Radiation pressure (corpuscular calculation)
Take 20 minutes to prepare this exercise.
Then, if you lack ideas to begin, look at the given clue and start searching for the solution.
A detailed solution is then proposed to you.
If you have more questions, feel free to ask them on the forum.
The energy \(E\) and the momentum \(\vec p\) of a photon are given respectively by :
\(E = h\nu = \frac{{hc}}{\lambda }\;\;\;\) and \(\;\;\;\vec p = \frac{E}{c}\;\vec u = \frac{h}{\lambda }\;\vec u\)
Where \(\vec u\) is a unit vector directed in the direction of propagation, \(h\) Planck's constant and \(c\) the speed of light in vacuum.
Assume the isotropic and monochromatic solar radiation, average wavelength \(\lambda\).
Earth, located on average at a distance \(d = 150 \;millions \; km\) from the Sun, receives solar radiation power per unit area equal to\(\Pi_S\).
We neglect the absorption of solar radiation by the atmosphere.
Question
Indicate how varies the power received per unit area with the distance \( r\) from the source to the receiver and literally calculate the total power \(P\) emitted by the Sun in the form of electromagnetic radiation.
Solution
The radiation emitted by the Sun being isotropic, the power \(P\) emitted by the Sun is also distributed over a sphere centered on the Sun and radius \(r\).
Therefore, received power per unit area at distance \(d\) is :
\(\Pi _S = P/4\pi d^{\;2}\)
Finally :
\(P = 4\pi {d^{\;2}}{\Pi _S} = {4,2.10^{26}}\,W\)
Question
Consider a spherical material particle, perfectly reflecting, center M, with radius \(a_0\) and uniform density \(\mu\), located at a distance \(r\) from the Sun.
The light from the sun reflects on it according to the laws of geometrical optics and it is assumed that all the incident light is reflected in the reverse direction as in the case of a plane mirror at normal incidence.
Determining the mean value of the force \(\vec F_R\) due to solar radiation experienced by the particle.
We express the literal result as the function of \(P\), \(a_0\) and \(r\).
Derive the pressure of solar radiation on the particle.
Solution
A photon of initial momentum :
\(\vec p_0=\frac {h}{\lambda} \;\vec u_0\)
Restart after reflection on the particle M in the opposite direction by yielding to this momentum :
\(\vec P_R = 2\vec p_0 =2 \frac {h}{\lambda} \;\vec u_0\)
During the time interval \(dt\), the light energy \(dE\) received by the particle :
\(dE = (P/4\pi {r^2})\,\pi a_0^2\,dt\)
Corresponds to the number of photons :
\(dN = dE/h\nu = dE/(hc/\lambda )\)
The impulsion \(d\vec P_R\) received by the particle during \(dt\) is then :
\(d{\vec P_R} = (dN)\,{\vec P_R} = (Pa_0^2/2c{r^2})\,dt\,{\vec u_0}\)
The average force exerted on the particle is therefore :
\({\vec F_R} = \frac{{d{{\vec P}_R}}}{{dt}} = \frac{{Pa_0^2}}{{2c{r^2}}}\,{\vec u_0}\)
The pressure is deduced :
\(p=\frac {F_R}{\pi a_0^2}=\frac {P}{2\pi cr^2}\)