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Calculating albedo and eccentricity

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  1. Apr 24, 2015 #1
    I have a three part question:

    Background: For a planet on an orbit with semi-major axis a and eccentricity e, the distance of closest approach to the Sun is r = a(1 − e) and the farthest approach is r = a(1 + e).

    (1) Assuming an albedo A = 0.2, estimate the temperature on Earth in equilibrium with irradiation from
    the Sun. Estimate the correction factor necessary due to the greenhouse effect to bring us up to a balmy 300 K.

    I don't know what to do.

    (2) Assuming that this correction factor does not change, how large an eccentricity could the Earth have before the temperature extremes reach the point where the Earth reaches either boiling or freezing point?

    Freezing:

    Based off the equation T=T⊙((1-A)/4)^0.25(R⊙/r)^0.5,

    273.15=5778((1-0.2)/4)^0.25(6.96*10^8/r)^0.5

    0.005=(6.96/r)

    r=1.39*10^11

    Boiling:

    373.15=5778((1-0.2)/4)^0.25(6.96*10^8/r)^0.5

    r=7.46*10^10

    These both seem reasonable to me except for one thing. The actual distance from the Sun to the Earth is 1.5*10^11. This means that the Earth is actually farther than the distance I calculated for the boiling part, which doesn't make sense.

    Is this the equation I should use, and is the work (and answer) correct? Or did I do something wrong?

    (3) If we define habitability as having a level of irradiation between these two extremes, consider the
    habitable zone around a lower mass star. Assuming circular orbits again, and the same greenhouse correction factor as above, where is the habitable zone around a 0.5M⊙ star, which has radius 0.5R⊙ and effective temperature 3700 K?


    Not sure where exactly to get started here.

    Do I use the equations

    L=4*pi*R⊙^2*stefan-boltzmann constant*T⊙^4
    Labs=((R⊙^2*stefan-boltzmann constant*T⊙^4*pi*R^2)/r^2)(1-A)?

    Not sure where the mass of the star fits in here.
     
  2. jcsd
  3. Apr 25, 2015 #2

    mfb

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    Staff: Mentor

    (1): You have the formula in (2), you just have to plug in numbers for the first part. And then find out which prefactor you need in the equation to get the actual average temperature.

    That comes from correction factor you have to include.

    (3): right
    You don't need it.
     
  4. Apr 25, 2015 #3
    Thanks for the reply.

    Not sure what you mean by prefactor. Would I just multiply the albedo times a value (i.e. x) and then solve for x? So:

    300=5778((1-0.2x)/4)^0.25(6.96*10^8/(1.5*10^11))^0.5

    ----> x=-1.75
     
  5. Apr 25, 2015 #4

    mfb

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    Staff: Mentor

    I'm sure you have a definition of the correction factor somewhere.
     
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