S2 orbit deviates from Keplerian ellipse

[David Levy]

How could it be that S2 orbit deviates from Keplerian ellipse and at what percentage?

In the following article it is stated that (It is believed that) it is due to stars, dark stellar...:

https://en.wikipedia.org/wiki/S2_(star)

"The motion of S2 is also useful for detecting the presence of other objects near to Sagittarius A*. It is believed that there are thousands of stars, as well as dark stellar remnants (stellar black holes, neutron stars, white dwarfs) distributed in the volume through which S2 moves. These objects will perturb S2's orbit, causing it to deviate gradually from the Keplerian ellipse that characterizes motion around a single point mass.^[9] So far, the strongest constraint that can be placed on these remnants is that their total mass comprises less than one percent of the mass of the supermassive black hole.^[10]

 

However, it is clear (to me) that stars orbit must obey to Keplerian ellipse .

I assume that this statement is due to our believe that S2 single point mass must be located at the SAG A*.

Hence, could it be that it deviates from Keplerian ellipse as we believe that it must orbit SAG A*.

In other words, if we eliminate this requirement, and verify S2 orbit based on pure Keplerian ellipse could it be that we might find that its single point mass is located at a different location than SAG A*?

Did we try it?

This information is very critical.

In one hand we claim that S2 orbit deviates from Keplerian ellipse.

So, we are well aware that there are some other forces which might have an impact on S2 orbit

However, on the other hand we use a none Keplerian ellipse orbit to estimate SAG A* mass.

https://en.wikipedia.org/wiki/Sagittarius_A*

• From the motion of star S2, the object's mass can be estimated as 4.1 million solar masses.^[3] (The corresponding Schwarzschild radius is 0.08 AU/12 million km/7.4 million miles; 17 times bigger than the radius of the Sun.)

How can we set SAG A* mass estimation on an orbit which is effected by (thousands of stars, as well as dark stellar remnants (stellar black holes, neutron stars, white dwarfs)?

Never the less, it is also stated that:

"the strongest constraint that can be placed on these remnants is that their total mass comprises less than one percent of the mass of the supermassive black hole"

So, does it mean that S2 orbit deviates from Keplerian ellipse by only 1%?

Hence, can we understand that by using pure Keplerian ellipse S2 single point mass deviates from SAG A* location by only 1%?

If so, how far the real S2 single point mass is located with regards to SAG A* location?

Edited April 1, 2017 by David Levy

[Strange]

However, it is clear (to me) that stars orbit must obey to Keplerian ellipse .

 

 

Only if they are orbiting a point mass.

 

 

 

In other words, if we eliminate this requirement, and verify S2 orbit based on pure Keplerian ellipse could it be that we might find that its single point mass is located at a different location than SAG A*?

 

If you believe that can be done, then go ahead and do it.

 

 

This information is very critical.

 

Why?

 

 

 

How can we set SAG A* mass estimation on an orbit which is effected by (thousands of stars, as well as dark stellar remnants (stellar black holes, neutron stars, white dwarfs)?

 

Because it introduces less than a 1% error, as you stated.

 

And don't forget the other stars.

 

 

So, does it mean that S2 orbit deviates from Keplerian ellipse by only 1%?

Hence, can we understand that by using pure Keplerian ellipse S2 single point mass deviates from SAG A* location by only 1%?

If so, how far the real S2 single point mass is located with regards to SAG A* location?

 

Why not read the references that the information comes from?

 

https://arxiv.org/abs/1203.2625

https://arxiv.org/abs/0810.4674

[Airbrush]

"...So, does it mean that S2 orbit deviates from Keplerian ellipse by only 1%?

Hence, can we understand that by using pure Keplerian ellipse S2 single point mass deviates from SAG A* location by only 1%?"

​It only means that all the thousands of stellar masses in the region through which S2 is orbiting Sag A add up to 1% of 4 million solar masses, or about 40,000 solar masses. These are all pulling on S2 from thousands of different directions. So it seems hard to tell exactly what stars are pulling on S2. S2 is a heavy star at 15 solar masses. In mid 2018 it will get accelerated up to over 3,000 miles per second, in a short amount of time, as it swings close to Sag A. That should tell us something interesting.

Edited April 1, 2017 by Airbrush

[Janus]

Objects will follow Keplerian ellipses as long as all of the mass they are orbiting is contained inside a radius smaller than the periapis of the orbit. This insures that the mass effecting the orbit remains a constant.

