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Earth's surface temp without atmosphere vs moon's

  1. Jun 3, 2007 #1
    Without an atmosphere, the earth's average surface temperature would be -18 C.

    The moon's surface temperature averages -23 C. Is this 5 degree C difference due to heat from the earth's core?

    (I'm not a student; I'm curious about the climate debate, and trying to figure some things out my own way.)


  2. jcsd
  3. Jun 3, 2007 #2
    I believe heat from the Earth's core is negligible compared to solar radiation. I'm not going to check your figures, but I expect the Moon would be colder due to its higher albedo i.e. the Moon reflects more sunlight than the Earth, and thus is not as hot.
  4. Jun 3, 2007 #3

    Claude Bile

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    The Earth would also receive more radiation because it is bigger, or more accurately, occupies more of the suns "field of view" as it were.

  5. Jun 3, 2007 #4
    likewise it will also radiate more

    I would bet good money that the two would cancel out. in fact I would imagine that ifthe albedos were the same the earth would be cooler than the moon due to the fact that surface area increases faster than cross-sectional area with respect o radius.
  6. Jun 4, 2007 #5


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    1 lunar day = 29 earth days (approx) so the temperature range on the moon surface would be much larger than on the earth surface. So the radiation from the "hot" parts of the moon surface would be greater than from the "hot" parts of the earth. Since radiation from the surface into "empty space" is proportional to T^4 (not a linear function of temperature), this would mean that relatively more energy was lost from the moon for the same average temperature.

    The radiation received from the sun is independent of the earth or moon temperature.

    So if everything else was the same, this nonlinear effect would make the "average" surface temperature lower on the moon.

    The different sizes of earth and moon make no difference to the radiation balance. The total surface area for outward radiation, and the projected (circular disk) area for inward radiation, are both proportional to r^2. And the heating and cooling effects only penetrate to a small depth compared with the radius.

    The heat from the earths core (which is partly generated by radioactivity) could make a difference.

    BTW, I haven't tried to put any numbers to the hypotheses about the effect of day length - it would be a nice exercise in numerical solution of a differential equation, though.
  7. Jun 5, 2007 #6
    So in the middle of winter, when the outside temperature is 0 C,
    if I have a cup of coffee at 100C and another at 50 C, the first will cool by 50 degrees a lot faster than the second

  8. Jun 5, 2007 #7


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    Yes - though the reason isn't the same as the point I was making about radiation.

    Ignoring radiation, assuming Newton's law of convection cooling (heat flux proportional to temperature difference), and assuming all the coffee in the cup stays at the same temperature because of internal convection, then cooling from 100 to 50 would take the same time as from 50 to 25 (and from 25 to 12.5, 12.5 to 6.25, etc...)
  9. Jun 5, 2007 #8


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    This is a very crude simulation of the radiation effect I was proposing. Imagine an object with high thermal conductivity so its internal temperature is uniform. Cooling is by radiation proportional to T^4. Heat is supplied at a constant rate during the "day". The graphs show the temperature cycling once steady state has been reached for two different "day lengths". For short days, the temperature changes are almost linear with time. For longer days, the temp rise slows down as the amount of heat radiated away increases. The average amount of heat input is the same but the mean temperature is lower for long "days".

    Attached Files:

  10. Jun 5, 2007 #9
    OK, I should have probably checked my figures before I said that. Albedo Earth = 0.36, Albedo Moon = 0.07.
  11. Jun 5, 2007 #10


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    The Moon's albedo is lower, meaning it absorbs more energy per unit area than Earth. All other things equal, the Moon should be warmer.

    Radius makes no difference. If radius accounted for the difference, then we would expect real strange answers for small rocks 1mm in radius.

    Solar energy absorbed = infrarad energy reradiated

    4\pi R^2 \sigma T^4 = \pi R^2 \left( {1 - A} \right)F
    where R is the radius of the Earth, sigma is the Boltzmann constant, (5.67e-1 W/m^2 K^04), T is temperature in Kelvins, F is flux at distance from the Sun, and r is Earth's distance to the Sun.

    Pi and R2 appear on both sides of the = sign and thus can be cancelled. So radius has nothing to do with it. Flux is derived by:
    L/4\pi r^2
    where L is the luminosity of the Sun (3.827e26 W).

    Combining these 2 equations to solve for T gives:
    T = \sqrt[4]{{\frac{{\left( {1 - A} \right)L}}{{16\sigma \pi r^2 }}}}

    Or in degrees Celcius:
    T_{^\circ C} = \sqrt[4]{{\frac{{\left( {1 - A} \right)L}}{{16\sigma \pi r^2 }}}} - 273.16^^\circ K

    So for the Earth:
    T_{^\circ C} = \sqrt[4]{{\frac{{\left( {1 - 0.36} \right)\,3.827 \times 10^{26} {\rm{W}}}}{{16 \times 5.67 \times 10^{ - 8} {\rm{W/m}}^2 \cdot {\rm{K}}^{ - 4} \pi \left( {{\rm{149597870691m}}} \right)^2 }}}} - 273.16^^\circ {\rm{K = }} - 24.2263113319468

    And for the Moon:
    T_{^\circ C} = \sqrt[4]{{\frac{{\left( {1 - 0.07} \right)\,3.827 \times 10^{26} {\rm{W}}}}{{16 \times 5.67 \times 10^{ - 8} {\rm{W/m}}^2 \cdot {\rm{K}}^{ - 4} \pi \left( {{\rm{149597870691m}}} \right)^2 }}}} - 273.16^^\circ {\rm{K = }} 0.152451815813947

    Hmmm... I get a warmer moon.
    Don't worry that your Earth temp and mine are off by a few degrees. Tinkering with the solar luminosity and albedo can account for this difference.

    According to Astrobiology by Jonathan I. Lunine, this formula
    So perhaps the Moon's slower rotation accounts for the difference.
    Last edited: Jun 5, 2007
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