Chapter 16, Problem 52A

To determine

The velocity of the galaxy relative to the Earth

Answer to Problem 52A

  $3\times {10}^6\text{}\text{m/s}$

Explanation of Solution

Given:

The actual frequency of the light, $f=486\text{nm}$

The observed frequency of the light, $f_{\text{obs}}=491\text{nm}$

Formula used:

Doppler effect: The observed frequency of light can be written in terms of actual frequency of light $\left(f\right)$ and the relative velocity of the source with respect to the observer $\left(v\right)$ as,

  $f_{\text{obs}}=f\left(1\pm \frac{v}{c}\right)$

Where $c$ is the speed of light.

Now, if the observer and the source are approaching each other, then the observed frequency is,

  $f_{\text{obs}}=f\left(1+\frac{v}{c}\right)$

And, if the observer and the source are receding away from each other, then the observed frequency is,

  $f_{\text{obs}}=f\left(1-\frac{v}{c}\right)$

Calculation:

The speed of light is known as, $c=3\times {10}^8\text{}\text{m/s}$

Thus for the given case, using the given values in above formula,

  $\begin{array}{c}{f}_{\text{obs}}=f\left(1+\frac{v}{c}\right)\\ 491\times {10}^9=486\times {10}^9\left(1+\frac{v}{3.0\times {10}^8}\right)\\ 1-\frac{v}{3.0\times {10}^8}=1.01\\ v=-3\times {10}^6\text{}\text{m/s}\end{array}$

Negative sign of above velocity denotes that the observer is moving away from the source, that is, the galaxy is moving away from the Earth.

Conclusion:

Hence, the relative velocity of galaxy with respect to the Earth is $3\times {10}^6\text{}\text{m/s}$ .