# Rotation curve of the Milky Way and the mass of Sagittarius A*

J. Enrique H. Ramírez
Department of Physics
ABC Federal University
Sao Paulo, Brazil

###### Abstract

In the present work we derive an analytical expression for the mass density of
an object with spherical symmetry, whose corresponding potential allows
obtaining a circular velocity around it that is in agreement with the observed
rotation curve of galaxies. The rotation curve of our galaxy is analyzed,
determining the properties of the central object, its radius and mass whose
value obtained is very close to that reported in the literature.

_K_ eywords MIlky Way $\cdot$ Galaxy rotation curve $\cdot$ Dark matter
$\cdot$ Sagittarius A* $\cdot$ Angular momentum

##  1 Introduction

The problem of the rotation curve of galaxies has generated various theories
and profile models ([3]), to explain the non-correspondence between the theory
and the observational data. The stars and galaxies when rotating around their
center should show a decay in their rotation speed as they move away from
their center. Many and different observations made show the opposite, as they
move away from the center, their speeds increase and then maintain an almost
constant value. The speed of rotation of the stars is much greater than what
is allowed, which does not correspond to the estimated mass of the galaxy,
calculated as the result of the masses of all the stars visible in it. The
most accepted theory refers to the presence of a non-visible mass (dark
matter), with the consideration of which it is possible to explain the non-
decaying shape of the rotation curve. In this work we will consider the
presence of this non-visible mass, whose density will be determined and which
obeys a certain law.

##  2 The model

We will build a density $\rho$ that allows us to explain its properties as
well as its behavior with the objects with which it interacts. We will deal,
in this model, with a central object without rotation and with spherical
symmetry.

The dimension $r$ of the object is obviously a magnitude to consider in the
expression of the density that we want to build. As daily experience shows us,
the density in large objects, such as planets and stars, is not constant, it
depends on which point in the interior of the object we refer to. If we are
close to the surface of the object, its density will be less than if we refer
to any point very close to its center. This leads us to consider that the
density is inversely proportional to the radius of the object, $\rho\propto
1/r^{n}$, where $n$ is an integer to be determined. As we are dealing with
problems related to gravitation, we will also consider a magnitude that
characterizes the interaction of the object of study with other objects around
it, the universal gravitational constant $G$. The other magnitude to consider,
which characterizes the internal properties of the object of study, is its
escape velocity, $v_{es}$.

Thus we can express the density as,

$\displaystyle\rho(r)=k\,G^{x}\,v_{es}^{y}\,r^{n},$ (1)

where $k$ is a dimensionless proportionality constant and $x$, $y$ and $n$
numbers to be determined. A dimensional analysis gives us that $x=-1$, $y=2$
and $n=2$, so,

$\displaystyle\rho(r)=k\frac{v_{es}^{2}}{G\,r^{2}}.$ (2)

As we known, the escape velocity of an object of mass $M$ and radius $a$ is
given by, $v_{es}^{2}=2\,M\,G/a$, thus, we can write the above equation as

$\displaystyle\rho(r)=k\frac{2\,M}{a\,r^{2}}.$ (3)

We will determine the proportionality constant $k$, knowing that the mass
contained in the volume $4/3\,\pi\,a^{3}$ is $M$. Thus, $k=1/8\,\pi$. In this
way we can finally express the density as

$\displaystyle\rho(r)=\frac{1}{3}\,\rho_{obj}\left(\frac{a}{r}\right)^{2},$
(4)

being $\rho_{obj}=3M/4\,\pi a^{3}$.

The mass corresponding to the distribution given by 4 is

$\displaystyle m(r)=M\,\frac{r}{a}.$ (5)

Note that for small distances from the center of the distribution, the density
given by 4 is large but the mass given by 5 is small, but for large distances
the opposite is true.

It should be noted that the above density obtained analytically is it very
similar to that used by [2], obtained through high-resolution N-body
simulations, to study the equilibrium density profiles of dark matter in
halos.

