Subfield: Mechanics


Question: Three non-collinear points $P_{1}, P_{2}$ and $P_{3}$, with known masses $m_{1}, m_{2}$ and $m_{3}$, interact with one another through their mutual gravitational forces only; they are isolated in free space and do not interact with any other bodies. Let $\sigma$ denote the axis going through the center-ofmass of the three masses, and perpendicular to the triangle $P_{1} P_{2} P_{3}$. What conditions should the angular velocities $\omega$ of the system (around the axis $\sigma$ ) and the distances:

$$
P_{1} P_{2}=a_{12}, \quad P_{2} P_{3}=a_{23}, \quad P_{1} P_{3}=a_{13}
$$

fulfill to allow the shape and size of the triangle $P_{1} P_{2} P_{3}$ unchanged during the motion of the system, i.e. under what conditions does the system rotate around the axis $\sigma$ as a rigid body?

Solution:
As the system is isolated, its total energy, i.e. the sum of the kinetic and potential energies, is conserved. The total potential energy of the points $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ in the inertial system (i.e. when there are no inertial forces) is equal to the sum of the gravitational potential energies of all the pairs of points $\left(\mathrm{P}_{1}, \mathrm{P}_{2}\right),\left(\mathrm{P}_{2}, \mathrm{P}_{3}\right)$ and $\left(\mathrm{P}_{1}, \mathrm{P}_{3}\right)$. It depends only on the distances $a_{12}, a_{23}$ and $a_{23}$ which are constant in time. Thus, the total potential energy of the system is constant. As a consequence the kinetic energy of the system is constant too. The moment of inertia of the system with respect to the axis $\sigma$ depends only on the distances from the points $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ to the axis $\sigma$ which, for fixed $a_{12}, a_{23}$ and $a_{23}$ do not depend on time. This means that the moment of inertia $I$ is constant. Therefore, the angular velocity of the system must also be constant:

$$
\omega=\text { const. }
\tag{1}
$$

This is the first condition we had to find. The other conditions will be determined by using three methods described below. However, prior to performing calculations, it is desirable to specify a convenient coordinates system in which the calculations are expected to be simple.

Let the positions of the points $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ be given by the vectors $\mathbf{r}_{1}, \mathbf{r}_{2}$ and $\mathbf{r}_{3}$. For simplicity we assume that the origin of the coordinate system is localized at the center of mass of the points $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ and that all the vectors $\mathbf{r}_{1}, \mathbf{r}_{2}$ and $\mathbf{r}_{3}$ are in the same coordinate plane, e.g. in the plane $(x, y)$. Then the axis $\sigma$ is the axis $z$.

In this coordinate system, according to the definition of the center of mass, we have:

$$
m_{1} \mathbf{r}_{1}+m_{2} \mathbf{r}_{2}+m_{3} \mathbf{r}_{2}=0
\tag{2}
$$


Consider the point $\mathrm{P}_{1}$ with the mass $m_{1}$. The points $\mathrm{P}_{2}$ and $\mathrm{P}_{3}$ act on it with the forces:

$$
\mathbf{F}_{21}=G \frac{m_{1} m_{2}}{a_{12}^{3}}\left(\mathbf{r}_{2}-\mathbf{r}_{1}\right), 
\tag{3}
$$
$$
\mathbf{F}_{31}=G \frac{m_{1} m_{3}}{a_{13}^{3}}\left(\mathbf{r}_{3}-\mathbf{r}_{1}\right) .
\tag{4}
$$

where $G$ denotes the gravitational constant.

In the inertial frame the sum of these forces is the centripetal force

$$
\mathbf{F}_{r 1}=-m_{1} \omega^{2} \mathbf{r}_{1}
$$

which causes the movement of the point $\mathrm{P}_{1}$ along a circle with the angular velocity $\omega$. (The moment of this force with respect to the axis $\sigma$ is equal to zero.) Thus, we have:

$$
\mathbf{F}_{21}+\mathbf{F}_{31}=\mathbf{F}_{r 1}
\tag{5}
$$

In the non-inertial frame, rotating around the axis $\sigma$ with the angular velocity $\omega$, the sum of the forces (3), (4) and the centrifugal force



$$
\mathbf{F}_{r 1}^{\prime}=m_{1} \omega^{2} \mathbf{r}_{1}
$$

should be equal to zero:

$$
\mathbf{F}_{21}+\mathbf{F}_{31}+\mathbf{F}_{r 1}^{\prime}=0
\tag{6}
$$

(The moment of this sum with respect to any axis equals to zero.)

