Try the SIMBAD database, it has rotation speeds as described in item 10 here:

http://simbad.u-strasbg.fr/Pages/guide/chG.htx

George Dishman

Perfect, thanks this gives me something to work with.

• Rotation of the Andromeda Nebula from a Spectroscopic Survey of Emission Regions, V. C. Rubin et. al, Astrophys.J. 159 (1970) 379-403
• Die Rotverschiebung von extragalaktischen Nebeln (In German), F. Zwicky, Helv.Phys.Acta 6 (1933) 110-127 and Gen.Rel.Grav. 41 (2009) 207-224
• No evidence for a dark matter disk within 4 kpc from the Galactic plane Astrophys.J. 724 (2010) L122-L126

Try this website for Rotation Curves and Mass Distribution in spiral galaxies:

http://www.ioa.s.u-tokyo.ac.jp/~sofue/h-rot.htm

Mohammed Naji Al Najm

Thanks, lots of good data there.

Would you also know a good source for luminous mass distribution?

SIMBAD is quite confusing...

Dear Steven Sesselmann

Thank you very much

I'll try to find you a good source for luminous mass distribution and and I'll tell you later!

I found the following velocity measurements on NGC224 here.

http://www.ioa.s.u-tokyo.ac.jp/~sofue/RC99/0224.dat

Can someone tell me if these velocities are absolute or tangential velocities adjusted for the velocity of NGC224 towards us?

It is my understanding that NGC224 is approaching us at approx. 300 km/s

The numbers are the tangential speed relative to the centre of the galaxy and have been corrected for both the overall motion of the galaxy towards us and the inclination of the disc.

NGC 224 is the Andromeda galaxy.

Thanks, I got the information I needed, and I think I can show where the missing mass is. Look at the attached image;

1) The green line is the actual rotation curve data from NGC224

2) The black dotted line is the expected Kepler orbital velocity

3) The Yellow line is rotation curve data for NGC224 (negative rotation)

4) The purple dotted line is inverse Kepler = √GM/r - √2GM/r

Wiki gives the NGC224 a mass of 1.5 x 10^12, but I had to increase it to 2.2 x 10^12 for the curves to fit.

The point I am trying to make here is that a different possibly better fit can be achieved by accepting negative orbital velocity. Since it appears we have arbitrarily chosen positive as a direction, why not accept negative.

As shown above, flipping the sign is not a symmetrical move, as we can't simply flip Kepler, but rather we must take the inverse.

v(orbital) = √GM/r - √2GM/r

Here I am assuming that escape velocity will result in an orbit of infinitely large radius, I then subtract the normal Keplerian orbital velocity to get the inverse.

Does this make sense?

SS: Does this make sense?

None whatsoever.

1) Using negative speeds makes no difference, the equation you used was:

vorbital = √(GM/r)

hence

M = (r vorbital)2 / G

Since the mass depends on the square of the speed, calling it positive or negative makes no difference.

2) The equation you have tried to use applies only for a single point mass, the galaxy is a disc with mass distributed through it, you need integrate the density over the disc.

SS: Here I am assuming that escape velocity will result in an orbit of infinitely large radius, I then subtract the normal Keplerian orbital velocity to get the inverse.

3) That is incorrect, to get the speed in the opposite direction, you simply negate the value which would give:

vorbital = -√(GM/r)

However, that's the wrong equation to start with, see point (2).

George Dishman

I'm sorry to be the one having to break the news to you, but classic orbital velocity doesn't make sense.

If orbital velocity decreases with increasing radius as in √GM/r, things ought to fall upwards towards a lower energy state !

...and escape velocity should be slower than orbital velocity.

Recall we are talking doppler velocity here (not Euclidean) which we already agreed is not the same.

Steven

Dear Steven,

SS: I'm sorry to be the one having to break the news to you, but classic orbital velocity doesn't make sense.

It does once you learn why it is that way. Your own lack of understanding is not a problem for the planets or stars.

SS: If orbital velocity decreases with increasing radius as in √GM/r, things ought to fall upwards towards a lower energy state !

You just need to learn some basics Steven, orbital mechanics tends to be counter-intuitive but it can be learnt.

• Mercury is closer to the Sun than Earth and it moves at 47 km/s.
• Earth is farther away and it moves at 30 km/s.

