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Chapter 38: Problem 44

An unpowered spaceship whose rest length is 350 meters has a speed \(0.82 c\) with respect to Earth. A micrometeorite, also with speed of \(0.82 c\) with respect to Earth, passes the spaceship on an antiparallel track that is moving in the opposite direction. How long does it take the micrometeorite to pass the spaceship as measured on the ship?

Short Answer

Expert verified
It takes approximately \(1.19 \times 10^{-6} \) seconds for the micrometeorite to pass the spaceship.

Step by step solution

01

Understand the relative velocity

Since the micrometeorite and spaceship are moving in opposite directions with the same speed \(0.82c\), the relative velocity between them is the sum of their speeds. Therefore, the relative velocity is \(0.82c + 0.82c = 1.64c\).
02

Account for Relativistic Effects

The velocity cannot exceed the speed of light. Apply the formula for relativistic addition of velocities: \[ u' = \frac{u+v}{1+\frac{uv}{c^2}} \] Here, both \(u\) and \(v\) are \(0.82c\). Substituting these values, we get: \[ u' = \frac{0.82c + 0.82c}{1+\frac{(0.82c)(0.82c)}{c^2}} = \frac{1.64c}{1+0.6724} = \frac{1.64c}{1.6724} \approx 0.981c \]
03

Calculate time in spaceship's frame

The rest length of the spaceship is 350 meters. Using the relativistic speed \( 0.981c \), the time needed to pass the spaceship length is given by: \[ t = \frac{L}{v_{rel}} = \frac{350 \text{ meters}}{0.981c} \] Knowing that \(c \approx 3 \times 10^8 \text{ meters/second}\), we get: \[ t = \frac{350}{0.981 \times 3 \times 10^8}\] \[ t \approx 1.19 \times 10^{-6} \text{ seconds}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

micrometeorite
Imagine a tiny rock zooming through space. This rock is called a micrometeorite. Micrometeorites are really small, often just a few millimeters wide.
But even though they are small, they can move extremely fast, sometimes thousands of kilometers per hour!
Because they move so fast, they can pose a danger to spacecraft and satellites.

In our problem, the micrometeorite is moving at 0.82 times the speed of light (\(0.82c\)). That's incredibly fast!
For comparison, the speed of light (\(c\)) is approximately 300,000 kilometers per second.
When two objects move at such high speeds, things work a bit differently than we're used to due to the rules of special relativity.
special relativity
Albert Einstein's theory of special relativity explains how objects behave at high speeds.
One crucial idea in this theory is that nothing can travel faster than the speed of light.

When speeds get close to the speed of light, the usual ways of adding velocities together don't work.
Instead, we use the relativistic addition of velocities formula:
\( u' = \frac{u+v}{1+\frac{uv}{c^2}} \).

In our problem, both the spaceship and the micrometeorite are moving at 0.82c, but in opposite directions.
By plugging those values into the formula, we figured out their relative velocity.
It turned out to be approximately 0.981c, not 1.64c as one might initially think.

This slower-than-expected combined speed shows how relativity tweaks our everyday understanding of motion.
relative velocity
Relative velocity is how fast one object appears to be moving when observed from another moving object.
This concept becomes very interesting in the realm of special relativity.

In our scenario, the spaceship and micrometeorite are moving in opposite directions with the same speed relative to Earth.
By considering their relative velocity, we needed to see how fast the micrometeorite looks to be moving from the spaceship's viewpoint.
Using the special relativity formula for velocity addition, we found this relative velocity to be 0.981c.
With this, we could then determine how long it takes for the micrometeorite to pass by the spaceship.

This involved calculating the time based on the spaceship's length and the determined relative velocity.
Ultimately, we found that it takes approximately 1.19 microseconds for the micrometeorite to pass the 350-meter-long spaceship.

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Most popular questions from this chapter

A pulse of protons arrives at detector \(\mathrm{D}\), where you are standing. Prior to this, the pulse passed through detector C, which lies 60 meters upstream. Detector C sent a light flash in your direction at the same instant that the pulse passed through it. At detector D you receive the light flash and the proton pulse separated by a time of 2 nanoseconds \(\left(2 \times 10^{-9} \mathrm{~s}\right)\). What is the speed of the proton pulse?

