Question #266247

A muon formed high in the Earth’s atmosphere is measured by an observer on the Earth’s surface to travel at speed 𝑣 = 0.990𝑐 for a distance of 4.60 km before it decays into an electron, a neutrino, and an antineutrino (πœ‡βˆ’ β†’ π‘’βˆ’ + 𝜈 + πœˆΜ…).  

(a) For what time interval does the muon live as measured in its reference frame?  

(b) How far does the Earth travel as measured in the frame of the muon? 


1
Expert's answer
2021-11-16T10:14:06-0500

Lorentz factor:\text {Lorentz factor:}

v=0.99cv = 0.99c

Ξ³=11βˆ’v2c2=11βˆ’0.992β‰ˆ7.09\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-0.99^2}}\approx7.09

for Earthbound observer:\text{for Earthbound observer:}

sE=4.6kms_E = 4.6km

vE=0.99cv_E= 0.99c

cβ‰ˆ3βˆ—108mcc \approx 3*10^8\frac{m}{c}

tE=sEvE=46000.99βˆ—3βˆ—108β‰ˆ1.54βˆ—10βˆ’5 st_E= \frac{s_E}{v_E}= \frac{4600}{0.99*3*10^8}\approx1.54*10^{-5}\ s

for muonbound observer:\text{for muonbound observer:}

tm=tEΞ³=1.54βˆ—10βˆ’57.09β‰ˆ2.18βˆ—10βˆ’6st_m = \frac{t_E}{\gamma}=\frac{1.54*10^{-5}}{7.09}\approx 2.18*10^{-6}s

sm=vtm=vβˆ—tEΞ³=sEΞ³β‰ˆ648.8ms_m = vt_m= v* \frac{t_E}{\gamma}=\frac{s_E}{\gamma}\approx648.8m

Answer:\text{Answer:}

a)1.54βˆ—10βˆ’5 sa)1.54*10^{-5}\ s

b)648.8 mb)648.8\ m


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS