A spaceship approaches the earth with a speed 0.50c. 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 the earth

61 bpm

Explanation:

By applying the relativistic Doppler Effect;

where;

frequency (f) = the heartbeat rate according to an observer that is at rest relative to the earth

= 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:

f = 60.62 bpm

f ≅ 61 bpm

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explanation:

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