Answer:
[tex]3.0\cdot 10^{-5} N/m^2[/tex]
Explanation:
The intensity of the radiation at the location of the satellite is given by:
[tex]I=\frac{P}{4\pi r^2}[/tex]
where
[tex]P=3.8\cdot 10^{26}W[/tex] is the power of the emitted radiation
[tex]4\pi r^2[/tex] is the area over which the radiation is emitted (the surface of a sphere), with
[tex]r=5.8\cdot 10^{10}m[/tex] being the radius of the orbit
Substituting,
[tex]I=\frac{3.8\cdot 10^{26}}{4\pi (5.8\cdot 10^{10})^2}=8989 W/m^2[/tex]
Now we can find the pressure of radiation, that for a totally absorbing surface (such as the satellite) is:
[tex]p=\frac{I}{c}[/tex]
where
[tex]c=3.0\cdot 10^8 m/s[/tex]
is the speed of light. Substituting,
[tex]p=\frac{8989}{3.0\cdot 10^8}=3.0\cdot 10^{-5} N/m^2[/tex]
