Question

A star of radius $R_S$ and surface temperature of $T_S$, irradiates a planet of radius $R_E$. The distance between the star and the planet is $R_{SE}$. The average temperature $T_E$ on the surface of the planet can be estimated from balance of energy flux between incoming and outgoing radiation. Assuming that the radiation from the star is uniformly distributed over the planet, select the correct option.

• A. $T_E \propto R_E^2$
• B. $T_E \propto R_{SE}^{-1}$
• C. $T_E$ depends on Stefan-Boltzmann constant.
• D. $T_E \propto R_S^{1/2}$

Answer 1

Answer: (D)

(D)

Answer 2

To determine the average temperature $T_E$ on the surface of the planet, we need to balance the energy flux between the incoming radiation from the star and the outgoing radiation from the planet.

1. Power radiated by the star ($P_S$):
   The star radiates energy from its entire surface area $4\pi R_S^2$ at a surface temperature $T_S$. According to the Stefan-Boltzmann law, the power radiated per unit area is $\sigma T_S^4$, where $\sigma$ is the Stefan-Boltzmann constant.
   Therefore, the total power radiated by the star is:
   $P_S = (\sigma T_S^4) \times (4\pi R_S^2)$

2. Intensity of radiation from the star at the planet's distance ($I$):
   The power $P_S$ radiated by the star spreads out uniformly over a sphere of radius $R_{SE}$ (the distance between the star and the planet).
   The intensity (power per unit area) of the star's radiation at the planet's orbit is:
   $I = \frac{P_S}{4\pi R_{SE}^2} = \frac{\sigma T_S^4 (4\pi R_S^2)}{4\pi R_{SE}^2} = \frac{\sigma T_S^4 R_S^2}{R_{SE}^2}$

3. Power absorbed by the planet ($P_{abs}$):
   The planet intercepts the star's radiation over its cross-sectional area, which is $\pi R_E^2$. Assuming the planet absorbs all incident radiation (i.e., it behaves as a perfect black body absorber, absorptivity = 1):
   $P_{abs} = I \times (\pi R_E^2) = \left( \frac{\sigma T_S^4 R_S^2}{R_{SE}^2} \right) \times (\pi R_E^2)$

4. Power radiated by the planet ($P_{rad}$):
   The planet radiates energy from its entire surface area $4\pi R_E^2$ at its average temperature $T_E$. Assuming it radiates as a perfect black body (emissivity = 1):
   $P_{rad} = (\sigma T_E^4) \times (4\pi R_E^2)$

5. Energy balance:
   At thermal equilibrium, the power absorbed by the planet must equal the power radiated by the planet:
   $P_{abs} = P_{rad}$
   $\frac{\sigma T_S^4 R_S^2 \pi R_E^2}{R_{SE}^2} = \sigma T_E^4 (4\pi R_E^2)$

6. Solve for $T_E$:
   We can cancel common terms from both sides of the equation: $\sigma$, $\pi$, and $R_E^2$.
   $\frac{T_S^4 R_S^2}{R_{SE}^2} = 4 T_E^4$
   Rearranging to solve for $T_E^4$:
   $T_E^4 = \frac{T_S^4 R_S^2}{4 R_{SE}^2}$
   Taking the fourth root of both sides:
   $T_E = \left( \frac{T_S^4 R_S^2}{4 R_{SE}^2} \right)^{1/4}$
   $T_E = T_S \left( \frac{R_S^2}{4 R_{SE}^2} \right)^{1/4}$
   $T_E = T_S \frac{(R_S^2)^{1/4}}{(4)^{1/4} (R_{SE}^2)^{1/4}}$
   $T_E = T_S \frac{R_S^{1/2}}{\sqrt{2} R_{SE}^{1/2}}$

Now, let's analyze the given options based on this derived relationship for $T_E$:

• (A) $T_E \propto R_E^2$: From the derivation, $R_E$ (radius of the planet) cancels out. Thus, $T_E$ does not depend on $R_E$. So, this option is incorrect.
• (B) $T_E \propto R_{SE}^{-1}$: From the derivation, $T_E \propto R_{SE}^{-1/2}$. So, this option is incorrect.
• (C) $T_E$ depends on Stefan-Boltzmann constant: The Stefan-Boltzmann constant $\sigma$ cancels out from both sides of the energy balance equation. Thus, $T_E$ does not depend on $\sigma$. So, this option is incorrect.
• (D) $T_E \propto R_S^{1/2}$: From the derivation, $T_E \propto R_S^{1/2}$. So, this option is correct.