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Chapter 22: Problem 56

Light of wavelength \(659.6 \mathrm{nm}\) is emitted by a star. The wavelength of this light as measured on Earth is \(661.1 \mathrm{nm} .\) How fast is the star moving with respect to Earth? Is it moving toward Earth or away from it?

Short Answer

Expert verified
Answer: The star is moving away from Earth with a relative speed of approximately 681 m/s.

Step by step solution

01

Calculate the change in observed wavelength

Using the given information, we can calculate the change in wavelength: \(\Delta \lambda = \lambda_{observed} - \lambda_{emitted} = 661.1 \,\text{nm} - 659.6 \,\text{nm} = 1.5 \,\text{nm}\) #Step 2: Use the Doppler effect formula for the change in wavelength#
02

Apply the Doppler effect formula

The Doppler effect formula for the change in wavelength due to relative motion is given by the following equation: \(\Delta \lambda = \frac{v}{c} \lambda_{emitted}\) where \(\Delta \lambda\) is the change in wavelength, \(v\) is the relative speed of the star with respect to Earth, \(c\) is the speed of light in vacuum, and \(\lambda_{emitted}\) is the wavelength of light emitted by the star. #Step 3: Solve for the speed of the star relative to Earth#
03

Calculate the relative speed

Rearrange the Doppler effect formula to solve for the speed \(v\): \(v = \frac{\Delta \lambda \cdot c}{\lambda_{emitted}}\) Now plug in the values for \(\Delta \lambda\), \(\lambda_{emitted}\), and \(c = 3.0 \times 10^8 \, \text{m/s}\) (converted to meters): \(v = \frac{(1.5 \times 10^{-9}\, \text{m}) \cdot (3.0 \times 10^8\, \text{m/s})}{(659.6 \times 10^{-9}\, \text{m})}\) #Step 4: Calculate the result#
04

Evaluate the expression

Calculate the relative speed of the star: \(v = \frac{(1.5 \times 10^{-9}\, \text{m}) \cdot (3.0 \times 10^8\, \text{m/s})}{(659.6 \times 10^{-9}\, \text{m})} \approx 681 \, \text{m/s}\) #Step 5: Determine the direction of motion#
05

Determine if the star is moving towards or away from Earth

Since \(\lambda_{observed} > \lambda_{emitted}\), the light has been redshifted, which means the star is moving away from Earth. #Solution# The star is moving away from Earth with a relative speed of approximately \(681 \,\text{m/s}\).

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Most popular questions from this chapter

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An unpolarized beam of light (intensity \(I_{0}\) ) is moving in the \(x\) -direction. The light passes through three ideal polarizers whose transmission axes are (in order) at angles \(0.0^{\circ}, 45.0^{\circ},\) and \(30.0^{\circ}\) counterclockwise from the \(y\) -axis in the \(y z\) -plane. (a) What is the intensity and polarization of the light that is transmitted by the last polarizer? (b) If the polarizer in the middle is removed, what is the intensity and polarization of the light transmitted by the last polarizer?
Verify that the equation \(I=\langle u\rangle c\) is dimensionally consistent (i.e., check the units).
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