Question

For each star, determine how its light would be shifted. Not all choices may be used, and some may be used more than once. A red dwarf moving away from Earth at 39.1 km/s
A yellow dwarf moving transversely at 15.1 km/s

A red giant moving towards Earth at 23.3 km/s

A blue dwarf moving away from Earth at 25.9 km/s

A red dwarf moving transversely at 14.1 km/s

Answer 1

Answer:

Explanation:

To calculate the red shift you use the following formula:

$z=\frac{1+vcos\theta/c}{\sqrt{1-v^2/c^2}}-1$

\tetha: angle between the observer and the motion of the body

v: speed of the body

c: speed of light

for motion with angle 90° (transversal motion):

$z=\sqrt{\frac{c+v}{c-v}}-1$

- A red dwarf moving away from Earth at 39.1 km/s :

$z=\sqrt{\frac{3*10^8m/s+39.1*10^3m/s}{3*10^8m/s-39.1*10^3m/s}}-1=1.3*10^{-4}$

- A yellow dwarf moving transversely at 15.1 km/s (angle = 90°):

$z=\frac{1+0}{\sqrt{1-(15.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.27*10^{-9}$

- A red giant moving towards Earth at 23.3 km/s (angle = 0°):

$z=\frac{1+(23.3*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(23.3*10^3m/s)^2/(3*10^8m/s)^2}}-1=7.76*10^{-5}$

- A blue dwarf moving away from Earth at 25.9 km/s$z=\frac{1+(25.9*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(25.9*10^3m/s)^2/(3*10^8m/s)^2}}-1=8.63*10^{-5}$

- A red dwarf moving transversely at 14.1 km/s

$z=\frac{1+0}{\sqrt{1-(14.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.11*10^{-9}$