W = σ T4where T is in kelvins and
σ = 5.6703 ×10-8 W m-2 K-4is the Stefan-Boltzmann constant. (A "black body" is an idealised object which radiates heat energy according to a certain simple physical model. The black body approximation holds quite well for most real objects.) Now let E represent total power radiated (= W × surface area), and r represent radius. We have:
Tplanet = (Wplanet / σ)1/4but:
Wplanet= Eplanet / (4 π rplanet2 )Now the planet intercepts a fraction of the sun's total radiation equal to the fraction of solid angle it subtends at the sun (i.e. the fraction of the whole "sky" seen from the sun that is covered by the planet). However, not all of the intercepted radiation is absorbed by the planet. A significant fraction, called the albedo α, is reflected by clouds, the atmosphere in general, and the surface. Another fraction &beta is absorbed by the atmosphere rather than the planet's surface (which is where we are interested in the temperature).
The amount of radiation actually absorbed by the planet's surface is equal to (1 - α - β) times the incident radiation, plus the amount of radiation emitted from the atmosphere in a downwards direction. The atmosphere has an inward-facing surface area equal to the planet's surface area, so this component is equal to 4 π rplanet2 σ Tatm4. For equilibrium, this sum must equal the total power radiated from the planet:
Eplanet = 4 π rplanet2 σ Tatm4 + (1 - α - β) Esun (π rplanet2 ) / (4 π d2 ),[equation 1] where d is the orbital distance of the planet. And:
Esun = 4 π rsun2 WsunNow consider the total outgoing radiation from the planet and atmosphere. This must equal (1 - α) times the incident radiation, so:
= 4 π rsun2 σ Tsun4
4 π rplanet2 σ ( Tatm4 + Tplanet4 ) = (1 - α) Esun[equation 2].
Solving equations 1 and 2 simultaneously for Tplanet gives:
Tplanet = Tsun √[rsun √(1 - α - β / 2) / 2d ]Note that the size of the planet is irrelevant.
For example: Plugging in
the radius of the sun = 6.963 ×108 mgives TEarth = 250 K, or -23 celsius.
Earth's average orbital distance = 1.496 ×1011 m
sun surface temperature = 6000 K
Earth's albedo = 31% = 0.31
absorption by Earth's atmosphere = 26% = 0.26
This is significantly colder than the true average temperature of the Earth's surface, which is 13 celsius (286 K). The difference is caused by the famous greenhouse effect. This is a warming effect caused by the existence of certain gases in the atmosphere (notably carbon dioxide, but also including methane and various artificial industrial gases). Such greenhouse gases are transparent to the visible portion of the electromagnetic spectrum, which corresponds to most of the incoming solar radiation, but strongly absorb infrared radiation, which is where most of the Earth's thermal radiation is emitted. The overall effect is to raise the Earth's equilibrium temperature to a somewhat higher value, where the increased thermal radiation compensates for the re-absorption of energy by the greenhouse gases.
The simplest way to account for greenhouse warming is to add a certain greenhouse effect temperature increase TG to the formula derived above:
Tplanet = TG + Tsun √[rsun √(1 - α - β / 2) / 2d ]For the Earth, TG = 36 K.
Note that the temperature derived is an average over all latitudes, longitudes, times of day, and seasons. Geographical and time variations from this average can be quite substantial - up to 30-40 K or even more.
Another couple of effects which can raise the surface temperature are latent "geo"thermal heat, and tidal forces. The Earth is hot inside because of the heat released by the slow decay of radioactive nuclides. Some of this heat can escape to the surface via vulcanism. The effect on the surface of the Earth is pretty small, except in close proximity to active vulcanism. (Volcanic eruptions can also have the opposite effect of reducing temperatures by injecting clouds of dust into the atmosphere and increasing atmospheric absorption.) Gas giant planets may still be hot internally because they have not yet had time to cool after their formation. This may the case for the gas giants in our solar system - which have been measured to radiate considerably more energy than they receive from the sun.
Tidal forces only effect a body orbiting very close to its primary. An example is the Jovian moon of Io, which is internally heated by the tidal stresses Jupiter causes in it, and is hence the most active volcanic body in the solar system.
Wplanet= Eplanet / (2 π rplanet2 )and the final formula comes out to:
Tplanet = TG + Tsun √[ (rsun / d) √((1 - α - β / 2) / 2) ]Using the figures from the Earth example above, if Earth was tidally locked the "hot" side would have an average temperature of TG + 298K = 334K, or 61 celsius, which is far too hot for humans.
