Problem

$\cdot$ An enemy spaceship is moving toward your …

02:21

Problem 15 Medium Difficulty

A A muon is created 55.0 $\mathrm{km}$ above the surface of the
earth (as measured in the earth's frame). The average life-
time of a muon, measured in its own rest frame, is 2.20$\mu \mathrm{s}$ ,
and the muon we are considering has this lifetime. In the
frame of the muon, the earth is moving toward the muon
with a speed of 0.9860$c$ . (a) In the muon's frame, what is its
initial height above the surface of the earth? (b) In the
muon's frame, how much closer does the earth get during
the lifetime of the muon? What fraction is this of the muon's
original height, as measured in the muon's frame? (c) In the
earth's frame, what is the lifetime of the muon? In the
earth's frame, how far does the muon travel during its life-
time? What fraction is this of the muon's original height in
the earth's frame?

Answer

$$
1.32 \times 10^{-5} \mathrm{s}
$$
$$
7.09 \%
$$

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Video Transcript

Okay, so in this problem, we know that, um, Yuan is created at a distance off. Let's put in here a distance. L off 50 five kilometers from the year. This is measure in the reference frame off the earth. We know that the proper time of the mule Let's go doubt the zero is 2.2 microseconds. And we know that the mule has a relative speed. Let's go. You off zero point ny 86 See? So this time is the proper time, Which means is measure in the reference frame off the mule brought for time. Okay, so the first item off this problem, I think a We have to calculate, uh, in the mules reference frame. What is the initial height above the surface of the earth? So we know that the immune is in I speed close to the speed of light. Therefore, he received the world different way. He sees the distance contract because of the relativistic effects. So since he's moving and he sees the distance contract, we know that the distance off the mule sees should be decent over our initial reference Observer sees contract by gamma, and this is equal to L Square, Root off one miners, You square c square all this in the square route and we have all this information here So we can say that l zero his equal 55 the multiplies one minus 0.9 86 square in this crab ridge. And this should be good too. Nine point 17 kilometers. So this is the height that the immune cease when he's created above the surface of the earth. Now let's go to the item B in ight nd we have to calculate in the mules frame How much closer does the year get during the lifetime of the mule? And what fraction in this When what fraction is this off the mules? Original height. Okay, all of these measures in the mules reference frame. So we know that the muse reference frame We have a proper time off 2.2 microseconds, and we already discover that he covers a distance l zero off nine point 17 kilometers. So the distance that the mewling covers in his lifetime is just let's put in here. That's called L one is the distance that the new one covers. And during his lifetime, so this is just going to be, too. They speed multiply by the appropriate time. So this is just, let's see, 2.2 times tend to the power off minor six. This is the time microseconds in here, the multiplies they speed in the speed is just zero point nine eight six see, and C is the speed of light, which is three times 10 to the power off. Eight. If we multiply all this, we're gonna get 651 meters. So this is a distance that the mule covers doing. He's appropriate time, and this is represent a fraction off. Let's put in here. This represents a fraction off 651 divided by nine points, 17 times 10 to the power off three, which is kilometers. So this represents a fraction off 7.1 were sent off its initial in need show height. So this is the answer to the second item. Uh, I can't see final right end of this problem. Okay, so in the final item, we have to calculate the proper time at the appropriate time to the lifetime off the mule. In the reference frame of the Earth we have to calculate the distance he travels during his lifetime. Let's go l one slash, and we have to calculate what fraction off the original height that that distance represents. So again, we know that the moon is under relativistic effect, so his time is dilated. So it's just going to be doubt. T equals gamma delta t zero. So this is just doubt t zero divided by one minus You square c square in this square route and we have here 2.2 times 10 to the power off minor six divided by one minus zero 0.9 86 square in the square root. And after calculating all this, we have a time measure in the reference frame of the year off one point 32 times 10 to the power off miners six minus five seconds. So this is the time our observer in the Earth measure for the mule and the distance he cover during this time is just l won't slash is it? It's just the time. Multiply by his speed. So this is simply one point 22 times 10 to the power of minors. Five. The multiplies the speed, which is 0.9 86 c and C is just three times 10. What if our off eight multiplying all this? We have 3.9 kilometers and the fraction off this initial height is just going to be tree point nine. Divided by 55 kilometres, this represents again seven point one percent of its initial height measure in the earth's frame. And that's the final answer. Thanks for watching.

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