[joefoxon]

Apparently it states on the Astronomy spec that on the delta lambda/lambda = v/c, you have to use the un-redshifted wavelength on the bottom, which would be the 480. I don't think it says that on the physics spec, so it may be ok to use 393.

[zabuzar]

(Original post by joefoxon)
Apparently it states on the Astronomy spec that on the delta lambda/lambda = v/c, you have to use the un-redshifted wavelength on the bottom, which would be the 480. I don't think it says that on the physics spec, so it may be ok to use 393.

what do you mean physics spec?
in the book it says : " where 'delta lambda' is the change of wavelength, and 'lambda' is the original wavelength"

[joefoxon]

The specification.

The original wavelength would be the one that wasn't redshifted, so the 480.

[zabuzar]

god, you just gave me a heart attack
wow
look at the question paper again, its in the first post, the 480.0nm is the spectrum from distant galaxy
it threw me off too that the redshifted one is so rounded and 'perfect' and the original isn't.

[joefoxon]

But the one as viewed from Earth WILL be redshifted, so you want to use the 480.

[Pachydermal]

if you look at the wording it says "spectrum of calcium as observed from a source on earth and from a distant galaxy" so the 393.4 one is the non redshifted one

[bdfzyt]

can we get 1 mark if used 480.....?

[zabuzar]

thank you
at least i used the correct values

[rikhilrai]

You have to use the 393.4 nm wavelength on the bottom of the equation d(lamda)/lamda = v/c.

For the galaxy question:

F = GMm/r^2
= (6.67x10^-11 . 10^41 x 10^41)/(4x10^22)^2

[joefoxon]

The source as observed from Earth was 393.4nm, the one from the galaxy was 480nm, you use the one from the galaxy, because it hasn't been redshifted.

[bdfzyt]

(Original post by joefoxon)
The source as observed from Earth was 393.4nm, the one from the galaxy was 480nm, you use the one from the galaxy, because it hasn't been redshifted.

I also used 480. But it looks that we were wrong...
I just wonder can we get at least 1 mark for that question..,

[rikhilrai]

(Original post by joefoxon)
The source as observed from Earth was 393.4nm, the one from the galaxy was 480nm, you use the one from the galaxy, because it hasn't been redshifted.

Umm ... 480 nm is the wavelength observed from a distant galaxy, so I think that would be the one that was red-shifted.

[joefoxon]

But we are observing the galaxy, not the Earth. And if you look at the galaxy from the galaxy. it won't be redshifted.

[rikhilrai]

Hey joefoxon,
Do you mind showing me your full working. I think I may be misunderstanding you.
Cheers

[bdfzyt]

In my opinion, the grade boundary is: A* 76, A 69, B62, C55...
Anyone agree with me?

[Pachydermal]

(Original post by joefoxon)
But we are observing the galaxy, not the Earth. And if you look at the galaxy from the galaxy. it won't be redshifted.

As i said a minute ago the wording was "the spectrum of calcium as observed from a source on the earth and from a distant galaxy" so you want to use the earth one as non redshifted

[rikhilrai]

(Original post by Pachydermal)
As i said a minute ago the wording was "the spectrum of calcium as observed from a source on the earth and from a distant galaxy" so you want to use the earth one as non redshifted

Yep I agree ... thanks!

[Jamiethevamp]

(Original post by zabuzar)
i got 1320Mpc
there was a decimal point and a digit after but i rounded to 3s.f.

I used the value i got in my calculator and got 1320.79 then rounded to 1321? but when i wrote down the value i got i wrote 6.60x10⁷, reckon i have still got the mark?

[joefoxon]

(Original post by joefoxon)
For the distance in MPc one I did:

(480-393.4)/480 = v/3x10^8

Got v ~ 5x10^7ms^-1

Then divide by 50 to get about 1200. Can't remember the exact values though

My working, for rikhilrai.

The wavelengths given are the one of viewing the galaxy from the Earth, and one from viewing the galaxy from just next to the galaxy. You need to use the one from next to the galaxy (480), because from that point of view, the light would not be redshifted. By the time it reaches the Earth, there will be a shift.

[just george]

(Original post by joefoxon)
My working, for rikhilrai.

The wavelengths given are the one of viewing the galaxy from the Earth, and one from viewing the galaxy from just next to the galaxy. You need to use the one from next to the galaxy (480), because from that point of view, the light would not be redshifted. By the time it reaches the Earth, there will be a shift.

hmm if you are right about this then that is officially the ****test wording i have ever seen :L