1 INTRODUCTION

Until recently, among the many double asteroids, there was known no a pair that: (i) would have double synchronous rotation, and (ii) wherein both components would have the same masses and sizes. Since tidal forces are inversely proportional to the cube of distances, tidal captures should be sought among pairs of asteroids close to each other. In the Solar System, systems of dwarf planets with double synchronous rotation are well known (Pluto-Charon [1], Eris-Dysnomia [2]) while the second point is not fulfilled for them since the masses of the components differ markedly. There are many examples of double asteroids with complete synchronous rotation both in the classical belt [3–6] and in the Kuiper belt, see, for example, [7] and the literature cited there, but also, in these systems, the condition (ii) is not fulfilled.

All the more curious is the fact that in 2005, one pair of asteroids with synchronous rotation and approximately the same masses and figures of both components was discovered [8]. This unique pair of asteroids, which aroused great interest among researchers [9, 10], was assigned number (190166) 2005 UP156. Doppler images of binary asteroid (190166) 2005 UP156 were obtained by the Arecibo radio telescope. Both components of this pair are elongated, almost identical in size, and their major axes are presumably aligned.

The origin and evolution of asteroid (190166) 2005 UP156 is of great interest, but no less important is the problem of the equilibrium figures of the asteroids in this pair. The very fact of the existence of double asteroid (190166) 2005 UP156 made it possible to pose and then solve in the final analytical form a complex problem about the equilibrium figures of asteroids in a state of close tidal interlocking. Here, besides its own gravity and centrifugal forces, one should take into account, at least in a tidal approximation, the attraction from the second ellipsoidal asteroid (Fig. 1). Previously, the problem of tidal capture of celestial bodies within the framework of the Roche problem was considered by astronomer Darwin while Chandrasekhar did this in more detail and in a specific mathematical form [11].

Fig. 1
figure 1

Cross sections of figures of a double asteroid. Both bodies have a congruent ellipsoidal shape and the same masses, and \({{a}_{1}}\) and \({{a}_{3}}\) are the semi-axes of the figure. Asteroids orbit around point \(O\) and the distance between the centers of mass \(O_{1}^{'}O_{2}^{'}\) equals to \(D\).

Full size image

Here, we continue to study the equilibrium figures of asteroids in a synchronous pair using the example of double asteroid (190166) 2005 UP156. Our method for solving hydrodynamic equilibrium equations is somewhat different from Chandrasekhar’s approach [11]. Our emphasis is on solving hydrodynamic equilibrium equations using the level surface method. In Section 2, the problem statement is given, the basic equilibrium equations are derived, and the spatial shape of both triaxial asteroids is found using an analytical-numerical method. In Section 3, this method is used to study double asteroid (190166) 2005 UP156. During the research, it was discovered that in the case of close tidal interlocking, the spin rotation of ellipsoidal equilibrium figures can occur not around the small axes, as is usually assumed, but around the middle axes of the ellipsoids. In Subsection 4.1, we discuss possible physical reasons for such rotation of asteroids. In Subsection 4.2, we substantiate the conclusion that, with the currently known parameters, the system of two asteroids (190166) 2005 UP156 is nonequilibrium.

2 PROBLEM FORMULATION AND DERIVATION OF THE BASIC EQUILIBRIUM EQUATIONS

According to observations [10], the asteroids of pair (190166) 2005 UP156 have approximately the same masses and a congruent (identical) elongated geometric shape, see Fig. 1.

The asteroids of this pair are sufficiently close to each other that the tidal humps that one asteroid produces on the other remain stationary in the rotating frame of reference. In combination with self-gravity and centrifugal force, tidal forces form the body’s equilibrium figure. The semimajor axes of the equilibrium figures lie approximately on the same line, and the asteroids always face each other with the same pointed ends. The period of axial and orbital rotation of asteroids is the same and equals \({{T}_{{{\text{rot}}}}} = {{40}^{{\text{h}}}_{.}}542\) [10].

