A navigational beacon in deep space broadcasts at a radio frequency of 50 MHz.A spaceship approaches the beacon with a relative velocity of 0.40c.The spaceship passes the beacon and departs from it with a DIFFERENT relative velocity.The beacon signal is now detected at a frequency of 40 MHz.What is the new velocity of the spaceship relative to the beacon?
A) 0.20c
B) 0.22c
C) 0.24c
D) 0.26c
E) 0.28c

Question

Quizplus · Accepted Answer

The frequency detected by the spaceship can be calculated using the formula:

f' = f(1 - v/c)

where f is the frequency of the beacon signal, v is the velocity of the spaceship relative to the beacon, c is the speed of light, and f' is the frequency detected by the spaceship.

Initially, the frequency of the beacon signal was 50 MHz. When the spaceship approaches the beacon with a relative velocity of 0.40c, the frequency detected by the spaceship is:

f' = 50 MHz (1 - 0.40c/c) = 30 MHz

When the spaceship departs from the beacon with a different relative velocity, the frequency detected by the spaceship is 40 MHz. We can use the same formula to find the new velocity of the spaceship relative to the beacon:

40 MHz = 50 MHz (1 - v/c)

1 - v/c = 0.8

v/c = 0.2

v = 0.2c

Therefore, the new velocity of the spaceship relative to the beacon is 0.2c or 0.22 times the speed of light. The best choice is B).

A Navigational Beacon in Deep Space Broadcasts at a Radio