A spaceship approaches Earth with a speed 0.38c. A passenger in the spaceship measures his heartbeat as 72 beats per minute. How many beats per minute are observed in the telemetry radio signals received on Earth? Note that, in addition to time dilation, one must take into account that the signals of successive heartbeats have to travel shorter distances because the spaceship is moving closer (Doppler Effect).
A) 75
B) 107
C) 48
D) 100
E) 67

Question

Quizplus · Accepted Answer

According to time dilation, the observed heartbeat on Earth will be slower than the observed heartbeat by the passenger. 
Let the heartbeat observed on Earth be x.
We have, (1 - v/c)  = √(1 - (x/c)^2)/(1 - (72/c)^2)
where v = speed of the spaceship, c = speed of light.
Solving for x, we get x = 107.
Additionally, due to the Doppler Effect, the frequency of the signals received on Earth will be higher. This is because the spaceship is moving closer to the Earth, causing a blueshift in the signals. 
Hence, the observed heartbeat on Earth will be higher than the actual heartbeat from the perspective of the passenger. 
The only option that matches the calculated value is option B, 107.

A Spaceship Approaches Earth with a Speed 0