# Thread: Specific Values for Kepler's Third Law

1. ## Specific Values for Kepler's Third Law

I have been formulating the motions of two bodies that travel in elliptical orbits due to the effects of each other's gravity and have come up with some interesting results. I originally posted this in The Grand Puzzle but got no response about this particular part of it so I decided to start a whole new thread. I am wondering if these observations are already known or if I may have discovered something new. I have looked through some websites to find out but have only found close approximations to the actual values. Some sites even say that they are not completely accurate.

Kepler's third law of motion states 4pi2a3=GMT2 for a body in orbit around another, so if M is the mass of the sun and (a) is the semi-major axis of a planet's orbit, then the period of its revolution (T) can be found with this formula. In this case, since GM/4pi2 is a constant for the sun, then a3/T2 is a constant for all planets that orbit it. Now, it is indeed already known that this law isn't quite accurate and a large error is found for larger planets, but I have yet to see the accurate version of this law. Another version says that for larger planets, one can use the sum of the masses in place of the mass of the sun to gain a more accurate result. This is as far as I have seen it taken. The actual formulas are so simple that I would have trouble believing that I am the first to come up with them. But if this is the case, why haven't I seen it and why are only the approximations available? Could the reason noone has responded to the original post be that nobody has found anything either?

The precise formulas which I have found that directly relate to Kepler's third law of motion are simply 4pi2(a1+a2)3=G(M1+M2)T2, pi2(Df+Dc)3=2G(M1+M2)T2, 4pi2a13=GM23T2/(M1+M2)2, and 4pi2a23=GM13T2/(M1+M2)2. More can be found in the original post.

Originally Posted by The Grand Puzzle
I began with two known masses which are travelling in parallel but opposing directions with a given distance between them which is perpendicular to their lines of motion. The difference in their velocities is also given and must be small enough so that the bodies will not escape each other and will instead produce closed, elliptical orbits. My initial observation was that the shape of the ellipses are identical for both bodies but that the dimensions are inversely proportional to the mass. I also noticed that a line drawn through the two bodies are only perpendicular to their lines of motion at their closest and furthest distances from each other and that the distance between these two points on the same orbit will describe the major axis of its ellipse.

Let's look at what is observed so far. Two bodies of masses M1 and M2 are orbitting each other and a line drawn between them will be perpendicular to their lines of motion at their closest and furthest distances from each other (Dc & Df). The sum of these two distances will equal the sum of the major axes of both orbits, so that 2(a1+a2)=Df+Dc, where (a) is the semi-major axis, equal to the distance from the center of the ellipse to its outermost point. The dimensions of each ellipse are inversely proportional to the mass of the body, so that the center of mass can be found at (1/M1)/(1/M1+1/M2)=(1/M1)/[(M1+M2)/M1M2]=M2/(M1+M2)=M2/MT of the distance from M1 to M2 or M1/MT from M2 to M1. If we are to use the center of mass as a frame of reference in order to stabilize the positions of the orbits around it, then the ratio of their individual velocities to the total difference in velocities at any given time will be directly related to the dimensions of each ellipse and must also be divided accordingly. If V is the difference in velocities at some point in their orbits, then V1=M2V/MT and V2=M1V/MT, so M1V1=M2V2.The semi-major axis of each ellipse is, of course, also proportional to its dimensions, so that we can find their lengths with a1=(Dc+Df)M2/2MT and a2=(Dc+Df)M1/2MT. Further observations show that Dc1Vc1=Df1Vf1=Vb1b1(MT/M2) and Dc2Vc2=Df2Vf2=Vb2b2(MT/M1), where Vc1 and Vf1 are the velocities for M1 at the closest and furthest distances and Vb1 is the velocity where the semi-minor axis (b1) crosses the ellipse on either side. One also finds that VcVf=Vb2 for each body, (Vb12/a1)+(Vb22/a2)=GMT/(a1+a2)2, and (Df/Dc)(Dc+Df)(Vf1+Vf2)2=2GMT (all quantities are positive).

