- Albedo
(common symbol α: see, e.g.,
Wikipedia:
Albedo:White-sky and black-sky albedo) is usually given as a
decimal fraction.
Albedo can be given in
The percentage form in which case it
is the decimal fraction
form times 100 %. This what the image uses.
In the analysis below, we use
symbol
α for
albedo and the
decimal fraction form---and
NOT the percentage form.
- The displayed land covers are as follows:
cirrus cloud,
crops,
desert,
forest,
ice,
meadows,
sand,
Savanna,
snow,
soil,
stratus cloud,
water.
- Effective temperature
is a fiducial-value
temperature for a
spherical body (e.g., star,
planet).
It is the temperature the
body would have if it radiated its
non-reflective luminosity
like an exact blackbody radiator.
If the body is approximately a
blackbody radiator, the
effective temperature
is approximately the actual surface temperature,
and so is a good characteristic or sort of average
temperature.
If the body is NOT approximately a
blackbody radiator, the
effective temperature is still often
a useful value for characterizing the body particularly for comparison to other bodies.
- We can do a derivation
of a useful formula
for the effective temperature
of planets.
- F = L/(4πr**2) : The formula
for flux from an isotropically radiating
star of
luminosity at
distance r from the star.
It is an inverse-square-law
formula
- P = πR**2*F : The formula for the
power captured by
a planet of
radius R at distance r from the
star.
- f = [P/(4πR**2)]*(1-α) = σ*T_eff**4 : The f is the
average power per unit area
absorbed into the
heat energy
of the planet---the
reflected
electromagnetic radiation (EMR)
is removed by the factor (1-α)---and
we have equated it to the
Stefan-Boltzmann law
for the flux radiated by
a blackbody radiator at
temperature T_eff
which is, of course, the
effective temperature of the
planet.
The
Stefan-Boltzmann constant
σ = 5.670367(13)*10**(-8) W/*m**2*K**4).
- Solving for T_eff gives
    T_eff = {[L/(16πr**2*σ)]*(1-α)}**(1/4)
or in a fiducial-value formula
    T_eff = ( 278.33000 K )*
(L/L_☉)**(1/4)*(r_⊕/r)**(1/2)*(1-α)**(1/4) ,
where
solar luminosity L_☉ = 3.828*10**26 W and
Earth's mean orbital radius
r_⊕ = 1.000001018 * 1 AU
(epoch J2000
and astronomical unit (AU) = 1.49597870700*10**11 m,
and where more than
significant figures
are given for the fiducial-value formula
to allow tests of reproducibility.
The formula agrees with the one given by
Wikipedia: Effective temperature: Planet.
- Using the above formula, we determine
the following
effective temperatures for
the
inner-Solar-System
planets:
Mercury ☿,
Venus ♀,
Earth ⊕,
Mars ♂.
_______________________________________________________________________
Table: Effective Temperatures for the Inner-Solar-System Planets
_______________________________________________________________________
Planet R_orbital_mean α_b T_eff T_eff T_(mean/fiducial)
α=0 α_b
(AU) (K) (K) (K)
_______________________________________________________________________
Mercury 0.387098 0.068 447.4 439.5 100 (night), 700 (day)
Venus 0.723332 0.90 327.3 184.0 740
Earth 1.000001018 0.306 278.3 254.0 288
Mars 1.523679 0.25 225.5 209.8 210
_______________________________________________________________________
Note: The α_b is
the bond albedo which is
the appropriate planetary albedo
for calculating effective temperature.
- The effective temperature
can only be a crude average temperature
if the planet has extreme day-night
temperature variation like
Mercury.
Recall the Mercury's
synodic day (i.e., solar day)
is 176 days which is almost twice
Mercury's
orbital period 87.9691 days
and almost 3 times its
sidereal rotation period 58.646 days
due
Mercury's 3:2
spin-orbit resonance.
It's a long cold night on Mercury
of ∼ 88 days.
- The effective temperature
departs increasingly from the actual average temperature
with increasing greenhouse effect.
Venus has an extreme
greenhouse effect
and Earth a moderate one.
- What would be planet's
average temperature with the
greenhouse effect turned off?
That depends on what other counterfactual
assumptions you make?
For example, if you turn off the
greenhouse effect
for Venus, but keep its
bond albedo α_b = 0.90,
then Venus will
have a very low average temperature
closely approximating the
effective temperature 184.0 K
given in the table above.
But Venus's
greenhouse effect
is caused by Venusian atmosphere
which also gives Venus its high
bond albedo.
Without the Venusian atmosphere,
Venus would probably have
a bond albedo similar to that
of the Moon 0.136.
Then Venus would have
probably have an average temperature
an average temperature
closely approximating the
effective temperature 327.3 K
given in the table above.
- Because the relatively short
Martian day = 25h,39m,35.244s)
(causing very small
thermal inertia)
very thin Martian atmosphere
(causing very little greenhouse effect)
and
relatively low bond albedo
(which roughly appoximates zero),
Mars's
actual average temperature
closely approximates the
effective temperatures
given in the table above.
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