##  3 Potential

Now we are going to consider the potential $\phi$ generated by this mass
distribution. We will consider that this density is not limited to the border
$a$ of the object, but that it extends beyond it. That is, the space that
surrounds our object of study is not empty. In this model we only consider the
mass distribution given by 4 and the vacuum, and that the object is not
rotating.

Solving the Poisson’s equation for this mass distribution, we found that

$\displaystyle\phi(r)=\frac{M\,G}{a}\,\left[-\ln(r/a)-c_{1}\,a/r+c_{2}\right],$
(6)

where by the boundary conditions $\phi(a)=-G\,M/a$ and
$\phi\prime(a)=-G\,M/a^{2}$ we found that $c_{1}=2$ and $c_{2}=1$, thus the
potential will be given by

$\displaystyle\phi(r)=\frac{M\,G}{a}\,\left[-\ln(r/a)-2\,a/r+1\right].$ (7)

##  4 Rotation Velocity

Now we can determine the rotation velocity $v_{rot}$ of some particle of mass
$m$ around our central object, considering that the space around it is not
empty but rather filled with an unknown substance whose density obeys the
distribution given by 4 .

The rotation velocity $v$ can be determined as,

$\displaystyle\frac{v^{2}}{r}=-\frac{d\phi}{dr},$ (8)

thus we find that,

$\displaystyle v=\sqrt{\frac{G\,M}{a}\left(1-\frac{2\,a}{r}\right)},$ (9)

whose graph is shown in (Fig.1).

Figure 1: The curve is valid for $r\geq 2a$ . As shown in the figure, the
circular velocity does not decrease and approaches to $\sqrt{G\,M/a}$ as
$r\to\infty$. Here we set $a=1$ and $GM=0.5$ for illustrative porpouses.

This simple but extremely useful expression will allow us to determine the
main characteristics of the celtral object, such as its mass $M$ and dimension
$a$, as we will see later.

For long distances ($r\gg 2a$),

$\displaystyle v\approx\sqrt{\frac{G\,M}{a}}\left(1-\frac{\,a}{r}\right).$
(10)

##  5 Fitting the Galaxy Rotation Curve and Mass of the Galactic Center

In this section we will show the rotation curves of the Milky Way and how to
find the fundamental characteristics of the central object, its mass($M$) and
radius($a$). The observational data used in this work were taken from [4].

We can rewrite equation 9 in a convenient way

$\displaystyle Y=B-A\,X$ (11)

where $Y=v^{2}$, $A=2G\,M$, $B=G\,M/a$ and $X=1/r$. Determining, from Eq. 11 ,
the constants $A$ and $B$ using the observational data, we were able to find
the mass $M$ and the radius $a$ of the central object.

In the case of our galaxy, the curve is shown in the Fig.2. This does not show
the linear behavior predicted by the model, because it considers that the
central object has a fixed mass $M$ and radius $a$. In the case of galaxies,
this mass varies, due, on the one hand, to the large number of objects between
the central object and the object that orbits and due to the non-visible mass,
whose density distribution is given by4 . However, we note that certain groups
of data have a linear behavior. For this reason, we split the data into five
groups, using the high degree of linearity in each of them as a criterion. The
first group from 1-4, the second from 5-21, the third from 22-41, the fourth
from 42-68 and the last from 69-99. The equations of the lines, according to
equation 11, are:

$Y_{1}=-1.1177\times 10^{25}+7.3641\times 10^{9}$,
$Y_{2}=-1.1543\times 10^{26}+2.3121\times 10^{10}$,
$Y_{3}=-2.3857\times 10^{26}+2.643\times 10^{10}$,
$Y_{4}=-9.3890\times 10^{26}+3.67\times 10^{10}$ and
$Y_{5}=-3.7650\times 10^{27}+6.3970\times 10^{10}$, respectively.