The direction of the orbits will be perpendicular to their centers where it crosses the a and b axes. These orbits sweep out equal areas in equal times from the focus closest to where the orbits are closest (Dc), so that area per time is a constant. If we designate the distance from the center of the orbit to this focus as f, then the area swept out per time for Vc within an extremely small time interval is [(a-f)VcTi/2]/Ti=(a-f)Vc/2. For Vf it is (a+f)Vf/2. For Vb it is [(f+VbTi)b/2-fb/2]/Ti=Vbb/2. And for the entire orbit it is piab/Trev. These values are all equal, so (a-f)Vc/2=(a+f)Vf/2=Vbb/2=piab/T. Since a2-f2=b2, these terms can be recombined to show that (piab/T)2=(Vbb/2)2=[(a-f)Vc/2][(a+f)Vf/2]=(a2-f2)VcVf/4=b2VcVf/4, which becomes (2pia/T)2=Vb2=VcVf. Much can be determined by the formulas we have obtained so far. Further recombining yields formulas such as a/b=(Vb/2)(1/Vc+1/Vf) and GM23/MT2a1=(2pia1/T)2. This last formula demonstrates that Kepler's third law for the ratio of the cube of the semi-major axis to the square of the orbittal period requires some minor adjusting since it will only approach a constant for a body which is much more massive than the other (M2>>M1), but a discrepency will be noticed for bodies whose masses are close in value. (Is this last formula known? I have not found anything with these same simple results but much that says that there is a discrepency between Kepler's third law and the real orbits of planets. Perhaps I have corrected for this. Please let me know.)

If we know the initial conditions for M1, M2, Df, and Vo, where Vo is the original difference in velocities between the bodies in parallel lines of motion with a perpendicular distance between them of Df,we can now describe their orbits in detail relative to their center of mass. We will determine the values for M1, but the values for M2 can of course be found using the same equations by substituting M1 for M2 and M2 for M1. These values can be determined as follows:

Dc=Df2Vo2/(2GMT-DfVo2)
Vf1=VoM2/MT
a1=(Dc+Df)M2/2MT
Vc1=GM23/MT2a1Vf1
Vb1=(GM23/MT2a1)1/2
b1=Df1Vf1M2/MTVb1
f1=(a12-b12)1/2
If I get no responses here, should I post a fuller version in another part of the forum (perhaps astronomy)?

Tensor,

I am still working on the binary problem and I am close to the results, but I need other decaying binary systems to compare it to. I have looked for some but have not found any with results quite like the one you gave me. It would have be be two bodies only that orbit in a single plane with the semi-axis and rate of decay of the semi-axis over time which are both expressed in units of distance. Can you recommend any specific sites?
Last edited by grav; 2006-Jun-17 at 03:31 AM.

2. Originally Posted by grav
Kepler's third law of motion states 4pi2a3=GMT2 for a body in orbit around another, so if M is the mass of the sun and (a) is the semi-major axis of a planet's orbit, then the period of its revolution (T) can be found with this formula. In this case, since GM/4pi2 is a constant for the sun, then a3/T2 is a constant for all planets that orbit it. Now, it is indeed already known that this law isn't quite accurate and a large error is found for larger planets, but I have yet to see the accurate version of this law. Another version says that for larger planets, one can use the sum of the masses in place of the mass of the sun to gain a more accurate result. This is as far as I have seen it taken. The actual formulas are so simple that I would have trouble believing that I am the first to come up with them. But if this is the case, why haven't I seen it and why are only the approximations available? Could the reason no has responded to the original post be that nobody has found anything either?
Kepler's original third law was just that the cube of the semimajor axis was proportional to the square of the period; the proportionality constant was worked out empirically. You can show pretty easily that, if the mass of the one body is negligible compared to the mass of the other, you get your first formula. The planets masses are fairly small compared to the Sun, but just as you note, for the more massive planets this isn't as close to being true, and the second formula you have gives a first-order correction for this effect. That's pretty common actually, since to solve a two body problem with Newtonian mechanics, you actually solve a one body problem with a reduced mass.

Of course, the solar system is not a two body system. There are many objects in the solar system, and they all perturb each other to some degree. You won't find any simple equations like the ones above that takes these perturbations into account, though, because as soon as you do that, the equations just aren't simple any more. In fact, you can show that there are no closed analytic solutions to even the three body problem; the only way to work it out is through numerical analysis ("ugly math"). That is, you figure out where everything is, determine what all the forces are, figure out the accelerations for everything, and move your simulation forward a little bit in time. Then you recalculate all the forces, and go forward another step. How precise you make your measurements and how small you make each step of time depends on how accurate you need your results to be and how much computing power you have available. And of course if you really need a lot of precision, Newton's laws aren't going to be sufficient, and you'll need to add in general relativistic corrections, especially for the innermost planets.