The half-life of stationary muons is measured to be \(1.6\) microseconds. Half of any initial number of stationary muons decays in one half-life. Cosmic rays colliding with atoms in the upper atmosphere of the Earth create muons, some of which move downward toward the Earth's surface. The mean lifetime of high- speed muons in one such burst is measured to be 16 microseconds. (a) Find the speed of these muons relative to the Earth. (b) Moving at this speed, how far will the muons move in one half-life? (c) How far would this pulse move in one half-life if there were no relativistic time stretching? (d) In the relativistic case, how far will the pulse move in 10 half-lives? (e) An initial pulse consisting of \(10^{8}\) muons is created at a distance above the Earth's surface given in part (d). How many will remain at the Earth's surface? Assume that the pulse moves vertically downward and none are lost to collisions. (Ninety-nine percent of the Earth's atmosphere lies below \(40 \mathrm{~km}\) altitude.)

The famous twin paradox is often introduced as follows: Two identical twins grow up together on Earth. When they reach adulthood, one twin zooms to a distant star and returns to find her stay-at-home sister much older than she is. Thus far no paradox. But Alexis Allen formulates the Twin Paradox for us: "The theory of special relativity tells us that all motion is relative. With respect to the traveling twin, the Earth-bound twin moves away and then returns. Therefore it is the Earthbound twin who should be younger than the 'traveling' twin. But when they meet again at the same place, it cannot possibly be that each twin is younger than the other twin. This Twin Paradox disproves relativity." The paradox is usually resolved by realizing that the traveling twin turns around. Everyone agrees which twin turns around, since the reversal of direction slams the poor traveler against the bulkhead of the decelerating starship, breaking her collarbone. The turnaround, evidenced by the broken collarbone, destroys the symmetry required for the paradox to hold. Good-bye Twin Paradox! Still, Alexis's father Cyril Allen has his doubts about this resolution of the paradox. "Your solution is extremely unsatisfying. It forces me to ask: What if the retro-rockets malfunction and will not fire at all to slow me down as I approach a distant star a thousand light-years from Earth? Then I cannot even stop at that star, much less turn around and head back to Earth. Instead, I continue moving away from Earth forever at the original constant speed. Does this mean that as I pass the distant star, one thousand light-years from Earth, it is no longer possible to say that I have aged less than my Earth-bound twin? But if not, then I would never have even gotten to the distant star at all during my hundred-year lifetime! Your resolution of the Twin Paradox is insufficient and unsatisfying." Write a half-page response to Cyril Allen, answering his objections politely but decisively.

A meter stick lies at rest in the rocket frame and makes an angle \(\phi^{\prime}\) with the \(x^{\prime}\) axis as measured by the rocket observer. The laboratory observer measures the \(x\) - and \(y\) -components of the meter stick as it streaks past. From these components the laboratory observer computes the angle \(\phi\) that the stick makes with his \(x\) axis. (a) Find an expression for the angle \(\phi\) in terms of the angle \(\phi^{\prime}\) and the relative speed \(v^{\text {rel }}\) between rocket and laboratory frames. (b) What is the length of the "meter" stick measured by the laboratory observer? (c) Optional: Why is your expression in part (a) different from equations derived in Problems 34 and \(35 ?\)

In the 24 th century the fastest available interstellar rocket moves at \(v=0.75 c .\) Mya Allen is sent in this rocket at full (constant) speed to Sirius, the Dog Star, the brightest star in the heavens as seen from Earth, which is a distance \(8.7\) ly as measured in the Earth frame. Assume Sirius is at rest with respect to Earth. Mya stays near Sirius, slowly orbiting around that Dog Star, for 7 years as recorded on her wristwatch while making observations and recording data, then returns to Earth with the same speed \(v=0.75 \mathrm{c}\). According to Earth-linked observers: (a) When does Mya arrive at Sirius? (b) When does Mya leave Sirius? (c) When does Mya arrive back at Earth? According to Mya's wristwatch: (d) When does she arrive at Sirius? (e) When does she leave Sirius? (f) When does she arrive back on Earth?
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