The reason is simple, Mercury is closer so the Sun pulls on it harder and it has to move faster to avoid falling inwards. For any radius, there is a particular speed where the pull of gravity is balanced by the so-called "centrifugal force" and the planet moves in a circular orbit at constant speed.

Similarly the International Space Station is in a low orbit and moves at 7.66 km/s while GPS satellites are much higher and move at 3.88 km/s.

The detail is a little more complex, we could talk about elliptical orbits, but you need to master the basic idea first.

George Dishman

From your reply I can see that you are still having difficulty understanding the difference between Euclidean rotation and that obtained by a doppler speedometer, it's not the same direction.

It appears, you and everyone else seem to be cheating, breaking the basic rules of science, you measure the velocity of a rocket blasting off into space with a speedometer which clearly reads -100 m/s and then at some arbitrary point you drop the negative sign because it doesn't fit your world view.

That's not science!

Steven,

SS: it's not the same direction.

That's right, but from your reply I can see that you are still having difficulty understanding the difference between radial motion, moving directly away from or towards the centre, and tangential motion which is at right angles to the radial direction.

SS: It appears, you and everyone else seem to be cheating, breaking the basic rules of science, ..

No, we just all know the difference between motion in a straight line versus motion round a circle and we don't get them confused as you are doing.

Have you found out what "clockwise" and "anti-clockwise" mean yet?

George Dishman

You are thinking in two dimensions, we don't live in flatland.

There are four possible vectors for orbital motion

CW - forwards

CW - backwards

CCW - forwards

CCW - backwards

Until you understand this we will never agree.

Here I have created a graphic to show more clearly how there are 4 ways an orbit can be observed in 3D space. Rotational direction DOES depend on your point of reference.

Kepler, Newton used the Euclidean plane which is essentially the same as infinity, and for most purposes this works just fine, except when it comes to measuring orbital velocity of stars in galaxies....

The problem with galaxies is that they are too far away, so it is not possible to measure the shift in position of stars with any accuracy against the fixed background stars.

Consequently astronomers look at the doppler shift in light spectra from stars in the galactic disc, and convert this to km per second speed.

The problem is that doppler measurements are observer referenced NOT infinity referenced, so this measurement yields NEGATIVE velocity.

There is no Hokus Pokus in my method, just common sense, and I believe this is the root of the DM problem.

Dear Steven,

Sorry for the slow response.

SS: Here I have created a graphic to show more clearly how there are 4 ways an orbit can be observed in 3D space.

It shows only two. You say in the text "Observer sees orbit as left to right" but that is incomplete, he sees the stars moving:

• left to right at the top of your cone
• right to left at the bottom (the arrow I added)
• bottom to top on the left and
• top to bottom on the right (not shown).

We call that combined motion "clockwise" as I have indicated.

SS: Rotational direction DOES depend on your point of reference.

Yes, for any observer on the far side of the galaxy, the motion would be the opposite of the above list and we call that "anti-clockwise".

SS: Kepler, Newton used the Euclidean plane which is essentially the same as infinity, ...

That is nonsense. Kepler used the orbital plane, Newton used 3D Euclidean space, and both work fine.

SS: The problem with galaxies is that they are too far away, so it is not possible to measure the shift in position of stars with any accuracy against the fixed background stars.

Yes.

SS: Consequently astronomers look at the doppler shift in light spectra from stars in the galactic disc, and convert this to km per second speed.

Correct.

SS: The problem is that doppler measurements are observer referenced NOT infinity referenced, so this measurement yields NEGATIVE velocity.

No, they always give the rate of motion along the line of sight, so it can be either positive or negative depending on a variety of parameters. I'll address that in the next post. First though, do you now understand what "clockwise" and "anti-clockwise" mean? That is important.

George Dishman

Call it whatever, my point is simple, a body can turn two ways, and either direction can be positive or negative depending where you observe it.

My point is that we call the two directions "clockwise" and "anticlockwise" because saying things like "left to right" is unclear, both the observers see "left to right" but one at the "top" and the other at the "bottom", and all those terms, left, right, top and bottom are terrestrial terms that have no meaning for say the Hubble telescope floating in space.

