# Given the following, calculate the total power radiated by the Sun. Calculate the rms values of the electric and magnetic fields of by the Sun. What is the net force of sunlight on Saturn?

## The sunlight that reaches Earth has an intensity of 1350 watts/square meter. The distance from the Earth to the Sun is 150 million kilometers and has a radius of 6400 km. Saturn is ten times as far away from the Sun and has a diameter that is nine times that of Earth.

Oct 8, 2016

$F = 4.7 \times {10}^{8} \text{N}$

#### Explanation:

The intensity of the radiation decays proportional to $\frac{1}{r} ^ 2$, where $r$ is the distance away from the source.

Therefore, at a distance 10 times the Earth-Sun distance, the intensity of the light reaching Jupiter is $\frac{1}{100}$ times that of the sunlight reaching the Earth.

I = frac{1350 "W/m"^2}{100} = 13.5 "W/m"^2

The power absorbed by a planet can be assumed to be proportional to it's cross-section area, as seen below.

The effective area of Jupiter is

A = pi * (9 xx 6400 "km")^2 = 1.0 xx 10^16 "m"^2

The total power absorbed by Jupiter is

$P = I \cdot A$

$= \left(13.5 {\text{W/m"^2) * (1.0 xx 10^16 "m}}^{2}\right)$

$= 1.4 \times {10}^{17} \text{W}$

Assuming that Jupiter is a black-body, the force experienced by Jupiter due to the in-elastic collision of photons is

$F = \frac{P}{c}$

= frac{1.4 xx 10^17 "W"}{3.0 xx 10^8 "m/s"}

$= 4.7 \times {10}^{8} \text{N}$

The magnetic field and electric field rms values depend on the point in space. Is it at Jupiter? Or is it near the surface of the Sun. In that case, the radius of the Sun should be given. However, if the intensity, $I$, at a particular point is known, ${E}_{\text{rms}}$ and ${B}_{\text{rms}}$ can be calculated by

${E}_{\text{rms}} = \sqrt{{\mu}_{0} I c}$

${B}_{\text{rms}} = \sqrt{\frac{{\mu}_{0} I}{c}}$

where ${\mu}_{0}$ is the permeability of vacuum.