An astronaut is traveling in a space vehicle moving at 0.520c relative to the Earth. The astronaut measures her pulse rate at 74.0 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
To solve this problem, we need to use the concept of time dilation, which relates the time interval experienced by an object in motion relative to an observer at rest.
(a) First, let's determine the pulse rate as measured by the Earth-based observer when the vehicle is moving at 0.520c relative to the Earth.
The time dilation formula is given by:
Δt' = Δt / γ
Where:
Δt' is the time interval experienced by the Earth-based observer,
Δt is the time interval experienced by the astronaut (measured pulse rate),
and γ (gamma) is the Lorentz factor, given by γ = 1 / sqrt(1 - v^2/c^2), where v is the velocity of the vehicle and c is the speed of light.
In this case, the vehicle is moving at a velocity of 0.520c, so v = 0.520c.
Let's plug in the values into the formula:
Δt' = 74.0 beats/min / γ
To find γ, we substitute the value of v:
γ = 1 / sqrt(1 - (0.520c)^2 / c^2)
= 1 / sqrt(1 - 0.2704)
= 1 / sqrt(0.7296)
= 1 / 0.855
≈ 1.170
Now, let's calculate the pulse rate as measured by the Earth-based observer:
Δt' = 74.0 beats/min / γ
= 74.0 beats/min / 1.170
≈ 63.23 beats/min
Therefore, the pulse rate measured by the Earth-based observer when the vehicle is moving at 0.520c is approximately 63.23 beats/min.
(b) Now, let's determine the pulse rate if the speed of the space vehicle were increased to 0.940c.
Using the same formula, we find:
γ = 1 / sqrt(1 - (0.940c)^2 / c^2)
= 1 / sqrt(1 - 0.8836)
= 1 / sqrt(0.1164)
≈ 3.188
The pulse rate as measured by the Earth-based observer would be:
Δt' = 74.0 beats/min / γ
= 74.0 beats/min / 3.188
≈ 23.19 beats/min
Therefore, if the speed of the space vehicle were increased to 0.940c, the pulse rate measured by the Earth-based observer would be approximately 23.19 beats/min.