Good Morning Robittybob1... What time is it where you are? lol
The formula I'm using V in is just the initial mass calc... using thee nearest published speeds I could find, closest to the center, of about 5ly.
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.Good Morning Robittybob1... What time is it where you are? lol
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??
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
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??
Please could you paste part of the sentence that made you think this mass was inside a 5 ly radius. I must be missing it?Here is where I got the intitial data to resolve the mass withing the 5ly radius
http://hubblesite.org/newscenter/archive/releases/2005/26/full/
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 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.
You got it right!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..
M=(7.47177024803790242104411460627691422509.kg X 10^38
Drop as many characters as you need to fit the digits in to your calculator, for mass, but the closer the better of course.
Please could you paste part of the sentence that made you think this mass was inside a 5 ly radius. I must be missing it?
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.
M = V ^ 2 * r / G derived from the orbital speed formula V = SQRT(GM/r) looks OK.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)
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.
The STIS observations of Andromeda are so precise that astronomers have eliminated all other possibilities for what the central, dark object could be. They also calculated that the black hole's mass is 140 million Suns, which is three times more massive than once thought.So these estimates keep on changing!
The STIS observations of Andromeda are so precise that astronomers have eliminated all other possibilities for what the central, dark object could be. They also calculated that the black hole's mass is 140 million Suns, which is three times more massive than once thought.So these estimates keep on changing!
I know, that's why we use what mosltly agrres with itself at present, which is the R= 5ly, 1000 kms, and the mass within that radius. The estimates for the actual Black Hole are derived by trying to guess the radius which is tiny, compared to our overall within the 5ly, where the blue stars are said to rotate.
I will try and help, but we just got some terrible news, and I have to stop at the moment. I will interject when I can, but have to drive to PA possibly in the next few hours. Sorry.
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.Let's just cut to the chase.
....
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.
@Scott - I am hoping you are asking Janus58 that question, for sorry I don't understand the question at all.Using the parameters I provided, what is the numeric. Time percentage found using the simple format of the time dilation formula?