You say " a body can turn two ways ... depending where you observe it" and that is correct and we call them "clockwise" and "anticlockwise", not negative or positive.

The other point is that, in your previous post you said "Here I have created a graphic to show more clearly how there are 4 ways an orbit can be observed in 3D space." but that is not true, as you now follow I hope, it shows only two directions for rotation, but it is a useful graphic.

Now if you have learnt that part. I'll go on to talk about a side view of the cone and relate that to Doppler but I don't want to do that if there is still any confusion in your mind about these terms for describing rotation.

George Dishman

Seems like we are not going to reach an easy agreement on this one..

An observer here is not an observer there and vice versa, so clockwise here is not clockwise there.

In 3D space you have to define 3 coordinates, so you have to define direction of rotation as observed from and observer at X,Y,Z.

Simplifying orbital motion to clockwise/anti clockwise means dropping one coordinate in favour of a 2D plane.

Further on the definitions I agree to simplify the signing to;

CW = (+v)

CCW = (-v)

Adding the extra coordinate we have ;

CW-Here (+v)

ACW-Here (-v)

CW There (+v)

ACW There (-v)

When you follow my regime you will see how doppler measured velocity (here) is opposite to how we normally describe orbital motion.

Dear Steven,

SS: Seems like we are not going to reach an easy agreement on this one..

There is no need to reach any agreement, I am simply telling you what the conventions are that astronomers use. Those conventions apply to the data you will find in all published papers and the databases you are using, so you need to understand them, not agree them.

SS: An observer here is not an observer there and vice versa, so clockwise here is not clockwise there.

Correct, in the diagram you drew, clockwise here is anticlockwise there, and vice versa.

SS: In 3D space you have to define 3 coordinates, so you have to define direction of rotation as observed from and observer at X,Y,Z. Simplifying orbital motion to clockwise/anti clockwise means dropping one coordinate in favour of a 2D plane.

Right, and that is a very important point we will come back to soon.

In a remote galaxy, the stars are generally moving round the centre in a plane with only a slight movement above or below the plane. Clockwise and anticlockwise refer to their motion projected onto that 2D plane.

SS: Further on the definitions I agree to simplify the signing to;

SS: CW = (+v)

SS: CCW = (-v)

No, there is no need for that and it is not how information in databases is provided. As you say, what is clockwise form 'here' is anticlockwise from 'there' so assigning positive or negative that way would be meaningless. The database numbers are always stated without any sign for that reason. Regarding your list of four possibilities, there are only two:

Now if you follow that, we can move on to spectroscopic (Doppler) measurements which can have positive and negative applied and I'll explain how that works. Please though, be sure you understand the preceding comments first.

George Dishman

Suppose we agree on the definitions now...

My hypothesis remains, the spectral doppler shift measurement method records a view from 'here', while the other method, observing a bodies drift relative the celestial sphere records the view from 'there'.

For all intents and purposes we may consider a rocket traveling with linear motion from here to there, we agree it has negative velocity from here and positive velocity to there. The anemometer exercise shows it's the same thing right?

Hi Steven,

I think we're making some progress.

SS: Suppose we agree on the definitions now...

It will certainly speed up the discussion. You can always convert to your own definitions separately but sticking with what everyone else uses will let you follow the published data.

SS: My hypothesis remains, the spectral doppler shift measurement method records a view from 'here', while the other method, observing a bodies drift relative the celestial sphere records the view from 'there'.

You are right about the first part but measuring speeds by observing stars moving against the background isn't really practical. What we need to discuss is the Doppler technique.

SS: For all intents and purposes we may consider a rocket traveling with linear motion from here to there, we agree it has negative velocity from here and positive velocity to there.

It's the other way round but don't worry, what we'll do works with either definition.

SS: The anemometer exercise shows it's the same thing right?

It will be relevant shortly. First we need to look at your cone but viewed from the side, I've added a sketch of that and marked yours. Your "here" is 'A', your "there" is 'B' and my sketch is drawn from location 'C'. You've drawn the cone with the galaxy rotating so we see it "face one". The motion of the stars round the galaxy doesn't move towards or away from us so there is no Doppler shift, we generally can't measure the speed of the stars in that case.

Can you see why there's a problem with this orientation? If so we can look at a different situation where the speed can be measured.

