Question

Assume that a certain location on the Earth reflects 28.0% of the incident sunlight from its clouds and surface.

Answer 1

(a) The radiation pressure is $5.9\cdot 10^{-6} Pa$

(b) The atmospheric pressure is greater by a factor of $1.72\cdot 10^{10}$

Explanation:

(a)

We have two different formulas to calculate the radiation pressure due to an electromagnetic wave:

- For a wave completely reflected, the radiation pressure is

$p=2\frac{I}{c}$

where I is the intensity and c the speed of light

- For a wave completely absorbed, the radiation pressure is instead

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

Here the intensity of the wave is

$I=1370 W/m^2$

However:

- 28.0% of this radiation is reflected, so the intensity associated to the reflected part of the wave is

$I_1 = (0.28)(1370)=383.6 W/m^2$

So, the radiation pressure due to this component is

$p_1 = 2\frac{I_1}{c}=2\frac{383.6}{3\cdot 10^8}=2.6\cdot 10^{-6} Pa$

- The other 72.0% of the radiation is absorbed, so the intensity associated to the absorbed part is

$I_2 = (0.72)(1370)=986.4 W/m^2$

So the radiation pressure due to this component is

$p_2 = \frac{I_2}{c}=\frac{986.4}{3\cdot 10^8}=3.3\cdot 10^{-6} Pa$

So, the total radiation pressure is

$p=p_1 + p_2 = 2.6\cdot 10^{-6}+3.3\cdot 10^{-6}=5.9\cdot 10^{-6} Pa$

(b)

The atmospheric pressure here is given and it is

$p=101 kPa = 101\cdot 10^3 Pa = 1.01\cdot 10^5 Pa$

So, we immediatly see that it is much larger than the radiation pressure calculated in part a), which was

$p'=5.9\cdot 10^{-6} Pa$

In particular, atmospheric pressure is greater by a factor of

$\frac{p}{p'}=\frac{1.01\cdot 10^5}{5.9\cdot 10^{-6}}=1.72\cdot 10^{10}$

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