A spaceship approaches earth with speed of 0.55c. A passenger in the ship measures his heartbeat as 70 beats per minute. What is his heartbeat rate according to an observer that is at rest relative to earth?
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Time dilation formula:
∆t= γ∆t_0
γ=(1 - (v^2/c^2))^-(1/2)
∆t_0 = time interval between two events in the frame where they occur at the same spatial coordinates.
∆t = time interval between the same two events in a frame travelling at a speed v relative to the first.
v = 0.55c
Using this information:
∆t_0 = 60/70
This is the length of time for one beat (in seconds).
γ =1.197
=> ∆t = 1.197(60/70) = 60/H
H = heart rate as observed on earth.
=> H = 70/1.197 = 58.48 beats per minute
i.e. slower. Hence time DIALATION.
Hope this is clear.
∆t= γ∆t_0
γ=(1 - (v^2/c^2))^-(1/2)
∆t_0 = time interval between two events in the frame where they occur at the same spatial coordinates.
∆t = time interval between the same two events in a frame travelling at a speed v relative to the first.
v = 0.55c
Using this information:
∆t_0 = 60/70
This is the length of time for one beat (in seconds).
γ =1.197
=> ∆t = 1.197(60/70) = 60/H
H = heart rate as observed on earth.
=> H = 70/1.197 = 58.48 beats per minute
i.e. slower. Hence time DIALATION.
Hope this is clear.