Optics question

  • #1
296
0
Hi guys

I'm flicking thorugh some past papers for an upcoming exam and came across this seemingly easy problem.

"The average intensity of solar radiation at the Earth is 1.4kW/m^2. Assuming the earth is a perfect absorber, calculate the force exerted by the radiation on the surface of the Earth. The mean radius of the earth is 6400 km"

Now I know that the radiation pressure on a perfect absorber is equal to the energy density, which is equal to the intensity divided by the speed of light.

[tex] P = U = \frac{I}{c} [/tex]

Given that P=F/A I get that the force should be.

[tex] F = \frac {IA}{c} = \frac {I 4\pi R^2}{c} [/tex]

Although it could be [tex] \frac {I 2\pi R^2}{c} [/tex]

As the sunlight only directly hits approximately half of the Earth's surface. But either way, when I plug in the numbers I don't get the answer my Lecturer gave of 600 MN.

I must be missing something, but I can't see it. If any of you guys can I'd appreciate it.:biggrin:
 

Answers and Replies

  • #2
Doc Al
Mentor
45,137
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Since the sunlight is essentially parallel, the area of interest is that of a circle (not a hemisphere or sphere).
 
  • #3
296
0
That makes sense, since I had noticed I was out by a factor of 4, I just couldn't come up with the physical meaning.

Thanks Doc Al
 

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