The conditions (5) and (6) are equivalent. They give the same vector equality:

$$
G \frac{m_{1} m_{2}}{a_{12}^{3}}\left(\mathbf{r}_{2}-\mathbf{r}_{1}\right)+G \frac{m_{1} m_{3}}{a_{13}^{3}}\left(\mathbf{r}_{3}-\mathbf{r}_{1}\right)+m_{1} \omega^{2} \mathbf{r}_{1}=0,
\tag{7'}
$$
$$
G \frac{m_{1}}{a_{12}^{3}} m_{2} \mathbf{r}_{2}+G \frac{m_{1}}{a_{13}^{3}} m_{3} \mathbf{r}_{3}+m_{1} \mathbf{r}_{1}\left(\omega^{2}-\frac{G m_{2}}{a_{12}^{3}}-\frac{G m_{3}}{a_{13}^{3}}\right)=0
\tag{7''}
$$

From the formula (2), we get:

$$
m_{2} \mathbf{r}_{2}=-m_{1} \mathbf{r}_{1}-m_{3} \mathbf{r}_{3}
\tag{8}
$$

Using this relation, we write the formula (7) in the following form:

$$
G \frac{m_{1}}{a_{12}^{3}}\left(-m_{1} \mathbf{r}_{1}-m_{3} \mathbf{r}_{3}\right)+G \frac{m_{1}}{a_{13}^{3}} m_{3} \mathbf{r}_{3}+m_{1} \mathbf{r}_{1}\left(\omega^{2}-\frac{G m_{2}}{a_{12}^{3}}-\frac{G m_{3}}{a_{13}^{3}}\right)=0,
$$

i.e.

$$
\mathbf{r}_{1} m_{1}\left(\omega^{2}-\frac{G m_{2}}{a_{12}^{3}}-\frac{G m_{3}}{a_{13}^{3}}-\frac{G m_{1}}{a_{12}^{3}}\right)+\mathbf{r}_{3}\left(\frac{1}{a_{13}^{3}}-\frac{1}{a_{12}^{3}}\right) G m_{1} m_{3}=0 .
$$

The vectors $\mathbf{r}_{1}$ and $\mathbf{r}_{3}$ are non-collinear. Therefore, the coefficients in the last formula must be equal to zero:

$$
\begin{gathered}
\left(\frac{1}{a_{13}^{3}}-\frac{1}{a_{12}^{3}}\right) G m_{1} m_{3}=0 \\
m_{1}\left(\omega^{2}-\frac{G m_{2}}{a_{12}^{3}}-\frac{G m_{3}}{a_{13}^{3}}-\frac{G m_{1}}{a_{12}^{3}}\right)=0 .
\end{gathered}
$$

The first equality leads to:

$$
\frac{1}{a_{13}^{3}}=\frac{1}{a_{12}^{3}}
$$

and hence,

$$
a_{13}=a_{12} \text {. }
$$

Let $a_{13}=a_{12}=a$. Then the second equality gives:

$$
\omega^{2} a^{3}=G M
\tag{9}
$$

where

$$
M=m_{1}+m_{2}+m_{3}
\tag{10}
$$

denotes the total mass of the system.



In the same way, for the points $\mathrm{P}_{2}$ and $\mathrm{P}_{3}$, one gets the relations:

a) the point $\mathrm{P}_{2}$ :

$$
a_{23}=a_{12} ; \quad \omega^{2} a^{3}=G M
$$

b) the point $\mathrm{P}_{3}$ :

$$
a_{13}=a_{23} ; \quad \omega^{2} a^{3}=G M
$$

Summarizing, the system can rotate as a rigid body if all the distances between the masses are equal:

$$
a_{12}=a_{23}=a_{13}=a,
\tag{11}
$$

the angular velocity $\omega$ is constant and the relation (9) holds.
As the system is isolated, its total energy, i.e. the sum of the kinetic and potential energies, is conserved. The total potential energy of the points $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ in the inertial system (i.e. when there are no inertial forces) is equal to the sum of the gravitational potential energies of all the pairs of points $\left(\mathrm{P}_{1}, \mathrm{P}_{2}\right),\left(\mathrm{P}_{2}, \mathrm{P}_{3}\right)$ and $\left(\mathrm{P}_{1}, \mathrm{P}_{3}\right)$. It depends only on the distances $a_{12}, a_{23}$ and $a_{23}$ which are constant in time. Thus, the total potential energy of the system is constant. As a consequence the kinetic energy of the system is constant too. The moment of inertia of the system with respect to the axis $\sigma$ depends only on the distances from the points $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ to the axis $\sigma$ which, for fixed $a_{12}, a_{23}$ and $a_{23}$ do not depend on time. This means that the moment of inertia $I$ is constant. Therefore, the angular velocity of the system must also be constant:

$$
\omega=\text { const. }
\tag{1}
$$

This is the first condition we had to find. The other conditions will be determined by using three methods described below. However, prior to performing calculations, it is desirable to specify a convenient coordinates system in which the calculations are expected to be simple.

Let the positions of the points $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ be given by the vectors $\mathbf{r}_{1}, \mathbf{r}_{2}$ and $\mathbf{r}_{3}$. For simplicity we assume that the origin of the coordinate system is localized at the center of mass of the points $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ and that all the vectors $\mathbf{r}_{1}, \mathbf{r}_{2}$ and $\mathbf{r}_{3}$ are in the same coordinate plane, e.g. in the plane $(x, y)$. Then the axis $\sigma$ is the axis $z$.

