91Ó°ÊÓ

The density of the spherical particle is$\mathrm{ÒÏ}=1.0×{10}^3{\text{kg/m}}^{\text{3}}$

The mass of the sun is$\text{M}=1.99×{10}^{30}\text{kg}$

The power radiation is $\text{P}=3.90×{10}^{26}\text{W}$

We can use the concept of Newton’s law of gravitation and radiation pressure.Both the forces will be the same for critical value but it will different for the radius of the particle. Due to the small radius of the particle, it will blow away by the radiation force of the sun.

Formula:

$\mathit{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathit{Ï€}{\mathit{r}}^{\mathbf{3}}$

$\mathit{ÒÏ}\mathbf{=}\frac{\mathbf{M}}{\mathbf{V}}$

${\mathit{F}}_{\mathbf{g}}\mathbf{=}\frac{\mathbf{G}\mathbf{M}\mathbf{m}}{{\mathbf{R}}^{\mathbf{2}}}$

$\mathit{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{A}}$

${\mathit{F}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{I}\mathbf{A}}{\mathbf{c}}$

$\mathit{A}\mathbf{=}\mathit{Ï€}{\mathit{r}}^{\mathbf{2}}$

If the radius is less than some critical radius $R$, the particle will be blown out of the solar system:

Let be the radius of the spherical particle, and its volume is

$V=\frac{4}{3}{\mathrm{Ï€°ù}}^3$

The expression of density is

$$\begin{gathered} \mathrm{ÒÏ}=\frac{M}{V} \\ M=\mathrm{ÒÏ}V \end{gathered}$$

$M=\mathrm{ÒÏ}\frac{4}{3}\mathrm{Ï€}{r}^3$

The gravitational force of the attraction of the sun on the particle is

$F_g=\frac{GMm}{R^2}$

where$R$is the distance between the particle and the sun.

$F_g=\frac{GM\left(\mathrm{ÒÏ}\frac{4}{3}\mathrm{Ï€}{r}^3\right)}{R^2}$

$F_g=\frac{4\mathrm{Ï€}GM\mathrm{ÒÏ}{r}^3}{3R^2}…\left(1\right)$

If$\text{P}$is the power output of the sun, then the intensity of the radiation is

$I=\frac{P}{A}$

The area of the sphere (sun) is

role="math" localid="1662973058834" $$\begin{gathered} A=4\mathrm{Ï€}{R}^2 \\ I=\frac{P}{4\mathrm{Ï€}{R}^2} \end{gathered}$$

When a surface intercepts the electromagnetic radiation, a force is exerted on the surface. If the radiation is absorbed by the surface, the force is

${\text{F}}_{\text{r}}=\frac{\text{IA}}{\text{c}}$

Here,$A$is the area of the surface perpendicular to the path of radiation. All the radiation passes through a particle of radius$r$, then the area of the particle is

$A={\mathrm{Ï€°ù}}^2$

$F_r=\frac{P}{4{\mathrm{Ï€¸é}}^2}×\frac{{\mathrm{Ï€°ù}}^2}{c}$

${\text{F}}_{\text{r}}=\frac{{\text{Pr}}^{\text{2}}}{{\text{4R}}^{\text{2}}\text{c}}…\left(2\right)$

The force is acting on the particle along the radius of the sun. From equations (1) and (2) the gravitational force and the radiation forces are inversely proportional to the $R^2$. If one of the forces increases, then the other can increase for all distances.

${\text{F}}_{\text{g}}$ is directly proportional to ${\text{r}}^{\text{3}}$ and ${\text{F}}_{\text{r}}$ is directly proportional to ${\text{r}}^{\text{2}}$. Due to the small radius of the particle, it blows away by the radiation force of the sun. If the two forces are equal, then the value of the radius is the critical value.

$\begin{array}{rcl}{F}_g& =& F_r\\ \frac{4\mathrm{Ï€}GM\mathrm{ÒÏ}{r}^3}{3R^2}& =& \frac{P{r}^2}{4R^{2}c}\\ r& =& \frac{3P}{16GMc\mathrm{ÒÏ}\mathrm{Ï€}}\end{array}$

The value of the critical radius:

$r=\frac{3P}{16GMc\mathrm{ÒÏ}\mathrm{Ï€}}$

$r=\frac{3×3.90×{10}^{26}\text{W}}{16×6.67×{10}^{-11}\text{N.}\frac{{\text{m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}×1.99×{10}^{30}\text{kg}×1.0×{10}^3\frac{\text{kg}}{{\text{m}}^{\text{3}}}×3×{10}^8\text{m/s}}$

$\text{r}=5.8×{10}^{-7}\text{m}$