How Is the Force of Sunlight Calculated on a Building's Roof?

[TLeo198]

Homework Statement

After filtering through the atmosphere, the Sun's radiation illuminates Earth's surface with an average intensity of 1.0 kW/m^2. Assuming this radiation strikes the 15-m x 45-m black, flat roof of a building at normal incidence, calculate the average force the radiation exerts on the roof.

Homework Equations

Dunno if this is all I need, but;
(1)I-average = u-average (energy density) x c (speed of light) = P-average (power) / A (area)
(2)P-average = F-average (force) x d (distance), where d = c (speed of light) x t (time), where time is 500 seconds, since that's how long it takes for light to reach the Earth from the sun.

The Attempt at a Solution

Since it's asking for average force, I remembered that P = Fxd. I knew I could find P-average by setting the intensity equal to P/A, in which A = 15m x 45m = 675 m^2, therefore getting a P-average of 675,000 Watts. I then went to the equation P = F x d, found d to be (c)(t) = (3 x 10^8)(500s) = 1.5 x 10^11 meters. Plugging that in, I calculated the average force to be about 4.5 x 10^-6 Newtons, but that answer does not coincide with the correct answer in the back of the book. Any comments? Any help is greatly appreciated

 

[mgb_phys]

One step at a time.
Find the area of the roof and from the intensity the total number of watts hititng the roof.
Then use the wavelength of peak emmision form the sun (or the middle of the visible) to work out how many photons this is. Note that since you are given watts this is the energy per second, so you can get photons/second.

Work out the momentum of a single photon of this wavelength and so the total momentum/second hitting the roof.