Answer:
The speed of the gas relative to the earth = 6.314 * [tex]10^{6} m/s[/tex]
Explanation:
Let the observed wavelength be [tex]\lambda_{0} = 1915 nm[/tex]
The wavelength of the source, [tex]\lambda_{s} = 1875 nm[/tex]
Let the speed of the gas relative to the earth be = v
The observed wavelength in relation to the source wavelength, according to doppler shift in relativity can be expressed as:
[tex]\lambda_{0} = \lambda_{s}\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }[/tex]
Where the speed of light, c = 3 * 10⁸m/s
[tex]\frac{\lambda_{0}}{\lambda_{s}} = \sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }\\(\frac{\lambda_{0}}{\lambda_{s}})^{2} = \frac{1-\frac{v}{c} }{1+\frac{v}{c} }\\(\frac{1915}{1875}) ^{2}( {1+\frac{v}{c}) = ({1-\frac{v}{c} }) \\[/tex]
[tex]1.043 +\frac{1.043v}{c} = 1 - \frac{v}{c} \\\frac{1.043v}{c} + \frac{v}{c} = 1-1.043\\2.043v = -0.043c\\v = \frac{-0.043c}{2.043} \\v = -0.02105c\\v = -0.02105 * 3 * 10^{8} \\v = -6314243.76 m/s\\The absolute value, v = 6314243.76 m/s\\v = 6.314 * 10^{6} m/s[/tex]
