the daily mean flux by averaging cos Z over a rotation period of the planet. This results in a nondimensional flux factor f, by which one multiplies the solar constant in order to obtain the daily mean solar radiation incident on each square meter of the planet's spherical surface. The daily average can be performed analytically, resulting in

= — [cos(^) cos(5) sin(ht) + sin(^) sin(5)ht] n where ht is determined by Eq. 7.8. This derivation of the daily average assumes that the length of the day is much less than the length of the year, so that 5 may be regarded as constant over the course of the day. If the length of the day is a significant fraction of the length of the year, as is be the case for nearly tide-locked planets like Mercury or Venus, the expression still gives the correct average along the latitude circle, but this average is no longer identical to the time average over a day.

During the equinoxes, 5 = 0 and f = cos(^)/n, independant of the obliquity. This agrees with the result we obtained earlier by direct geometrical reasoning. At other times of year, the daily mean flux is governed by two competing factors: the varying length of day, which tends to produce higher fluxes near the summer pole, and the average zenith angle, which tends to produce high fluxes near the subsolar latitude (which remains near the Equator if the obliquity is not too large). The latitude where the maximum daily mean insolation occurs is always between the subsolar latitude 5 and the summer pole. For 5 = 0 the maximum occurs at the Equator, and a little numerical experimentation shows that the latitude of the maximum increases to about 43.4o when 5 = 23.4o (and similarly, with reversal of signs, in the Southern hemisphere). For larger 5, the length-of-day effect wins out over the slant angle effect at the pole, and the maximum occurs at the summer pole itself. This state of affairs just barely happens at the solstice for the present obliquity of Earth and Mars; as a result, the summer hemisphere solstice insolation is fairly uniform in these two cases. It is also useful to note that the daily mean insolation at the summer pole exceeds the daily mean insolation at the Equator when |5| > 17.86o.

To obtain a general appreciation of the seasonal cycle, recall that 5 varies from —7 during the southern hemisphere summer solstice to 7 at the northern hemisphere summer solstice, taking on a value of zero at the equinox which lies between the two solstices. Consider a planet with uniform albedo, so that the absorbed solar radiation is determined by the distribution of incident solar radiation. Suppose further that the thermal response time is long enough to average out the diurnal cycle, but short compared to the length of the year. If the obliquity is below 23.4o, the "hot spot" starts some distance poleward of the Equator in the Southern hemisphere, moves to the Equator as the equinox is approached, and then migrates a similar distance into the Northern hemisphere as the Northern summer solstice is aproached. If the obliquity is greater than 23.4o, the hot spot starts at the South Pole, discontinously jumps to —43.4o at the point in the season

Fall

Fall

Spring

Figure 7.4:

Spring

Figure 7.4:

where the subsolar latitude crosses — 23.4o, smoothly migrates throught the Equator and on to 43.4o when the subsolar latitude approaches 23.4o, and then discontinuously jumps to the North Pole. Note that in either case, the hot spot crosses the Equator twice per year, at the equinoxes; the two solstices are the coldest times at the Equator. The climate at the Equator has a periodicity that is half the planet's year.

It only remains to express 5 as a function of the position of the planet in its orbit. The planet is spinning like a top, and if there are no torques acting on the planet (an assumption we will relax later) its angular momentum is conserved. Hence the rotation axis keeps a fixed orientation relative to the distant stars throughout the year. This is why Polaris is the Northern Hemisphere pole star all year around. Let k be the angle describing the position of the planet, as shown in Figure 7.4. We shall adopt the convention that k = 0 occurs at the Northern Hemisphere summer solstice. We shall refer to k as the season angle, but it is more commonly (and more obscurely) referred to as the longitude of the sun. In our case, we have defined the longitude of the sun relative to the Northern Hemisphere summer solstice, but other choices are also common, for example defining it relative to the Northern winter solstice or the Spring equinox. When discussing the progression through the seasonal cycle on planets other than Earth, the season angle is almost universally used to describe where the planet is in its cycle, since this description obviates the need to make up names for months for each planet. If we project the rotation axis onto the plane of the ecliptic (i.e. the plane containing the planet's orbit), then the angle made by this vector with PS is equal to k. The rotation axis projected onto the plane of the ecliptic acts like the hand of a clock, which rotates around the clock face once per year, though at a non-uniform rate if the orbit is not perfectly circular.

Let n be the unit normal vector to the plane of the ecliptic, and na be the unit vector in the direction of the rotation axis. Introduce a new cartesian coordinate system with x pointing along PS, z pointing along n, and y perpendicular to the two in a right-handed way. Then na = (cos(k) sin(Y), sin(K) sin(Y), cos(y)) and the latitude of the sun is the complement of the angle between na and the x axis, whence

Was this article helpful?

## Post a comment