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Originally Posted by grav
Tensor,

I am still working on the binary problem and I am close to the results, but I need other decaying binary systems to compare it to. I have looked for some but have not found any with results quite like the one you gave me. It would have be be two bodies only that orbit in a single plane with the semi-axis and rate of decay of the semi-axis over time which are both expressed in units of distance. Can you recommend any specific sites?
Grav, I appologize for missing this. I can't give you any specific sites, but below are a list of Neutron-neutron star binaries. A google search should provide you with the particulars of the binary:

B1913+16
B1534+12
J1756-2251
J0737-3039
J1518+4904
J1811-1736
J1829+2456

I'm not sure why they two systems have to be similar. You should be able to use the same equations for both (GR does) The only differences should be the masses, the orbital elelments. Those should give you the data needed to plug into the equations and match their observed rate of insprial.

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GR corrections are only of interest with regard to the orbit of Mercury. Old fashioned Newtonian gravity works very well for all the other planets. The asteroid belt [and Kuiper cloud] has too little mass [or are too distant] to be of any relevance.

5. Originally Posted by Thanatos
GR corrections are only of interest with regard to the orbit of Mercury. Old fashioned Newtonian gravity works very well for all the other planets. The asteroid belt [and Kuiper cloud] has too little mass [or are too distant] to be of any relevance.
No, there are measurable general relativistic corrections for both Venus and Earth, although they are smaller than that for Mercury. Whether you need them or not just depends on how precise you want to be.

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Originally Posted by Tensor
Grav, I appologize for missing this. I can't give you any specific sites, but below are a list of Neutron-neutron star binaries. A google search should provide you with the particulars of the binary:

B1913+16
B1534+12
J1756-2251
J0737-3039
J1518+4904
J1811-1736
J1829+2456

I'm not sure why they two systems have to be similar. You should be able to use the same equations for both (GR does) The only differences should be the masses, the orbital elelments. Those should give you the data needed to plug into the equations and match their observed rate of insprial.
The Jodrell Bank Observatory Pulsar page may be a good place to start; not only does it have some interesting articles, but the publications of observatory members include many that seem to have direct bearing on your quest, grav.

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Originally Posted by Nereid
The Jodrell Bank Observatory Pulsar page may be a good place to start; not only does it have some interesting articles, but the publications of observatory members include many that seem to have direct bearing on your quest, grav.
Doh! I forgot about that site. I had it bookmarked, but then had to get a new computer and evidently missed a few when I changed over. Now I got it back again. Thanks Nereid.

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HI Grav

I am interested in your post on Keplers laws with masses that are somewhat proportionalto each other.

You listed these as exact

"The precise formulas which I have found that directly relate to Kepler's third law of motion are simply
4pi2(a1+a2)3=G(M1+M2)T2,
pi2(Df+Dc)3=2G(M1+M2)T2,
4pi2a13=GM23T2/(M1+M2)2, and
4pia23=GM13/(M1+M2)2. More can be found in the original post."

It is hard to tell the superscripts, expecially in the last two equations. Would ^be better ?

Also, you may find my post on a uniform expansion interesting. Especially post number 14. Base on a uniform expansion I generate Keplers third law, and produce a model in which the principles of conservation of energy and momentium are the result of geometry.

Snowflake

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Originally Posted by snowflakeuniverse

Also, you may find my post on a uniform expansion interesting. Especially post number 14. Base on a uniform expansion I generate Keplers third law, and produce a model in which the principles of conservation of energy and momentium are the result of geometry.

Snowflake
Then you ought to be able to reproduce the observations on the Neutron star binaries then, right? I look forward to your calculations.

10. Thank you Tensor and Nereid, but I still can't find compatible binary systems. Maybe I just don't know how to read the data correctly. But I have learned that others have come up with similar results to my own concerning the speed of gravity. I originally thought that the aberration of gravity might cancel out the effects of virtual relativity, but alas they only add to it. Virtual relativity depends on the speed of the source and aberration on the observer. The overall effect is an angle of observervance which depends on the relative velocities of both bodies to each other and the speed of gravity itself. It is not the same as the analogy of rain aberration, however, but depends on more relativistic effects where the hypotenuse is the speed of gravity vector (not the side of a right angle) and the base is the sum of velocities. Because of the lack of data on my part, perhaps we could try something else. If your time permits, you could give me some data of one or two binary systems where you already know the rate of decay (don't tell me which ones, of course) and I will tell you the predicted rate of decay using what information I have obtained so far using only the B1913+16 system. The data can include the masses of each, their closest and furthest distances from each other and/or the velocities at these points, and/or the orbittal period. You can then compare the results to your observed decay.

