Question

a. At Earth's distance of 1 AU from the Sun, how many times less bright is the light from the Sun than at Earth's distance?

b. What about at Neptune's distance of 30.05 AU?

Answer 1

1 AU = 1.496 x 10^11 m

The inverse square law says that a light source decreases in intensity by the inverse square of the distance. Double the distance to a lightsource, the intensity of the light (l for luminosity) goes down to l=1/2^2 or 1/4. Or 25%.

(a) Suppose 'I' is the intensity of light at the sun.

So, intensity of light at the earth, I' = I x 1 / (1.496 x 10^11)^2 = 4.468 x 10^-23 [I]

So, sun light is 4.468 x 10^-23 times dimmer at the earth than that of sun.

(b) Neptune's distance = 30.05 AU

So, compute the intensity of sun light at Neptune comparative to earth,

This is -

I" = [I' x 1 / (30.05)^2]x100 = 0.1086 x I' %

So, for Neptune, the sun light is 0.1086% intense than that of the intensity of sun light at the earth.