Interplanetary space contains many small particles referred to as interplanetary dust. Radiation pressure from the sun sets a lower limit on the size of such dust particles. To see the origin of this limit, consider a spherical dust particle of radius R and mass density r. (a) Write an expression for the gravitational force exerted on this particle by the sun (mass M) when the particle is a distance r from the sun. (b) Let L represent the luminosity of the sun, equal to the rate at which it emits energy in electromagnetic radiation. Find the force exerted on the (totally absorbing) particle due to solar radiation pressure, remembering that the intensity of the sun’s radiation also depends on the distance r. The relevant area is the cross-sectional area of the particle, not the total surface area of the particle. As part of your answer, explain why this is so. (c) The mass density of a typical interplanetary dust particle is about 3000 kg/m3 Find the particle radius R such that the gravitational and radiation forces acting on the particle are equal in magnitude. The luminosity of the sun is . Does your answer depend on the distance of the particle from the sun? Why or why not? (d) Explain why dust particles with a radius less than that found in part (c) are unlikely to be found in the solar system. [Hint: Construct the ratio of the two force expressions found in parts (a) and (b).]

The dust particle is far a distance r from the sun with mass density and radius R. The dust particle is in the shape of a sphere. So, its volume is$\mathrm{V}=\frac{4}{3}{\mathrm{\pi R}}^3$

The mass of the particle is m and it is related to the volume where the densityis the mass per unit volume. So, we can get an expression for m by

$$\begin{gathered} \mathrm{p}=\frac{\mathrm{m}}{\mathrm{V}} \\ \mathrm{m}=\mathrm{pV} \\ \mathrm{m}=\mathrm{p}\frac{4}{3}{\mathrm{\pi R}}^3 \end{gathered}$$

To find the force of gravity of the Sun on the dust particle, we use Newton's general law

$$\begin{gathered} \mathrm{F}=\mathrm{G}\frac{\mathrm{mM}}{{\mathrm{r}}^2} \\ =\mathrm{G}\frac{\mathrm{M}}{{\mathrm{r}}^2}\mathrm{p}\frac{4}{3}{\mathrm{\pi R}}^3 \end{gathered}$$

The intensity I of the output light is the power per unit area. The average power output of the light L is related to the intensity and the area of the spread spot by

$\mathrm{l}=\frac{\mathrm{L}}{\mathrm{A}}=\frac{\mathrm{L}}{4{\mathrm{\pi r}}^2}$

The wave exerts an average force ${\mathrm{F}}_{\mathrm{rad}}$per unit area and this is the radiation pressure Prad and it is the average value of dp/dt divided by the area. So, the radiation pressure of the wave that totally absorbed is given by equation

${\mathrm{p}}_{\mathrm{rad}}=\frac{\mathrm{l}}{\mathrm{c}}=\frac{\mathrm{L}/4{\mathrm{\pi r}}^2}{\mathrm{c}}=\frac{\mathrm{L}}{4{\mathrm{\pi r}}^2\mathrm{c}}$

For the area of the sphere, we can get the force that exerted on that dust particle by the radiation by the next equation

$$\begin{gathered} {\mathrm{F}}_{\mathrm{rad}}={\mathrm{p}}_{\mathrm{rad}}\mathrm{A} \\ =\frac{\mathrm{L}}{4{\mathrm{\pi r}}^2\mathrm{c}}\left({\mathrm{\pi R}}^2\right) \end{gathered}$$

A condition given that the gravitational and radiation forces acting on the particle are equal in magnitude$\mathrm{F}={\mathrm{F}}_{\mathrm{rad}}$ .Substitute the values we have

$$\begin{gathered} \mathrm{G}\frac{\mathrm{M}}{{\mathrm{r}}^2}\mathrm{p}\frac{4}{3}{\mathrm{\pi R}}^3=\frac{{\mathrm{LR}}^2}{4{\mathrm{r}}^2\mathrm{c}} \\ \mathrm{R}=\frac{3\mathrm{L}}{16\mathrm{\pi pMGc}} \\ =\frac{3\left(3.9\times {10}^{26}\right)}{16\mathrm{\pi }\left(1.99\times {10}^{30}\right)\left(6.67\times {10}^{-11}\right)\left(3\times {10}^8\right)} \\ =1.94\times {10}^{-7}\mathrm{m} \end{gathered}$$

Therefore, the radius R of the dust is$1.94\times {10}^{-7}\mathrm{m}$

The ratio of the two force expressions of F and ${\mathrm{F}}_{\mathrm{rad}}$that found in above steps is

$$\begin{gathered} \frac{{\mathrm{F}}_{\mathrm{rad}}}{\mathrm{F}}=\frac{\mathrm{L}/4\mathrm{c}}{\mathrm{G}{\displaystyle \frac{\mathrm{M}}{{\mathrm{r}}^2}}\mathrm{p}{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3} \\ =\frac{3\mathrm{L}}{16\mathrm{\pi cpMG}}\left(\frac{1}{\mathrm{R}}\right) \end{gathered}$$

As shown by this expression, the force of the radiation ${\mathrm{F}}_{\mathrm{rad}}$is inversely proportional to the radius of the dust${\mathrm{F}}_{\mathrm{rad}}\infty \frac{1}{\mathrm{R}}$So, if the radius of the dust is less than in part (c), the radiation force will be larger than the gravitational force. This leads to force the dust particles out of the solar system.