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Hi Tensor
Regarding rotating neutron binaries, this is link is a good general description of the complexity of the problem.
http://www.sissa.it/ap/RelAstro/Binary.html

To explain how binaries rotate faster and faster and eventually merge, all I do is keep adding some dark matter to the binaries. Is that assumption ok? I am not sure how the dark matter is added, but it must be there because the evidence is there, they are pulling themselves closer and closer and spinning faster and faster as a consequence of the increased gravitational attraction.

How is that any different than assuming that there is dark matter keeping galaxies and groups of galaxies together?

You can see my calculations that produce the inverse square laws, establish the principles of conservation of energy and momentum, and lay a foundation for a Unified field theory, all because of a geometric expansion of spacetime. The model also makes predictions that conform to observation. It is all there for your review. If you want to pick on me, or challenge me, start there.

Snowflake

12. Thanks, snowflakeuniverse. You may have unknowingly helped to solve part of the binary problem with your link. It begins with a reference to point masses, which is how gravitational bodies can usually be viewed. But I realized that the the spin of a body might affect its aberration. The side that is closest to another body may turn in the opposite direction of the direction of revolution, cancelling the aberration for that side. That is to say, every atom in the body must really be accounted for when it comes to gravity, so a body that is spinning in a direction opposite its revolution should cancel most of its aberration. For example, this is the case with the moon. The same side always faces the Earth, so that it turns at the same angular velocity as its orbital velocity, making its orbit very stable. I have reason to believe that aberration is also related to the eccentricity of the orbits, so that circular orbits are also very stable.

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Originally Posted by snowflakeuniverse
Hi Tensor
Regarding rotating neutron binaries, this is link is a good general description of the complexity of the problem.
http://www.sissa.it/ap/RelAstro/Binary.html

To explain how binaries rotate faster and faster and eventually merge, all I do is keep adding some dark matter to the binaries. Is that assumption ok?
It might be ... except for the fact that there is, observationally, essentially no DM in the MW (thin) disk, which is where most (all?) binary pulsars are to be found.
I am not sure how the dark matter is added, but it must be there because the evidence is there, they are pulling themselves closer and closer and spinning faster and faster as a consequence of the increased gravitational attraction.
Or, GR (and Hulse and Taylor) are right - gravitational radiation accounts for the inspiral of binary pulsars, to within the limits of the observational data.
How is that any different than assuming that there is dark matter keeping galaxies and groups of galaxies together?
Hugely.

The two are quite different, in terms of the supporting observations (I really must try harder to find the time to write the relevant posts to that thread).

Suffice it to say that the scales are hugely different, the physical processes also different, and the data unequivocal.
You can see my calculations that produce the inverse square laws, establish the principles of conservation of energy and momentum, and lay a foundation for a Unified field theory, all because of a geometric expansion of spacetime. The model also makes predictions that conform to observation. It is all there for your review. If you want to pick on me, or challenge me, start there.

Snowflake
Actually, this is explicitly a violation of the BAUT Rules ("thread hijacking"). As a long time BAUT member, we expect that you would be aware of these rules, snowflakeuniverse, so I have no choice but to warn you (next time you will be banned).

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Originally Posted by grav
Thank you Tensor and Nereid, but I still can't find compatible binary systems. Maybe I just don't know how to read the data correctly. But I have learned that others have come up with similar results to my own concerning the speed of gravity. I originally thought that the aberration of gravity might cancel out the effects of virtual relativity, but alas they only add to it. Virtual relativity depends on the speed of the source and aberration on the observer. The overall effect is an angle of observervance which depends on the relative velocities of both bodies to each other and the speed of gravity itself. It is not the same as the analogy of rain aberration, however, but depends on more relativistic effects where the hypotenuse is the speed of gravity vector (not the side of a right angle) and the base is the sum of velocities. Because of the lack of data on my part, perhaps we could try something else. If your time permits, you could give me some data of one or two binary systems where you already know the rate of decay (don't tell me which ones, of course) and I will tell you the predicted rate of decay using what information I have obtained so far using only the B1913+16 system. The data can include the masses of each, their closest and furthest distances from each other and/or the velocities at these points, and/or the orbittal period. You can then compare the results to your observed decay.
I think the pulsar to watch is PSR J0737-3039B ("the double pulsar"). This page has references to papers published on it.

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HI grav
Im a bit confused, you said
"The side that is closest to another body may turn in the opposite direction of the direction of revolution, cancelling the aberration for that side."

