As a comet swings around the Sun, ice on the comet’s surface vaporizes, releasing trapped dust particles and ions. The ions, because they are electrically charged, are forced by the electrically charged solar wind into a straight ion tail that points radially away from the Sun (Fig. 33-39). The (electrically neutral) dust particles are pushed radially outward from the Sun by the radiation force on them from sunlight. Assume that the dust particles are spherical, have density$\mathbf{\text{3.5}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{3}}}{\mathbf{\text{\hspace{0.17em}kg/m}}}^{\mathbf{\text{3}}}$ , and are totally absorbing

.(a) What radius must a particle have in order to follow a straight path, like path 2 in the figure?

(b) If its radius is larger, does its path curve away from the Sun (like path 1) or toward the Sun (like path 3)?

Density is, .$\rho =3500{\text{\hspace{0.17em}kg/m}}^{\text{3}}$

Newton's law of universal gravitation states that a particle attracts every other particle in the universe with a force, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. And radiation pressure depends on the intensity and speed of light.

Formulae are as follows:

Radiation force $\left(F_r\right)$is given by,

$F_r=\frac{IA}{c}$

Gravitation force $\left(F_g\right)$ is given by,

$F_g=\frac{G{M}_{s}m}{r^2}$

Where, is the radiation force, $F_g$is the gravitation force, l is the intensity, A is the area,C is the speed of light, $M_s$is the mass of the Sun, m is the mass of the dust particles, and G is the gravitational constant.

Here, radiation force and gravitation force balance each other.

$F_r=F_g$

Let's find these two forces first.

Radiation force $\left(F_r\right)$is given by,

$F_r=\frac{IA}{c}$

Substitute the values in the above expression, and we get,

$\begin{array}{l}{F}_r=\frac{P_s}{4\pi r^2}\times \frac{\pi R^2}{c}\\ F_r=\frac{P_s{R}^2}{4r^{2}c}\end{array}$ (2)

Here, R is the radius of the particle.$P_s$is the pressure of the sun.

Gravitation force$\left(F_g\right)$:

$F_g=\frac{G{M}_{s}m}{r^2}$

We know that $m=\frac{4\pi R^3\rho }{3}$.

Substitute the values in the above expression, and we get,

$F_g=\frac{G{M}_s\frac{4\pi R^3\rho }{3}}{r^2}$ $\left(2\right)$

Now equating equations 1 and 2 as;

$\frac{P_s{R}^2}{4r^{2}c}=\frac{G{M}_s\frac{4\pi R^3\rho }{3}}{r^2}$

Rearrangingtheabove equation for R :

$R=\frac{3P_s}{16\pi c\rho G{M}_s}$

Substitute the values in the above expression, and we get,

$\begin{array}{l}R=\frac{3(3.9\times {10}^{26})}{16\pi (3\times {10}^8)\times (3.5\times {10}^3)\times (6.67\times {10}^{-11})\times (1.99\times {10}^{30})}\\ R=1.7\times {10}^{-7}\text{\hspace{0.17em}m}\end{array}$

Hence, the radius is $R=1.7\times {10}^{-7}\text{\hspace{0.17em}m}$

$F_g$Varies with$R^3,$and$F_r$varies with$R^2$.

Therefore, the magnitude of gravitational force is greater than the radiation force.

Hence, the path will be curved, and the direction of the path will be towards the Sun.