We consider the equilibrium figures of the asteroids in this pair, assuming that both asteroids have the same ellipsoidal shape:

$$\frac{{x_{1}^{2}}}{{a_{1}^{2}}} + \frac{{x_{2}^{2}}}{{a_{2}^{2}}} + \frac{{x_{3}^{2}}}{{a_{3}^{2}}} = 1,$$
(1)

and their masses are the same: \({{M}_{1}} = {{M}_{2}} = M.\) Rotation (spin and orbital) occurs at the same angular velocity \({{\Omega }}\) around axes with index 3. We introduce two systems of rectangular Cartesian coordinates. We orient system \(O{{x}_{1}}{{x}_{2}}{{x}_{3}}\) with the origin at the center of mass of the first (left) body so that the axis \(O{{x}_{1}}\) was directed to the center of the second body while axis \(O{{x}_{3}}\) was directed parallel to rotation axis \({{\Omega }}\). Origin of second coordinate system \(O{\kern 1pt} '{\kern 1pt} x_{1}^{'}x_{2}^{'}x_{3}^{'}\) is compatible with the center of mass of the second (right) asteroid, its axes will be oriented in the same way as axes \(O{{x}_{1}}{{x}_{2}}{{x}_{3}}\) relative to the first mass. Both coordinate systems are related by the relations:

$${{x}_{1}} + x_{1}^{'} = D,\quad {{x}_{2}} = - x_{2}^{'},\quad {{x}_{3}} = x_{3}^{'}.$$
(2)

In the rotating reference system at angular velocity \({{\Omega }}\), the balance of forces for the mass elements of the first body is described by the equation:

$$\frac{{\partial p}}{{\partial {{x}_{i}}}} = \rho \frac{\partial }{{\partial {{x}_{i}}}}\left\{ {\varphi + \varphi {\kern 1pt} ' + \frac{{{{{{\Omega }}}^{2}}}}{2}\left[ {{{{\left( {{{x}_{1}} - \frac{D}{2}} \right)}}^{2}} + x_{2}^{2}} \right]} \right\}.$$
(3)

Here, term \(\frac{{{{{{\Omega }}}^{2}}}}{2}\left[ {{{{\left( {{{x}_{1}} - \frac{D}{2}} \right)}}^{2}} + x_{2}^{2}} \right]\) is the potential of centrifugal forces acting on the test point when the asteroid rotates relative to the center of mass. (3) also includes \(\rho\) is the density, which we consider constant, \(p\left( x \right)\) is the pressure, and \(\varphi \left( x \right)\) is the potential on the internal point of the first body from itself:

$$\varphi \left( {{{x}_{i}}} \right) = \pi G\rho \left( {I - {{A}_{1}}x_{1}^{2} - {{A}_{2}}x_{2}^{2} - {{A}_{3}}x_{3}^{2}} \right),$$
$${{A}_{i}} = {{a}_{1}}{{a}_{2}}{{a}_{3}}\int\limits_0^\infty {\frac{{du}}{{\left( {a_{i}^{2} + u} \right)\Delta \left( u \right)}}} ,$$
(4)
$$\Delta \left( u \right) = \sqrt {\left( {a_{1}^{2} + u} \right)\left( {a_{2}^{2} + u} \right)\left( {a_{3}^{2} + u} \right)} {\kern 1pt} .$$

Term \(\varphi {\kern 1pt} '\left( x \right)\) in formula (3) is the potential for trial point \({{x}_{i}}\) from the second (external) body. In the integral form, this external potential of a homogeneous triaxial ellipsoid that is written in the coordinate system \(O{\kern 1pt} '{\kern 1pt} x_{1}^{'}x_{2}^{'}x_{3}^{'}\) is set by the formula:

$$\begin{gathered} \varphi {\kern 1pt} '(x_{i}^{'}) = \pi G\rho {{a}_{1}}{{a}_{2}}{{a}_{3}} \\ \times \;\int\limits_\lambda ^\infty {\left( {1 - \frac{{x{{{_{1}^{'}}}^{2}}}}{{a_{1}^{2} + u}} - \frac{{x{{{_{2}^{'}}}^{2}}}}{{a_{2}^{2} + u}} - \frac{{x{{{_{3}^{'}}}^{2}}}}{{a_{3}^{2} + u}}} \right)} \frac{{du}}{{\Delta \left( u \right)}}. \\ \end{gathered} $$
(5)

