Respuesta :
Answer:
a) The radiation pressure on the Earth is 6.065x10⁻⁶Pa
b) The atmospheric pressure is 1.665x10¹⁰ greater than the pressure exerted by the radiation.
Explanation:
Given:
I = intensity of solar radiation = 1368 W/m²
Earth reflects 33%, therefore Earth absorbs 67%
P = pressure = 101 kPa = 1.01x10⁵Pa
c = speed of light = 3x10⁸m/s
Questions:
a) Find the radiation pressure on the Earth, in pascals, at the location where the Sun is straight overhead, P = ?
b) State how this quantity compares with normal atmospheric pressure at the Earth's surface
a) To solve, it is necessary to calculate the pressure exerted by both the reflected light and the light that is absorbed, in this way:
The pressure exerted by the reflected light:
[tex]P_{1} =\frac{2*0.33*I}{c} =\frac{2*0.33*1368}{3x10^{8} } =3.01x10^{-6} Pa[/tex]
The pressure exerted by the absorbed light:
[tex]P_{2} =\frac{0.67*I}{c} =\frac{0.67*1368}{3x10^{8} } =3.055x10^{-6} Pa[/tex]
The radiation pressure on the Earth:
Pt = P₁ + P₂ = 3.01x10⁻⁶ + 3.055x10⁻⁶ = 6.065x10⁻⁶Pa
b) Comparing with normal atmospheric pressure
[tex]Ratio=\frac{P_{atm} }{P_{t} } =\frac{1.01x10^{5} }{6.065x10^{-6} } =1.665x10^{10}[/tex]
According to this result, the atmospheric pressure is 1.665x10¹⁰ greater than the pressure exerted by the radiation.
