Question

how does the change in position of distant galaxies compared to the change in position of nearby galaxies

Answer 1

Are you talking about red and blue shift/

Answer 2

I dont believe i am @Astrophysics

Answer 3

red shift is spoken about a few questions later
perhaps you could help with that instead?

Answer 4

Sure, I guess I just don't get this question :P

Answer 5

ok let me see the oher question @Astrophysics

Answer 6

well perhaps this one first, lets see if this one is simple enough without context.

Answer 7

Does the fact that we see all galaxies receding from us require that our own galaxy is in the center of the universe? Explain.

Answer 8

No, there is no centre in the universe. And down goes human ego :P it's not that galaxies are receding, but everything is receding from each other, the whole universe is expanding, I don't really know how much detail you want but I think this should give you the idea and you can fill in the gaps.

Answer 9

Actually that should be just enough detail! @Astrophysics

Answer 10

There is a famous balloon analogy

Answer 11

go on!

Answer 12

Here watch this video it's 2 minutes long it can explain better than I can on here https://www.youtube.com/watch?v=i1UC6HpxY28

Answer 13

ok i will. Stick around please? See if you are able to understand like 2 others without context? but yes, i am going to watch that video right now. Give me a moment! @Astrophysics

Answer 14

Yes, it did clarify a bit more, thank you so much @Astrophysics

Answer 15

are you able to stick around for one or two more? because besides these two questions, the rest are going to be ridiculously hard.

Answer 16

Go ahead and ask

Answer 17

lets see if this makes sense...

Answer 18

When the redshift is much less than one, we can use a simple formula to convert the redshift into a recession speed, v, with the formula below:

\[z=\frac{ v }{ c }\]

Recall that c= 300,000 km/s is the speed of light. What is the recession speed of the galaxy Zax in our example?

Answer 19

I think you require more context to answer this question tbh... :/

Answer 20

Ok so we're solving for v, what is the value of z?

Answer 21

yes, so it would appear i have to solve the question prior to this one to get z

Answer 22

What is galaxy zax, you did not give enough information

Answer 23

i think galaxy zax is a made up galaxy in this book... it says made up... this book is so ridiculous.. would you like to see the previous question so you can understand a little more ?

Answer 24

I understand that, but it requires the example, what was the example

Answer 25

Yes, give as much information as you can

Answer 26

z is the redshift, you should have wavelengths to solve for z

Answer 27

In this "toy universe" you measured the distances at two different times and calculated the speed of each "galaxy." In the real Universe, we measure the speeds and distances of galaxies at a single time: the present. To find the speed of a galaxy at an instant in time, astronomers rely on the Doppler shift.

Light emitted by an object in motion along the line of sight for an observer will undergo a Doppler shift of its wavelength \[\lambda \] By measuring the amount of the shift, we find the redshift "z" of the object using the formula

\[z=\frac{ \lambda obs -\lambda0}{ \lambda0 }\]

When z is a positive number, is the line shifted to longer or shorter wavelengths?

Say we observe the (made up!) galaxy Zax and identify a particular spectral line, for example a hydrogen line that has a wavelength of \[\lambda0 = 656.3 nm\] when the emitting gas is at rest in the laboratory. If we observe the spectral line in galaxy Zax at a wavelength of \[\lambda0bs = 673.8 nm\] what is the redshift "z" of Zax

Answer 28

@Astrophysics

Answer 29

So thats everything it says above...

Answer 30

if its too complicated or still not enough information, let me know and i have a more than likely simpler one to ask

Answer 31

I'm helping someone right now, I'll get back to you in a second, but that looks good!

Answer 32

ok take your time!

Answer 33

Hey, so did you calculate z then given all the info?

Answer 34

\[\huge z = \frac{ \lambda _{obs}- \lambda_0 }{ \lambda_0 }\]

Answer 35

0.0266646350754228

Answer 36

thats what i got.

Answer 37

i mean i dont think that could be wrong ? it was subtracting and then dividing

Answer 38

@Astrophysics

Answer 39

Yes, so it's 0.0267

Answer 40

Now we can use the formula \[z = \frac{ v }{ c }\] solve for v, what do you get?

Answer 41

hang on before we get to that

Answer 42

when z is a positive number (in this case it is) the line is shifted to longer wavelengths correct?

Answer 43

pretty sure thats correct, just need to verify

Answer 44

If it's positive it will be redshifted

Answer 45

Meaning it's moving away from us

Answer 46

and redshifted means longer wavelengths right?

Answer 47

reason i ask is because that question is hidden inside of that question haha

Answer 48

Yes red is longer

Answer 49

You can think of a rainbow, ROYGBIV

Answer 50

Red being the longest

Answer 51

ok great !

Answer 52

ok lets move on to the next part, lets see

Answer 53

z=v/c

Answer 54

so....

Answer 55

0.0267=

Answer 56

huh... how do i solve for v?

Answer 57

we have z and c but i dont know how to get v

Answer 58

You should really practice with your algebra, as it's very important especially if you're taking astronomy.

So we have