Here, \(\lambda\) is the ellipsoidal coordinate of the trial point, which is the largest (for an external point, positive) root of the cubic equation:

$$\frac{{x{{{_{1}^{'}}}^{2}}}}{{a_{1}^{2} + \lambda }} + \frac{{x{{{_{2}^{'}}}^{2}}}}{{a_{2}^{2} + \lambda }} + \frac{{x{{{_{3}^{'}}}^{2}}}}{{a_{3}^{2} + \lambda }} = 1$$
(6)

while \({{\Delta }}\left( u \right)\) is given in (4).

In this problem, we take into account the external potential at the points of the left ellipsoid in the tidal approximation, and with the required accuracy (in the quadratic approximation) we will write:

$$\begin{gathered} \varphi {\kern 1pt} '(x_{1}^{'},x_{2}^{'},x_{3}^{'}) \approx \varphi {\kern 1pt} '\left( {D,0,0} \right) + ({{x}_{1}} - D) \\ \times \;{{\left. {\frac{{\partial \varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x_{1}^{'}}}} \right|}_{{D,0,0}}} + \frac{1}{2}{{({{x}_{1}} - D)}^{2}}{{\left. {\frac{{{{\partial }^{2}}\varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x{{{_{1}^{'}}}^{2}}}}} \right|}_{{D,0,0}}} \\ + \;\frac{1}{2}x_{2}^{2}{{\left. {\frac{{{{\partial }^{2}}\varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x{{{_{2}^{'}}}^{2}}}}} \right|}_{{D,0,0}}} + \frac{1}{2}x_{3}^{2}{{\left. {\frac{{{{\partial }^{2}}\varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x{{{_{3}^{'}}}^{2}}}}} \right|}_{{D,0,0}}}. \\ \end{gathered} $$
(7)

Substituting this tidal potential \(\varphi {\kern 1pt} '{\kern 1pt} (x_{1}^{'},x_{2}^{'},x_{3}^{'})\) into equilibrium equation (3) and again taking into account relations between the coordinates (2), we obtain:

$$\begin{gathered} \frac{{\partial p}}{{\partial {{x}_{i}}}} = \rho \frac{\partial }{{\partial {{x}_{i}}}}\left\{ {\varphi + \frac{{{{\Omega }^{2}}}}{2}\left( {x_{1}^{2} + x_{2}^{2}} \right) + \frac{1}{2}\sum\limits_{j = 1}^3 {x_{j}^{2}{{{\left. {\frac{{{{\partial }^{2}}\varphi '}}{{\partial x{{{_{2}^{'}}}^{2}}}}} \right|}}_{{D,0,0}}}} } \right. \\ \left. {\mathop - \limits_{\mathop {}\limits_{}^{} }^{} \;{{x}_{1}}\left[ {{{{\left. {\frac{{\partial \varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x_{1}^{'}}}} \right|}}_{{D,0,0}}} + \frac{{{{\Omega }^{2}}}}{2}D} \right]} \right\}. \\ \end{gathered} $$
(8)

Since we are interested in forces, we will further determine the potential up to a constant, so constant term \(\frac{{{{{{\Omega }}}^{2}}}}{8}{{D}^{2}}\) in (8) was immediately rejected.

For a pair of rotating gravitating asteroids to be in relative equilibrium, the distance between their centers must remain unchanged, so the angular velocity of rotation will satisfy the equation:

$${{\left. {{{\Omega }^{2}} = - \frac{2}{D}\frac{{\partial \varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x_{1}^{'}}}} \right|}_{{D,0,0}}}.$$
(9)

We explain: only if condition (9) is satisfied, the expression in square brackets on the right side of (8) will vanish and, importantly, linear term \({{x}_{1}}\) disappears in the equilibrium equation. Then, equilibrium equation (8) takes the form:

$$\frac{{\partial p}}{{\partial {{x}_{i}}}} = \rho \frac{\partial }{{\partial {{x}_{i}}}}\Phi \left( {x,{{\Omega }^{2}}} \right),\quad \left( {i = 1,2,3} \right).$$
(10)

