Someone plans to float a small, totally absorbing sphere 0.500 m above an isotropic point source of light,so that the upward radiation force from the light matches the downward gravitational force on the sphere. The sphere’s density is 19.0 g/cm3, and its radius is 2.00 mm. (a) What power would be required of the light source?

Respuesta :

Answer:

468449163762.0812 W

Explanation:

m = Mass = [tex]\rhoV[/tex]

V = Volume =[tex]\dfrac{4}{3}\pi r^3[/tex]

r = Distance of sphere from isotropic point source of light = 0.5 m

R = Radius of sphere = 2 mm

[tex]\rho[/tex] = Density = 19 g/cm³

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

A = Area = [tex]\pi R^2[/tex]

I = Intensity = [tex]\dfrac{P}{4\pi r^2}[/tex]

g = Acceleration due to gravity = 9.81 m/s²

Force due to radiation is given by

[tex]F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}[/tex]

According to the question

[tex]F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W[/tex]

The power required of the light source is 468449163762.0812 W

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