INTERESTED IN THE EXPLANATION NOT THE ANSWER 11 Fleet Admira

INTERESTED IN THE EXPLANATION, NOT THE ANSWER!!!

11. Fleet Admiral Pike on the starship USS Excalibur, which has suffered battle damage due to an encounter with a Romulan warship, is attempting to contact Captain Kirk on the starship USS Enterprise in order to warn him about the presence of the Romulan ship. Due to the battle damage Admiral Pike must use an old style FM radio transmitting at a frequency of 100 MHz in order to contact Kirk. The USS Excalibur is traveling at 15% of the speed of light and the USS Enterprise is following behind at 20% of the speed of light. At what frequency must Lieutenant Uhura tune the FM radio on board the USS Enterprise in order to receive Pike’s critical communication before the ship is destroyed? A. 95 MHz B. 105 MHz C. 115 MHz D. 120 MHz E. 100 MHz

Solution

speed of the ship (Pike-sender) = 0.15c

speed of ship behind (Kirk-reciever) = 0.2c

frequency transmitted f = 100 MHz.

This is the case of relativistic Doppler effect.

Here the source (Pike) is moving at speed of 0.15c

The observer (Kirk) is moving at a speed of 0.2c

Source and observer are moving in the same direction.

The observer is approaching the source at a speed of v = 0.05c

Let us consider form the observer’s point of view. The source is emitting wave fronts. Let there be two wave fronts separated by a time interval T’ in observer’s reference frame S’, By the time the second wave front is emitted the source has moved a distance of vT’, from the observer’s point of view as the source is approaching the observer at a speed of v(0.05c).

The distance between two successive wave fronts as measured by the observer is

l’ = cT’ –vT’

The observed frequency f’ =c/l’ = c/(c-v)T’

T’ is the time measured by the observer’s frame of reference.

Proper time T is the time measured in source frame

T’ = T/(1-v²/c²)

f’ = c(1-v²/c²)/(c-v)T

1/T is the source frequency

f’ = f(1+v/c)/ (1-v/c)

putting the values for v and f we get

observed frequency

f’ = 100MHz*sqrt(1+0.05)/sqrt(1-0.05)

   = 105.13 MHz

105 MHz