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Chapter 37: Problem 11

A spaceship whose rest length is \(280 \mathrm{~m}\) has a speed of \(0.94 c\) with respect to a certain reference frame. A micrometeorite, also with a speed of \(0.94 c\) in this frame, passes the spaceship on antiparallel track. How long does it take this object to pass the ship as measured on the ship?

Short Answer

Expert verified
It takes approximately 3.19 \times 10^{-7} seconds.

Step by step solution

01

Calculate the Relative Velocity

First, calculate the relative velocity between the spaceship and the micrometeorite. Since both objects are moving at a speed of \(0.94c\) in opposite directions, their relative velocity \(v_{rel}\) can be calculated using the relativistic velocity addition formula: \[ v_{rel} = \frac{v_1 + v_2}{1 + \frac{v_1 \cdot v_2}{c^2}} \]Substitute \(v_1 = 0.94c\) and \(v_2 = 0.94c\):\[ v_{rel} = \frac{0.94c + 0.94c}{1 + \frac{(0.94c)(0.94c)}{c^2}} = \frac{1.88c}{1 + 0.8836} \approx \frac{1.88c}{1.8836} \approx 0.9989c \]
02

Calculate the Lorentz Contracted Length

Next, calculate the length of the spaceship in the frame of the micrometeorite, which experiences length contraction due to its relative speed. The contracted length \(L\) is given by:\[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]where \(L_0 = 280\, \text{m}\) and \(v = 0.94c\):\[ L = 280 \sqrt{1 - (0.94)^2} = 280 \times \sqrt{1 - 0.8836} = 280 \times \sqrt{0.1164} \approx 280 \times 0.3415 \approx 95.6 \, \text{m} \]
03

Calculate Time for Meteorite to Pass Ship

To find the time that the micrometeorite takes to pass the spaceship, use its speed relative to the ship and the contracted length calculated in the previous step. The time \( t \) is given by:\[ t = \frac{L}{v_{rel}} \]Plugging in the values:\[ t = \frac{95.6 \text{ m}}{0.9989c} \]We know \(c = 3.0 \times 10^8 \text{ m/s}\), so:\[ t = \frac{95.6}{0.9989 \times 3.0 \times 10^8} \approx \frac{95.6}{2.9967 \times 10^8} \approx 3.19 \times 10^{-7} \text{ seconds} \]

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Key Concepts

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

Lorentz contraction
When dealing with objects moving at high speeds, such as a spaceship traveling near the speed of light, their measured length changes according to the principles of relativity. This phenomenon is known as Lorentz contraction. In this case, the length of the moving object becomes shorter in the direction of motion compared to its rest length, the length measured at rest.
The formula for Lorentz contraction is given by:\[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]where:
  • \(L_0\) is the rest length of the object.
  • \(v\) is the velocity of the object.
  • \(c\) is the speed of light.
For the spaceship in our exercise, its rest length is 280 m. Moving at a velocity of 0.94c, the contracted length is calculated to be approximately 95.6 m. This shows how the spaceship would appear shorter to an observer due to its high speed.
velocity addition
In the realm of relativity, adding velocities isn't as simple as traditional arithmetic suggests, especially when dealing with speeds close to the speed of light. The relativistic velocity addition formula adjusts for these high-speed conditions.
The formula is:\[ v_{rel} = \frac{v_1 + v_2}{1 + \frac{v_1 \cdot v_2}{c^2}} \]Here:
  • \(v_1\) and \(v_2\) are the velocities of the two objects as measured in the same reference frame.
  • \(c\) is the speed of light.
In the given exercise, both the spaceship and micrometeorite travel at 0.94c but in opposite directions. Using the formula, their relative velocity is calculated to be approximately 0.9989c, which is nearly the speed of light. This calculation is crucial because it helps determine the time it takes for the micrometeorite to pass the spaceship.
relative velocity
Relative velocity refers to the velocity of one object as observed from another. In scenarios involving high speeds, like those in our exercise with the spaceship and micrometeorite, accurately determining relative velocities requires considering the effects of relativity.
When both objects move in opposite directions each at 0.94c, their relative velocity isn't simply twice their speed due to relativistic effects. Instead, using the velocity addition formula, we find the relative velocity to be a very close 0.9989c. Understanding relative velocity is essential because it influences other calculations, such as time taken for the micrometeorite to pass the spaceship.
This concept ensures that when objects move at significant fractions of the speed of light, calculations remain accurate and reliable, preventing unrealistic outcomes.
time dilation
Time dilation is a concept in relativity where time is observed to run differently for objects in relative motion. For objects moving at speeds approaching the speed of light, time appears to pass more slowly compared to stationary observers. This difference becomes critical when measuring durations in high-speed contexts.
In our exercise, we want to determine how long it takes for a micrometeorite to pass by a moving spaceship. Here, the spaceship perceives a different flow of time due to its velocity and relational speed with the micrometeorite. By incorporating both Lorentz contraction and relative velocity, we calculate the passing time to be approximately 3.19 脳 10-7 seconds.
Understanding time dilation is key in high-speed scenarios, ensuring that we accurately predict time-based events as they diverge from typical low-speed observations.

