The solar constant at the mean position of the Earth is 1.40 times 103 W.m-2; th

Question

The solar constant at the mean position of the Earth is 1.40 times 103 W.m-2; the mean distance from the Earth to the Sun is 1.496 times 106 km. What is the total power developed by the Sun? Using this result calculate the surface temperature of the Sun. The radius of the Earth is 6.98 times 105 km and Stefan's constant is 5.67 times 10-8 in SI units. Using the solar constant given above, calculate the mean total power intercepted by the Earth. The radius of the Earth is 6400km. Use the above result to calculate the surface temperature of the Earth in the absence of the atmosphere and assuming that the surface is totally black, that is the albedo is zero.

Explanation / Answer

The strategy is much more applicable to Mars than to the Earth. Mars is a much simpler planet. The assumptions to this strategy are as follows: 1. The planet is thermally dormant and generates no heat of its own 2. The planet is sufficiently conductive and always has a uniform temperature 3. Greenhouse effects on the planet are non-existent 4. Surface of the planet can be treated as a gray body 5. The parent star radiates as a black body Black body radiation power emitted by the parent star: Q_dot_s = sigma*As*Ts^4 Irradiative flux received at orbit of the planet, in terms of orbit surface area Ao: q"_o = Q_dot_s/Ao Amount of this irradiative flux received as incident radiation power by the planet: Q_dot_in = q"_o*Adisc Consolidate previous formulas: Q_dot_in = sigma*Adisc*As*Ts^4/Ao Now let's consider the planet radiating its heat, as a black body, to the cosmic background. Even if it were a gray body and we applied a planetary emissivity to the equation, it would cancel out. The only way for which it wouldn't cancel out is if its solar spectrum absorptivity weren't equal to its cold-spectrum emissivity, in which it wouldn't be a gray body. The planet radiates like a sphere: Q_dot_out = sigma*Ap*T^4 Equate heat in = heat out as per conservation of energy and thermally dormant assumption: sigma*Ap*T^4 = sigma*Adisc*As*Ts^4/Ao Cancel Stefan-Boltzmann constant: Ap*T^4 = Adisc*As*Ts^4/Ao Now find the areas in terms of dimensions: Radius of planet: Rp Orbital radius of planet: r Radius of parent star: Rs Hence areas are: Adisc = pi*Rp^2 Ap = 4*pi*Rp^2 Ao = 4*pi*r^2 As = 4*pi*Rs^2 Substitute: 4*pi*Rp^2*T^4 = pi*Rp^2*4*pi*Rs^2*Ts^4/(4*pi*r^2) Cancel out all the pi's: 4*Rp^2*T^4 = Rp^2*4*Rs^2*Ts^4/(4*r^2) Cancel Rp^2 from both sides: 4*T^4 = 4*Rs^2*Ts^4/(4*r^2) Cancel the 4's: 4*T^4 = Rs^2*Ts^4/(r^2) Solve for T: T^4 = Rs^2*Ts^4/(4*r^2) Resulting expression: T = Ts*sqrt(Rs/(2*r)) Data for Earth and the sun: Ts:=5778 K; Rs:=0.004652 AU; r:=1 AU; Result: T = 278.7 Kelvin

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