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Electromagnetic radiation from a star is observed with an earth-based telescope. The star is moving away from the earth at a speed of \(0.520 c\). If the radiation has a frequency of \(8.64 \times 10^{14} \mathrm{~Hz}\) in the rest frame of the star, what is the frequency measured by an observer on earth?

Short Answer

Expert verified
The frequency measured by an observer on Earth is \(4.84 \times 10^{14} Hz\).

Step by step solution

01

Understand the Doppler Effect for Light

The Doppler effect for light, unlike sound, does not depend on the medium. The formula is \[ f' = f \times \sqrt{\frac{1 + \frac{v}{c}}{1 - \frac{v}{c}}}\], where f' is the observed frequency, f is the source frequency, v is the speed of the source relative to the observer, and c is the speed of light.
02

Substitute the given values into the formula

Substitute the given values into the formula. Given that \(f = 8.64 \times 10^{14} Hz\), \(v = 0.520c\), and \(c = 1\) (in natural units where c is used as the unit of speed, often used in Special Relativity problems), the formula becomes \[ f' = 8.64 \times 10^{14} Hz \times \sqrt{\frac{1 + 0.520}{1 - 0.520}} \]
03

Calculation

Calculate the value of f' using these inputs. The result will be the observed frequency on Earth.

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Key Concepts

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

Electromagnetic Radiation
Electromagnetic radiation encompasses a broad range of wavelengths and frequencies, from radio waves to gamma rays. These waves are vital in astronomy because they enable us to observe distant celestial objects like stars, planets, and galaxies. When dealing with light from distant stars, such as in our original exercise, we're often focusing on the visible spectrum, a small section of the entire electromagnetic spectrum.
By observing electromagnetic radiation, scientists can gather essential information about stars, including their composition, temperature, motion, and even potential planets orbiting them. This broad spectrum of waves travels through space at the speed of light, helping us understand the universe.
In the case of the Doppler Effect for Light, which is part of electromagnetic radiation, the apparent frequency of light changes due to the relative motion between the light source (like a star) and the observer (such as on Earth). This phenomenon plays a key role in astronomy as it helps in studying how stars and galaxies move.
Special Relativity
Special relativity, introduced by Albert Einstein, revolutionized our understanding of space, time, and motion. Its core principles include the constancy of the speed of light in a vacuum and the idea that the laws of physics are the same in all inertial frames.
One of the essential consequences of special relativity is time dilation and length contraction, which become significant when dealing with objects moving at speeds close to the speed of light. For our exercise, special relativity provides the framework for understanding how the observed frequency of electromagnetic radiation changes due to the relative motion between the observer and the source through the relativistic Doppler effect.
In our problem, the star's movement away from Earth at a speed close to a significant fraction of the speed of light requires us to apply a relativistic correction to calculate the frequency observed. This ensures that our measurements align with the principles of relativity, accounting for the high velocities involved.
Frequency Shift
Frequency shift refers to the change in the observed frequency of a wave when the source of the wave moves relative to the observer. This shift can result in a higher or lower frequency depending on whether the source is moving towards or away from the observer. In the context of light waves, this concept is closely associated with the Doppler Effect.
A frequency shift can manifest as a redshift or a blueshift. If an object in space, like the star in our problem, moves away from us, the light it emits shifts to lower frequencies and longer wavelengths, a phenomenon known as redshift. Conversely, if the object approaches, we observe a blueshift, where the frequency increases.
Calculating the frequency shift requires an understanding of the source's velocity relative to the observer. In our mathematical solution, substituting given values into the Doppler effect formula for light helps us find the new frequency that an Earth-based observer would measure.

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Most popular questions from this chapter

A proton (rest mass \(1.67 \times 10^{-27} \mathrm{~kg}\) ) has total energy that is 4.00 times its rest energy. What are (a) the kinetic energy of the proton; (b) the magnitude of the momentum of the proton; (c) the proton's speed?

The distance to a particular star, as measured in the earth's frame of reference, is 7.11 light-years ( 1 light-year is the distance that light travels in \(1 \mathrm{y}\) ). A spaceship leaves the earth and takes \(3.35 \mathrm{y}\) to arrive at the star, as measured by passengers on the ship. (a) How long does the trip take, according to observers on earth? (b) What distance for the trip do passengers on the spacecraft measure?

In the earth's rest frame, two protons are moving away from each other at equal speed. In the frame of each proton, the other proton has a speed of \(0.700 c\). What does an observer in the rest frame of the earth measure for the speed of each proton?

A source of electromagnetic radiation is moving in a radial direction relative to you. The frequency you measure is 1.25 times the frequency measured in the rest frame of the source. What is the speed of the source relative to you? Is the source moving toward you or away from you?

Compute the kinetic energy of a proton (mass \(\left.1.67 \times 10^{-27} \mathrm{~kg}\right)\) using both the nonrelativistic and relativistic expressions, and compute the ratio of the two results (relativistic divided by nonrelativistic) for speeds of (a) \(8.00 \times 10^{7} \mathrm{~m} / \mathrm{s}\) and (b) \(2.85 \times 10^{8} \mathrm{~m} / \mathrm{s}\)

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