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Ronch
1 month ago
11

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
Physics
1 answer:
Ostrovityanka [3.2K]1 month ago
8 0

Complete Question

A satellite in geostationary orbit transmits data using electromagnetic radiation, positioned 35,000 km above the Earth's surface, with an isotropic power output of 1 kW (although actual satellite antennas transmit with less power but more directionally).

Now, imagine this satellite relays television signals and you have a satellite TV receiver with a circular dish of radius R that focuses the electromagnetic energy from the satellite onto a receiver with a surface area of 5 cm2.

What radius R must this dish have to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

Assuming your house is situated directly beneath the satellite (as calculated previously), that the dish reflects all incoming signals to the receiver without any losses in the reception process, and that the dish has a curvature where R refers to its projection perpendicular to the direction of the incoming signal.

Present your answer in centimeters, rounded to two significant figures.

Answer:

 The radius  of  the dish is R = 18cm

Explanation:

  The question indicates that

     The orbital radius is  = R = 35,000km = 35,000 *10^3 m

    The power output is  P = 1 kW = 1000W

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

    The receiver's area is   A_R = 5cm^2

Overall, the intensity of the dish can be mathematically expressed as

         I = \frac{P}{A}

Here, A represents the area of the orbit, described as the surface area of a sphere, thus calculated as

          A = 4 \pi r^2

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

              =1.5395*10^{16} m^2

Ultimately, substituting the values into the intensity equation

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

            = 6.5*10^ {-14}W/m2

 Therefore, the intensity received by the dish can be mathematically evaluated as

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

  With c symbolizing the speed of light which has a constant value of c = 3.0*10^8 m/s

              \epsilon_o is the permittivity of free space, valued at 8.85*10^{-12} N/m

              E_D indicates the electric field present on the dish

Assuming no loss, the intensity from the satellite equals the intensity hitting the receiver dish

      Setting the electric field intensity as the subject of the formula

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

Inputting values

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

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

The power incident on the dish reflects what is delivered to the receiver

                P_D = P_R

Where P_D symbolizes the power incident on the dish expressed mathematically as

              P_D = I_d A_d

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

And  P_R is the incident power on the dish represented as

                 P_R = I_R A_R

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

Equating the two gives

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

  Solving for R yields

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

Substituting values gives

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

                     R = 18cm

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