Here,

$$\begin{gathered} \Phi \left( {x,{{\Omega }^{2}}} \right) = \varphi \left( x \right) + \frac{1}{2}x_{1}^{2}\left( {{{{\left. {{{\Omega }^{2}} + \frac{{{{\partial }^{2}}\varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x{{{_{1}^{'}}}^{2}}}}} \right|}}_{{D,0,0}}}} \right) \\ + \;\frac{1}{2}x_{2}^{2}\left( {{{{\left. {{{\Omega }^{2}} + \frac{{{{\partial }^{2}}\varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x{{{_{2}^{'}}}^{2}}}}} \right|}}_{{D,0,0}}}} \right) + \frac{1}{2}x_{3}^{2}{{\left. {\frac{{{{\partial }^{2}}\varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x{{{_{3}^{'}}}^{2}}}}} \right|}_{{D,0,0}}} \\ \end{gathered} $$
(11)

is the total (gravitational plus centrifugal) potential inside the considered ellipsoid.

Next, we find from (6) the auxiliary derivative:

$$\frac{{\partial \lambda }}{{\partial x_{1}^{'}}} = \frac{{2x_{1}^{'}}}{{\left( {a_{1}^{2} + \lambda } \right)\sum\limits_1^3 {\frac{{x_{i}^{'2}}}{{{{{\left( {a_{i}^{2} + \lambda } \right)}}^{2}}}}} }}.$$
(12)

Since the ellipsoidal coordinate at the point \(\left( {D,0,0} \right)\) is \(\lambda = {{D}^{2}} - a_{1}^{2},\) then formula (12) produces:

$${{\left. {\frac{{\partial \lambda }}{{\partial x_{1}^{'}}}} \right|}_{{D,0,0}}} = 2D.$$
(13)

Now differentiating potential (5) and taking into account (13), we find that

$${{\left. {\frac{{\partial \varphi {\kern 1pt} '{\kern 1pt} }}{{\partial {{x}_{1}}}}} \right|}_{{D,0,0}}} = - 2\pi G\rho {{\bar {A}}_{1}}D;\quad {{\left. {\frac{{\partial \varphi {\kern 1pt} '{\kern 1pt} }}{{\partial {{x}_{2}}}}} \right|}_{{D,0,0}}} = {{\left. {\frac{{\partial \varphi {\kern 1pt} '{\kern 1pt} }}{{\partial {{x}_{3}}}}} \right|}_{{D,0,0}}} = 0;$$
$${{\left. {\frac{{{{\partial }^{2}}\varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x_{1}^{2}}}} \right|}_{{D,0,0}}} = - 2\pi G\rho {{\bar {A}}_{1}} + \frac{{4\pi G\rho {{a}_{1}}{{a}_{2}}{{a}_{3}}}}{{\Delta \left( \lambda \right)}}$$
$$ = 2\pi G\rho \left( {{{{\bar {A}}}_{2}} + {{{\bar {A}}}_{3}}} \right) > 0;$$
(14)
$${{\left. {\frac{{{{\partial }^{2}}\varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x_{2}^{2}}}} \right|}_{{D,0,0}}} = - 2\pi G\rho {{\bar {A}}_{2}} < 0;$$
$${{\left. {\frac{{{{\partial }^{2}}\varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x_{3}^{2}}}} \right|}_{{D,0,0}}} = - 2\pi G\rho {{\bar {A}}_{3}} < 0.$$