George Dishman

All good so far, I turned my cone around a little an d made the orbital plane oblique, so it is now an elliptical conic section.

To point out the obvious an observer at 'B' the cone would show the wider celestial sphere at 'A'.

This might also be one way to understand the gyroscopic effect, for if rotation was just a matter of semantics we should be able to rotate a spinning wheel freely in all directions, but this is not the case is it?

Okay, lets continue..., astronomers record light spectra and measure the doppler shift from earth based observatories.

Over to you...

Steven, that's excellent.

SS: To point out the obvious an observer at 'B' the cone would show the wider celestial sphere at 'A'.

Yes, no problem. It looks that way but only because I kept the image the same size as yours, if B were the same distance from the galaxy, it would appear the same size. We're only going to use observations from A though.

SS: .. we should be able to rotate a spinning wheel freely in all directions, but this is not the case is it?

We can spin a gyroscope in any direction but it has angular momentum so it is hard to change the axis of the spin.

SS: Okay, lets continue..., astronomers record light spectra and measure the doppler shift from earth based observatories.

Right. You have shown the motion on one side of the cone with a white arrow, I've added a second white arrow showing the motion of stars on the far side.

Now the Doppler technique tells us how rapidly the source is moving towards or away from us here but doesn't say anything about motion across the sky against the background. To deal with that we can decompose the white arrows into two motions, one is around the cone as you had in your previous diagrams, I've shown that with green arrows. The other component is the radial motion away from us for the arrow you drew and towards us when the star is on the far side.

Spectroscopic measurements tell us the value of the red arrows but say nothing about the green, OK?

Note that one is towards us and one away from us so one will be positive and the other negative no matter which way you define positive.

If you're happy with this part, the next post should explain what is in the databases and following one will link back to the anemometer analogy.

George Dishman

Yes, all clear we agree on this...

That's great. The attached diagram shows the configuration of the Andromeda galaxy, NGC 224. The disc is at about 77 degrees to our line of sight. If the speed of the edge of the disc is V, can you see that the component of the green line that is directed towards us would be V cos(77o).

Of course it would be positive on one side and negative on the other, let's say the Doppler values are:

Vpos = +V cos(77o)

Vneg = -V cos(77o)

It doesn't matter which way you define positive, towards or away, that only changes whether positive is on the left or right as we look at it.

What we are interested in is the value of V so we can calculate that as:

V = (Vpos - Vneg) / (2 cos(77o) )

That is the number you'll find in the databases.

George Dishman

Thank you for the details, and yes, this is more or less how I understand the method. Didn't know it was 77˚ but that makes sense.

So from these measurements we now have the speed, but we still don't have the direction. How do we prove which way it's turning?

Hi Steven,

SS: Didn't know it was 77˚.

You don't need to, the database values will usually be corrected for inclination.

We also don't need to know which way it's turning, the speed versus radius alone tells you the mass profile, job done.

However, it's trivial to work out the rotation, the side where the received frequency is higher is moving towards us while it is lower on the side that is moving away.

George Dishman

[GD: We don't need to know which way it's turning, the speed versus radius alone tells you the mass profile, job done.]

George, we already established that there is such a thing as clockwise and anti-clockwise, but now you are saying it doesn't matter.

When did you last gaze at the full moon?

In one day our moon has moved around 2πr/28 while you standing on earth have moved 2πr/1, therefore if we call movement against the celestial sphere as positive, the moons movement relative to you is negative.

This is relativity (ordinary garden variety, not special or general).

Please think about this very carefully before you tell me it doesn't matter.

Dear Steven:

SS: George, we already established that there is such a thing as clockwise and anti-clockwise, but now you are saying it doesn't matter.

That is correct.

SS: .. if we call movement against the celestial sphere as positive ..

We don't call movement against the celestial sphere either positive or negative, we've already agreed this too. Doppler speeds are towards or away from us, not tangential to the orbit. Think about a moon of Mars instead of ours. When it first appears over the horizon, it is moving towards us, when it disappears on the other side of Mars it is moving away. One speed is positive, the other negative. To get the speed of the moon relative to Mars, put those two values into the equation:

V = (Vpos - Vneg) / (2 cos(inclination) )

You can't use that technique for our own Moon because we're inside the orbit and Doppler is nearly zero all the time, but you can find the speed another way very easily.

