A spaceship approaches Earth with a speed $0.50$ c. A passenger in the spaceship measures his heartbeat as 70 beats per minute. What is his heartbeat rate according to an observer that is at rest relative to Earth?
A) 75 beats per minute
B) 69 beats per minute
C) 65 beats per minute
D) 73 beats per minute
E) 61 beats per minute

Question

Quizplus · Accepted Answer

Due to time dilation, the heartbeat rate measured by the observer on Earth will be slower than the rate measured by the passenger. The time dilation factor is $\gamma = \frac{1}{\sqrt{1-(v^2/c^2)}}$, where $v = 0.50c$. Calculating this gives $\gamma \approx 1.15$. The rate observed on Earth is $70/\gamma \approx 61$ beats per minute.

A Spaceship Approaches Earth with a Speed 0.500.500.50 C a Passenger in the Spaceship Measures His Heartbeat as Passenger

A Spaceship Approaches Earth with a Speed $0.50$ C a Passenger in the Spaceship Measures His Heartbeat as Passenger