Why shouldn't gravitional drag betwen the two rotating neutron stars cause them to rotate in sync with the same rotation, clock wise = clock wise, counter clock wise= counter clockwise.

(again sorry for stepping on you thread, but I got dragged in)
Snowflake

16. Originally Posted by grav
The precise formulas which I have found that directly relate to Kepler's third law of motion are simply 4pi2(a1+a2)3=G(M1+M2)T2
That formula appears in many textbooks. Notice that the distance is actually a1+a2, which is the distance between the two bodies, not the semi-major axis of one of their orbits.

17. Originally Posted by Grey
Kepler's original third law was just that the cube of the semimajor axis was proportional to the square of the period; the proportionality constant was worked out empirically. You can show pretty easily that, if the mass of the one body is negligible compared to the mass of the other, you get your first formula. The planets masses are fairly small compared to the Sun, but just as you note, for the more massive planets this isn't as close to being true, and the second formula you have gives a first-order correction for this effect. That's pretty common actually, since to solve a two body problem with Newtonian mechanics, you actually solve a one body problem with a reduced mass.
The reduced mass formulas look like they would probably amount to the same thing as some of the formulas I came up with if introduced into Kepler's original laws. However, if that's the case, then why are the masses for the binary system 1913+16 measured so far off? It appears to be one of the best known systems, yet the mathematics is incorrect (as far as I can tell). It's been bugging me for a while now. The masses of the stars are given as 1.441Msun and 1.387 Msun, which totals 2.828 Msun. But according to the formula (mine) where pi2(Df+Dc)3=2G(M1+M2)T2, the total mass should be 2.8324 Msun. This formula uses those factors which can be measured best, and it should then be a simple procedure to find the masses (at least the total mass), yet the numbers are very far off. Does anyone know why this is so?

(I am using the orbital period (T) of 7.751939106 hours, a periastron separation (Dc) of 746,600 km, and an apastron separation (Df) of 3,153,600 km)

18. Which values are you using for G, and the mass of the Sun?

PS: Might also want to know which value of pi, too! When I use G=6.672x10-11, MS=1.99x1030, and pi=3.14, I get 2.828.
Last edited by hhEb09'1; 2006-Jul-18 at 07:59 PM.

19. Originally Posted by grav
The reduced mass formulas look like they would probably amount to the same thing as some of the formulas I came up with if introduced into Kepler's original laws. However, if that's the case, then why are the masses for the binary system 1913+16 measured so far off? It appears to be one of the best known systems, yet the mathematics is incorrect (as far as I can tell). It's been bugging me for a while now. The masses of the stars are given as 1.441Msun and 1.387 Msun, which totals 2.828 Msun. But according to the formula (mine) where pi2(Df+Dc)3=2G(M1+M2)T2, the total mass should be 2.8324 Msun. This formula uses those factors which can be measured best, and it should then be a simple procedure to find the masses (at least the total mass), yet the numbers are very far off. Does anyone know why this is so?

(I am using the orbital period (T) of 7.751939106 hours, a periastron separation (Dc) of 746,600 km, and an apastron separation (Df) of 3,153,600 km)
I don't know for certain, but my guess would be that with two stellar mass objects at a remarkably small distance, you can't just use Newton's equations for gravity. You'd have to use general relativity instead. After all, you can't get Mercury's orbit right using Newtonaian mechanics, and it's about two orders of magnitude farther from the Sun than the separation here.

20. Originally Posted by hhEb09'1
Which values are you using for G, and the mass of the Sun?

PS: Might also want to know which value of pi, too! When I use G=6.672x10-11, MS=1.99x1030, and pi=3.14, I get 2.828.
I used about the same values, where G=6.6725985*10-11, MS=1.9891*1030, and, of course, pi=3.14159265. If you found 2.828 MS for a result, what formula did you use to find this?

21. Originally Posted by Grey
I don't know for certain, but my guess would be that with two stellar mass objects at a remarkably small distance, you can't just use Newton's equations for gravity. You'd have to use general relativity instead. After all, you can't get Mercury's orbit right using Newtonaian mechanics, and it's about two orders of magnitude farther from the Sun than the separation here.
This seems like it might be the case. If so, it might help determine a formula for the rate of orbital decay and/or the transmission time of gravity and/or its aberration. Would you or anyone else happen to know the formula for how the masses are originally determined (or a link)?

