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Chapter 14: Problem 35

A spaceship, moving away from Earth at a speed of \(0.900 c,\) reports back by transmitting at a frequency (measured in the spaceship frame) of \(100 \mathrm{MHz}\). To what frequency must Earth receivers be tuned to receive the report?

Short Answer

Expert verified
The Earth receivers must be tuned to approximately 22.9 MHz.

Step by step solution

01

Understanding the problem

We need to find the frequency received on Earth from a spaceship moving away at a speed of \(0.900c\), where \(c\) is the speed of light. The frequency emitted from the spaceship is \(100 \text{ MHz}\). This problem involves the relativistic Doppler effect.
02

Relativistic Doppler Effect Formula

The relativistic Doppler effect formula for a source moving away is given by:\[f_r = f_s \sqrt{\frac{1-\beta}{1+\beta}} \]where \(f_r\) is the received frequency, \(f_s\) is the source frequency, and \(\beta = \frac{v}{c}\), with \(v\) being the velocity of the spaceship.
03

Plug in the known values

Given \(f_s = 100 \text{ MHz}\) and \(\beta = 0.900\), we substitute these into the formula:\[f_r = 100 \times 10^6 \sqrt{\frac{1-0.900}{1+0.900}} \]
04

Solve the expression under the square root

Calculate the expression under the square root:\[\sqrt{\frac{1-0.900}{1+0.900}} = \sqrt{\frac{0.100}{1.900}} = \sqrt{0.05263} \approx 0.229 \]
05

Calculate the received frequency

Multiply the emitted frequency by the calculated factor:\[f_r = 100 \times 10^6 \times 0.229 \approx 22.9 \times 10^6 \approx 22.9 \text{ MHz} \]
06

Conclusion

The frequency that the receivers on Earth must be tuned to in order to receive the report is approximately \(22.9 \text{ MHz}\).

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Key Concepts

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

Frequency Shift in Relativistic Doppler Effect
The concept of frequency shift is vital when discussing the relativistic Doppler effect. It describes how the observed frequency of waves changes due to the motion of the source and observer. In simpler terms, when an object like a spaceship approaches or moves away from us, the waves it emits (such as light or sound) appear at different frequencies.
This change in frequency is what we call a frequency shift. For a source moving away, like a spaceship traveling away from Earth, the frequency of its emitted signals appears lower than it actually is.
This phenomenon occurs due to the stretching of waves over increased distance. Calculations for such scenarios use the relativistic Doppler effect formula, which factors in relative velocity and the speed of light. Understanding this shift is crucial for various applications, including astronomy and telecommunications.
**Key Points to Remember**:
  • Frequency shifts help determine object speed and direction.
  • Lower observed frequencies imply the source is moving away.
  • The relativistic Doppler effect formula is essential to calculate these shifts.
Speed of Light as a Cosmic Constant
The speed of light, symbolized by the letter \(c\), is one of the most fundamental constants in physics. Its value in a vacuum is approximately \(3 \, \times \, 10^8 \, \text{m/s}\). It acts as a cosmic speed limit, meaning nothing can travel faster than light.
This constant is crucial in our calculations related to the relativistic Doppler effect. For instance, when determining frequency shifts, the fraction involving speed \(v\) over \(c\) helps us understand how much frequencies will change due to motion.
The speed of light's invariance makes it a cornerstone of Einstein's theory of relativity. Every observer, regardless of their motion, measures light's speed as \(c\).
**Remember:**
  • The speed of light is a constant \(c = 3 \, \times \, 10^8 \, \text{m/s}\).
  • It plays a critical role in the relativistic formulas used in physics.
  • Nothing can exceed the speed of light in a vacuum.
Understanding Spaceship Velocity
Spaceship velocity significantly impacts the frequency shift observed due to the relativistic Doppler effect. In our exercise, the spaceship travels at \(0.900c\), meaning it moves at 90% the speed of light. This high velocity is why the frequency seen from Earth is significantly lower than the one emitted aboard the spaceship.
When calculating the Doppler shift, velocity's role is captured through the parameter \(\beta = \frac{v}{c}\). Here, \(v\) refers to the spaceship's speed, and \(c\) is the speed of light. For practical purposes, understanding this parameter helps in computing observed frequencies for spaceships or other quick-moving objects.
**Key Takeaways**:
  • Higher velocities lead to more significant frequency shifts.
  • \(\beta\) represents the fraction of the speed of light a spaceship travels.
  • Calculating \(\beta\) is pivotal for determining the relativistic Doppler effect.

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Most popular questions from this chapter

An iron anchor of density \(7870 \mathrm{~kg} / \mathrm{m}^{3}\) appears \(200 \mathrm{~N}\) lighter in water than in air. (a) What is the volume of the anchor? (b) How much does it weigh in air?

Space cruisers \(A\) and \(B\) are moving parallel to the positive direction of an \(x\) axis. Cruiser \(A\) is faster, with a relative speed of \(v=0.900 c,\) and has a proper length of \(L=200 \mathrm{~m}\). According to the pilot of \(A\), at the instant \((t=0)\) the tails of the cruisers are aligned, the noses are also. According to the pilot of \(B,\) how much later are the noses aligned?

An object hangs from a spring balance. The balance registers \(30 \mathrm{~N}\) in air, \(20 \mathrm{~N}\) when this object is immersed in water, and \(24 \mathrm{~N}\) when the object is immersed in another liquid of unknown density. What is the density of that other liquid?

Suppose that your body has a uniform density of 0.95 times that of water. (a) If you float in a swimming pool, what fraction of your body's volume is above the water surface? Quicksand is a fluid produced when water is forced up into sand, moving the sand grains away from one another so they are no longer locked together by friction. Pools of quicksand can form when water drains underground from hills into valleys where there are sand pockets. (b) If you float in a deep pool of quicksand that has a density 1.6 times that of water, what fraction of your body's volume is above the quicksand surface? (c) Are you unable to breathe?

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 \(90^{n}\) 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 \(c\) )
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