Question

Consider a perfectly reflecting mirror oriented so that solar radiation of intensity i is incident upon, and perpendicular to, the reflective surface of the mirror. if the mirror has surface area a, what is frad, the magnitude of the average force due to the radiation pressure of the sunlight on the mirror

Answer 1

The answer is Frad = 2IA/c.

Answer 2

To solve this problem we need to know the following:

the mass of the sun, Msun, 2.0 * 10^30 kg

the intensity of sunlight as a function of the distance R from the sun,

Isun(R) = 3.2 * 10^25 * 1/R2 (W/m2), and

the gravitational constant G = 6.67 * 10^-11 m3/(kg⋅s2)

Suppose that the mirror described in Part A is initially at rest a distance R away from the sun.

What is the critical value of area density for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun?

Using the formula from Part A, find the force due to solar radiation:

Fr = 2IA/c

And compare to the force due to gravity:

Fg = G * m1 * m2 / R2

Fr = Fg

2IA/c = G * m1 * m2 / R2

Substitute in for I and the mass of the sun:

2 * (3.2 * 10^25 * 1/R2) * A / c = G * (2.0*10^30) * m2 / R2

The R2s cancel:

2 * (3.2 * 10^25) * A / c = G * (2.0 * 10^30) * m2

Now solve for m/A (where m = m2, the mass of the solar sail):

m/A = 2 * (3.2 * 10^25) / (c * G * (2.0*10^30))

m/A = 2 * (3.2 * 10^25) / ((3.0 * 10^8) * (6.67 * 10^-11) * (2.0 * 10^30))

mass/area = 0.00160 kg/m2