Question

A muon has a lifetime 2×10^-6 in it's rest frame. It is created 100 km above the earth moves towards it at a speed of 2.97 x 10^8 m/s. At what altitude does it decay?

Answer 1

The lifetime of the muon for the observer on Earth is

$\displaystyle t = \frac{\tau}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{2 \cdot 10^{-6}}{\sqrt{1-(2.97/2.998)^2}}=14.667 \cdot 10^{-6}$

The distance muon can travel before the decay is

$L = v \cdot t = 2.97 \cdot 10^8 \cdot 14.667 \cdot 10^{-6} = 4356\; m.$

Therefore, it decays at the altitude of 4356m.

Answer: 4356 m.