Question

41. What is the velocity (speed AND direction) relative to Earth of a star with a spectral 1 normally seen at 625 nm that is now at 690 nm? (speed of light = 300,000 km/sec) A) 31,200 km/sec TOWARD Earth B) 31,200 km/sec AWAY from Earth C) 10.4 km/sec TOWARD Earth D) 10.4 km/sec AWAY from Earth

Answer 1

Answer:

V = 31200000 m/s = 31200km/s

Explanation:

when star recedes away from earth its wavelength will appear to increae

$\frac{\Delta \lambda}{\lambda_o} =\frac{v}{c}$

where c is speed of light

$\Delta \lambda$ is change in wavelength

$\lambda_o$ is wavelength of star = 625 nm

therefore we have

solving for v

$v = \frac{\Delta \lambda}{c}{\lamda_0}$

$v = \frac{(690-625)\times 10^{-9}\times 3\times 10^8}{625\times 10^{-9}}$

V = 31200000 m/s = 31200km/s