If this is not the case, such as when you have mass spread out in a spherical volume that encompasses the entire orbit, the mass effecting the orbit varies over the orbit. This has an effect on the parameters of the orbit. For instance, ratio of periapis to apoapis velocities will deviate from those predicted for a Kepler orbit of the measured eccentricity. (An extreme case would occur if the mass were evenly distributed throughout the spherical volume. Then you would get an orbit were the orbital velocities at the ends of the major axis would be the same and the highest velocities would be at the points at the ends of the minor axis.)

A non-Keplerian orbit means the object does not orbit the geometric focus of a ellipse and has different parameters. You can't "fix" this by just shifting the assumed position of the point mass.

 

The 1% limit is imposed by how much the orbit deviates from the perfect ellipse. In other words, the vast majority of the mass must reside in Sag A, while a small part is spread out so the amount of it closer to Sag A than S2 at different points of S2's orbit changes over the course of S2's orbit, thus slightly changing the mass effecting its orbit from point to point.

[David Levy]

For better understanding let's use the solar system as an example.

Especially - Moon, Earth and Sun.

So, let's isolate that system.

However, in this example we will shut down Earth light.

Hence, all we can see is Moon and Sun.

We know the total mass of the Moon, but we are requested to calculate the Sun mass.

We get full information about the moon orbit.

We see that the moon orbits the Sun in one year, however, its movement deviates from a perfect Keplerian ellipse.

It orbits the Sun with minor changes – a little bit up, a little bit down, sometimes it moves fast sometimes slow.

Technically, we can ignore that vital data and could get into simple calculation that the total mass of the sun is as big as Earth mass.

Is it correct?

 

We can easily claim:

 

Objects will follow Keplerian ellipses as long as all of the mass they are orbiting is contained inside a radius smaller than the periapis of the orbit. This insures that the mass effecting the orbit remains a constant.

 

So, yes, the moon orbit the Sun in one year.

Could it be that the moon orbits the Sun with only 1% deviation from a perfect ellipse?

 

 

The 1% limit is imposed by how much the orbit deviates from the perfect ellipse. In other words, the vast majority of the mass must reside in Sag A, while a small part is spread out so the amount of it closer to Sag A than S2 at different points of S2's orbit changes over the course of S2's orbit, thus slightly changing the mass effecting its orbit from point to point.

 

Could it be that the Earth mass is deviated by only 1% from the Sun mass?

Hence, we first must find the temporary location of the Moon's single point mass (which is – Earth), based on perfect Keplerian ellipse.

It is quite difficult as we only see a moon deviation from perfect Keplerian ellipse.

However, based on our ultra high powerful commuters, we should verify the temporary location of that Moon's single point mass (Earth), and even calculate its expected mass.

 

Then we should try to verify if that Moon's single point mass (Earth) is orbiting the Sun in perfect Keplerian ellipse.

If the answer is positive, then we can easily calculate the real mass of the Sun.

Conclusion –

There is no direct way to calculate a mass of an object which isn't in a perfect Keplerian ellipse orbit.

Any deviation from Keplerian ellipse is critical.

With regards to S2

We must first calculate the temporary location of S2 single point mass and its total mass.

Than we have to verify if that S2 single point mass orbits the SAG A* in a perfect Keplerian ellipse.

If so, we can easily extract the real mass of SAG A*.

Edited April 1, 2017 by David Levy

[David Levy]

Please look at the following S2 orbit diagram

http://www.universetoday.com/wp-content/uploads/2010/08/nature01121-f2.22.jpg

It is quite clear that S2 is moving in and out the from Keplerian ellipse

However, the time resolution is quite poor. (There is about one spot per year).

In the same token please look at the following moon orbit around the Sun:

https://en.wikipedia.org/wiki/Talk%3AOrbit_of_the_Moon#/media/File:Earth_orbit.svg

We also see that it moves in and out from the Keplerian ellipse.

So, if we could get better time resolution of S2 orbit diagram, we might get better understanding about its single point of mass.

