# Spaceship relativity problem

1. Sep 5, 2010

### PhMichael

1. The problem statement, all variables and given/known data

A spaceship is moving with a velocity $$0.8c\hat{x}$$ towards a planet. At some instant it explodes into two pieces such that the rest mass of the first piece is one thirds the original rest mass while the rest mass of the second piece is half the original one.
The problem is 1D such that the first piece moves to the left while the second one moves to the right.
Find the velocity of the second piece.

2. The attempt at a solution

I assume the rest mass of the spaceship is $$M_{0}$$.

Energy conservation:

$$\frac{M_{0}c^{2}}{\sqrt{1-0.8^{2}}}=\gamma_{1}(\frac{M_{0}}{3})c^{2}+\gamma_{2}(\frac{M_{0}}{2})c^{2}$$

(1):

$$\frac{5}{3}=\frac{1}{3 \sqrt{1-(v_{1}/c)^{2}}} + \frac{1}{2 \sqrt{1-(v_{2}/c)^{2}}}$$

Momentum conservation:

$$\frac{M_{0}(0.8c)}{\sqrt{1-0.8^{2}}} = -\gamma_{1}(\frac{M_{0}}{3})v_{1}+\gamma_{2}(\frac{M_{0}}{2})v_{2}$$

(2):

$$\frac{4}{3}c= - \frac{v_{1}}{3 \sqrt{1-(v_{1}/c)^{2}}} + \frac{v_{2}}{2 \sqrt{1-(v_{2}/c)^{2}}}$$

However, these equation are too hard to solve so there must be another way to look at this.
Help! =)

2. Sep 5, 2010

### PhMichael

Re: relativity

anyone?! .. this problem is freaking me out :pulling my hair out:

3. Sep 5, 2010

### jcsd

Re: relativity

Hi, perhaps make things a little easier for yourself by 1) starting in the orignal rest frame of the ship set and 2) set c=1 then just change back to the frame you want and put c back in at the end.

Just remember that

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

Last edited: Sep 5, 2010
4. Sep 5, 2010

### PhMichael

Re: relativity

what do you mean by this:

?

Is it to set the velocity of the ship to be zero?

5. Sep 5, 2010

### jcsd

Re: relativity

Sorry I deleted a bit. I meant start in the original frame of the ship (i.e. the frame travelling at 0.8c to one we want to get our answers in). TBH not really necessary, it's just how I started the problem myself.

6. Sep 5, 2010

### PhMichael

Re: relativity

Sorry, but I still don't get you ... what do you mean by "start in the original frame of the ship"? How will it change my "insolvable" equations that are written above?

7. Sep 5, 2010

### jcsd

Re: relativity

Anyway this is what I got (in the frame of the ship):

$$\frac{\gamma_1}{3} + \frac{\gamma_2}{2} = 1$$

therefore:

1) $$\gamma_2 = 2- \frac{2\gamma_1}{3}$$

for the momentum:

$$\frac{\gamma_1 v_1}{3} + \frac{\gamma_2 v_2}{2} = 0$$

therefore:

2)$$\frac{\sqrt{{\gamma_1}^2 -1}}{3} + \frac{\sqrt{{\gamma_2}^2 -1}}{2} = 0$$

Then 1 into 2 find the Lorentz factors, convert back to right frame and put c back in.

Last edited: Sep 5, 2010
8. Sep 5, 2010

### PhMichael

Re: relativity

first of all, thanks alot!

The answer for this solution is:

$$v_{1}=0.633c$$ and $$v_{2}=0.479c$$

while the correct answer should be:

$$0.853c$$

what could be the mistake?

9. Sep 5, 2010

### PhMichael

Re: relativity

nevermind ... I've obtained the right answer ...

THANKS ALOT!!

Last edited: Sep 5, 2010
10. Sep 5, 2010

### PhMichael

Re: relativity

By solving the equations in the rest frame of the original spaceship we got:

1) $$\gamma_2 = 2- \frac{2\gamma_1}{3}$$
2)$$\frac{\sqrt{{\gamma_1}^2 -1}}{3} + \frac{\sqrt{{\gamma_2}^2 -1}}{2} = 0$$

$$\to v_{1}=0.633c$$ and $$v_{2}=0.479c$$

and by solving the "regular" equations (from the 1st post in this thread) we'd get:

1) $$\frac{5}{3}=\frac{1}{3 \sqrt{1-(v_{1}/c)^{2}}} + \frac{1}{2 \sqrt{1-(v_{2}/c)^{2}}}$$

2) $$\frac{4}{3}c= - \frac{v_{1}}{3 \sqrt{1-(v_{1}/c)^{2}}} + \frac{v_{2}}{2 \sqrt{1-(v_{2}/c)^{2}}}$$

$$\to v_{1}=0.665c$$ and $$v_{2}=0.912c$$

My question is: What's the relationship between these results? i.e. between both $$v_{1}'s$$ and $$v_{2}'s$$.

11. Sep 5, 2010

### vela

Staff Emeritus
Re: relativity

You can use the relativistic velocity addition formula to relate the velocities from one inertial frame to those in another.

12. Sep 5, 2010

### PhMichael

Re: relativity

I tried to relate them by using the velocity transformation formula:

$$u_{x}' = \frac{u_{x}-v}{1-\frac{vu_{x}}{c^{2}}}$$

where $$v=0.8c$$ and $$u_{x}$$ is the corresponding velocity ... however, nothing worked :D

13. Sep 5, 2010

### vela

Staff Emeritus
Re: relativity

Your speeds in the planet's rest frame are wrong. I actually don't find a solution where one piece moves to the left in that frame.

14. Sep 6, 2010

### PhMichael

Re: relativity

for the sake of my insight ;), can you please tell me how you've reached this coclusion?

15. Sep 6, 2010

### vela

Staff Emeritus
Re: relativity

I had Mathematica solve your equations in the planet's frame. It produced two solutions, neither of which match yours, and in both cases, the sign of the velocities indicate both pieces move to the right.

Also, look at the velocities you found in the ship's rest frame. The speed of the first piece is less than 0.8c, so when you transform to the planet's rest frame, it has to be moving to the right.

Last edited: Sep 6, 2010