Astronomy 11 - Fall 1998 (S.T. Myers)

Solutions to Problem Set #3

Solutions:

  1. The easiest thing to do is to scale from 1 AU

    F(r) = F( 1 AU ) [ r / 1 AU ]-2 = 1370 W/m2 [ r / 1 AU ]-2

    which gives F=50.7 W/m^2 at r=5.2 AU. Note that direct computation gives

    F = Lsun / (4r2)

    which agrees within our rounding of the solar constant 1370 W/m^2. Thus, to get our 500 W of power, we need an area of around

    Area = P / F 500 W / 50 W/m2 = 10 m2

    or a panel size of 3.3m x 3.3m! This is pretty big (10ft on a side) and is why Galileo carries an onboard nuclear power source.

    A flat panel oriented toward the Sun absorbs over the forward half of its surface area only, while it emits thermal radiation over both sides, so the power equilibrium is given by

    Pin = Fin L2 =
    Pout = T4 2L2
    => T = [ Fin/2 ]1/4 = 145 K.

    Note if you forgot the other side of the panel, you would have gotten just the subsolar temperature

    Tss = [ Fin/ ]1/4

    which we often use for our solar system in the form

    Tss = [ Lsun / (4 r2 ]1/4 = 394 K [ r / 1 AU ]-1/2.

  2. A spherical planet absorbs over its cross-section (a flat disk)

    Pin = Fin r2 ( 1 - A ) = Tss4 r2 ( 1 - A )

    which includes the correction for reflection with albedo A. The planet emits over its full surface area

    Pout = Tsurf4 4r2 ( 1 - G )

    where we have included the greenhouse factor G. In equilibrium Pin=Pout and thus

    Tsurf = Tss 2-1/2 [ ( 1 - A ) / ( 1 - G ) ]1/4

    where for our solar system

    Tsurf = 279 K [ r / 1 AU ]-1/2 [ ( 1 - A ) / ( 1 - G ) ]1/4

    using the scaling from problem 1. If A=G, we get the relation appropriate to the Earth

    Teq = Tss 2-1/2 = 279 K [ r / 1 AU ]-1/2

    which we call the "equilibrium temperature". Thus, for our habitable zone, we want

    279 K [ rinner / 1 AU ]-1/2 = 373 K => rinner = [279/373]2 = 0.56 AU
    279 K [ router / 1 AU ]-1/2 = 173 K => router = [279/173]2 = 2.58 AU

    where these are the inner and outer radii of the zone respectively. The habitable zone thus includes Venus (0.7 AU), Earth (1 AU) and Mars (1.5) AU. Note that the asteroids are generally not included, like Ceres (2.8 AU) though some Earth or Mars-crossing asteroids will be.

  3. The hottest point of the sunlit side of the Moon, directly beneath the Sun, is the subsolar point. If the lunar soil (or whatever object we were interested in) were perfectly absorbing blackbody, it would then have the subsolar temperature for 1 AU of 394 K! As it is, the lunar surface albedo is A=0.07 on average, so

    T = 394 K [ 1 - 0.07 ]1/4 = 387 K

    or nearly 114 C, which is above the boiling point! This is why astronauts were air-conditioned white colored space suits. The average temperature of the Earth was 279 K as mentioned above, or only 6 C, and thus the peak temperature of the moon is about 100 degrees higher!

  4. At r = 0.723 AU the subsolar temperature is 463 K and the equilibrium temperature is 328 K. Thus, our surface temperature should be given by

    Tsurf = Tss 2-1/2 [ ( 1 - A ) / ( 1 - G ) ]1/4 = Teq [ ( 1 - A ) / ( 1 - G ) ]1/4

    which for the measured surface temperature of 700 K and albedo A=0.76 gives

    ( 1 - G ) = ( 1 - 0.76 ) [328/700]4 = 0.0116 => G = 0.988

    and so Venus is so blanketed by clouds and greenhouse gases that only 1% of the surface radiation escapes! Zow!

  5. The two possibilities correspond to

    [A]
    Like Neptune A=0.62, G=A=0.62
    [B]
    Like Pluto A=0.5 G=0

    (a) We have from Problem 2

    T = 279 K [ 77.2 ]-1/2 [ ( 1 - A ) / ( 1 - G ) ]1/4 = 31.6 K [ ( 1 - A ) / ( 1 - G ) ]1/4

    so for our cases:

    [A]
    T = 31.6 K
    [B]
    T = 26.6 K

    (b) We use the Wien's Law

    max = 2.898 x 10-3 m [ T / 1 K ]-1 = 2898 µm / T(K)

    which gives

    [A]
    max = 92 µm
    [B]
    max = 109 µm

    which are in the far-infrared part of the spectrum.

    (c) We now assume R = 24800 km (Neptune, appendix A3), and the total blackbody luminosity is just flux times area

    L = 4r2 Tsurf4 = 4.37 x 1014 W [ T / 31.6 K ]4

    so

    [A]
    L = 4.37 x 1014 W
    [B]
    L = 2.19 x 1014 W

    The flux at the Earth (r=77.2-1=76.2 AU at closest approach) is

    F = L / (4r2 = 2.7 x 10-13 W/m2 [ L / 4.37 x 1014 W ]

    so we have

    [A]
    F = 2.7 x 10-13 W/m2
    [B]
    F = 1.3 x 10-13 W/m2

    which is very faint.

    (d) You should draw a skinny triangle with base D, height r, and apex angle theta, which approximates an arc of length D on a circle of radius r:

    D / r = (4.96x104 km)/(1.14x1010 km) = 4.35 x 10-6 rad x 206265 arcsec/rad = 0.9 arcsec

    which is just barely resolvable by most ground-based telescopes.

smyers@nrao.edu   Steven T. Myers