Mass, energy problem in relativity

[zimo]

Homework Statement

A $$\pi$$+ meson whose rest energy is 140 MeV is created 100km above sea level in the Earth's atmosphere. The pi+ meson has a total energy of 1.5X10^5 MeV and is moving vertically downward. If it disintegrates 2x10^-8 s after it's creation as determined in its own frame of reference, at what altitude above sea level does the disintegration occur?

Homework Equations

energy conservation.

The Attempt at a Solution

I used energy const. to get the velocity of the particle, then attempted to calculate it's displacement which is approximately 6 km and I get approx. 94 km.
the correct answer is given to be 93.6, which is not a small error in my opinion.

 

[CompuChip]

It looks like you did the right thing, coming so close to the answer. Also, you say you got "approx. 94" while it should be 93.6, so how approx is your approx Did you round (too much) in intermediate steps?

If that doesn't help, please post your calculation.

 

[zimo]

I came down to V=0.9999c
since x=V*t=0.9999c*2*10^-8=~5.99524958
so 100-x is ~94.0047504
maybe the time is need to be measure relative to the Earth frame of reference or something?
e.g. t(earth)=gamma*2x10^-8, where gamma=1/sqrt(1-v^2/c^2)

 

[BenjF]

Hi zimo,

maybe the time is need to be measure relative to the Earth frame of reference or something?
e.g. t(earth)=gamma*2x10^-8, where gamma=1/sqrt(1-v^2/c^2)

Exactly. You have to "convert" the time to the observer's referential (i.e. Earth referential) , since the distance is measured in the latter.

Cheers!

 

[zimo]

Exactly. You have to "convert" the time to the observer's referential (i.e. Earth referential) , since the distance is measured in the latter.

but gamma=~70.712 so Earth time is 0.00000141424891
so the total distance is 0.9999c*0.00000141424891=423.938 ...

what is going on!?

 

[BenjF]

Your value of gamma is not correct. You haven't probably kept enough decimals in V.

 

[zimo]

Your value of gamma is not correct. You haven't probably kept enough decimals in V.

These are my steps to get gamma:

by definition: gamma=1 / sqrt(1-v^2/c^2)
1 / sqrt(1 - 0.9999^2) = 1/ 0.014141782 =~ 70.712

 

[BenjF]

Your definition of gamma is correct, and so are your steps to calculate it, but I wonder how you have exactly derived V. This value of V/c=0.9999 seems a bit too small for me.

 

[zimo]

please, help me here, I'm feeling lost... would you mind?

 

[Dick]

These are my steps to get gamma:

by definition: gamma=1 / sqrt(1-v^2/c^2)
1 / sqrt(1 - 0.9999^2) = 1/ 0.014141782 =~ 70.712

Why are you using a rather approximate v to compute gamma? Didn't you use gamma to find v to begin with? How DID you get v?