Inactive Tutor answered • 15h
This a great question. The answer is not significantly, since your maximum tangential velocity on Earth is 1674.4 Km/h which is 465 m/s, which by everyday standards is blisteringly fast, but to the universe, it's nothing. Compared to the speed of light thats, 0.00000154773 times the speed of light. However, in very precise experiments, yes, it's important to factor it in.
The redshift formula is z = (λobserved - λemmited)/(λemmited), where the λ (lambda) is the wavelength, so λemitted is the wavelength of the light when it is emitted from the source and λobserved is the wavelength of the light when it's observed. An equivalent form of the equation in terms of velocity is z= √((1+v/c)/(1-v/c)) - 1, where z is the redshift factor, v is the velocity of the object relative to the observer, and c is the speed of light. I know it looks complicated, but v/c is the part where velocity comes in; let's examine that and see what changes when accounting for Earth's spin. First off, v/c just tells you how many times the speed of light you're going. So if v = 0.99c, or 99 percent the speed of light, 0.99c/c just gives you 0.99. Or if you're using meters per second lets say v = 149,896,229 m/s and c = 299,792,458 m/s; 149,896,229 / 299,792,458 is just 0.5. So you're going 0.5 times the speed of light. But let's say we include Earth's spin here, so v = 149,896,229 m/s + 465m/s = 149896694, which divided by the speed of light gives you 0.50000155. So it's only important to factor in Earth's rotation if your equipment is precise enough; if not, your margin of error far exceeds the contribution on the order of 10-6
