Question: The average intensity of the solar radiation that strikes normally on a surface just outside Earth’s atmosphere is 1.4kw/m².(a) What radiation pressure P_r is exerted on this surface, assuming complete absorption?(b) For comparison, find the ratio of P_r to Earth’s sea-level atmospheric pressure, which is$\mathbf{\text{1.0}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{5}}}\mathbf{\text{\hspace{0.17em}Pa}}$

The average intensity of solar radiation is,

$I=1.4\times {10}^3{\text{\hspace{0.17em}W/m}}^{\text{2}}$

Earth’s sea level atmospheric pressure is,

$P_e=1.0\times {10}^5\text{\hspace{0.17em}Pa}$

Radiation pressure depends upon the intensity of radiation and the speed of light. Radiation pressure is the mechanical pressure exerted upon any surface due to the exchange of momentum between the object and the electromagnetic field. This includes the momentum of light or electromagnetic radiation of any wavelength that is absorbed, reflected, or otherwise emitted by matter on any scale.

Formulae are as follows:

Radiation pressure (P_r):

$p_r=\frac{I}{c}$

Where,$I-\text{intenisity}\left(\frac{\text{W}}{{\text{m}}^{\text{2}}}\right)$, P_r is the radiation pressure, c is the speed of light.

To find the radiation pressure P_r, we can use the formula as,

$p_r=\frac{I}{c}$

Substitute the values in the above expression, and we get,

$\begin{array}{l}{p}_{\text{r}}=\frac{1400{\text{\hspace{0.17em}W/m}}^{\text{2}}}{3\times {10}^8\text{\hspace{0.17em}m/s}}\\ p_{\text{r}}=4.7\times {10}^{-6}\cdot (\frac{1{\text{\hspace{0.17em}W/m}}^{\text{2}}}{1\text{\hspace{0.17em}m/s}}\times \frac{1\text{\hspace{0.17em}N}\cdot \text{m/s}}{1\text{\hspace{0.17em}W}})\\ p_{\text{r}}=4.7\times {10}^{-6}{\text{\hspace{0.17em}N/m}}^{\text{2}}\end{array}$

Therefore, the radiation pressure is

$p_{\text{r}}=4.7\times {10}^{-6}{\text{\hspace{0.17em}N/m}}^{\text{2}}$

The ratio of pressure P_r to Earth’s sea level atmospheric pressure can be calculated as,

$\begin{array}{l}\frac{p_{\text{r}}}{P_e}=\frac{4.7\times {10}^{-6}}{1\times {10}^5}\\ \frac{p_{\text{r}}}{P_e}=4.7\times {10}^{-11}\end{array}$

Therefore, the ratio of pressure P_r to Earth’s sea level atmospheric pressure is $\frac{p_{\text{r}}}{P_e}=4.7\times {10}^{-11}$.