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, we need to find the average force associated with the radiation pressure on a solar energy collecting panel of an earth-orbiting satellite. Here are the steps to arrive at the solution:

### Step 1: Understand the relationship between pressure, force, and area
We know that pressure (P) is defined as force (F) per unit area (A):
\[ P = \frac{F}{A} \]
Rearranging this gives us:
\[ F = P \times A \]

### Step 2: Determine the radiation pressure
The radiation pressure (P_radiation) can be calculated using the solar constant (S) and the speed of light (c). The formula for radiation pressure when the radiation is completely absorbed is:
\[ P_{radiation} = \frac{S}{c} \]
Where:
- \( S = 1.4 \, kW/m^2 = 1.4 \times 10^3 \, W/m^2 \)
- \( c = 3 \times 10^8 \, m/s \) (speed of light)

### Step 3: Calculate the radiation pressure
Substituting the values into the formula:
\[ P_{radiation} = \frac{1.4 \times 10^3 \, W/m^2}{3 \times 10^8 \, m/s} \]
Calculating this gives:
\[ P_{radiation} = \frac{1.4 \times 10^3}{3 \times 10^8} = 4.67 \times 10^{-6} \, Pa \]

### Step 4: Calculate the average force
Now that we have the radiation pressure, we can calculate the average force using the area of the solar panel (A = 5 m²):
\[ F = P_{radiation} \times A \]
Substituting the values:
\[ F = 4.67 \times 10^{-6} \, Pa \times 5 \, m^2 \]
Calculating this gives:
\[ F = 2.335 \times 10^{-5} \, N \]

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