An earth- orbiting satellite has a solar energy collecting panel with total area `5 m^2` If solar radiations are perpendicular and completely absorbed, the average force associated with the radiation pressure is (Solar constant `= 1.4 k Wm^(-2)` )

Options

A. `2.33xx10^(-3)N`
B. `2.33xx10^(-4)N`
C. `2.33xx10^(-5)N`
D. `2.33xx10^(-6)N`

Answer

To solve the problem of finding the average force associated with the radiation pressure on a solar energy collecting panel of an earth-orbiting satellite, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Area of the solar panel, \( A = 5 \, m^2 \)
- Solar constant (energy flux), \( S = 1.4 \, kW/m^2 = 1.4 \times 10^3 \, W/m^2 \)
- Speed of light, \( c = 3 \times 10^8 \, m/s \)

2. **Calculate the Radiation Pressure:**
The radiation pressure \( P \) can be calculated using the formula:
\[ P = \frac{S}{c} \]
Substituting the values:
\[ P = \frac{1.4 \times 10^3 \, W/m^2}{3 \times 10^8 \, m/s} \]
\[ P = \frac{1.4 \times 10^3}{3 \times 10^8} = 4.67 \times 10^{-6} \, Pa \]

3. **Calculate the Force:**
The force \( F \) associated with the radiation pressure can be calculated using the formula:
\[ F = P \times A \]
Substituting the values:
\[ F = 4.67 \times 10^{-6} \, Pa \times 5 \, m^2 \]
\[ F = 2.335 \times 10^{-5} \, N \]

4. **Final Answer:**
The average force associated with the radiation pressure is:
\[ F \approx 2.33 \times 10^{-5} \, N \]