Is the Relativistic Doppler Shift Formula Correctly Applied in this Problem?

[Solarmew]

I was just wondering if this problem is right and I'm missing something or ...
http://www.astronomy.ohio-state.edu/~ryden/ast143/ps3_soln.pdf
Cuz I'm reading the GRE book and it says here to use the relativistic Doppler shift formula. So the problem in the link would need to be solved like

500/700 =√[(1-β)/(1+β)]
25/49 = (1-β)/(1+β)
24/49 = 74/49 β
β = 13/37
v = 9.7 E4

isn't that how you're supposed to do it? >.> ...
i tried doing the problem in the book using the method from the link and didn't get the right answer +.+

 

[PeterDonis]

The PDF you linked to is using a non-relativistic approximation for the Doppler shift formula--which, as your calculation shows, is actually not a very good approximation for this problem, since the speed is high enough to be an appreciable fraction of the speed of light. That's why the PDF's answer is about 12 percent lower than yours; that's the error involved in using the non-relativistic approximation for this speed.

 

[Solarmew]

oooooh, ok, thanks :3

 

[PAllen]

I was just wondering if this problem is right and I'm missing something or ...
http://www.astronomy.ohio-state.edu/~ryden/ast143/ps3_soln.pdf
Cuz I'm reading the GRE book and it says here to use the relativistic Doppler shift formula. So the problem in the link would need to be solved like

500/700 =√[(1-β)/(1+β)]
25/49 = (1-β)/(1+β)
24/49 = 74/49 β
β = 13/37
v = 9.7 E4

isn't that how you're supposed to do it? >.> ...
i tried doing the problem in the book using the method from the link and didn't get the right answer +.+

Your fourth, and last steps are wrong. Just check your algebra for the 4th step. For the last I have no idea what you did but it is a nonsensical answer given that mph are desired. The rest is ok.

I think this should be moved to a homework forum, and have indicated such to the mentors.

[Correct answer should be about 2.17 E8 mph]

 

[PeterDonis]

Just check your algebra for the 4th step.

I think he meant to write 12/37 instead of 13/37. Although 13/37 is a better number if you're using the urban dictionary.

For the last I have no idea what you did but it is a nonsensical answer given that mph are desired.

Yes, but the answer makes sense if it's km/sec and beta is 12/37. That was what I assumed when I responded.

 

[Solarmew]

yes, i meant 12/37 X3 ... and also yes on km/sec :]
also I'm a she >.> ... not that it matters <.< ...