Here, we indicated

$${{\bar {A}}_{i}} = {{a}_{1}}{{a}_{2}}{{a}_{3}}\int\limits_{{{D}^{2}} - a_{1}^{2}}^\infty {\frac{{du}}{{\left( {a_{i}^{2} + u} \right)\Delta \left( u \right)}}} ;$$
$$\Delta \left( \lambda \right) = D\sqrt {\left( {a_{2}^{2} + \lambda } \right)\left( {a_{3}^{2} + \lambda } \right)} $$
(15)
$$ = D\sqrt {\left( {a_{2}^{2} - a_{1}^{2} + {{D}^{2}}} \right)\left( {a_{3}^{2} - a_{1}^{2} + {{D}^{2}}} \right)} $$

and took into account that to check coefficients \({{\bar {A}}_{i}}\), the identity must hold (which is not difficult to prove):

$${{\bar {A}}_{1}} + {{\bar {A}}_{2}} + {{\bar {A}}_{3}} = \frac{{2{{a}_{1}}{{a}_{2}}{{a}_{3}}}}{{\Delta \left( \lambda \right)}}.$$
(16)

Multiplying hydrostatic equilibrium equation (10) by \(x_{i}^{2}\;\left( {i = 1,2,3} \right)\) and integrating over the volume of the first figure, we obtain tensor virial equations of relative equilibrium [11]:

$${{W}_{{11}}} + \left( {{{\Omega }^{2}} + {{{\left. {\frac{{{{\partial }^{2}}\varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x_{1}^{2}}}} \right|}}_{{D,0,0}}}} \right){{I}_{{11}}}$$
$$ = {{W}_{{22}}} + \left( {{{\Omega }^{2}} + {{{\left. {\frac{{{{\partial }^{2}}\varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x_{2}^{2}}}} \right|}}_{{D,0,0}}}} \right){{I}_{{22}}}$$
(17)
$$ = {{W}_{{33}}} + \left( {{{{\left. {\frac{{{{\partial }^{2}}\varphi {\kern 1pt} '{\kern 1pt} }}{{\partial x_{3}^{2}}}} \right|}}_{{D,0,0}}}} \right){{I}_{{33}}} = - \Pi {\kern 1pt} .$$

Here, \({{W}_{{ij}}}\) and \({{I}_{{ij}}}\) are the components of the gravitational energy tensor and the figure inertia tensor, respectively. Considering this figure to be a homogeneous triaxial ellipsoid with surface (1), we have:

$$\begin{gathered} {{W}_{{11}}} = - 2\pi G\rho {{A}_{1}}{{I}_{{11}}},\quad {{W}_{{22}}} = - 2\pi G\rho {{A}_{2}}{{I}_{{22}}}, \\ {{W}_{{33}}} = - 2\pi G\rho {{A}_{3}}{{I}_{{33}}}, \\ \end{gathered} $$
(18)
$${{I}_{{11}}} = \frac{1}{5}Ma_{1}^{2},\quad {{I}_{{22}}} = \frac{1}{5}Ma_{2}^{2},\quad {{I}_{{33}}} = \frac{1}{5}Ma_{3}^{2}.$$
(19)

Taking these formulas into account, equations (17) can be also written as equalities for the weights of channels in an ellipsoid [12, 13]:

$$\begin{gathered} \frac{{{{p}_{0}}}}{{2\pi G\rho }} = a_{1}^{2}\left( {{{A}_{1}} - \frac{{{{\Omega }^{2}}}}{{2\pi G\rho }} - {{{\bar {A}}}_{2}} - {{{\bar {A}}}_{3}}} \right) \\ = a_{2}^{2}\left( {{{A}_{2}} + {{{\bar {A}}}_{2}} - \frac{{{{\Omega }^{2}}}}{{2\pi G\rho }}} \right) = a_{3}^{2}\left( {{{A}_{3}} + {{{\bar {A}}}_{3}}} \right). \\ \end{gathered} $$
(20)

Combining equations (17), we find:

$$\begin{gathered} {{\Omega }^{2}}\left( {{{I}_{{11}}} + {{I}_{{22}}}} \right) = 2{{W}_{{33}}} - {{W}_{{11}}} - {{W}_{{22}}} \\ - \;\left( {{{I}_{{11}}} + 2{{I}_{{33}}}} \right){{\left. {\frac{{{{\partial }^{2}}\varphi }}{{\partial x_{1}^{2}}}} \right|}_{{D,0,0}}} - \left( {{{I}_{{22}}} + 2{{I}_{{33}}}} \right){{\left. {\frac{{{{\partial }^{2}}\varphi }}{{\partial x_{2}^{2}}}} \right|}_{{D,0,0}}}. \\ \end{gathered} $$
(21)