The distance from the barycentre of the Earth/Moon system to the centre of the Moon is 382,700 km so the orbit has a circumference of 2πr which is 2,404,600 km.

It takes 27.322 days or 2,360,600 seconds for the Moon to complete an orbit so its speed is 2404600/ 2360600 = 1.02 km/s.

If it was orbiting in the opposite direction but with the same radius and period, it would be travelling at the same speed.

That speed tells us the mass of the Earth regardless of the direction of the orbit.

George Dishman

[GD: You can't use that technique for our own Moon because we're inside the orbit..]

Technically we are inside any orbit which lies above us, regardless of how far away it is or at which angle, by "above us" I mean it would require energy to reach the orbit from here.

You see the magnitude of the velocity may change with the cosine angle, but not the direction.

Every moving body must have a positive and negative velocity, but convention has it that we usually only speak of one. When I serve you a tennis ball to you it has negative and positive velocity, but we never talk about the negative.

For most things it's okay to simplify velocity to positive speed, because it is measurable, but in the special case of galaxy rotation we can't actually measure the positive velocity because it's too far away, so we are stuck measuring the negative.

That's not a problem, as we can fix this in the equation by deducting the escape velocity to give us the opposite velocity. v = √GM/r - √2GM/r

The Earth/Moon rotation is like the hour hand and the minute hand on a clock, the earth turns 27.322 times faster than the moon so with respect to the earths rotation (our observation point) the moon therefore moves backwards at around -26 km/s

As you can see, the backwards rotation is NOT equal to the forwards rotation.

I am not trying to change any laws of physics here, just applying the ones we have.

Dear Steven,

SS: Technically we are inside any orbit which lies above us, regardless of how far away it is or at which angle, by "above us" I mean it would require energy to reach the orbit from here.

No, that is wrong. Earth is always at a greater distance from Mars than its moons (Phobos and Deimos) so we are outside their orbits.

SS: You see the magnitude of the velocity may change with the cosine angle, but not the direction.

The cosine factor is a common factor, you multiply the component of the velocity in the orbital plane of the galaxy by that to get the value in the plane in which both we and the centre of the galaxy lie.

SS: Every moving body must have a positive and negative velocity, but convention has it that we usually only speak of one.

No, again that is wrong. When scientists talk of a moving body, they may define velocity using three components in the x, y and z directions, or perhaps r, ϴ, ϕ for polar coordinates, and then each of the three can be positive or negative.

If the velocity has components (vx, vy, vz) then the speed is given by:

V = +√(vx2 + vy2 + vz2)

SS: When I serve you a tennis ball to you it has negative and positive velocity, but we never talk about the negative.

Your thinking is one-dimensional. The tennis ball has negative radial velocity when approaching me (distance is decreasing) and positive velocity after I return it (distance increasing).

SS: For most things it's okay to simplify velocity to positive speed, because it is measurable, but in the special case of galaxy rotation we can't actually measure the positive velocity because it's too far away, so we are stuck measuring the negative.

Why are you saying this nonsense, you were doing really well until this post. Just 2 days ago I said:

• The attached diagram shows the configuration of the Andromeda galaxy, NGC 224. The disc is at about 77 degrees to our line of sight. If the speed of the edge of the disc is V, can you see that the component of the green line that is directed towards us would be V cos(77o).
• Of course it would be positive on one side and negative on the other, let's say the Doppler values are:
• Vpos = +V cos(77o)
• Vneg = -V cos(77o)
• It doesn't matter which way you define positive, towards or away, that only changes whether positive is on the left or right as we look at it.

You replied:

SS: Thank you for the details, and yes, this is more or less how I understand the method. Didn't know it was 77˚ but that makes sense.

Now you are contradicting yourself. I think you need to go back to that previous post and make sure you really understand it.

SS: That's not a problem, as we can fix this in the equation by deducting the escape velocity to give us the opposite velocity. v = √GM/r - √2GM/r

No, that is completely wrong, I have no idea where you say that. The first part is correct

v = √(GM/r)

hence

M = rv2/G

https://www.astronomynotes.com/gravappl/s8.htm

SS: The Earth/Moon rotation is like the hour hand and the minute hand on a clock, the earth turns 27.322 times faster than the moon so with respect to the earths rotation (our observation point) the moon therefore moves backwards at around -26 km/s

We don't care how fast the Earth is turning, only how long it takes the Moon to complete one orbit. Divide the circumference by the time and you get the speed (not velocity !!)