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Most popular questions from this chapter

An electron of \(\beta=0.999987\) moves along the axis of an evacuated tube that has a length of \(5.00 \mathrm{~m}\) as measured by a laboratory observer \(S\) at rest relative to the tube. An observer \(S^{\prime}\) who is at rest relative to the electron, however, would see this tube moving with speed \(v(=\beta c)\). What length would observer \(S^{\prime}\) measure for the tube?

An alpha particle with kinetic energy \(7.70 \mathrm{MeV}\) collides with an \({ }^{14} \mathrm{~N}\) nucleus at rest, and the two transform into an \({ }^{17} \mathrm{O}\) nucleus and a proton. The proton is emitted at \(90^{\circ}\) to the direction of the incident alpha particle and has a kinetic energy of \(4.44 \mathrm{MeV}\). The masses of the various particles are alpha particle, \(4.00260 \mathrm{u},{ }^{14} \mathrm{~N}, 14.00307 \mathrm{u} ;\) proton, \(1.007825 \mathrm{u}\); and \({ }^{17} \mathrm{O}, 16.99914 \mathrm{u}\). In \(\mathrm{MeV}\), what are (a) the kinetic energy of the oxygen nucleus and (b) the \(Q\) of the reaction? (Hint: The speeds of the particles are much less than \(c\).)

A meter stick in frame \(S^{\prime}\) makes an angle of \(30^{\circ}\) with the \(x^{\prime}\) axis. If that frame moves parallel to the \(x\) axis of frame \(S\) with speed \(0.95 c\) relative to frame \(S\), what is the length of the stick as measured from \(S\) ?

Bullwinkle in reference frame \(S^{\prime}\) passes you in reference frame \(S\) along the common direction of the \(x^{\prime}\) and \(x\) axes, as in Fig. 37-9. He carries three meter sticks: meter stick 1 is parallel to the \(x^{\prime}\) axis, meter stick 2 is parallel to the \(y^{\prime}\) axis, and meter stick 3 is parallel to the \(z^{\prime}\) axis. On his wristwatch he counts off \(10.0 \mathrm{~s}\), which takes \(30.0 \mathrm{~s}\) according to you. Two events occur during his passage. According to you, event 1 occurs at \(x_{1}=33.0 \mathrm{~m}\) and \(t_{1}=22.0 \mathrm{~ns}\), and event 2 occurs at \(x_{2}=53.0 \mathrm{~m}\) and \(t_{2}=62.0 \mathrm{~ns} .\) According to your measurements, what is the length of (a) meter stick 1, (b) meter stick 2, and (c) meter stick 3? According to Bullwinkle, what are \((\mathrm{d})\) the spatial separation and \((\mathrm{e})\) the temporal separation between events 1 and 2 , and (f) which event occurs first?

A rod lies parallel to the \(x\) axis of reference frame \(S\), moving along this axis at a speed of \(0.892 c\). Its rest length is \(1.70 \mathrm{~m}\). What will be its measured length in frame \(S ?\)
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