Through them we were able to determine the corresponding masses, being them
$M_{1}=8.44\times^{34}\,kg$,$M_{2}=8.72\times^{35}\,kg$,$M_{3}=1.80\times^{36}\,kg$
and $M_{4}=7.09\times^{36}\,kg$, $M_{5}=2.84\times^{37}\,kg$, whose average is
$M_{SgrA^{*}}=7.65\times 10^{36}\,kg$, value very close to that reported in
[1] (3.85 million solar masses).

The corresponding radius are: $a_{1}=7.55\times 10^{14}\,m$ ,
$a_{2}=2.51\times 10^{15}\,m$ , $a_{3}=4.59\times 10^{15}\,m$ ,
$a_{4}=1.27\times 10^{16}\,m$ , $a_{5}=2.94\times 10^{16}\,m$ whose mean value
is $a_{SgrA^{*}}=9.98\times 10^{15}\,m$ . The observational and fitted curves
are shown in 3 .

$M_{SrgA^{*}}$ and $a_{SrgA^{*}}$ are average values. The best approximation
for these is privided by the two objects closest to the galactic center that
orbit it. On the other hand, the best approximation to the mass and dimension
of the galaxy ($M_{galaxy}$ and $R_{galaxy}$) come from the two most distant
objects that orbit it.

##  6 Angular Momentum Conservation

Since the Lagrangian of the system does not depend explicitly on $\theta$ the
angular momentum $L$ is conserved. It happens that after a certain distance
the speed of rotation remains almost constant, but not the position of the
orbiting object, which increases. It may be asked on account of what angular
momentum is conserved. Let’s see,

$\displaystyle L=m\,v\,d,$ (12)

being $m$ the mass of the star, $v$ its speed of rotation and $d$ its distance
from the galactic center. Since $L$ is constant then, the mass must vary
according to the equation

$\displaystyle
m(r)=\frac{m_{0}\,v_{0}\,r_{0}}{r\,\sqrt{\frac{GM}{a}\left(1-2\,a/r\right)}}.$
(13)

As the star moves away from the center of rotation, it loses mass and with it,
energy. The mass loss mechanism occurs by the well-known nuclear process that
occurs in stars, in addition to the one exposed by conservation of angular
momentum.

##  7 Conclusion

The data obtained by the model was possible assuming the presence of a non-
visible distribution of matter, outside the limits of the central object. The
Eq. 9 is in correspondence with the observational data as it does not present
a decay and it is valid for spiral galaxies or others whose dependence $1/r$
vs $v^{2}$ is given by the Eq. 11 . Since the speed of rotation of the stars
depends on the mass of the galaxy contained in a radius $r$, several rotation
curves are obtained, each of which will reproduce the observational data in a
certain range. A more realistic model should include the rotation of the
central object, this would allow a better fit of the model with the
observational data.

## References

  * [1] Abuter, R., Amorim, A., Bauböck, M., Berger, J., Bonnet, H., Brandner, W., Clénet, Y., Du Foresto, V. C., de Zeeuw, P., Deen, C., et al. Detection of orbital motions near the last stable circular orbit of the massive black hole sgra.  Astronomy & Astrophysics 618 (2018), L10.
  * [2] Navarro, J. F., Frenk, C. S., and White, S. D. A universal density profile from hierarchical clustering.  The Astrophysical Journal 490, 2 (1997), 493.
  * [3] Sofue, Y. Rotation curve of the milky way and the dark matter density.  Galaxies 8, 2 (2020), 37.
  * [4] Sofue, Y., Tutui, Y., Honma, M., Tomita, A., Takamiya, T., Koda, J., and Takeda, Y. Central rotation curves of spiral galaxies.  The Astrophysical Journal 523, 1 (1999), 136.

Figure 2: The figure shows the graph of $1/r$ vs $v^{2}$ according to the data
show in Fig. 4. Because the mass varies with $r$, it does not show a linear
dependence according to the proposed model. Figure 3: The figure shows the
rotation curve obtained from the observational data (red circles) and the best
fit according to our model (blue asterisks). $M_{3}=1.80\times 10^{36}\,Kg$
and $a=4.59\times 10^{15}\,m$ were used. Figure 4: Rotation curve of the Milky
Way