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Planetary Orbit from Albedo?

  1. Jul 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Assume that a planet can have an atmosphere if the escape speed of the planet is 6 times larger than the thermal speed of the molecules in the atmosphere (also known as the root-mean-square molecular velocity). Suppose a hypothetical object having the same mass and radius as Mercury, and an albedo of 0.1, orbits the Sun at just the right location for this condition to be met. Assume that its atmosphere is made of up of carbon dioxide. What is the radius of the this object's orbit around the Sun?

    2. Relevant equations

    v_esc = sqrt(2GM/R)
    M = M_Merc = 3.30 x 10^23 kg
    R = R_Merc = 2.439 x 10^6 m
    albedo = 0.1

    v_esc = 6 v_rms = 6 sqrt (3kT/M_CO2); M_CO2 = 44(1.67e-27) = 7.348e-26

    3. The attempt at a solution

    I equated v_esc and 6 v_rms:

    sqrt(2GM/R) = 6 sqrt(3kT/M_CO2)

    and solved for T, getting me a theoretical maximum temperature of 889.87K.

    After this I'm completely stuck. I don't know where albedo kicks in, and I don't know how to get the radius of an orbit around the Sun from the rms velocity or the escape velocity.

    Any help would be appreciated. Thanks!
     
  2. jcsd
  3. Jul 14, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    Once you have the surface temperature you don't need velocities any more. This the astronomy part: if a planet absorbs .1 of the incoming light and acts as a perfect blackbody emitter for infrared, how does its equilibrium temperature depend on the incoming radiation (which then depends on the orbital radius)?
     
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