Question

At what distance from the sun is the intensity of sunlight nine times the value at the earth? (the average earth–sun separation is 1.496 1011 m.)

Answer 1

Sunlight is type of EM wave. The intensity of EM follows inverse square law.

`(1)/( n^(2) )`
This means that the intensity at a distance that is n times greater will be n^2 times smaller.
Now we set this equation to be equal to 9 and we solve it for n:

`9= (1)/( n^(2) ) \\ \\ 9 n^(2) =1 \\ \\ n^(2) = (1)/(9) \\ \\ n= (1)/(3)`

The intensity is 9 times greater at a distance that is equal to a 1/3 of Earth's distance:

`(1)/(3) *1.496* 10^(11) m=4.99* 10^(10) m`