SS: As you can see, the backwards rotation is NOT equal to the forwards rotation.

Again that is wrong, the circumference divided by the time taken for an orbit is the same whichever direction you measure the distance. In fact there is no direction associated with the circumference of a circle.

SS: I am not trying to change any laws of physics here, just applying the ones we have.

You are failing to understand the difference between speed and velocity, and you tried to subtract the escape velocity from the equation above for no reason. You problem is in not applying the maths of the laws correctly.

George Dishman

[GD:No, that is wrong. Earth is always at a greater distance from Mars than its moons (Phobos and Deimos) so we are outside their orbits.]

Not sure if that's right, all the planets are orbiting the sun's potential energy well, and Mars is further from the sun than Earth, so it's moons are also at higher potential than us. You might argue that we are outside the orbit of Mercury, to which I would agree, but not the Moons of Mars.

[GD Said: V = +√(vx2 + vy2 + vz2) ]

Thanks, no arguments with Pythagoras...

[GD: Your thinking is one-dimensional....]

My point here is, the one dimensional thought experiment can be extended to three dimensional rotation without violating the principle. We already showed this with our anemometer.

[GD:......Now you are contradicting yourself. I think you need to go back to that previous post and make sure you really understand it.....]

What I meant here is that one can use doppler spectral shift to measure speed using the method you describe, but when it comes to determining the direction we need to establish if the body is at a higher or lower potential than us. In the case of another galaxy outside the Milky Way, we can assume it must be at higher potential and my assumption of the orbital direction is therefore negative.

[GD: No, that is completely wrong, I have no idea where you say that...]

Orbital velocity is the well known and accurate v = √GM/r

In my opinion this equation is only accurate when the orbital motion is measured against the celestial sphere or from orbiting body to infinity.

Sidenote:

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Our world is like an onion, with layers upon layers of rotating shells, (like a tornado) both speed and direction of this rotation is a function of the relative potential to you the observer. Imagine finding yourself sucked up into a tornado, spinning around somewhere in the middle. This is essentially how we live on the surface of the earth, the liquid stuff below us swirls forward and the stuff above us swirls backwards.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

When we are measuring from the surface of the earth up towards a galaxy at much higher potential than us with doppler spectra, we are measuring the motion from below, so the speed is not the same and the direction flips the other way.

But now we also have to flip Kepler's equation the other way as well, otherwise we will be calculating the speed on the wrong side.

My solution is based on the following assumption;

If √GM/r gives us the external orbital velocity with respect to the celestial sphere, then √GM/r - (escape velocity) should give us the internal velocity with respect to centre of rotation.

To prove why the internal and external speeds are not the same, we probably have to get deep into GR, I don't think we need to go there because my argument seems to work without it.

I understand exactly how you see it, and I am trying to open your eyes and show you how a tiny assumption like orbital speed direction may have ballooned out to a problem where 70% of the Universe is missing.

Not sure about you, but I don't believe in a God, I no longer believe in Santa Claus, and I certainly don't believe we have lost 70% of the Universe. I am 100% sure the Galaxies know what they are doing.

Dear Steven,

Merry Christmas,

[GD:No, that is wrong. Earth is always at a greater distance from Mars than its moons (Phobos and Deimos) so we are outside their orbits.]

SS: Not sure if that's right, all the planets are orbiting the sun's potential energy well, and Mars is further from the sun than Earth, so it's moons are also at higher potential than us. You might argue that we are outside the orbit of Mercury, to which I would agree, but not the Moons of Mars.

We are outside the orbits of Mercury and Venus but inside the orbit of Mars around the Su. However, Phobos and Deimos don't orbit the Sun, they are in the gravitational well of Mars and we are always more distant from Mars than them so we are outside their orbits.

[GD: Your thinking is one-dimensional....]

SS: My point here is, the one dimensional thought experiment can be extended to three dimensional rotation without violating the principle. We already showed this with our anemometer.

The anemometer can be likened to the case of Phobos and Deimos orbiting Mars. The planet acts as the axle with the moons being the cups.

