Answer:
[tex]v=0.0198c, c=3.0\ \ \times 10^8\ m/s[/tex]
Explanation:
The signal observed on Earth is 2.0% greater than frequency observed by the source.
#This is a blueshift and the source is moving towards point of observation.
#For Doppler effect, observed frequency is given by:
[tex]V_o=v_s\sqrt{\frac{1\beta}{1-\beta}}\ \ \ \ \ ...i[/tex]
where [tex]v_s[/tex] is the frequency emitted by source and [tex]\beta=\frac{v}{c}, \ \ \[/tex]. v is +ve when emission source is approaching observer.
[tex]\frac{v_o}{v_s}=1.02\ \ \ \ \ ....ii[/tex]
#we equate the two equations:
[tex]\sqrt{\frac{1+\beta}{1-\beta}}=1.02\\\\\sqrt{\frac{1+\beta/c}{1-\beta/c}}=1.02\\\\\#take \ root \ on \ both \ sides:\\\\1.0404=\frac{c+v}{c-v}\\\\\\1.0404c-1.0404v=c+v\\\\0.0404c=2.0404v\\\\v=0.0198c, c=3.0\ \ \times 10^8\ m/s[/tex]
Hence, the relative velocity of the galaxy relative to earth is [tex]v=0.0198c, c=3.0\ \ \times 10^8\ m/s[/tex]