The "cold" side of an airless, tidally locked planet would have a temperature approaching absolute zero (-273 celsius), but can never actually reach this figure. In practice, some heat will flow through the planet itself from the hot side. If there is an atmosphere, wind circulation will act to distribute the heat. The coldest temperature must be above the condensation point of the atmosphere, otherwise the gases would liquefy or freeze on to the surface. The drop in atmospheric pressure would suck more of the atmosphere towards the cold area, until most or all of the atmosphere is frozen out, leaving an airless world. Even if this fate is avoided, cold side of a tide-locked world will still be intolerably cold, probably -100 celsius or more, unless the atmosphere is very dense.
Also note that the average temperature of the hot side is likely to be distributed such that the centre of the hot side (where the sun is always directly overhead) will be hotter than the edges (where the sun is always on the horizon).
In fact, if we assume there is no atmosphere to circulate heat, nor any other appreciable heat circulation mechanism, then we can calculate the variation of temperature with latitude. (Here we define latitude such that the terminator, where the sun is on the horizon, is at 0 degrees, and the hot "pole", where the sun is permanently directly overhead, is at 90 degrees.) Note that no atmosphere means no atmospheric absorption and no greenhouse effect.
In this case, the incident radiation power on a unit area patch of planet surface at latitude θ is
Eunit = (1 - α) σ Tsun4 rsun2 sinθ / d2and the power radiated by the unit area is
Eunit = σ Tunit4Equating these and solving for the unit area temperature gives (this equation is shown in both graphics and text formats):
Tplanet(θ) = Tsun √[rsun √( (1 - α) sinθ) / d ]
This makes the hot pole √2 times as hot as the mean temperature of a rotating planet. This drops off as a fourth root of sin(latitude) function, reaching absolute zero at the terminator. Obviously it's not going to get quite that cold there in reality, but the terminator will be pretty much the same temperature as the night side - only a few degrees above absolute zero.
Atmosphere will even out some of this temperature variation, most importantly warming up the terminator and dark side areas with heat flow from the hot side, but also cooling down the hot pole somewhat.
What this calculation does give for a planet with an atmosphere is an upper limit to the temperature of the hot pole - namely only √2 times as hot as the mean temperature of a rotating planet. Of course in Kelvins, multiplying a temperature like 13 Celsius (286 K) by √2 gives 131 C (404 K). So an earthlike planet which was tide-locked might have a hot pole well above the boiling point of water - but not hot enough to melt lead.
The root-mean-square velocity of a gas particle is roughly √(3kT/m) where k is Boltzmann's constant, and T and m are the gas temperature and particle mass. For oxygen this is about 28 √(T) in SI units. For Earth the escape velocity is about 11,000 metres per second, which corresponds to 162,000 Kelvin for oxygen. If the air was this temperature, we would lose most of our oxygen within a matter of minutes. Luckily for us, the air never gets this hot!
This RMS velocity is a sort of average gas particle velocity, so there will be some gas particles with substantialy higher and lower velocities. What this means is that even though the gas temperature may be well below that required to make the RMS velocity near escape velocity, there will still be a few particles with velocities high enough to escape the atmosphere. If the difference between RMS and escape velocities is large, the gas will trickle away very slowly, taking billions of years to make any substantial difference. This is the case for oxygen on Earth.
For hydrogen the temperature comes to about 10,000 Kelvin. The air never gets anywhere near this hot either... at ground level. But up in the ionosphere, where it is heated by the full blast of the sun's radiation and particle streams, it gets close. So all of our hydrogen has leaked away on a timescale that is short compared to the age of the Earth.
Basically, if you want to avoid having your nice oxygen-nitrogen atmosphere leak away into space, you can make the air as hot as you please, as long as you don't intend to make it hot enough to ionise the atmosphere. If this was the case you certainly wouldn't want to live there anyway.
On planets substantially smaller than Earth (think Mercury, or the moon) the escape velocity is small enough that any gas will leak away in a relatively short time (compared to the planet's lifetime). On a Mars size world, again, so long as you don't ionise the atmosphere you're pretty safe.
Some particular cases, assuming our own Earth and sun, are noteworthy:
For information, star temperatures range from about 4000K for red dwarfs and red giants, to about 40 000K for blue supergiants. Stellar radii are listed in the following table. Interpolate between these values for stars of intermediate spectral types.
|Main Sequence Stars|
|M I, M II||100+||7.0×1010|