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Chapter 26: Problem 71

Muons are created by cosmic-ray collisions at an elevation \(h\) (as measured in Earth's frame of reference) above Earth's surface and travel downward with a constant speed of \(0.990 c .\) During a time interval of \(1.5 \mu \mathrm{s}\) in the rest frame of the muons, half of the muons present at the beginning of the interval decay. If one fourth of the original muons reach Earth before decaying, about how big is the height \(h ?\)

Short Answer

Expert verified
Answer: The approximate height of the point where muons are created above Earth's surface is \(1.58 \times 10^{6} m\).

Step by step solution

01

Understand time dilation

We are given the decay time of muons in their rest frame, and we need to find the decay time of muons relative to the observer on Earth. We can do this by finding the time dilation factor, given by \(\gamma = \frac{1}{\sqrt{1 - v^{2}/c^{2}}}\). Here, \(v\) is the speed of muons, and \(c\) is the speed of light in a vacuum.
02

Calculate time dilation factor and proper time

Using the given muon speed \(v = 0.990c\): \(\gamma = \frac{1}{\sqrt{1 - (0.990c)^{2}/c^{2}}} = \frac{1}{\sqrt{1 - 0.9801}} = \frac{1}{\sqrt{0.0199}} \approx 7.088\) Now, we can calculate the proper time for the observer on Earth, \(t_{e} = \gamma t_{m}\), where \(t_{m} = 1.5 \mu s\): \(t_{e} = 7.088 \times 1.5 \mu s = 10.632 \mu s\)
03

Calculate distance muons travel in Earth's frame of reference

Now that we have the proper time in the Earth's frame (observer's frame), we can determine the distance muons travel in this time. We are given that the muons travel with a constant speed of \(0.990c\), so the distance traveled is: \(d = v \cdot t_{e} = 0.990c \cdot 10.632 \mu s \approx 10.525 c \mu s\)
04

Find the height \(h\)

Given that only one-fourth of the original muons reach the Earth before decaying, the distance traveled by the muons can be considered as the half-decay distance: \(h = \frac{1}{2} (10.525 c \mu s) \approx 5.262 c \mu s\)
05

Convert the height to meters

Now we convert the height to meters by multiplying it with the speed of light in a vacuum (\(c\)), which is approximately \(3\times10^8 m/s\): \(h = 5.262 \times 10^{-6} s \times 3\times10^8 \frac{m}{s} = 1.5786 \times 10^{6} m\) So, the height of the point where muons are created above Earth's surface is approximately \(1.58 \times 10^{6} m\).

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Most popular questions from this chapter

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