a) ![1.51\cdot 10^{-14} W/m^2](https://tex.z-dn.net/?f=1.51%5Ccdot%2010%5E%7B-14%7D%20W%2Fm%5E2)
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
![T=12 h = 12 h \cdot (3600 s/h)=43,200 s](https://tex.z-dn.net/?f=T%3D12%20h%20%3D%2012%20h%20%5Ccdot%20%283600%20s%2Fh%29%3D43%2C200%20s)
So the angular frequency is
![\omega=\frac{2\pi}{T}=\frac{2\pi}{43200 s}=1.45\cdot 10^{-4} rad/s](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D%3D%5Cfrac%7B2%5Cpi%7D%7B43200%20s%7D%3D1.45%5Ccdot%2010%5E%7B-4%7D%20rad%2Fs)
The gravitational force between the satellite and Earth provides the centripetal force that keeps the satellite in orbit:
![G\frac{Mm}{r^2}=m \omega^2 r](https://tex.z-dn.net/?f=G%5Cfrac%7BMm%7D%7Br%5E2%7D%3Dm%20%5Comega%5E2%20r)
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,
![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](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BGM%7D%7B%5Comega%5E2%7D%7D%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%286.67%5Ccdot%2010%5E%7B-11%7D%29%285.98%5Ccdot%2010%5E%7B24%7D%20kg%29%7D%7B%281.45%5Ccdot%2010%5E%7B-4%7Drad%2Fs%29%5E2%7D%7D%3D2.67%5Ccdot%2010%5E7%20m)
However, the distance of the satellite from Earth's surface is actually this value minus the radius of the Earth:
![h=r-R=2.67\cdot 10^7 m-6.37\cdot 10^6 m=2.30\cdot 10^7 m](https://tex.z-dn.net/?f=h%3Dr-R%3D2.67%5Ccdot%2010%5E7%20m-6.37%5Ccdot%2010%5E6%20m%3D2.30%5Ccdot%2010%5E7%20m)
The waves travel in a downward hemisphere, so the area of the surface of propagation is
![A=2 \pi h^2 = 2\pi (2.30\cdot 10^7 m)^2=3.32\cdot 10^{15} m^2](https://tex.z-dn.net/?f=A%3D2%20%5Cpi%20h%5E2%20%3D%202%5Cpi%20%282.30%5Ccdot%2010%5E7%20m%29%5E2%3D3.32%5Ccdot%2010%5E%7B15%7D%20m%5E2)
And since the power is
P = 50.0 W
We can now calculate the intensity:
![I=\frac{P}{A}=\frac{50.0 W}{3.32\cdot 10^{15} m^2}=1.51\cdot 10^{-14} W/m^2](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BP%7D%7BA%7D%3D%5Cfrac%7B50.0%20W%7D%7B3.32%5Ccdot%2010%5E%7B15%7D%20m%5E2%7D%3D1.51%5Ccdot%2010%5E%7B-14%7D%20W%2Fm%5E2)
b)
, 0.077 s
The average intensity of an electromagnetic wave is given by
![I=\frac{1}{2}c \epsilon_0 E^2](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B1%7D%7B2%7Dc%20%5Cepsilon_0%20E%5E2)
where
c is the speed of light
is the vacuum permittivity
E is the amplitude of the electric field
Solving for E, we find
![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](https://tex.z-dn.net/?f=E%3D%5Csqrt%7B%5Cfrac%7B2I%7D%7Bc%5Cepsilon_0%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%281.51%5Ccdot%2010%5E%7B-14%7D%20W%2Fm%5E2%29%7D%7B%283%5Ccdot%2010%5E8%20m%2Fs%29%288.85%5Ccdot%2010%5E%7B-12%7D%20F%2Fm%29%7D%7D%3D3.37%5Ccdot%2010%5E%7B-6%7D%20V%2Fm)
The amplitude of the magnetic field is given by
![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](https://tex.z-dn.net/?f=B%3D%5Cfrac%7BE%7D%7Bc%7D%3D%5Cfrac%7B3.37%5Ccdot%2010%5E%7B-6%7D%20V%2Fm%7D%7B3.0%5Ccdot%2010%5E8%20m%2Fs%7D%3D1.12%5Ccdot%2010%5E%7B-14%7DT)
And since the waves travels at speed of light and the distance is h, the time thay take to reach the receiver is
![t=\frac{h}{c}=\frac{2.30\cdot 10^7 m}{3.0\cdot 10^8 m/s}=0.077 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bh%7D%7Bc%7D%3D%5Cfrac%7B2.30%5Ccdot%2010%5E7%20m%7D%7B3.0%5Ccdot%2010%5E8%20m%2Fs%7D%3D0.077%20s)
c) ![5.03\cdot 10^{-23}Pa](https://tex.z-dn.net/?f=5.03%5Ccdot%2010%5E%7B-23%7DPa)
In case of a perfect absorber, the radiation pressure due to an electromagnetic wave is given by
![p=\frac{I}{c}](https://tex.z-dn.net/?f=p%3D%5Cfrac%7BI%7D%7Bc%7D)
where
I is the intensity of the wave
c is the speed of light
Here we have
![I=1.51\cdot 10^{-14} W/m^2](https://tex.z-dn.net/?f=I%3D1.51%5Ccdot%2010%5E%7B-14%7D%20W%2Fm%5E2)
![c=3.0\cdot 10^8 m/s](https://tex.z-dn.net/?f=c%3D3.0%5Ccdot%2010%5E8%20m%2Fs)
So the average pressure is
![p=\frac{1.51\cdot 10^{-14} W/m^2}{3.0\cdot 10^8 m/s}=5.03\cdot 10^{-23}Pa](https://tex.z-dn.net/?f=p%3D%5Cfrac%7B1.51%5Ccdot%2010%5E%7B-14%7D%20W%2Fm%5E2%7D%7B3.0%5Ccdot%2010%5E8%20m%2Fs%7D%3D5.03%5Ccdot%2010%5E%7B-23%7DPa)
d) 0.190 m
The receiver must be tune on the same wavelength of the emitter.
The wavelength of an electromagnetic wave is
![\lambda=\frac{c}{f}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bc%7D%7Bf%7D)
where
c is the speed of light
f is the frequency
The frequency of the waves emittted by the satellite is
![f=1575.42 MHz=1575.42\cdot 10^6 Hz](https://tex.z-dn.net/?f=f%3D1575.42%20MHz%3D1575.42%5Ccdot%2010%5E6%20Hz)
so the wavelength is
![\lambda=\frac{3.0\cdot 10^8 m/s}{1575.42\cdot 10^6 Hz}=0.190 m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B3.0%5Ccdot%2010%5E8%20m%2Fs%7D%7B1575.42%5Ccdot%2010%5E6%20Hz%7D%3D0.190%20m)