# Solution to the Galaxy Rotation Problem, without Dark Matter

Discussion in 'Alternative Theories' started by Scott Myers, Feb 2, 2013.

1. ### Scott MyersNewbieRegistered Senior Member

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Good Morning Robittybob1... What time is it where you are? lol

3. ### Robittybob1BannedBanned

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Scott could you write that formula in the style used in a macros or Excel. (Visual Basic has gaps (spaces) between each variable and action like T0 = Tf * (1 - (2 * G * M / (r * c ^ 2))) ^ 0.5
so if you could write out the formula and list the values I'll include that part into the macro somewhere??

5. ### Robittybob1BannedBanned

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It was 5:00 AM when I started so I'm listening to BBC World News at the same time as well as having my breakfast. I nearly gave up on this project yesterday.
Sometimes I get up at 3:00 AM do a bit and go back to bed for a while.

7. ### Scott MyersNewbieRegistered Senior Member

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I think this is the correct format and initial values I used just for the mass Calc

(1,000,000)X(1,000,000)X(5.271)/(6.67398X10^11)=Mass in KG

8. ### Scott MyersNewbieRegistered Senior Member

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Raidius I left in LY sorry

Mass in KG =(1,000,000)X(1,000,000)X(4.98664452 x 10^16)/(6.67398X10^11)

9. ### Scott MyersNewbieRegistered Senior Member

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T0 = Tf * (1 - (2 * 6.67398X10^11* 7.47177024803790242104411460627691422509 x10^38 / (4.98664452 x 10^16 * 299 792 458 ^ 2))) ^ 0.5

I do not know what Tf should be, and I don't see where you are using the square root of the entire phrase on the other side og tf, but I'm sure it's there somehow. I'm not understanding where you get the ^0.5 is that your square root? I'm lost in that, obviously..

10. ### Scott MyersNewbieRegistered Senior Member

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I intially tried to use seconds per year (31,556,925.9936) as tf, but my answers were so ridiculous we would have to adjust the velocity of the blue stars to well beyound the speed of light, so see what you come up with. We may just have to step away from the black hole a bit? I don't know yet. Maybe I just did it all wrong.

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12. ### Robittybob1BannedBanned

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The scientist always use the Julian calender year of 31557600 seconds. If you used this does it come out better? In Wikipedia on GTD they give the example for the Sun and the Earth, were you able to work those out before starting with the very very complex situation of Andromeda?
I thought I'll see if I can do the easy one first just to see if I got the formula right.

13. ### Robittybob1BannedBanned

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You got it right!
You have to count the brackets, they must balance (e.g. ((----))) would be wrong) the inner most brackets hold together the "divided by" bit (4.98664452 x 10^16 * 299 792 458 ^ 2)
The second set is around the whole fraction bit (2 * 6.67398X10^11* 7.47177024803790242104411460627691422509 x10^38 / (4.98664452 x 10^16 * 299 792 458 ^ 2))
The 3rd set is around the bit that has to be square rooted
(1 - (2 * 6.67398X10^11* 7.47177024803790242104411460627691422509 x10^38 / (4.98664452 x 10^16 * 299 792 458 ^ 2)))
and the way of doing a square root is to the power of 0.5.
Tf is just 1, but if were looking from an area of time dilation this Tf value would take that into account.

You are making it hard on yourself by including so many digits in mass number.

14. ### Aqueous Idflat Earth skepticValued Senior Member

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Not quite. You only get to keep the number of digits of precision that can be accurately measured. For this type of discussion you can at best probably only justify one or two digits. There isn't any advantage to carrying more digits since these are all rough estimates at best.

15. ### Scott MyersNewbieRegistered Senior Member

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I did not find it within some of this other documents I have been pasting. I came accross one a while back that estimated the 5ly radius for the blue stars, and I saw one that claimed 4.9 ly, Assuming that we (all astronomers) are using the same calcs for orbital calculations, I merely figured out how far the blue stars will travel in one year, and solved that then for pie, then split that in half for the radius. This seemed better than using a vaugue number published. To use the higher estimate I calced this simple way, is the more conservative, given the effect a larger radius will have on GTD, I prefer to use the higher number instead of the shortest estimate of 4.9 lys, since to use 4.9 would be padding my answers, or helping my cause, if you will.

I'll serch for my original findings. I jotted some of these numbers down a while ago, just haven't found them again. Calciong the radiu7s by using their distance traveled in the total orbit, should be equal to what we will find elswhere.

16. ### Scott MyersNewbieRegistered Senior Member

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Drop them we will. I have just wanted to reduce any introduced error, but I realized each individual KG will not make or break the experiment

17. ### Robittybob1BannedBanned

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M = V ^ 2 * r / G derived from the orbital speed formula V = SQRT(GM/r) looks OK.

