The sun delivers `10^3W//m^2` of electromagnetic flux to the earth's surface, which is at a distance of `1.5xx10^(11)m` from the sun. The whole incident electromagnetic flux is absorbed by the earth.
The radiation force on the roof is

Options

A. `2.67xx10^-4N`
B. `5.34xx10^-4N`
C. `2.33xx10^-4N`
D. `1.33xx10^-4N`

Explanation

To solve the problem of finding the radiation force on the roof due to the electromagnetic flux from the Sun, we can follow these steps:

### Step 1: Understand the Given Data
We are given:
- Electromagnetic flux (intensity) delivered by the Sun, \( I = 10^3 \, \text{W/m}^2 \)
- Distance from the Sun to the Earth, \( d = 1.5 \times 10^{11} \, \text{m} \)

### Step 2: Calculate the Total Power Incident on the Earth
The total power \( P \) incident on the Earth can be calculated using the formula:
\[ P = I \times A \]
where \( A \) is the area of the Earth that is exposed to the sunlight. The area \( A \) can be approximated as the cross-sectional area of the Earth, which is a circle:
\[ A = \pi r^2 \]
where \( r \) is the radius of the Earth. The average radius of the Earth is approximately \( r \approx 6.4 \times 10^6 \, \text{m} \).

Calculating the area:
\[ A = \pi (6.4 \times 10^6)^2 \approx 1.28 \times 10^{14} \, \text{m}^2 \]

Now, substituting the values to find total power:
\[ P = 10^3 \, \text{W/m}^2 \times 1.28 \times 10^{14} \, \text{m}^2 \approx 1.28 \times 10^{17} \, \text{W} \]

### Step 3: Calculate the Radiation Force
The radiation force \( F \) can be calculated using the formula:
\[ F = \frac{P}{c} \]
where \( c \) is the speed of light, approximately \( c \approx 3 \times 10^8 \, \text{m/s} \).

Substituting the values:
\[ F = \frac{1.28 \times 10^{17} \, \text{W}}{3 \times 10^8 \, \text{m/s}} \approx 4.27 \times 10^8 \, \text{N} \]

### Final Answer
The radiation force on the roof is approximately \( 4.27 \times 10^8 \, \text{N} \).

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