Question

The surface temperature of the Sun is ${T_0}$ , and it is at an average distance ‘d’ from a planet. The radius of the Sun is R. What is the temperature at which the planet radiates the energy?

Answer 1

Hint: In this question, we need to determine the temperature at which the planet radiates the energy such that the surface temperature of the Sun is ${T_0}$ and it is ‘d’ distance away from the planet. For this, we will use the relation between the power radiated by the energy source, the temperature at which the power is radiated and surface area of the body.

Complete step by step answer:
Let, the radius of the planet be ‘r’ and the temperature at which the planet radiates energy be ${T_1}$.
The radius of the Sun is R, and the surface temperature of the Sun is ${T_0}$
Energy received by the planet is the product of the flux and the area of the surface through which the energy is radiated. Mathematically, ${E_{received}} = \phi A$ where $\phi $ is the flux and ‘A’ is the area.
Here, the flux is given as $\phi = \sigma {T_1}^4$ and the area is given as $A = 4\pi {d^2}$ as the average distance of the planet from the Sun is ‘d’.
So, the energy received by the planet is given as:
$
{E_{received}} = \phi A \\
\Rightarrow{E_{received}} = \left( {\sigma {T_1}^4} \right)\left( {4\pi {d^2}} \right) - - - - (i) \\
 $
Similarly, the energy radiated by the planet has been given as
$
{E_{radiated}} = \phi A \\
\Rightarrow{E_{radiated}} = \left( {\sigma {T_0}^4} \right)\left( {4\pi {R^2}} \right) - - - - (ii) \\
 $
According to the question, at Thermal Equilibrium:
Absorbed power = Radiated Power. So, ${E_{received}} = {E_{radiated}}$
Substituting the values from the equation (i) and (ii) in the equation ${E_{received}} = {E_{radiated}}$ to determine the temperature at which the planet radiates the energy.
$
{E_{received}} = {E_{radiated}} \\
\Rightarrow\left( {\sigma {T_1}^4} \right)\left( {4\pi {d^2}} \right) = \left( {\sigma {T_0}^4} \right)\left( {4\pi {R^2}} \right) \\
\Rightarrow{T_1}^4{d^2} = {T_0}^4{R^2} \\
\Rightarrow{T_1}^4 = \dfrac{{{T_0}^4{R^2}}}{{{d^2}}} \\
\therefore{T_1} = \sqrt[4]{{\dfrac{{{T_0}^4{R^2}}}{{{d^2}}}}} \\
 $
Hence, the temperature at which the planet radiates the energy such that the surface temperature of the Sun is ${T_0}$ and it is ‘d’ distance away from the planet is $\sqrt[4]{{\dfrac{{{T_0}^4{R^2}}}{{{d^2}}}}}$.

Note: It is interesting to note here that, at thermal equilibrium the energy (or the power) received by a body equals to the energy (or the power) radiated by the body.