A fellow astronaut passes by you in a spacecraft traveling a

A fellow astronaut passes by you in a spacecraft traveling at a high speed. The astronaut tells you that his craft is 21.9 m long and that the identical craft you are sitting in is 16.6 m long. According to your observation, how long is your craft? According to your observation, how long is the astronaut\'s craft? According to your observation, what is the speed of the astronaut\'s craft relative to your craft? A deep-space vehicle moves away from the Earth with a speed of 0.770c. An astronaut on the vehicle measures a time interval of 2.70 s to rotate her body through 1.00 rev as she floats in the vehicle. What time interval is required for this rotation according to an observer on the Earth?

Solution

4 .
Since both spacecraft are identical, then each astronaut sees his own spacecraft as being the same length. Each astronaut sees his own spacecraft as 21.9 meters long since the relative velocity of the astronaut to his own spacecraft is zero.

Since each astronaut is traveling at the same speed relative to each other (each astronaut sees himself as standing still and the other astronaut zipping by), then they each see the other\'s spacecraft to be 16.6 meters long, since there is a relative velocity between the astronaut and the other spacecraft. The question is, what is that relative velocity?

So now, the final question is, how fast does something need to be traveling to appear to contract by a factor of 16.6 / 21.9 = 0.75799

Let:

L_rest = length at rest = 21.9 m
L_atspeed = length at speed = 16.6 m
c = 2.998*10^8 m/sec (speed of light in vacuum)

From relativity (see link below on length contraction due to relative velocity)

L_atspeed = L_rest * sqrt(1 - v^2 / c^2)

The only unknown in the equation above is v.
Rearrange the equation algebraically to solve for v.

I calculate a velocity of:
V = 1.95*10^8 m/sec

5) Gamma for 0.770c is 1 / sqrt(1- (0.705)^2) = 1.4728

So 2.7* 1.4728 = 3.977 (to 3 sig figs).