Edited April 1, 2017 by David Levy

[Enthalpy]

Non-point mass distribution does result in non-Keplerian orbits. This is commonly used for observation satellites in sun-synchronous orbits, which are possible because the Earth isn't spherical but has a bulge at the Equator.

https://en.wikipedia.org/wiki/Sun-synchronous_orbit

[Strange]

 

Please look at the following S2 orbit diagram

http://www.universetoday.com/wp-content/uploads/2010/08/nature01121-f2.22.jpg

It is quite clear that S2 is moving in and out the from Keplerian ellipse

 

 

Is it?

Could it be that the Earth mass is deviated by only 1% from the Sun mass?

 

Why would it be? The mass inside the Earth's orbit is constant and almost entirely the mass of the Sun.

 

The rest of you post doesn't make much sense. I cannot work out what you are trying to say.

[David Levy]

The rest of you post doesn't make much sense. I cannot work out what you are trying to say.

 

O.K.

 

Gravity is gravity.

It works for all and in the same way.

if you are moon, Earth, Sun, S2 or even BH, you have to obey to gravity law.

 

In order to understand this statement, let's start with the following explanation about Earth/Moon Orbit and their centre of mass.

 

 

http://www.animations.physics.unsw.edu.au/jw/gravity.htm

 

Orbits and the centre of mass

It is a common simplification to say that 'the moon's orbit is a circle about the earth': the gravitational attraction of the moon towards the earth provides the centripetal force for the moon's orbit. From Newton's third law, of course, the moon attracts the earth with a force of equal magnitude, which accelerates the earth. So the earth and the moon each trace (almost perfectly) circular orbits about their common centre of mass. The sketched below illustrates this, but is not to scale.

• 

As you can see, the earth and the moon each trace (almost perfectly) circular orbits about their common centre of mass.

This center of mass orbits the Sun.

Hence, the Earth and the Moon orbits their common center of mass, while this virtual point of center mass orbits the Sun.

So, this virtual common center of Earth/Moon mass point orbits the Sun in a perfect Keplerian ellipse. (Although some people would assume that the Earth orbits the sun in a perfect Keplerian ellipse - and that is a mistake),

However, the Earth is significantly heavier than the Moon, therefore, this Common center of mass is located in its radius.

Therefore, it isn't so big mistake to assume that the Earth orbits the sun in a almost perfect Keplerian ellipse.

Never the less, the moon itself doesn't orbit the Sun in a perfect Keplerian ellips.

The only way to verify this issue is by monitoring its orbital path.

So, assuming that we can't see the Earth, by monitoring the Noon orbital path we can easily verify that it doesn't orbit the Sun in a perfect Keplerian ellips.

If we ignore that vital information, we could assume that the total mass of that Earth/Moon common center of mass is as big as the moon itself.

So we have neglected completely the Earth mass.Therefore, if we will try to extract the Sun mass directly out of the Moon mass and its none perfect Keplerian ellips, we would make a severe mistake which is proportional to the Earth/moon mass ration.

 

That is not an error of 1%, but a huge error!!!

 

Don't forget that the distance between the Moon/Earth is about 1.5 Sec of light and the Earth/Sun is above 8 Minutes of light.

Therefore, just by looking on the Moon/Sun orbit it might be almost an impossible mission to verify that the moon orbit around the Sun isn't a perfect Keplerian ellips.

However, there is one more issue which can help us in this verification.

It is the moon speed.

As it orbits the Common center of mass, we should easily verify that there are picks in its speed.

If we can detect those picks, we could calculate the time duration that it takes the moon to complete one cycle around that virtual Earth/Moon common of mass.

If we can verify the radius of that cycle, we can extract the value of Earth mass.

With this info, we can easily get the real mass of the Sun.

So that was all about our solar system.

In the same way we have to solve S2 Orbital path around the SMBH.

As its orbit isn't a perfect Keplerian ellips it is clear that there must be at least one level of common center of mass.

First we must monitor S2 speed and see if there are picks in that speed.

Each pick indicates that S2 had completed one cycle around its common center of mass.

If we can verify the radius of that orbit we can extract the total mass of the object which it orbits.

Then, we have to look carefully on that center of mass and verify if it orbits the SMBH is perfect Keplerian ellips.

If it does, than by simple calculation of S2 and the other object mass, we can calculate the real mass of SMBH.

Is it clear by now?

Edited April 3, 2017 by David Levy

[swansont]

 

As you can see, the earth and the moon each trace (almost perfectly) circular orbits about their common centre of mass.

This center of mass orbits the Sun.

