Question

The force experienced by the Earth due to solar radiation is $9 \times 10^9$, N. Find the radiation pressure on Earth. (Radius of Earth R = $6.4 \times 10^6$, m)

Hint:

A highly common mistake is using the total surface area of the sphere ($4\pi R^2$). Always remember that parallel incoming light rays only ever "see" the flat circular shadow ($\pi R^2$) of a spherical object!

Answer 1

Step 1: Understanding the Concept:
Radiation pressure is defined exactly as the physical force exerted by electromagnetic radiation per unit area on a given target.
For a massive spherical body like the Earth absorbing radiation from a distant source (the Sun), the effective intercepting area is not the total surface area of the sphere, but rather its flat projected cross-sectional area.
Step 2: Key Formula or Approach:
The defining formula for pressure $P$ based on applied force $F$ and intercepting area $A$ is $P = \frac{F}{A}$.
The effective intercepting cross-sectional area of a sphere of radius $R$ is exactly a circle: $A = \pi R^2$.
Step 3: Detailed Explanation:
The total force exerted by the solar radiation is given directly as $F = 9 \times 10^9\text{ N}$.
The given radius of the Earth is $R = 6.4 \times 10^6\text{ m}$.
First, calculate the effective cross-sectional area intercepting the light:
\[ A = \pi R^2 \] \[ A = \pi (6.4 \times 10^6)^2 \] \[ A = \pi (40.96 \times 10^{12})\text{ m}^2 \approx 1.2868 \times 10^{14}\text{ m}^2 \] Now, directly substitute the force and the calculated area into the radiation pressure formula:
\[ P = \frac{9 \times 10^9}{1.2868 \times 10^{14}} \] \[ P \approx 6.99 \times 10^{-5}\text{ N/m}^2 \] Step 4: Final Answer:
The calculated radiation pressure acting on the Earth is approximately $6.99 \times 10^{-5}\text{ N/m}^2$.