Question

Using Hubble’s law, find the wavelength of the $590-nm$ sodium line emitted from galaxies

(a)$2.00\times {10}^{6}ly$,
(b) $2.00\times {10}^{8}ly$, and (c) $2.00\times {10}^{9}ly$ away from the
Earth.

Answer 1

First, we calculate $v=HR$, using $H=22\times {10}^{-3}m/s\cdot ly$, and then we

use the result of Problem , ${\lambda }^{\prime }=\lambda \sqrt{\frac{1+v/c}{1-v/c}}$, and $c=2.998\times {10}^{8}m/s$, to

calculate the wavelength emitted by the galaxy.
(a) $v=(22\times {10}^{-3}m/s\cdot ly)(2.00\times {10}^{6}ly)=4.4\times {10}^{4}m/s,$
${\lambda }^{\prime }=\lambda \sqrt{\frac{1+v/2.998\times {10}^{8}m/s}{1-v/c}}=\lambda \sqrt{\frac{1+(4.4\times {10}^{4}m/s)/(2.998\times {10}^{8}m/s)}{1-(4.4\times {10}^{4}m/s)/(2.998\times {10}^{8}m/s)}}$
$=\left(590nm\right)\sqrt{\frac{1+0.0001468}{1-0.0001468}}=590.09nm$
(b) $v=(22\times {10}^{-3}m/s\cdot ly)(2.00\times {10}^{8}ly)=4.4\times {10}^{6}m/s,$
$=\left(590nm\right)\sqrt{\frac{1+0.01468}{1-0.01468}}=599nm$
(c) $v=(22\times {10}^{-3}m/s\cdot ly)(2.00\times {10}^{9}ly)=4.4\times {10}^{7}m/s,$
$=\left(590nm\right)\sqrt{\frac{1+0.1468}{1-0.1468}}=684nm$