Question

Even at such low fractions of the speed of light these stars are moving quite quickly (thousands of kilometers a second) compared to the standard velocity dispersion in the milky way galaxy (which is around 300km/s). suppose the star was actually moving toward the earth at a more reasonable 300km/s. what wavelength λ would the 656.46-nm line move to? use 299,792km/s for the speed of light. express your answer in nanometers to five significant figures.

Answer 1

According to Doppler Effect, an observer at rest will perceive a shift in the wavelength or frequency of the radiation emitted by a source in movement.This shift is given by the formula:
$\frac{ \lambda - \lambda_{0} }{ \lambda_{0} } = \frac{-v}{c}$

where:
$\lambda$ = observed wavelength
$\lambda_{0}$ = wavelength at rest
v = speed of source (positive if towards the observer, negative if away from the observer)
c = speed of light

Therefore, we can solve for the observed wavelength:
$\lambda = \lambda_{0} (\frac{-v}{c}) + \lambda_{0} \\ \lambda = \lambda_{0} (1 - \frac{v}{c})$

Substituting the given data:
$\lambda = 656.46 (1 - \frac{300}{299792})$
= 655.80 nm

Hence, the observed wavelength of the line would be 655.80 nm. Note that this value is smaller than the one at rest, which means that we have a blue-shift, as expected for an approaching source.