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Radiation pressure

  1. Oct 19, 2005 #1
    Ok, i got a problem that reads as followed.

    Suppose that a perfectly reflecting circular mirror is initially at rest a distance R away from the sun and is oriented so that the solar radiation is incident upon, and perpendicular to, the plane of the mirror. What is the critical value of mass/area for which the radiation pressure exactly cancels out the force due to gravity?

    Ok so lets start with the given:, I know mass of the sun is 2.0 x 10^30 kg
    intensity of sunlight as a function of the distance R from the sun = 3.2x10^25(1/R^2) (w/m^2)
    and the gravitational constant is 6.67x 10^-11

    Radiation pressure ecerted on a perfectly reflecting surface is P= 2S/c where C is the speed of light and S is the poynting vector? I know the answer is going ot be mass/area in which the mass is mass of the sun. The answer will also be in kg/m^2. Now i am not sure how to setup this problem.
  2. jcsd
  3. Oct 19, 2005 #2

    Andrew Mason

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    The counter-force to radiation force is the weight of the reflecting surface (mass of mirror x acceleration due to gravity ([itex]F=mGM_{sun}/r^2 [/itex]). In terms of pressure this is:

    [tex]P = F/A = \frac{\rho*Ad*GM_{sun}}{Ar^2} = \frac{\rho*d*GM_{sun}}{r^2}[/tex]

    Equating the two:

    [tex]P = \Phi_E/c = \frac{\rho*d*GM_{sun}}{r^2}[/tex]

    where [itex]\Phi_E = \frac{E}{4\pi r^2}[/itex] is the energy flux (E/A)

    You should be able to work it out from that.
    Last edited: Oct 19, 2005
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