Chapter 16.2, Problem 19PP

To determine

The distance of the galaxy that it will move relative to Earth.

Whether the galaxy is moving toward or away from Earth.

Answer to Problem 19PP

The galaxy would move at a speed of $7.02\times {10}^6{\text{ms}}^{\text{-1}}$ relative to the Earth and the astronomer and the galaxy are moving away from each other.

Explanation of Solution

Given: The observed wavelength of the oxygen spectral line, ${\lambda }_{\text{obs}}=525\text{nm}$

And the laboratory value of the wavelength, $\lambda =513\text{nm}$

Formula used:

From the above measures it is clear that observed wavelength is greater than the actual wavelength. So, one can use the Doppler shift formula.

Since observed wavelength is greater than the laboratory value of the wavelength, ${\lambda }_{\text{obs}}\rangle \lambda$ . So, the quantity will be positive. It indicates that the astronomer and the galaxy are moving away from each other.The formula to be used here is

  $\left({\lambda }_{\text{obs}}-\lambda \right)=\pm \frac{v}{c}\lambda$

  $v=\frac{\left({\lambda }_{\text{obs}}-\lambda \right)\times c}{\lambda }$  (1)

Here, $v$ is the speed and $c=3\times {10}^8{\text{ms}}^{\text{-1}}$ is the speed of the light in vacuum.

Calculation: Substitute $525\text{nm}$ for ${\lambda }_{\text{obs}}$ , $513\text{nm}$ for $\lambda$ , $3\times {10}^8{\text{ms}}^{\text{-1}}$ for $c$ in the equation (1)

  $\begin{array}{c}v=\frac{\left(525-513\right)\times 3\times {10}^8}{513}{\text{ms}}^{\text{-1}}\\ =\frac{12\times 3\times {10}^8}{513}{\text{ms}}^{\text{-1}}\\ \boxed{v=7.02\times {10}^6{\text{ms}}^{\text{-1}}}\end{array}$

Conclusion: Hence, the galaxy would move at a speed of $7.02\times {10}^6{\text{ms}}^{\text{-1}}$ relative to the Earth and the astronomer and the galaxy are moving away from each other.