1. Pepsi24chevy

65
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. Andrew Mason

6,850
The counter-force to radiation force is the weight of the reflecting surface (mass of mirror x acceleration due to gravity ($F=mGM_{sun}/r^2$). In terms of pressure this is:

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

Equating the two:

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

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

You should be able to work it out from that.
AM

Last edited: Oct 19, 2005