Answer:
61 bpm
Explanation:
By applying the relativistic Doppler Effect;
[tex]f= f_o} \sqrt{1-\frac{v^2}{c}[/tex]
where;
frequency (f) = the heartbeat rate according to an observer that is at rest relative to the earth
[tex]f_o[/tex] = initial frequency of the heart beat = 70 beats per minutes (70 bpm)
v = speed of the spacecraft = 0.50 c
c = speed of the light
Replacing our values in the above equation; we have:
[tex]f= 70} \sqrt{1-\frac{0.5c^2}{c}[/tex]
[tex]f= 70bpm} \sqrt{1-{0.5^2}[/tex]
f = 60.62 bpm
f ≅ 61 bpm
