Relative velocity of Electromagnetic waves

[Eng67]

I have a problem that I cannot get the correct answer.

An Astronomer observes electromagnetic waves emmitted by oxygen atoms in a distant galaxy that have a frequency of 5.71 *10^14 Hz. On earth, oxygen atoms emit waves that have a freq. of 5.841 *10^14. What is the relative velocity of the galaxy with respect to the astronomer on earth.

Vrel = [(F(observed) - F(Source))/(2*F(Source))] * c

Vrel = [(5.710*10^14 - 5.841*10^14)/(2*5.841*10^14)] * (3.00 *10^8)

answer = 3.3641 *10^6 m/s


What is wrong with my calculations?

The correct answer is 6.724 * 10^6 m/s

Thanks

 

[Galileo]

Notice your answer is off by a factor of 1/2. Recheck the formula you used and you'll see there should be a 2 in the denominator.

 

[HallsofIvy]

By the way- your title, "Relative velocity of Electromagnetic waves", is very misleading! Obviously the velocity of electromagnetic waves (light) is, relative to anything, the speed of light, c. What you see different from another observer is the frequency which is what allows you to calculate the velocity of the source relative to you.

 

[Eng67]

Vrel = [(F(observed) - F(Source))/(2*F(Source))] * c

This formula already has the 2 in the denominator. To get the correct answer, It would need to be 4*F(Source)?

I cannot find anything to support this.

 

[Galileo]

Vrel = [(F(observed) - F(Source))/(2*F(Source))] * c
This formula already has the 2 in the denominator. To get the correct answer, It would need to be 4*F(Source)?
I cannot find anything to support this.

I mistyped, obviously I meant there shouldn't be a two in the denominator. Check your reference.

 

[Eng67]

Thanks.

I finally found the correct reference for this formula. Many sources have this listed differently.