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 density3.5×103 kg/m3 , 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)?

Short Answer

Expert verified

a .R=1.7×107 m

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

Step by step solution

01

Step 1: Given data

Density is, .ρ=3500 kg/m3

02

Determining the concept

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 (Fr)is given by,

Fr=IAc

Gravitation force (Fg) is given by,

Fg=GMsmr2

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

03

(a) Determining the radius a particle has in order to follow a straight path, like the path  2 in the figure

Here, radiation force and gravitation force balance each other.

Fr=Fg

Let's find these two forces first.

Radiation force (Fr)is given by,

Fr=IAc

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

Fr=Ps4πr2×πR2cFr=PsR24r2c (2)

Here, R is the radius of the particle.Psis the pressure of the sun.

Gravitation force(Fg):

Fg=GMsmr2

We know that m=4πR3ρ3.

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

Fg=GMs4πR3ρ3r2 (2)

Now equating equations 1 and 2 as;

PsR24r2c=GMs4πR3ρ3r2

Rearrangingtheabove equation for R :

R=3Ps16πcρGMs

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

R=3(3.9×1026)16π(3×108)×(3.5×103)×(6.67×1011)×(1.99×1030)R=1.7×107 m

Hence, the radius is R=1.7×107 m

04

 Step 4: (b) Determining the path curve away from the Sun (like path 1) or toward the Sun (like path 3, does it) if its radius is larger.

FgVaries withR3,andFrvaries withR2.

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.

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Figure 33-32 shows four long horizontal layers A–D of different materials, with air above and below them. The index of refraction of each material is given. Rays of light are sent into the left end of each layer as shown. In which layer is there the possibility of totally trapping the light in that layer so that, after many reflections, all the light reaches the right end of the layer?

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