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28P

Expert-verifiedFound in: Page 1002

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/m ^{2}.(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 ${\mathbf{\text{1.0}}}{\mathbf{\times}}{{\mathbf{\text{10}}}}^{{\mathbf{\text{5}}}}{\mathbf{\text{\hspace{0.17em}Pa}}}$**

- ${p}_{\text{r}}=4.7\times {10}^{-6}\text{\hspace{0.17em}N}/{\text{m}}^{2}$
- $\frac{{p}_{r}}{{P}_{e}}=4.7\times {10}^{-11}$

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 $\frac{{p}_{\text{r}}}{{P}_{e}}=4.7\times {10}^{-11}$.

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