Question

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99537 c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 40.0 km .
QUESTION 1: As measured by the scientist, how much time does it take the particle to travel the 40.0 km to the surface of the earth?
QUESTION 2: Use the length-contraction formula to calculate the distance from where the particle is created to the surfaceof the earth as measured in the particle's frame.
QUESTION 3: In the particle's frame, how much time does it take the particle to travel from where it is created to the surface of the earth?

Answer 1

Answer:

Q1)  Time taking by particle to travel the 40 km wrt. earth = $1.34\times10^{-6}$ sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = $1.285\times10^{-5}$ sec.

Explanation:

Given :

Speed of particle wrt. earth $v=0.99537c$

Distance between where particle is created and earth surface = 40 km

we know that,

⇒       $v = \frac{x}{t}$

Where $x = 40\times10^{3} m$, $v = 0.99537c$, we know speed of light $c = 3 \times10^{8}$

∴      $t = \frac{x}{v}$

         $= \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }$

      $t = 1.34\times10^{-6} sec$

∴ Thus, time taking by particle to travel the 40 km wrt. earth $t = 1.34\times10^{-6} sec$

According to the lorentz transformation,

⇒    $l = l_{o} \sqrt{1-\frac{v^2}{c^2} }$

Where $l =$ improper length, $l_{o} =$proper length (distance measured wrt. rest frame) = 40 km

     $l = 40 \sqrt{1-\frac{v^2}{c^2 }$

     $l = 40 \times 0.096$

     $l = 3.84$ km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

   $\Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }$

Where $\Delta t =$ improper time (wrt. earth frame time) $=1.34\times10^{-6} sec$ ,  $\Delta t _{o} =$ proper time (wrt. particle frame).

 $1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}$

 $\Delta t_{o} = 1.285 \times10^{-5} sec$

Thus, the time taking by particle to travel from where it is created to the surface of the earth $= 1.285 \times10^{-5} sec$.