Substituting into (21) the values from (14), (18), and (19), after reductions by \(\frac{2}{5}\pi G\rho M,\) we obtain the equation for the square of the angular velocity:

$$\begin{gathered} {{\Omega }^{2}}\left( {1 + n_{{12}}^{2}} \right) = {{A}_{1}} + n_{{12}}^{2}{{A}_{2}} - 2n_{{13}}^{2}{{A}_{3}} \\ - \;(1 - n_{{12}}^{2}){{{\bar {A}}}_{2}} - (1 + 2n_{{13}}^{2}){{{\bar {A}}}_{3}}. \\ \end{gathered} $$
(22)

The following notations for dimensionless quantities are adopted here:

$${{\Omega }^{2}} \equiv \frac{{{{\Omega }^{2}}}}{{2\pi G\rho }};\quad {{n}_{{12}}} = \frac{{{{a}_{2}}}}{{{{a}_{1}}}};\quad {{n}_{{13}}} = \frac{{{{a}_{3}}}}{{{{a}_{1}}}}.$$
(23)

On the other hand, according to (7) and the first formula from (14), in this problem, there must be:

$$\frac{{{{\Omega }^{2}}}}{{2\pi G\rho }} = 2{{\bar {A}}_{1}}.$$
(24)

From a dynamic point of view, this means that the angular velocity of rotation of bodies in a pair is determined by tidal coefficient \({{\bar {A}}_{1}}\). Substituting \(\frac{{{{{{\Omega }}}^{2}}}}{{2\pi G\rho }}\) from (24) to the left side of (22), after some transformations, we obtain the main equation:

$$\begin{gathered} 2\left( {1 + n_{{12}}^{2}} \right){{{\bar {A}}}_{1}} + (1 - n_{{12}}^{2}){{{\bar {A}}}_{2}} + (1 + 2n_{{13}}^{2}){{{\bar {A}}}_{3}} \\ = {{A}_{1}} + n_{{12}}^{2}{{A}_{2}} - 2n_{{13}}^{2}{{A}_{3}}, \\ \end{gathered} $$
(25)

which should be considered as an implicit equation linking the relations of the semi-axes of the equilibrium figure of the ellipsoid: \({{n}_{{12}}} = \frac{{{{a}_{2}}}}{{{{a}_{1}}}}\) and \({{n}_{{13}}} = \frac{{{{a}_{3}}}}{{{{a}_{1}}}}.\)

3 NUMERICAL CALCULATIONS FOR BINARY ASTEROID (190166) 2005 UP156

First of all, we note that there is some uncertainty in the parameters of asteroid (190166) 2005 UP156 obtained from observations. Thus, when determining the size of each of the asteroids of this pair in [9], the diameter length is estimated at \(d \approx \) 1.05 ± 0.02 km, and in [10] a noticeably lower value is given \(d \approx \) 900 m. Even less certain in [10] is the estimate of the distance between the centers of mass of asteroids \(D.\) In [9], this important parameter of the system was not assessed in any way, but in [10], unfortunately, there is some confusion between the concepts of the semi-major axis of the orbit and total distance \(D\). In this situation, as a first approximation, we took the value of \(D = \) 5.4 km. We also pay attention to the fact that in [10], the asteroid density value is estimated by the value of \(\rho \approx \) 1.6 g/cm3, for other works, \(\rho \approx \) 1.8 g/cm3.

In the calculations, we used the following parameters of double asteroid (190166) 2005 UP156:

$$\begin{gathered} {{a}_{1}} \approx 450~\;{\text{m}},\quad {{a}_{2}} \approx 300\;{\text{m}},\quad \frac{{{{a}_{1}}}}{{{{a}_{2}}}} = 1.5, \\ D \approx 12{{a}_{{\text{1}}}},\quad {\text{and}}\quad {{T}_{{{\text{rot}}}}} \approx {{40}^{{\text{h}}}_{.}}542. \\ \end{gathered} $$
(26)

