IB Physics Climate Model

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The following questions are taken from the book The Physics of Atmospheres by the late Sir John Houghton (3rd edition, 2007, Cambridge University Press).

To make a crude estimate of the surface temperature of a planet we can equate the solar radiation it absorbs with the infrared radiation it emits using the equation

In this equation the power of the radiation emitted by a planet of radius a (assumed to be emitting as a perfect black body) is equal to the average power absorbed by the planet having an albedo A at a distance R from the Sun (R is measured in terms of Earth distances from the Sun, for the Earth, R = 1 ). F is the solar constant.

  1. In the equation when the planet absorbs the solar radiation we use an area of 𝜋a2 but when the planet emits radiation we use an area of 4𝜋a2. Why is this? (The rays from the Sun are effectively parallel when they reach the planet and we use the area perpendicular to these rays which is the area of a circle, the cross sectional area that the rays "see". When the planet emits radiation it does so perpendicular (radially) outwards to its surface and so we use the surface area of a sphere)
  2. Given that F = 1370 W m-2 and for the Earth R = 1, A = 0.30, find the effective surface temperature of the Earth. (256 K). This result is lower than average temperature of 288 K. Why?
  3. Find the effective surface temperature for Venus given that R = 0.72 and A = 0.77. (227 K)
  4. Find the effective surface temperature for Mars given that R = 1.52 and A = 0.15. (216 K)
  5. Find the effective surface temperature for Jupiter using this model given that R = 5.20 and A = 0.58. The approximate measured surface temperature is 130 K. The difference is due to the energy generated inside Jupiter. (98 K)