/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 A HeNe laser onboard a spaceship... [FREE SOLUTION] | 91Ó°ÊÓ

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A HeNe laser onboard a spaceship moving toward a remote space station emits a beam of red light toward the space station. The wavelength of the beam, as measured by a wavelength meter on board the spaceship, is \(632.8 \mathrm{nm}\). If the astronauts on the space station see the beam as a blue beam of light with a measured wavelength of \(514.5 \mathrm{nm},\) what is the relative speed of the spaceship with respect to the space station? What is the shift parameter \(z\) in this case?

Short Answer

Expert verified
Answer: The relative speed of the spaceship with respect to the space station is approximately \(1.881 * 10^8 ms^{-1}\), and the shift parameter \(z \approx -0.1869\).

Step by step solution

01

Convert wavelengths to frequencies.

Using the formula \(c = \lambda f\), where \(c\) is the speed of light (\(3 * 10^8 ms^{-1}\)), \(\lambda\) is the wavelength, and \(f\) is the frequency, we can convert the given wavelengths to their corresponding frequencies. Let's calculate the emitted and observed frequencies. Emitted wavelength (\(\lambda\)): \(632.8 nm\) = \(632.8 * 10^{-9} m\) Emitted frequency (\(f\)): \(f = \frac{c}{\lambda} = \frac{3 * 10^8 ms^{-1}}{632.8 * 10^{-9} m} = 4.738 * 10^{14} Hz\) Observed wavelength (\(\lambda'\)): \(514.5 nm\) = \(514.5 * 10^{-9} m\) Observed frequency (\(f'\)): \(f' = \frac{c}{\lambda'} = \frac{3 * 10^8 ms^{-1}}{514.5 * 10^{-9} m} = 5.834 * 10^{14} Hz\)
02

Use the Doppler effect formula to find the relative velocity.

Now that we have the emitted and observed frequencies, we can use the Doppler effect formula to find the relative velocity of the spaceship with respect to the space station: \(f' = \frac{f}{1+\frac{v}{c}}\) Rearranging for \(v\): \(v = c (\frac{f}{f'} - 1)\) Substitute the values of \(f, f', c\): \(v = 3 * 10^8 ms^{-1} (\frac{4.738 * 10^{14} Hz}{5.834 * 10^{14} Hz} - 1)\) \(v \approx -1.881 * 10^8 ms^{-1}\) The negative sign indicates that the spaceship is moving towards the space station. Therefore, the relative speed of the spaceship with respect to the space station is approximately \(1.881 * 10^8 ms^{-1}\).
03

Compute the shift parameter \(z\).

Finally, we will compute the shift parameter \(z = \frac{\Delta\lambda}{\lambda}\), where \(\Delta\lambda = \lambda'-\lambda\). \(\Delta\lambda = 514.5 nm - 632.8 nm = -118.3 nm\) \(z = \frac{-118.3 nm}{632.8 nm} \approx -0.1869\) The shift parameter \(z \approx -0.1869\).

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Key Concepts

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

Relative Velocity
When we talk about relative velocity in the context of the Doppler Effect, we're referring to the speed at which an object is moving in relation to another. Here, it's the speed of the spaceship compared to the space station. The Doppler Effect in physics helps us understand how the frequency of a wave changes when the source and observer are in motion relative to each other.
In the exercise, we determined the relative speed by using the Doppler Effect formula:
  • The emitted frequency of the laser light was calculated from the spaceship's perspective.
  • The observed frequency was determined as the spaceship approached the space station.
  • The change in frequency gives insight into the spaceship's speed compared to the space station.
The formula used was:\[f' = \frac{f}{1+\frac{v}{c}}\]By rearranging, the velocity \(v\) was calculated, showing how quickly the spaceship moves towards the space station. A negative value here suggests that the distance between the spaceship and the space station is decreasing, hence getting closer.
Wavelength Shift
Wavelength shift happens when there is a change in the observed wavelength due to the relative motion of the source and the observer - a result of the Doppler Effect. For this problem, the red light emitted from the spaceship appeared blue to the astronauts on the space station.
Here's how it works:
  • The original wavelength of the laser emitted from the spaceship was 632.8 nm.
  • Upon reaching the space station, this wavelength was observed as 514.5 nm, a shorter wavelength associated with blue light.
These changes in wavelength are quantified as a Wavelength Shift. It's crucial in astrophysics for determining the movement and speed of astronomical objects relative to Earth.
The difference in wavelength, \(\Delta \lambda\), is computed by:\[\Delta \lambda = \lambda' - \lambda\]In our problem, this wavelength shift highlights the spaceship's quick approach towards the space station, evidenced by the shift towards the blue end of the spectrum.
Frequency Conversion
Frequency conversion is a critical step in resolving problems involving the Doppler Effect. To understand how waves change from one form (or frequency) to another, we use the speed of light equation:\(c = \lambda f\).
Here’s the breakdown:
  • Start with the emitted wavelength \(\lambda\) on the spaceship: 632.8 nm.
  • Convert this into its frequency \(f\) by rearranging the formula, which gives us \(f = \frac{c}{\lambda}\).
  • Next, take the observed wavelength\(\lambda'\) of 514.5 nm as seen by the station astronauts.
  • Convert this to the observed frequency \(f'\) using the same formula: \(f' = \frac{c}{\lambda'}\).
This process translates the problem into terms of frequency, which can then be used in the Doppler Effect equation to find the relative velocity. This step is invaluable because it mathematically bridges the shift in wavelength (color change from red to blue) to actual speeds.

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Most popular questions from this chapter

Radar-based speed detection works by sending an electromagnetic wave out from a source and examining the Doppler shift of the reflected wave. Suppose a wave of frequency \(10.6 \mathrm{GHz}\) is sent toward a car moving away at a speed of \(32.0 \mathrm{~km} / \mathrm{h}\). What is the difference between the frequency of the wave emitted by the source and the frequency of the wave an observer in the car would detect?

A square of area \(100 \mathrm{~m}^{2}\) that is at rest in the reference frame is moving with a speed \((\sqrt{3} / 2) c\). Which of the following statements is incorrect? a) \(\beta=\sqrt{3} / 2\) b) \(\gamma=2\) c) To an observer at rest, it looks like another square with an area less than \(100 \mathrm{~m}^{2}\) d) The length along the moving direction is contracted by a factor of \(\frac{1}{2}\)

In the age of interstellar travel, an expedition is mounted to an interesting star 2000.0 light-years from Earth. To make it possible to get volunteers for the expedition, the planners guarantee that the round trip to the star will take no more than \(10.000 \%\) of a normal human lifetime. (At that time the normal human lifetime is 400.00 years.) What is the minimum speed the ship carrying the expedition must travel?

In mechanics, one often uses the model of a perfectly rigid body to model and determine the motion of physical objects (see, for example, Chapter 10 on rotation). Explain how this model contradicts Einstein's special theory of relativity.

An astronaut in a spaceship flying toward Earth's Equator at half the speed of light observes Earth to be an oblong solid, wider and taller than it appears deep, rotating around its long axis. A second astronaut flying toward Earth's North Pole at half the speed of light observes Earth to be a similar shape but rotating about its short axis. Why does this not present a contradiction?

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