Relativistic Length Contraction in Three-Spacecraft Chase Scenario

[vachan]

Three enemy spacecraft have been causing trouble in the asteroid belt. They always travel in a line, evenly spaced apart, attempting to chase down local spacecraft to steal their goods. The local asteroid colonists have decided to set a trap to capture these three spacecraft . They'll get them to chase one of their fastest ships into an asteroid with a large hole in it and, once the three enemy ships are inside, close two giant trapdoors on each side of the asteroid to catch them. These spacecraft all travel close to the speed of light so the locals will have to take relativity into account. Intelligence about the enemy spacecraft reveals that, in their reference frame, they always travel 90 m behind their teammate, each spacecraft is 10 m in length, and their maximum velocity is 90% the speed of light (relative to the asteroids). The asteroid tunnel is only 215 m in length. In this problem we will analyze whether the locals will be able to capture the enemy spacecraft after taking into account relativity.

If the spacecraft are traveling at 90% the speed of light, what is the total length of the three- spacecraft team as observed from the asteroid?


My attempt was putting the number in the L=Lo( 1/\sqrt{}1-v2/c2

Then i add them up... but wasnt right... anyone has idea?!? please!

 

[HallsofIvy]

What do you mean by "the number" and "add them up"? Are you calculating contraction in the length of each spaceship separately? Are you calculating the contraction in the space between the ships? There are a total of 5 length numbers here: three ships and two distances between them. But it would be simpler to take the total distance, from the head of the first ship to the tail of the last ship and calculate the contraction of that.

Oh, and notice that I am talking about contraction. Since v< c, 1- v^2/c^2< 1 and dividing by it makes L larger than Lo. Is that what you want?

 

[vachan]

im not so sure how you do that... could you please show me?!