Question

An astronaut is traveling in a space vehicle moving at 0.490c relative to the Earth. The astronaut measures her pulse rate at 66.0 beats per minute. Signals generated by the astronaut's pulse are radioed to the Earth when the vehicle is moving in a direction perpendicular to the line that connects the vehicle with an observer on the Earth. (Due to vehicle's path there will be no Doppler shift in the signal.)(a) What pulse rate does the Earth-based observer measure? beats/min(b) What would be the pulse rate if the speed of the space vehicle were increased to 0.940c?beats/min

Answer 1

To solve the exercise it is necessary to use the reciprocal Lorenz factor of the special theory of relativity with the pulse rate.

$\Xi = Pulse_{astronaut} * \alpha$

Where,

$\Xi =$ Pulse rate measured by earth observed

$\alpha =$ Reciprocal Lorenz Factor

$\alpha = \sqrt{1-\frac{v^2}{c^2}}$

Here v is the velocity and c the light speed.

PART A) For the given values we can replace it at the previous equation,

$\Xi = Pulse_{astronaut} *\sqrt{1-\frac{v^2}{c^2}}$

$\Xi = 66beats/min * \sqrt{1-\frac{(0.49c)^2}{c^2}}$

$\Xi = 57.5beats/min$

PART B ) If the speed of vehicle is 0.94c:

$\Xi = Pulse_{astronaut} *\sqrt{1-\frac{v^2}{c^2}}$

$\Xi = 66beats/min * \sqrt{1-\frac{(0.94c)^2}{c^2}}$

$\Xi = 22.5beats/min$