Question

An astronaut at rest on earth has a heartbeat rate of 70 beats/min. what will this rate be when she is traveling in a spaceship at 0.90c as measured (a) by an observer also in the ship and (b) by an observer at rest on the earth?

Answer 1

(a) Rate as measured by an observer in the same ship will be same because as an observer in ship will not be moving relative to the astronaut, therefore their relative speed is zero and hence rate must be same.

(b) To calculate the rate measured by an observer at rest on the Earth we use the formula time dilation in relativity

`\gamma =(1)/(1-(v^2)/(c^2))`

Here, v is the speed of astronaut and c is the speed of light.

Substituting the given values in above formula we get

`\gamma =\frac{1}{\sqrt{1-(0.90c^2)/(c^2) } }`

`\gamma =2.29`

As one heartbeat takes (1/70)min = (6/7)s = 0.857 s in appropriate time, at that point to the observer on Earth it takes (2.29)(0.857 s) = 1.96 sec, and in this way her heartbeat is 30 beats/min.