22. Originally Posted by grav
I used about the same values, where G=6.6725985*10-11, MS=1.9891*1030, and, of course, pi=3.14159265. If you found 2.828 MS for a result, what formula did you use to find this?
I used your formula. When I use those values, I get your result. I'm a little dubious about the precision of your G, though. Where did you get so many figures?

23. Originally Posted by hhEb09'1
I used your formula. When I use those values, I get your result. I'm a little dubious about the precision of your G, though. Where did you get so many figures?
But my formula yields 2.8324 MS.

I got the value for G from an old high school physics book. I do notice, however, that the values of constants appear slightly different in different books. But I have gotten used to this one. I don't even remember it usually. But I punch the numbers for it into a calculator so often that it becomes automatic. I actually had to punch it in to retrieve its value before I could reply to your post.

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Originally Posted by grav
This seems like it might be the case. If so, it might help determine a formula for the rate of orbital decay and/or the transmission time of gravity and/or its aberration. Would you or anyone else happen to know the formula for how the masses are originally determined (or a link)?
Grav, you might try here. It is an early paper on PSR1913+16. Check section V (Discussion) for how the mass was computed. Please note the date on the paper (1976). Some of the measured results have been refined based on better meaurements since then.

25. Originally Posted by grav
But my formula yields 2.8324 MS.
Right! That's what I got using your values. Using the other values, I got a different answer.
I got the value for G from an old high school physics book.
I'm not sure what all the currently accepted values are, but I'm pretty sure that G is not known that precisely.

26. Originally Posted by hhEb09'1
Which values are you using for G, and the mass of the Sun?

PS: Might also want to know which value of pi, too! When I use G=6.672x10-11, MS=1.99x1030, and pi=3.14, I get 2.828.
Originally Posted by hhEb09'1
Right! That's what I got using your values. Using the other values, I got a different answer.
Okay. I see what you are saying now. When I used the "shortened" values you posted, I got Mtotal=2.8285 MS. That is probably the explanation right there. Sorry I didn't know what you meant before. I would never have thought the few extra digits would make that big of a difference. But it is probably also due to the exponential values as well, making the difference even larger than one would think. Wouldn't that mean, though, that the given values of the masses are indeed way off?

<<<<<>>>>>

Tensor, thank you also for your response. I will check that out, too.

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Originally Posted by hhEb09'1
Right! That's what I got using your values. Using the other values, I got a different answer.I'm not sure what all the currently accepted values are, but I'm pretty sure that G is not known that precisely.
A quick look at NIST yields the following value

6.6742(10) x 10-11 m3 kg-1 s-2

28. Originally Posted by Fortis
A quick look at NIST yields the following value

6.6742(10) x 10-11 m3 kg-1 s-2
It is interesting that this is so far off from the value I obtained. But it appears it is the value for pi (3.14, when shortened) that makes the biggest difference. The smaller value for G would actually bring the value for the total mass closer to my result.

I also learned something, which will answer hhEb09'1's question as well. It appears that the value I was using, G=6.6735985*10-11 should actually read 6.67259(85)*10-11. I always thought that the digits in parentheses meant that those digits may not be precisely known, but that they were still part of the value. The link you posted showed otherwise. I guess that's what you get when you try to learn this stuff on your own. I just thank goodness for this forum. I had never discussed my ideas with anyone before now and you are all so helpful. Heck, I never even used a computer until a year ago. It's about time I caught up with the rest of the world.

29. Originally Posted by Tensor
Grav, you might try here. It is an early paper on PSR1913+16. Check section V (Discussion) for how the mass was computed. Please note the date on the paper (1976). Some of the measured results have been refined based on better meaurements since then.
Thank you for the link. It appears (in section V) that they do indeed use the formula which I "discovered", the third formula shown at the beginning of this thread, 4pi2a13=GM23T2(M1+M2)2. Some of the values shown in this earlier paper seem to have been updated since then, as you said, but apparently the values for the masses weren't updated with them.

30. Originally Posted by grav
It is interesting that this is so far off from the value I obtained. But it appears it is the value for pi (3.14, when shortened) that makes the biggest difference. The smaller value for G would actually bring the value for the total mass closer to my result.
Anytime you have a product/quotient, and one value changes in the nth significant figure, the answer will also. I was only being facetious about using 3.14, but it appears you've found your answer in that paper.
I always thought that the digits in parentheses meant that those digits may not be precisely known, but that they were still part of the value.
The gravitational constant is one of the least well-measured physical constants there is. Experiments in various labs seem to pin it down further, but the results vary.

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