Hence, the Earth and the Moon orbits their common center of mass, while this virtual point of center mass orbits the Sun.

So, this virtual common center of Earth/Moon mass point orbits the Sun in a perfect Keplerian ellipse. (Although some people would assume that the Earth orbits the sun in a perfect Keplerian ellipse - and that is a mistake),

However, the Earth is significantly heavier than the Moon, therefore, this Common center of mass is located in its radius.

Therefore, it isn't so big mistake to assume that the Earth orbits the sun in a almost perfect Keplerian ellipse.

Never the less, the moon itself doesn't orbit the Sun in a perfect Keplerian ellips.

The only way to verify this issue is by monitoring its orbital path.

So, assuming that we can't see the Earth, by monitoring the Noon orbital path we can easily verify that it doesn't orbit the Sun in a perfect Keplerian ellips.

If we ignore that vital information, we could assume that the total mass of that Earth/Moon common center of mass is as big as the moon itself.

So we have neglected completely the Earth mass.Therefore, if we will try to extract the Sun mass directly out of the Moon mass and its none perfect Keplerian ellips, we would make a severe mistake which is proportional to the Earth/moon mass ration.

 

That is not an error of 1%, but a huge error!!!

 

 

...

 

Is it clear by now?

 

 

The problem you describe does not exist. You can determine the sun's mass from the orbit without knowing the moon's mass. As long as the mass is small compared to the mass of the central body, it can be ignored.

[Strange]

Never the less, the moon itself doesn't orbit the Sun in a perfect Keplerian ellips.

 

 

True.

http://www.math.nus.edu.sg/aslaksen/teaching/convex.html

 

 

 

If we ignore that vital information, we could assume that the total mass of that Earth/Moon common center of mass is as big as the moon itself.

 

Not sure what you mean by this.

 

If we could (somehow) only see the moon and not the Earth then we would know three things:

 

1. From the orbital period we would know the mass of the Sun.

2. From the shape of the moon's orbit (not exactly an ellipse) we would know that there was another mass present.

3. By measuring the departure of the moon's orbit from a perfect ellipse, we could work out the relative mass of the moon and its invisible partner. (Edit: or, at least, work out the mass of the Earth)

 

This is basically how planets like Neptune and some extra-solar planets were found.

 

 

 

Therefore, if we will try to extract the Sun mass directly out of the Moon mass and its none perfect Keplerian ellips, we would make a severe mistake which is proportional to the Earth/moon mass ration.

 

Can you explain why that is?

 

 

 

Therefore, just by looking on the Moon/Sun orbit it might be almost an impossible mission to verify that the moon orbit around the Sun isn't a perfect Keplerian ellips.

 

If the deviation is too small to measure, then can it have a significant effect? Please show your working.

 

 

In the same way we have to solve S2 Orbital path around the SMBH.

As its orbit isn't a perfect Keplerian ellips it is clear that there must be at least one level of common center of mass.

First we must monitor S2 speed and see if there are picks in that speed.

 

And, as far as I can tell, this is what has been done to estimate the extra mass of other matter that it passes through.

Edited April 3, 2017 by Strange

[David Levy]

O.K.

 

So we agree that the Earth is invisible.

We can only see the Moon and the Sun.

We know the total mass of the Moon, but we need to verify the mass of the Sun.

We also know that moon's orbit around the Sun isn't a perfect Keplerian ellips

 

 

 

1. From the orbital period we would know the mass of the Sun.

2. From the shape of the moon's orbit (not exactly an ellipse) we would know that there was another mass present.

3. By measuring the departure of the moon's orbit from a perfect ellipse, we could work out the relative mass of the moon and its invisible partner. (Edit: or, at least, work out the mass of the Earth)

 

1. No, you can't calculate the Sun mass directly from the moon orbital period without extract the Earth Mass first

2. - That is correct.

3 - Can you please elaborate how can you do it? How the departure of the moon's orbit from a perfect ellipse can give you indication about Earth mass?

 

 

With Regards to S2.

 

You claim correctly: "2. From the shape of the moon's orbit (not exactly an ellipse) we would know that there was another mass present".

However, S2 has no perfect Keplerian ellips, than it is clear indication that there is another mass present.

So, why do we ignore that vital info?

The problem you describe does not exist. You can determine the sun's mass from the orbit without knowing the moon's mass. As long as the mass is small compared to the mass of the central body, it can be ignored.

