Special Relativity: Length Contraction Problem

[peroAlex]

Hello! I have a small problem with a task professor gave us. I tried many options (you will see below) but I cannot seem to get the right solution. Any advice or guideline how to solve this would be really helpful. In advance I thank you for helping me.

Homework Statement

Our professor of physics has sense of humor, so he represented this task: starship Enterprise captain measures his ship to be $4500$ meters long. Enterprise passes Earth with velocity of $0.8 c_0$. In opposite direction, starship Galactica flies by with velocity of $0.9 c_0$. Compute how long will Enterprise appear to Galactica's captain.

Homework Equations

Pretty obvious, this task will implement length contraction formula $$ L = \sqrt{1 - \frac{v^2}{c_0^2}} L_0 $$. Also, according to solutions, final result should be $684$ meters.

The Attempt at a Solution

OK, so I began with computing Enterprise's velocity according to Galactica. Using $v_e' = \frac{0.8c_0 - 0.9c_0}{1 - \frac{0.9c_0 \cdot 0.8c_0}{c_0^2}} = 0.35714c_0$ I though I should just simply insert this into length contraction formula. It returned $3608.1$ meters.
Now I decided to use slightly different procedure. I used $L = \sqrt{1 - \frac{v_e' v_{galactica}}{c_0^2}} L_0$ but it returned $3706.9$ meters.

At this point I lost all hope. I really wish someone would be able to help me with this one.

 

[PeroK]

OK, so I began with computing Enterprise's velocity according to Galactica. Using v′e=0.8c0−0.9c01−0.9c0⋅0.8c0c20=0.35714c0 v_e' = \frac{0.8c_0 - 0.9c_0}{1 - \frac{0.9c_0 \cdot 0.8c_0}{c_0^2}} = 0.35714c_0 I

The two ships must have a relative velocity of close to $c$. How can $0.357c$ be right?

 

[peroAlex]

The two ships must have a relative velocity of close to $c$. How can $0.357c$ be right?

I used equation $v_e' = \frac{v_e - v_g}{1 - \frac{v_e v_g}{c_0^2}}$ to determine velocity of Enterprise according to Galactica.. Same equation appeared in previous example and in our textbook so I assumed it must be correct. It returned $v_e' = 1.07068 \cdot 10^8 m/s$. I think I must be missing an important step in all of this but I can't seem to find it.

 

[PeroK]

I used equation $v_e' = \frac{v_e - v_g}{1 - \frac{v_e v_g}{c_0^2}}$ to determine velocity of Enterprise according to Galactica.. Same equation appeared in previous example and in our textbook so I assumed it must be correct. It returned $v_e' = 1.07068 \cdot 10^8 m/s$. I think I must be missing an important step in all of this but I can't seem to find it.

Forget relativity for a moment. If two cars are coming at you from opposite directions at $80km/h$ and $90km/h$ respectively. The relative speed of the two cars is?

a) $170km/h$

b) $10km/h$