If an airplane flies at 300 m/s around the planet, and the earth\'s circumferenc

Question

If an airplane flies at 300 m/s around the planet, and the earth's circumference is about 4 times 10^7m, calculate the time dilation for an around-the-world trip. Ignore the earth's rotation and gravity. In the rest frame of the sample, the number of muons at time t is given by N(t) = N_0 e^-t/tau, where N_0 is the number at t=0, and tau is the mean lifetime (or "half life") of 2.2 mu s. A) Assume the muons are produced at a height of 5.0 km, and head toward the ground as a speed of 0.91c. What fraction will reach the ground? B) what fraction would reach the ground if classical mechanics were correct?

Explanation / Answer

3. given v = 300 m/s

so gamma = sqroot(1 - v^2/c^2) where c = 3*10^8 m/s

gamma = sqroot(1 - 10^-12) = 0.9999999

now circumference, l = 4*10^7 m

time, t = l/v = 4*10^7/300 = 133,333 s

time dialation = t/gamma - t = 8*10^-5 s

4. N(t) = No*e^(-t/T)

T = 2.2*10^-6 s

speed of muons, v = 0.91 c

distance, d = 5000 m

classical time taken by muons to reach earth, to = d/v = 5000/0.91*3*10^8 = 5.4945*10^-5 s

time taken in frame of muons = t' = to*sqroot(1 - (v/c)^2) = to*sqroot(1 - 0.91^2) = to/2.4119 =

so fraction reqaching earth, N/No = e^(-2.278*10^-5/2.2*10^-6) = 3.302*10^-5

if classical mechanics was correct

N/No = e^(-5.4945*10^-5/2.2*10^-6) = 1.423*10^-11

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