 

How can we do it?

If we can only see the moon and the Sun, how can we extract the Sun mass without the information about the moon mass? (Please remember - there is no info at all about the Earth!)

Edited April 3, 2017 by David Levy

[swansont]

 

1. No, you can't calculate the Sun mass directly from the moon orbital period without extract the Earth Mass first

 

 

Why not?

 

How can you do it?

If we can only see the moon and the Sun, how can we extract the Sun mass without the information about the moon mass? (Please remember - there is no info at all about the Earth!)

 

You set the mass of the moon to zero in the equation. It has a negligible effect on the answer. Your measurements likely don't have enough precision for this to matter, and even if they do, it matters at the parts per million level.

[David Levy]

Why not?

You set the mass of the moon to zero in the equation. It has a negligible effect on the answer.

 

It seems to me that you assume that the Earth mass is well known.

However, in this example we have no info about the earth mass.

Hence, how can we calculate the sun mass without any info about Earth or moon mass?

[swansont]

 

It seems to me that you assume that the Earth mass is well known.

However, in this example we have no info about the earth mass.

Hence, how can we calculate the sun mass without any info about Earth or moon mass?

 

Answer me this:

 

What is the ratio of the earth mass to the sun's mass?

 

Where does the mass of the earth show up in the orbital equation that would allow you to find the sun's mass? (copy/paste, or even posting a link to the equation should suffice. Something so we can be sure we're looking at the same thing)

[David Levy]

Why not?

 

Very simple:

 

Moon Mass - 7.342×10²² kg

Sun Mass – 1.98855×10³⁰ kg

Earth mass = 5.97219 * 10 ²⁴ kg

Distance : 149,600,000 Km

Please use the following formula to calculate the gravity force between Earth/Sun.

https://www.easycalculation.com/physics/classical-physics/newtons-law.php

 

I have got:

F= 3.5922175 ×10²² N

Now change it to calculate M1 mass based on M2.

However, instead of using the mass of the Earth, please use Moon mass.

The outcome is that the Sun mass is:

1.6409299575189575e+24

That is a severe mistake!!!

 

Therefore, we can't calculate the real value of the Sun without verifying first the Earth mass.

Edited April 3, 2017 by David Levy

[swansont]

 

Very simple:

 

Moon Mass - 7.342×10²² kg

Sun Mass – 1.98855×10³⁰ kg

Earth mass = 5.97219 * 10 ²⁴ kg

Distance : 149,600,000 Km

Please use the following formula to calculate the gravity force between Earth/Sun.

https://www.easycalculation.com/physics/classical-physics/newtons-law.php

 

I have got:

F= 3.5922175 ×10²²N

Now change it to calculate M1 mass based on M2.

However, instead of using the mass of the Earth, please use Moon mass.

The outcome is that the Sun mass is be:

1.6409299575189575e+24

That is a severe mistake!!!

 

Therefore, we can't calculate the real value of the Sun without verifying first the Earth mass.

 

 

Is that the only way to do the calculation? No. Nobody asked for the force.

 

https://en.wikipedia.org/wiki/Gravitational_two-body_problem

 

Scroll down to "for elliptical orbits" for the equation

 

The orbital parameters needed for an elliptical orbit are the semimajor axis, a, the orbital period T, and u (mu), which is G(m₁ + m₂)

 

So we can solve for m₁ + m₂

 

Since one of those is very small, we can determine the sun's mass to a precision of several parts in 10^6 or to the precision of the other parameters, whichever is larger.

[David Levy]

Is that the only way to do the calculation? No. Nobody asked for the force.

https://en.wikipedia.org/wiki/Gravitational_two-body_problem

Scroll down to "for elliptical orbits" for the equation

The orbital parameters needed for an elliptical orbit are the semimajor axis, a, the orbital period T, and u (mu), which is G(m₁ + m₂)

So we can solve for m₁ + m₂

Since one of those is very small, we can determine the sun's mass to a precision of several parts in 10^6 or to the precision of the other parameters, whichever is larger.

 

It is stated:

 

"The gravitational two-body problem concerns the motion of two point particles that interact only with each other, due to gravity. This means that influences from any third body are neglected."

So, if S2 orbit was a perfect Keplerian ellips, than yes, we could use this formula.