There was compiled a calculation program using the formulas specified in Section 2. The calculations were aimed to find the third semi-axis of the ellipsoid \({{a}_{3}},\) volume of each asteroid \(V = \frac{4}{3}\pi {{a}_{1}}{{a}_{2}}{{a}_{3}},\) and then estimate their density \(\rho \). To do this, it was necessary to calculate both the coefficients of the internal potential of the ellipsoid \(\left( {{{A}_{1}},{{A}_{2}},{{A}_{3}}} \right)\) and tidal potential coefficients \(\left( {{{{\bar {A}}}_{1}},{{{\bar {A}}}_{2}},{{{\bar {A}}}_{3}}} \right)\) from (15). Since important parameter \(D\) of the system was given in [10] with some uncertainty, we carry out calculations for different values of this quantity in the interval:

$$7 \leqslant \frac{D}{{{{a}_{1}}}} \leqslant 12.$$
(27)

The main calculation data are given in Table 1 and shown in Fig. 2.

Table 1.   Properties of ellipsoidal models in binary asteroid (190166) 2005 UP156. Designations of all quantities are given in the text
Full size table
Fig. 2.
figure 2

Dependence of the density of matter in double asteroid (190166) 2005 UP156 on the D/a1 normalized distance between the centers of mass.

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4 DISCUSSION

There are several noteworthy features in the above obtained results.

4.1 On the Issue of Rotation of Ellipsoids Around the Middle Axis

In the second column of Table 1, the calculation results of third semi-axis \({{a}_{3}}\) in asteroid pairs (190166) 2005 UP156 are given for various distances \(D\) between their centers of mass. As can be seen, the third semi-axis is larger than the second semi-axis:

$${{a}_{3}} > {{a}_{2}},$$
(28)

and the difference of \({{a}_{3}} - {{a}_{2}}\) when calculating in the interval for \(D \in \left( {12{-} 7} \right)\) depends very little on the distance between asteroids (it decreases by \(3\% \) only). This means that the spin rotation of ellipsoidal equilibrium figures occurs not around the small axes, as is usually assumed in the theory of equilibrium figures, but around the middle axes of the ellipsoids.

In this regard, we note the following. As is known, classical Maclaurin spheroids and triaxial Jacobi ellipsoids always rotate around the minor axis. However, these are figures of relative balance. Under certain conditions, ellipsoids with internal flows can also rotate around the middle axes. A note about such ellipsoids is in Chandrasekhar’s book [11, p. 174]. Besides, ellipsoids with rotation around the middle axis, which exist only with sufficiently strong countercurrents, were discussed in detail in Kondratyev’s monographs [12, 13, p. 372]. Here, we will limit ourselves to only brief explanations.

Inside a liquid gravitating ellipsoid, there is internal velocity field \(u\left( {{{x}_{i}}} \right)\) that is linear in coordinates, rotor, from which \(\zeta = \operatorname{curl} u\) does not depend on coordinates. We consider equilibrium figures, in which vectors \(\zeta \) and \({{\Omega }}\) lie on rotation axis \(O{{x}_{3}}.\) These figures are called Riemann S-ellipsoids [11]. If the ratio of the absolute values of these vectors is denoted by \(f = \frac{\zeta }{{{\Omega }}},\) then for ellipsoids with countercurrents, \(f < 0,\) and the Coriolis force on the particle will be directed towards the center of the figure. Considering channels of a single cross section along the principal semi-axes, we find their “weights” (these are integrals of the total force on the liquid particle). The “weights” of these channels in the stationary state of the ellipsoid must be equal to each other. It is important for us to compare the contribution to the “weight” of the second channel from the Coriolis forces and the centrifugal force. From physical considerations, it is clear that in countercurrent flows, the Coriolis force is directed towards the centrifugal force. Moreover, at sufficiently strong countercurrents, the Coriolis force will both balance the centrifugal force and under the condition [11]:

$$f = - \frac{{2 - e_{{12}}^{2}}}{{1 \pm {{e}_{{12}}}}} < - \frac{1}{2}$$
(29)

surpass it. It turns out that at \(f < - \frac{1}{2}\), each sequence of equilibrium figures has at least one ellipsoid rotating around the middle axis.