[GD:......Now you are contradicting yourself. I think you need to go back to that previous post and make sure you really understand it.....]

SS: What I meant here is that one can use doppler spectral shift to measure speed using the method you describe, but when it comes to determining the direction we need to establish if the body is at a higher or lower potential than us.

No, potential doesn't matter, to determine the direction, you compare the speed on the two sides of the orbit. On one side the body is approaching us, on the other side it is receding. Add those to the motion of the central body and the total is what Doppler tells us.

SS: In the case of another galaxy outside the Milky Way, we can assume it must be at higher potential

No we can't but it is irrelevant anyway. Most galaxies are moving away from us, a few move towards us, like Andromeda for example, but that motion adds to the Doppler on both sides of the orbit so when we calculate vpos - vneg, it cancels out automatically.

[GD: No, that is completely wrong, I have no idea where you say that...]

SS: Orbital velocity is the well known and accurate v = √GM/r In my opinion this equation is only accurate when the orbital motion is measured against the celestial sphere or from orbiting body to infinity.

No, it is only valid when measured relative to the location of mass M.

SS: My solution is based on the following assumption; If √GM/r gives us the external orbital velocity with respect to the celestial sphere

The assumption is wrong, the orbital velocity is relative to the body that creates the orbit.

SS: then √GM/r - (escape velocity) should give us the internal velocity with respect to centre of rotation.

Sorry, that is nonsense. You can't just invent equations like that with no regard for the physics.

SS: I understand exactly how you see it, ..

Your answers suggest you don't, you are simply making assumptions that contradict basic physics.

SS: Not sure about you, but I don't believe in a God, I no longer believe in Santa Claus, ...

On the contrary you are inventing random equations that have no more connection to reality than Santa Claus. The existence of dark matter is proven by the motion of galaxies in clusters, by the amount of deuterium in nebulae, the existence of small galaxies billions of years ago when the shouldn't have had enough time to form and the Euclidean geometry of the universe on large scales, none of which can be explained by your suggestion.

George Dishman

Happy 360 degrees.

It‘s come down to my gut feeling vs. your hardcore physics. Allthogh I think you are cheating by relocating the fiducial from earth to the centre of a galaxy thousands of light years away and denying that the new fiducial is inside the orbit.

The experiment I would like you to concider is looking at the moon with your telescope, as the earth turns, which way does the moon dissapear out of view. (Assuming you dont have a tracking telescope)

compare this to the direction the moons orbital direction.

Dear Steven,

SS: I think you are cheating by relocating the fiducial from earth to the centre of a galaxy thousands of light years away and denying that the new fiducial is inside the orbit.

To get v = √GM/r, you set the centrifugal force f=mv2/r equal to the gravitational force, f=GMm/r2. In those equations, r is measured from the centre and v is relative to the centre of the distant galaxy, not from us, so you have no choice.

That is of course at the centre of the orbit, it is we who are outside it so all our spectroscopic Doppler measurements have to be converted in order to use that equation. The equations we had before are:

v = (Vpos - Vneg) / (2 cos(inclination) )

M = v2r/G

George Dishman

Correct me if I am wrong, but the velocity in , f=mv2/r refers to the euclidean velocity no?

If so this doesn‘t prove the case for doppler.

How about my question anout the moon?

Dear Steven,

SS: Correct me if I am wrong, but the velocity in , f=mv2/r refers to the euclidean velocity no?

It refers to the speed, not the velocity. Euclidean geometry is a good enough approximation.

However, I should repeat that the above equation is only valid for a central mass, it is fine for the moons of Mars, planets orbiting the Sun and the Moon orbiting Earth but for a galaxy which is thin in comparison to its diameter, you have to integrate over the disc.

SS: If so this doesn‘t prove the case for doppler.

What's to prove?

SS: How about my question anout the moon?

The speed v is the circumference of the Moon's orbit divided by the time it takes, which way it appears to move is of no relevance at all. I already did that calculation as an example for you a few posts back.

Remember galaxies are too far away for us to see any lateral movement, we can only use Doppler to measure the component of the speed towards or away from us. The motion of the telescope along that direction can be subtracted out of course but it is tiny compared to the speed of the stars in the galaxy.