Question

please help

2. Now we are going to look at problem 1 again, but this time we are going to set the spaceship as the inertial stationary frame, meaning that the Earth is defined as the moving frame and traveling away from you. 1B:.8740 ; P = 2.058yr a. Using your answer from Ib as the speed of the moving frame, V, what is the speed of the Earth with respect to your spaceship? Using a Galilean Transformation equation for velocity, define and calculate Vx, Vx', and V given that we are now defining the spaceship as the inertial stationary frame. The first answer has been given to you, but you still need to fill in the numerical value. Show all work. (If you did not get an answer you trust for lb, use V = 0.900c for this question.) E.g) vx = speed of the Earth with respect to spaceship = (speed you calculate) b. From the stationary reference frame of the spaceship, the Earth travelled a distance of 4.37 light-years away from the spaceship at the speed you calculated from 2a. How long does it take the Earth to reach its destination, as measured by the clock on the spaceship? i. Time of round-trip back to the spaceship? c. How long does the trip of the Earth take as measured by the clock on Earth? i. Now who aged more during the trip?

Answer 1

a.

By Galilean transformation:

z = - (U-V)

where
T
is the position coordinate of the primed frame, x is the position coordinate of the unprimed frame, $v_x'-v_x$ is the relative velocity between the two frames and t is the time elapsed.

We will take the frame of the spaceship to be the stationary frame(unprimed). Therefore, x and $v_x$ is always 0 (with respect to itself, it is at the origin of the frame and at rest).

The distance travelled by the earth wrt the spaceship (it is moving in the opposite direction to the motion of the spaceship so it will be negative).

Plugging it all into the transformation relation:

$-0.874c\times 2.058=0-(0-v_x')\times2.058$

$-0.874c\times 2.058=v_x'\times2.058$

$v_x'=-0.874c$

This is the speed of the earth with respect to the spaceship.

b.

Time taken by the earth to cover the distance of 4.37 lightyears:

Time taken for the round trip:

c.

By time dialation, time elapsed on earth:

$2t'=2t \gamma\\ \implies 2t'=\frac{10}{\sqrt{1-\frac{0.874^2c^2}{c^2}}}=20.579\ years$

Therefore the earth or the residents of earth aged more. This isn't strictly correct (it's basically a form of the twin paradox) however the problem asks for it.