Question

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius RRR which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2cm2. How large does the radius RRR of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/mmV/m at the receiver? For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius RRR refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal

Answer 1

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.

How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.

Give your answer in centimeters, to two significant figures.

Answer:

 The radius  of  the dish is $R = 18cm$

Explanation:

  From the question we are told that

     The radius of the orbit is  = $R = 35,000km = 35,000 *10^3 m$

    The power output of the power is  $P = 1 kW = 1000W$

   The electric vector amplitude is given as $E = 0.1 mV/m = 0.1 *10^{-3}V/m$

    The area of thereciever  is   $A_R = 5cm^2$

Generally the intensity of the dish is mathematically represented as

         $I = \frac{P}{A}$

Where A is the area orbit which is a sphere so this is obtained as

          $A = 4 \pi r^2$

              $= (4 * 3.142 * (35,000 *10^3)^2)$

              $=1.5395*10^{16} m^2$

  Then substituting into the equation for intensity

          $I_s = \frac{1000}{1.5395*10^{16}}$

            $= 6.5*10^ {-14}W/m2$

 Now the intensity received by the dish can be mathematically evaluated as

              $I_d = \frac{1}{2} * c \epsilon_o E_D ^2$

  Where c is thesped of light with a constant value  $c = 3.0*10^8 m/s$

              $\epsilon_o$ is the permitivity of free space  with a value  $8.85*10^{-12} N/m$

              $E_D$ is the electric filed on the dish

So  since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish

      Now making the eletric field intensity the subject of the formula

                  $E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }$

substituting values

                 $E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }$

                       $= 7*10^{-6} V/m$

The incident power on the dish is what is been reflected to the receiver

                $P_D = P_R$

Where $P_D$ is the power incident on the dish which is mathematically represented as

              $P_D = I_d A_d$

                   $= \frac{1}{2} c \epsilon_o E_D^2 (\pi R^2)$

And  $P_R$ is the power incident on the dish which is mathematically represented as

                 $P_R = I_R A_R$

                       $= \frac{1}{2} c \epsilon_o E_R^2 A_R$

Now equating the two

                $\frac{1}{2} c \epsilon_o E_D^2 (\pi R^2) = \frac{1}{2} c \epsilon_o E_R^2 A_R$

   Making R the subject we have

                   $R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }$

Substituting values

                   $R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }$

                     $R = 18cm$