Q56E - University Physics with Modern Physics

Question

How fast (as a percentage of light speed) would a star have to be moving so that the frequency of the light we receive from it is 10.0% higher than the frequency of the light it is emitting? Would it be moving away from us or toward us? (Assume it is moving either directly away from us or directly toward us.)

Step-by-Step Solution

Answer

(v=0.095c) the star must be moving toward us because the frequency received from it is greater than the frequency emitted by it.

Step 1:

If the frequency of the star's light is $f_s$, then the frequency of the light received from that star will be $\left(1.1f_s\right)$. The relationship that describes the frequency of the received light is:

$f_r=f_S\sqrt{\frac{c+v}{c-v}}$

Replacing $f_r$ with $\left(1.1f_s\right)$;

$\begin{array}{l}1.1 f_s= f_s\sqrt{\frac{c+V}{c-v}}\\ 1.1 f_s=\sqrt{\frac{c+V}{c-v}}\end{array}$

Step 2:

On putting the value;

  $$\begin{gathered} {\left(1.1\right)}^2 =\frac{c+V}{c-v} \\ c+V={\left(1.1\right)}^{2}c-{\left(1.1\right)}^{2}v \\ v+{\left(1.1\right)}^{2}v={\left(1.1\right)}^{2}c-c \\ v=\frac{\left(\left(1.1^2\right)-1\right)}{\left(\left(1.1^2\right)+1\right)}c \end{gathered}$$           

So, the value is

$$\begin{gathered} v=\frac{\left(\left(1.1^2\right)-1\right)}{\left(\left(1.1^2\right)+1\right)}c \\ v=0.0095c \end{gathered}$$

Hence, the star must be moving toward us because the frequency received from it is greater than the frequency emitted by it.