Respuesta :
a) [tex]1.51\cdot 10^{-14} W/m^2[/tex]
In order to calculate the average intensity, we need to calculate the distance between the satellite and Earth's surface.
The satellite makes two orbits per day, so the period is
[tex]T=12 h = 12 h \cdot (3600 s/h)=43,200 s[/tex]
So the angular frequency is
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{43200 s}=1.45\cdot 10^{-4} rad/s[/tex]
The gravitational force between the satellite and Earth provides the centripetal force that keeps the satellite in orbit:
[tex]G\frac{Mm}{r^2}=m \omega^2 r[/tex]
where
G is the gravitational constant
M is the Earth's mass
m is the satellite mass
R is the distance between the Earth's center and the satellite
Solving for r,
[tex]r=\sqrt[3]{\frac{GM}{\omega^2}}=\sqrt[3]{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24} kg)}{(1.45\cdot 10^{-4}rad/s)^2}}=2.67\cdot 10^7 m[/tex]
However, the distance of the satellite from Earth's surface is actually this value minus the radius of the Earth:
[tex]h=r-R=2.67\cdot 10^7 m-6.37\cdot 10^6 m=2.30\cdot 10^7 m[/tex]
The waves travel in a downward hemisphere, so the area of the surface of propagation is
[tex]A=2 \pi h^2 = 2\pi (2.30\cdot 10^7 m)^2=3.32\cdot 10^{15} m^2[/tex]
And since the power is
P = 50.0 W
We can now calculate the intensity:
[tex]I=\frac{P}{A}=\frac{50.0 W}{3.32\cdot 10^{15} m^2}=1.51\cdot 10^{-14} W/m^2[/tex]
b) [tex]3.37\cdot 10^{-6} V/m, 1.12\cdot 10^{-14}T[/tex], 0.077 s
The average intensity of an electromagnetic wave is given by
[tex]I=\frac{1}{2}c \epsilon_0 E^2[/tex]
where
c is the speed of light
[tex]\epsilon_0[/tex] is the vacuum permittivity
E is the amplitude of the electric field
Solving for E, we find
[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(1.51\cdot 10^{-14} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=3.37\cdot 10^{-6} V/m[/tex]
The amplitude of the magnetic field is given by
[tex]B=\frac{E}{c}=\frac{3.37\cdot 10^{-6} V/m}{3.0\cdot 10^8 m/s}=1.12\cdot 10^{-14}T[/tex]
And since the waves travels at speed of light and the distance is h, the time thay take to reach the receiver is
[tex]t=\frac{h}{c}=\frac{2.30\cdot 10^7 m}{3.0\cdot 10^8 m/s}=0.077 s[/tex]
c) [tex]5.03\cdot 10^{-23}Pa[/tex]
In case of a perfect absorber, the radiation pressure due to an electromagnetic wave is given by
[tex]p=\frac{I}{c}[/tex]
where
I is the intensity of the wave
c is the speed of light
Here we have
[tex]I=1.51\cdot 10^{-14} W/m^2[/tex]
[tex]c=3.0\cdot 10^8 m/s[/tex]
So the average pressure is
[tex]p=\frac{1.51\cdot 10^{-14} W/m^2}{3.0\cdot 10^8 m/s}=5.03\cdot 10^{-23}Pa[/tex]
d) 0.190 m
The receiver must be tune on the same wavelength of the emitter.
The wavelength of an electromagnetic wave is
[tex]\lambda=\frac{c}{f}[/tex]
where
c is the speed of light
f is the frequency
The frequency of the waves emittted by the satellite is
[tex]f=1575.42 MHz=1575.42\cdot 10^6 Hz[/tex]
so the wavelength is
[tex]\lambda=\frac{3.0\cdot 10^8 m/s}{1575.42\cdot 10^6 Hz}=0.190 m[/tex]