Besides, the equilibrium figures of ellipsoidal stellar systems can also stably rotate (due to the anisotropy of the velocity dispersion) around the average axes [14, 15].

In the double asteroid problem considered here, the role of the force weakening the centrifugal force along the second channel is played by the tidal force from the second body. We consider the weights of the second and third channels according to formula (20):

$$\frac{{{{p}_{0}}}}{{2\pi G\rho }} = a_{2}^{2}\left( {{{A}_{2}} + {{{\bar {A}}}_{2}} - \frac{{{{\Omega }^{2}}}}{{2\pi G\rho }}} \right) = a_{3}^{2}\left( {{{A}_{3}} + {{{\bar {A}}}_{3}}} \right).$$
(30)

This formula implies that tidal force \({{\bar {A}}_{2}}\) from the second body actually weakens centrifugal force \(\left( { - \frac{{{{{{\Omega }}}^{2}}}}{{2\pi G\rho }}} \right)\).

4.2 On Nonstationarity of a System of Two Asteroids (190166) 2005 UP156

Table 1 and Fig. 2 show the calculation results of the density values in the asteroids of pair (190166) 2005 UP156 for various distances between their centers of mass. As can be seen, the found density values are very different from the approximate estimates given in [9, 10]. We recall that, in [10], the estimate is \(\rho \approx \) 1.6 g/cm3, which is consistent with the fact that asteroid (190166) 2005 UP156 was classified as an S-type, the density of which usually does not exceed \(\rho \approx \) 3 g/cm3. However, our calculations produce density values acceptable for S-type asteroids only for distances of \(D{\text{/}}{{a}_{1}} \approx 7,\) which does not agree with that a value of \(D{\text{/}}{{a}_{1}} \approx 12\), which is given in [10].

To confirm these calculations, we provide a simple example. We consider both asteroids to be homogeneous spheres with radii \(r\) average in volume. At the semi-axes of the ellipsoid found by us (in units \({{a}_{1}}\)), \({{a}_{1}} = 1,\) \({{a}_{2}} = \frac{2}{3},\) and \({{a}_{3}} \approx 0.82\), the average radius will be equal to \(r \approx 0.81766{{a}_{1}}\). From Keplerian formula.

$${{\Omega }^{2}} \approx \frac{{2GM}}{{{{D}^{3}}}},\quad \rho = \frac{{3{{\Omega }^{2}}{{D}^{3}}}}{{8\pi G{{r}^{3}}}},$$
(31)

after substituting here above found \(r,\) as well as \({{\Omega }} = \frac{{2\pi }}{{40.542h}}\) known from observations, and the distance of \(D \approx 12{{a}_{1}}\) between the centers of mass given in [10], we obtain:

$$\rho \approx 10.48\;{\text{g/c}}{{{\text{m}}}^{{\text{3}}}}.$$
(32)

This density value is close to our estimates (see Table 1) and does not fit into the density range for S-type asteroids. Thus, either an estimate of \(D \approx 12{{a}_{1}}\) given in [10] is incorrect (overestimated) or the system of asteroid (190166) 2005 UP156 turns out to be significantly nonequilibrium.

5 CONCLUSIONS

In this study, we posed and solved the problem of equilibrium figures of two liquid masses in a state of tidal interlocking. In this system, two conditions must be met: (i) complete synchronous rotation, and (ii) both bodies must have the same masses and congruent ellipsoidal surfaces. For each figure, besides own gravity and centrifugal forces, the tidal approximation takes into account the attraction from the second body. The spatial form of equilibrium figures as triaxial ellipsoids is determined using the analytical-numerical method. This method was used to study double asteroid (190166) 2005 UP156, which approximately satisfies the initial conditions of the problem. The spin rotation of ellipsoidal equilibrium figures was established to occur not around the small axes, as is usually assumed, but around the middle axes of the ellipsoids. The question of the physical reasons leading to the rotation of ellipsoids around the middle axis was briefly discussed. The study showed that at the parameters known today, the system of two asteroids (190166) 2005 UP156 turns out to be nonequilibrium.