The robotic rovers exploring Mars at a distance 15 AU astron

The robotic rovers exploring Mars, at a distance 1.5 AU (astronomical units; 1 AU is the distance of the earth from the sun) from the sun, get a significant fraction of their energy from solar panels that provide about 140 W when the sun is high.

Suppose the rovers were on Europa, a moon of Jupiter that is well worth a closer look. Europa is 5.2 A.U. from the sun. What maximum power could the solar panels provide at this distance?

Solution

We know luminosity L is inversely proportional to square of the distance(r).

i.e., L \' / L = ( r / r\' ) ²

Given r = 1.5 AU

         r \' = 5.2 AU

        L = 140 W

       L \' = ?

Substitute values you get , L \' / 140 = (1.5 / 5.2) ²

                                                    = 0.2884 ²

                                                    = 0.08321

                                             L \' = 0.08321 (140)

                                                 = 11.649 W