18. ### Robittybob1BannedBanned

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[/QUOTE][/QUOTE]
from http://hubblesite.org/newscenter/archive/releases/2005/26/full/

19. ### Scott MyersNewbieRegistered Senior Member

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[/QUOTE]

from http://hubblesite.org/newscenter/archive/releases/2005/26/full/

20. ### Janus58Valued Senior Member

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Let's just cut to the chase.

Newton's orbital velocity equation is this:

$V = \sqrt{\frac{GM}{r}}$

Gravitational time dilation is

$T_o = T_f \sqrt{1- \frac{2GM}{rc^2}}$

Thus if you want to know the orbital velocity measured by a far removed observer including time dilation, you use:

$V=\sqrt{\frac{GM}{r}} \sqrt{1- \frac{2GM}{rc^2}}$

What you want is the difference in M between that calculated by Newton and that calculated by time dilation for an object observed to have an orbital velocity of V at a orbital distance of r, or this relationship

$\sqrt{\frac{GM_k}{r}} = \sqrt{\frac{GM_R}{r}} \sqrt{1- \frac{2GM_R}{rc^2}}$

Here Mk is the mass determined by Newton and MR the mass if we include time dilation.

solving for MR:

$M_R = \left ( 1 \pm \sqrt{1-\frac{8GM_k}{rc^2}} \right ) \frac{rc^2}{4G}$

Now we just A values for Mk and r

http://www.spacetelescope.org/news/heic0512/

gives an estimated black hole mass of 140 million solar masses and a value of V of 1000 km/s.

This would give an r of 1.964 ly (1.859e16m)

So now we plug these values into our equation.

Note that there are two possible answers, as the equation has a + or - symbol.

One answer comes out to 140,003,113 solar masses. IOW, not much different from the Kepler determined mass.

The other answer is 6.2993e12 (6.3 trillion)solar masses!

If we use the 5.27 ly distance and 376 million solar masses, we get

376,008,374 solar masses (still not much of a difference) and

16.899 trillion solar masses.

The larger values are thousands of times larger than the entire estimated mass of the galaxy.

But which ones are correct, the larger or the smaller? To determine that, we need more data points.

So let's take the first example with possible values of 140,003,113 solar masses and 6.29 trillion solar masses.

We'll calculate what orbital velocity we should see for a star orbiting at 4 ly.

This gives 700.666 km sec if we use the 149,003,113 Solar mass

and

106,048.252 km/sec if we use the 6.3 trillion Solar mass. This is over 1/3 c and some 106 times faster than the star orbiting at just under 2 ly. The further star would appear to have the greater orbital velocity. The reason is that the time dilation factor goes from 0.0103 at 2 ly to only 0.714 at 4 ly with the larger mass.

If we use the masses from the other example and a star 10 ly away, we get close the same answers; the small mass gives 736 km/sec and the larger 106,000 km/sec.

IOW, to determine which "corrected mass" is right, we look at the pattern of orbital velocities of stars in the vicinity of the black hole at different distances. If we see a pattern that shows the velocities falling off with distance, we know that the "corrected mass" is the one that does not vary much from the predicted Kepler mass.

If we see the velocities increasing with distance, we know that the "corrected mass" is many times larger than the Kepler mass.

What we do see is the first case, the velocities of stars near the black hole fall off with distance. This means that the Time dilation "corrected mass" is only very slightly larger than the Kepler mass even near super massive black holes where the effect of time dilation would be the greatest.

21. ### Scott MyersNewbieRegistered Senior Member

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Using the parameters I provided, what is the numeric. Time percentage found using the simple format of the time dilation formula?

22. ### Robittybob1BannedBanned

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That was very helpful and I could only see 1 typo error in the whole post. ("This gives 700.666 km sec if we use the 149,003,113 Solar mass" was that supposed to be 140,003,113?) I sense Janus58 is a person who really knows what he/she is talking about.

OK so time dilation does not account for a sufficient part of the needed corrected mass (I know this poorly worded, for time dilation does not change mass). I was previously working on a project looking at correcting G rather than mass. If G varied in a function depending on distance from the central black hole maybe then we wouldn't need dark matter or adjusted mass. Have you ever considered G varying?

In a reasonably simple way (at similar level in your "cut to the chase post") could you explain how the addition of DM to the galaxy overcomes this problem?
Thank you for helping us out.

23. ### Robittybob1BannedBanned

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@Scott - I am hoping you are asking Janus58 that question, for sorry I don't understand the question at all.