Answer
(a) The average power on the telescope is $7.31 \times 10^{-22}~W$
(b) The average power on the Earth's surface is $1.28 \times 10^{-12}~W$
(c) $E_{rms} = 1.94\times 10^{-12}~V/m$
$B_{rms} = 6.47\times 10^{-21}~T$
Work Step by Step
(a) We can find the average power on the telescope:
$I = \frac{P}{A}$
$P = I~A$
$P = I~(\pi~r^2)$
$P = (1.0\times 10^{-26}~W/m^2)(\pi)~(152.5~m)^2$
$P = 7.31 \times 10^{-22}~W$
The average power on the telescope is $7.31 \times 10^{-22}~W$
(b) We can find the average power on the Earth's surface, which can be modeled as a circular disk:
$I = \frac{P}{A}$
$P = I~A$
$P = I~(\pi~r^2)$
$P = (1.0\times 10^{-26}~W/m^2)(\pi)~(6.38\times 10^6~m)^2$
$P = 1.28 \times 10^{-12}~W$
The average power on the Earth's surface is $1.28 \times 10^{-12}~W$
(c) We can find $E_{rms}$:
$E_{rms}^2 = \frac{I}{c~\epsilon_0}$
$E_{rms} = \sqrt{\frac{I}{c~\epsilon_0}}$
$E_{rms} = \sqrt{\frac{1.0\times 10^{-26}~W/m^2}{(3.0\times 10^8~m/s)~(8.85\times 10^{-12}~C^2/N~m^2)}}$
$E_{rms} = 1.94\times 10^{-12}~V/m$
We can find $B_{rms}$:
$B_{rms} = \frac{E_{rms}}{c}$
$B_{rms} = \frac{1.94\times 10^{-12}~V/m}{3.0\times 10^8~m/s}$
$B_{rms} = 6.47\times 10^{-21}~T$
