Question

Problem 4 (25 Points) A spaceship passes the earth at t=t’=0 with relative velocity v. At time t1 on earth clocks, a super space ship leaves the earth with relative velocity V>v to catch up with the first one. This will happen at earth time t2, where vt2 = V(t2-t1) or t2=V t1/(V-v).

(a) What does the clock on the slow spaceship read when the earth clock reads t1 ?

(b) If viewed from the slow spaceship, how far from the slow spaceship is the earth at time t1 ?

(c) What does the clock on the slow spaceship register when the ship is overtaken ?

(d) In the frame of reference of the slow spaceship, how much time has elapsed since the pursuit started ? (e) In the same reference frame, how large a distance was covered in the pursuit ?

Answer 1

S frame=Earth

S' frame=slower ship

Also, I'll work in c=1 units.

(a) Follow Lorentz transformation

$t'=\gamma(t-vx)$

where

$\gamma=\frac{1}{\sqrt{1-v^2}}$

Given:

$t=t_1,x=vt_1$

so that

$t_1'=\gamma t_1(1-v^2)=t_1\sqrt{1-v^2}$

(b)

Distance the ship covers according to the time it itself measures

$vt_1\sqrt{1-v^2}$

(c) The event overtake takes place in S frame at

$t=t_2,x=vt_2$

as before

$t_2'=t_2\sqrt{1-v^2}=\frac{Vt_1\sqrt{1-v^2}}{V-v}$

(d)

$\Delta t'=t_2'-t_1'=\frac{vt_1\sqrt{1-v^2}}{V-v}$

(e)

$\frac{v^2t_1\sqrt{1-v^2}}{V-v}$