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Physics calculation- does anyone understand the solution to this question?

Physics calculation- does anyone understand the solution to this question?

  • Physics calculation- speed of a moving star--? The yellow line emitted by the helium discharge tube in the laboratory has a wavelenght of 587 nm. The same yellow line in the helium spectrum of a star has a wavelenght of 590 nm. What can you deduce about the motion of the star? Calculate the speed of the moving star. (speed of light= 3.00 x 10^8 m s^-1) This is the official solution but I dont follow it!! any explanations?? thanks :) Sorry cant put the solution here but heres the link :) http://www.examinations.ie/archive/markingschemes/2010/LC021ALP000EV.pdf (page 8 of the solutions!! thanks)

  • Answer:

    Without looking at the answer: It changed because of doppler effect. The fact that the wavelength became longer tells me that the star is moving away from earth. The relative change is 3 nm out of 587 nm. So the speed is given by v/c = 3/587 or v = (3/587)*3 x 10^8 m/sec = 1.53 x 10^6 m/sec. This formula is an approximation by the way, suitable only for small doppler shifts. But this is small (3 is a small fraction of 587) so we're fine. Now looking at the solution: OK, they used the same formula in slightly different form. u = c * (590/587 - 1) = c*(590 - 587)/587 = c * (3/587) They just computed it differently. 590/587 - 1 = 1.0051 - 1 = 0.0051. c * 0.0051 = 1.53 x 10^6 m/sec Which part confuses you?

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Without looking at the answer: It changed because of doppler effect. The fact that the wavelength became longer tells me that the star is moving away from earth. The relative change is 3 nm out of 587 nm. So the speed is given by v/c = 3/587 or v = (3/587)*3 x 10^8 m/sec = 1.53 x 10^6 m/sec. This formula is an approximation by the way, suitable only for small doppler shifts. But this is small (3 is a small fraction of 587) so we're fine. Now looking at the solution: OK, they used the same formula in slightly different form. u = c * (590/587 - 1) = c*(590 - 587)/587 = c * (3/587) They just computed it differently. 590/587 - 1 = 1.0051 - 1 = 0.0051. c * 0.0051 = 1.53 x 10^6 m/sec Which part confuses you?

Randy P

that looks confusing. I also have some physics questions, and if you can answer them that would be nice :)

You don't really need to do any calculation to determine what's happening qualitatively. If the star were at rest it would give off the spectral line due to helium at the same wavelength. If it's different it must be "doppler" shifted. If it's shorter than the discharge tube... the wave is compressed and moving toward you. If longer it's stretched and moving away from you. So the longer wavelength says the star is moving away from you. Now... how fast? That will require some calculation.

Cletus

Geese, my dog can answer that one. Your yeller line up dare is a real pain!

Charlie D. Noble

You don't really need to do any calculation to determine what's happening qualitatively. If the star were at rest it would give off the spectral line due to helium at the same wavelength. If it's different it must be "doppler" shifted. If it's shorter than the discharge tube... the wave is compressed and moving toward you. If longer it's stretched and moving away from you. So the longer wavelength says the star is moving away from you. Now... how fast? That will require some calculation.

Cletus

No. People have theories that there is a particle that exists that causes gravity. There is complex math and logic that support the existence of such particle (google higgsbozan particle i think?). There are some people who believe that gravity is a force that is exerted by something in another dimension and that is why we can't find it. We are able to predict it so well because it seems to always follow the same law (gmm/r^3 anyone). there is a team of people around the world measureing gravity to 10 or more decimal places and to date there has not been any unexpected jumps that I know of.

zella

Geese, my dog can answer that one. Your yeller line up dare is a real pain!

Charlie D. Noble

that looks confusing. I also have some physics questions, and if you can answer them that would be nice :)

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