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An astronaut traveling at 0.90c, with respect to Earth, measures his pulse and finds it to be 70 beats per minute.
a) Calculate the time required for one pulse to occur, as measured by the astronaut.
b) Calculate the time required for one pulse to occur, as measured by an Earth-based observer.
c) Calculate the astronaut’s pulse, as measured by an Earth-based observer.
d) What effect, if any, would increasing the speed of the spacecraft have on the astronaut’s pulse as measured by the astronaut and by an Earth-based observer? Why?
this is my solution
a. the time for one pulse to occur in the astronauts frame is 1 minute/70 beats per minute or 1/70 minute or 60/70 seconds which in decimal form is 0.857 seconds.
b. the formula for time dilation is part of the Lorentz transformation is:
t = t0/(1 - (v^2/c^2))^1/2
= 60/70/[1 - (.81/1)]^1/2
= 60/70[.19]^1/2
= 60/70(0.435889894354067)
= 0.37 seconds
So they are about .37 seconds apart form the frame of reference of the Earthling
c. Pulse is 1/.37 seconds or 2.677 beats per seconds which is:
2.67651689515656 x 60 = 160.6 beats per minute
d. As the speed of the astronaut increases the astronauts pulse will also increase from the frame of reference of the Earthling. As v approaches c the denominator or the Lorentz transformation approaches 0 so the whole thing goes to infinity.
is that right?
a) Calculate the time required for one pulse to occur, as measured by the astronaut.
b) Calculate the time required for one pulse to occur, as measured by an Earth-based observer.
c) Calculate the astronaut’s pulse, as measured by an Earth-based observer.
d) What effect, if any, would increasing the speed of the spacecraft have on the astronaut’s pulse as measured by the astronaut and by an Earth-based observer? Why?
this is my solution
a. the time for one pulse to occur in the astronauts frame is 1 minute/70 beats per minute or 1/70 minute or 60/70 seconds which in decimal form is 0.857 seconds.
b. the formula for time dilation is part of the Lorentz transformation is:
t = t0/(1 - (v^2/c^2))^1/2
= 60/70/[1 - (.81/1)]^1/2
= 60/70[.19]^1/2
= 60/70(0.435889894354067)
= 0.37 seconds
So they are about .37 seconds apart form the frame of reference of the Earthling
c. Pulse is 1/.37 seconds or 2.677 beats per seconds which is:
2.67651689515656 x 60 = 160.6 beats per minute
d. As the speed of the astronaut increases the astronauts pulse will also increase from the frame of reference of the Earthling. As v approaches c the denominator or the Lorentz transformation approaches 0 so the whole thing goes to infinity.
is that right?