The sun having surface temperature `Ts` radiates like a black body. The rasius of sun is `Rs` and earth is at a distance R from the surface of sun. Earth absorbs radiations falling on its surface from sun only and is at constant temperature T. If radiations falling on earth's surface are alomost parallel and earth also radiates like a blackbody, then

Options

A. `T = T_(s) sqrt(R_(s)/(2R))`
B. `T = Ts`
C. `T = (T_(s))/(2) sqrt((R_(s)/(R ))`
D. `T = T_(s) sqrt((R_(s)/(R ))`

Explanation

To solve the problem, we will use the principles of black body radiation and the Stefan-Boltzmann law. Here’s a step-by-step solution:

### Step 1: Understand the situation
The sun radiates energy as a black body, and the Earth, which is at a distance \( R \) from the surface of the sun, absorbs this radiation. The Earth is also treated as a black body and is at a constant temperature \( T \).

### Step 2: Use the Stefan-Boltzmann Law
The power emitted by a black body is given by the Stefan-Boltzmann law:
\[ P = \sigma A T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T \) is the temperature.

### Step 3: Calculate the power emitted by the Sun
The surface area of the Sun \( A_s \) is:
\[ A_s = 4 \pi R_s^2 \]
Thus, the power emitted by the Sun \( P_s \) is:
\[ P_s = \sigma A_s T_s^4 = \sigma (4 \pi R_s^2) T_s^4 \]

### Step 4: Calculate the intensity of radiation at the Earth
The intensity \( I \) of the radiation at a distance \( R \) from the Sun is given by the power per unit area:
\[ I = \frac{P_s}{4 \pi R^2} = \frac{\sigma (4 \pi R_s^2) T_s^4}{4 \pi R^2} = \frac{\sigma R_s^2 T_s^4}{R^2} \]

### Step 5: Calculate the power absorbed by the Earth
The Earth has a radius \( r \), and its projected area \( A_e \) is:
\[ A_e = \pi r^2 \]
The power \( P_e \) absorbed by the Earth is:
\[ P_e = I \cdot A_e = \left( \frac{\sigma R_s^2 T_s^4}{R^2} \right) \cdot \pi r^2 \]

### Step 6: Set the power absorbed equal to the power emitted
At thermal equilibrium, the power absorbed by the Earth equals the power emitted by the Earth:
\[ P_e = \sigma A_e T^4 \]
Substituting for \( P_e \):
\[ \left( \frac{\sigma R_s^2 T_s^4}{R^2} \right) \cdot \pi r^2 = \sigma (\pi r^2) T^4 \]

### Step 7: Simplify the equation
We can cancel \( \sigma \) and \( \pi r^2 \) from both sides:
\[ \frac{R_s^2 T_s^4}{R^2} = T^4 \]

### Step 8: Solve for \( T \)
Rearranging gives:
\[ T^4 = \frac{R_s^2 T_s^4}{R^2} \]
Taking the fourth root:
\[ T = \left( \frac{R_s^2}{R^2} \right)^{1/4} T_s = \frac{R_s^{1/2}}{R^{1/2}} T_s \]

### Final Expression
Thus, the temperature of the Earth is given by:
\[ T = T_s \left( \frac{R_s}{R} \right)^{1/2} \]