Question

1. An astronaut at rest on Earth has a heartbeat rate of 56 beats/min. When the astronaut is traveling in a spaceship at 0.96c, what will this rate be as measured by an observer also in the ship and an observer at rest on Earth?

(a) an observer also in the ship

(b) an observer at rest on Earth

2.Spaceship R is moving to the right at a speed of 0.433c with respect to Earth. A second spaceship, L, moves to the left at the same speed with respect to Earth. What is the speed of L with respect to R?

Answer 1

(1) (a) The first observer is also in the same ship as the astronaut. So, he will measure the same heartbeat rate means, 56 beats/min. as the two are in the same reference frame.

(b) In the second case, an observer on the earth will measure the astronaut's time to be dilated by a factor

t(observed) = t(rest)/Sqrt[1-(v/c)^2]

= t(rest)/Sqart[1-(0.96c/c)^2] = 3.57 t(rest)

This means an Earth observer will measure the time between heart beats to be 3.57 times longer than will be measured by the astronaut on the ship

Therefore, if the astronaut measures 56 beats in 1 minute,

The observer at rest on Earth we will measure 56 / 3.57 = 15.7 beats / min. (Answer)