However, S2 orbit isn't a perfect one. It is also stated clearly that there are some other gravity forces on S2:

https://en.wikipedia.../wiki/S2_(star)

"The motion of S2 is also useful for detecting the presence of other objects near to Sagittarius A*. It is believed that there are thousands of stars, as well as dark stellar remnants (stellar black holes, neutron stars, white dwarfs) distributed in the volume through which S2 moves.

Hence, it seems to me that this formula is not relevant to S2 case.

How can we claim in one hand that "This means that influences from any third body are neglected",

While on the other hand it is stated that: "housands of stars, as well as dark stellar remnants (stellar black holes, neutron stars, white dwarfs) distributed in the volume through which S2 moves"?

How can it work together?

 

Therefore, I would consider it as a mistake.

Edited April 3, 2017 by David Levy

[swansont]

We're talking about the moon, earth and sun at the moment. Until your misunderstanding is cleared up, there's no point in going back to the other discussion.

[Strange]

Hence, how can we calculate the sun mass without any info about Earth or moon mass?

 

 

What is the formula for orbital period?

 

Clue: https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body

[David Levy]

We're talking about the moon, earth and sun at the moment. Until your misunderstanding is cleared up, there's no point in going back to the other discussion.

 

 

O.K.

Let's focus on Moon /Earth / Sun Example

 

In our example we have only eliminated the light on Earth.

However, it is still there. Its gravity has a severe impact on Moon / Sun orbit.

Therefore, it is clear to me that we can't use this formula for this example as it is stated: "This means that influences from any third body are neglected",

Edited April 3, 2017 by David Levy

[Strange]

 

 

O.K.

Let's focus on Moon /Earth / Sun Example

 

In our example we have only eliminated the light on Earth.

However, it is still there. Its gravity has a severe impact on Moon / Sun orbit.

Therefore, it is clear to me that we can't use this formula for this example as it is stated: "This means that influences from any third body are neglected",

 

 

 

Here are three simple questions requiring very little mathematical ability. You should be able to answer all three.

 

1. What is the orbital period of the moon around the sun?

 

2. What is the relationship between orbital period and mass?

 

3. From 1 and 2 (and nothing else) what is the mass of the sun?

 

After that, we can look into how to determine the mass of the Earth...

Edited April 3, 2017 by Strange

[David Levy]

Thanks

 

That is clear

https://imagine.gsfc.nasa.gov/features/yba/CygX1_mass/gravity/sun_mass.html

 

So,we can calculate Sun mass directly from Earth/Sun orbit without any info about Earth or moon mass.

 

However, it must be perfect Keplerian ellipse.

How that impact S2?

[swansont]

 

 

O.K.

Let's focus on Moon /Earth / Sun Example

 

In our example we have only eliminated the light on Earth.

However, it is still there. Its gravity has a severe impact on Moon / Sun orbit.

Therefore, it is clear to me that we can't use this formula for this example as it is stated: "This means that influences from any third body are neglected",

 

 

How much error is introduced by this? Because it's clear to me that this isn't a problem unless you need precision at the 10^-6 level.

[Janus]

Thanks

 

That is clear

https://imagine.gsfc.nasa.gov/features/yba/CygX1_mass/gravity/sun_mass.html

 

So,we can calculate Sun mass directly from Earth/Sun orbit without any info about Earth or moon mass.

 

However, it must be perfect Keplerian ellipse.

How that impact S2?

The difference between the Moon-Earth example and S2 is that for the Moon, the total mass effecting its trajectory at an given moment is a constant. And while the Moon's actual trajectory relative to the Sun doesn't perfectly follow the Keplerian Eclipse it does vary around one. (average out the path of the moon and you get a Keplerian Eclipse.

 

 

With S2, the mass effecting its trajectory at any given point of its orbit is not a constant, as this image demonstrates:

 

The red circle is Sag A, the larger pink circle is the extra mass distributed around Sag A. The Blue ellipse is S2's orbit. At point A,the mass effecting S2's trajectory is that of Sag A plus that part of the extra mass closer to Sag A than it is (enclosed by the innermost yellow circle. At either of the points B it is Sag A plus the mass enclosed by the next yellow circle. The same holds for points C and D. Thus as S2 moves in and out from Sag A in its orbit, the mass effecting it changes. This causes a departure from a Keplerian Eclipse in a number of ways.