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Found in: Page 1002

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# 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 Pr is exerted on this surface, assuming complete absorption?(b) For comparison, find the ratio of Pr to Earth’s sea-level atmospheric pressure, which is ${\mathbf{\text{1.0}}}{\mathbf{×}}{{\mathbf{\text{10}}}}^{{\mathbf{\text{5}}}}{\mathbf{\text{\hspace{0.17em}Pa}}}$

1. ${p}_{\text{r}}=4.7×{10}^{-6}\text{\hspace{0.17em}N}/{\text{m}}^{2}$
2. $\frac{{p}_{r}}{{P}_{e}}=4.7×{10}^{-11}$
See the step by step solution

## Step 1: Given data

The average intensity of solar radiation is,

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

Earth’s sea level atmospheric pressure is,

${P}_{e}=1.0×{10}^{5}\text{\hspace{0.17em}Pa}$

## Step 2: Determining the concept

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:

${p}_{r}=\frac{I}{c}$

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

## Step 3: (a) Determining theradiation pressure Pr is exerted on this surface, assuming complete absorption

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

${p}_{r}=\frac{I}{c}$

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

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

${p}_{\text{r}}=4.7×{10}^{-6}{\text{\hspace{0.17em}N/m}}^{\text{2}}$

## Step 4: (b) Determining theratio of Pr to Earth’s sea-level atmospheric pressure, which is 1.0×105 Pa for comparison.

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

$\begin{array}{l}\frac{{p}_{\text{r}}}{{P}_{e}}=\frac{4.7×{10}^{-6}}{1×{10}^{5}}\\ \frac{{p}_{\text{r}}}{{P}_{e}}=4.7×{10}^{-11}\end{array}$

Therefore, the ratio of pressure Pr to Earth’s sea level atmospheric pressure is $\frac{{p}_{\text{r}}}{{P}_{e}}=4.7×{10}^{-11}$.