Question

The total electromagnetic power emitted by the sun is 3.8 × 1026 W. What is the radiation pressure on a totally absorbing satellite at the orbit of Mercury, which has an orbital radius of 5.8 × 1010 m?

Answer 1

`3.0\cdot 10^(-5) N/m^2`

Step-by-step explanation:

The intensity of the radiation at the location of the satellite is given by:

`I=(P)/(4\pi r^2)`

where

`P=3.8\cdot 10^(26)W` is the power of the emitted radiation

`4\pi r^2` is the area over which the radiation is emitted (the surface of a sphere), with

`r=5.8\cdot 10^(10)m` being the radius of the orbit

Substituting,

`I=(3.8\cdot 10^(26))/(4\pi (5.8\cdot 10^(10))^2)=8989 W/m^2`

Now we can find the pressure of radiation, that for a totally absorbing surface (such as the satellite) is:

`p=(I)/(c)`

where

`c=3.0\cdot 10^8 m/s`

is the speed of light. Substituting,

`p=(8989)/(3.0\cdot 10^8)=3.0\cdot 10^(-5) N/m^2`