In this coordinate system, according to the definition of the center of mass, we have:

$$
m_{1} \mathbf{r}_{1}+m_{2} \mathbf{r}_{2}+m_{3} \mathbf{r}_{2}=0
\tag{2}
$$


At the beginning we find the moment of inertia $I$ of the system with respect to the axis $\sigma$. Using the relation (2), we can write:

$$
0=\left(m_{1} \mathbf{r}_{1}+m_{2} \mathbf{r}_{2}+m_{3} \mathbf{r}_{3}\right)^{2}=m_{1}^{2} \mathbf{r}_{1}^{2}+m_{2}^{2} \mathbf{r}_{2}^{2}+m_{3}^{2} \mathbf{r}_{3}^{2}+2 m_{1} m_{2} \mathbf{r}_{1} \mathbf{r}_{2}+2 m_{1} m_{3} \mathbf{r}_{1} \mathbf{r}_{3}+2 m_{3} m_{2} \mathbf{r}_{3} \mathbf{r}_{2} .
$$

Of course,

$$
\mathbf{r}_{i}^{2}=r_{i}^{2} \quad i=1,2,3
$$

The quantities $2 \mathbf{r}_{i} \mathbf{r}_{j}(i, j=1,2,3)$ can be determined from the following obvious relation:

$$
a_{i j}^{2}=\left|\mathbf{r}_{i}-\mathbf{r}_{j}\right|^{2}=\left(\mathbf{r}_{i}-\mathbf{r}_{j}\right)^{2}=\mathbf{r}_{i}^{2}+\mathbf{r}_{j}^{2}-2 \mathbf{r}_{i} \mathbf{r}_{j}=r_{i}^{2}+r_{j}^{2}-2 \mathbf{r}_{i} \mathbf{r}_{j} .
$$

We get:

$$
2 \mathbf{r}_{i} \mathbf{r}_{j}=r_{i}^{2}+r_{j}^{2}-a_{i j}^{2}
$$

With help of this relation, after simple transformations, we obtain:

$$
0=\left(m_{1} \mathbf{r}_{1}+m_{2} \mathbf{r}_{2}+m_{3} \mathbf{r}_{3}\right)^{2}=\left(m_{1}+m_{2}+m_{3}\right)\left(m_{1} r_{1}^{2}+m_{2} r_{2}^{2}+m_{3} r_{3}^{2}\right)-\sum_{i<j} m_{i} m_{j} a_{i j}^{2}
$$

The moment of inertia $I$ of the system with respect to the axis $\sigma$, according to the definition of this quantity, is equal to

$$
I=m_{1} r_{1}^{2}+m_{2} r_{2}^{2}+m_{3} r_{3}^{2}
$$

The last two formulae lead to the following expression:

$$
I=\frac{1}{M} \sum_{i<j} m_{i} m_{j} a_{i j}^{2}
$$

where $M$ (the total mass of the system) is defined by the formula (10).

In the non-inertial frame, rotating around the axis $\sigma$ with the angular velocity $\omega$, the total potential energy $V_{\text {tot }}$ is the sum of the gravitational potential energies

$$
V_{i j}=-G \frac{m_{i} m_{j}}{a_{i j}} ; \quad i, j=1,2,3 ; i<j
$$

of all the masses and the potential energies



$$
V_{i}=-\frac{1}{2} \omega^{2} m_{i} r_{i}^{2} ; \quad i=1,2,3
$$

of the masses $m_{i}(i=1,2,3)$ in the field of the centrifugal force:

$$
\begin{aligned}
& V_{t o t}=G \sum_{i<j} \frac{m_{i} m_{j}}{a_{i j}}-\frac{1}{2} \omega^{2} \sum_{i=1}^{3} m_{i} r_{i}^{2}=G \sum_{i<j} \frac{m_{i} m_{j}}{a_{i j}}-\frac{1}{2} \omega^{2} I=G \sum_{i<j} \frac{m_{i} m_{j}}{a_{i j}}-\frac{1}{2} \omega^{2} \frac{1}{M} \sum_{i<j} m_{i} m_{j} a_{i j}^{2}= \\
& =-\sum_{i<j} m_{i} m_{j}\left(\frac{\omega^{2}}{2 M} a_{i j}^{2}+\frac{G}{a_{i j}}\right)
\end{aligned}
$$

i.e.

$$
V_{t o t}=-\sum_{i<j} m_{i} m_{j}\left(\frac{\omega^{2}}{2 M} a_{i j}^{2}+\frac{G}{a_{i j}}\right)
$$

A mechanical system is in equilibrium if its total potential energy has an extremum. In our case the total potential energy $V_{t o t}$ is a sum of three terms. Each of them is proportional to:

$$
f(a)=\frac{\omega^{2}}{2 M} a^{2}+\frac{G}{a}
$$

The extrema of this function can be found by taking its derivative with respect to $a$ and requiring this derivative to be zero. We get:

$$
\frac{\omega^{2}}{M} a-\frac{G}{a^{2}}=0
$$

It leads to:

$$
\omega^{2} a^{3}=G M \quad \text { or } \quad \omega^{2} a^{3}=G\left(m_{1}+m_{2}+m_{3}\right) \text {. }
$$

We see that all the terms in $V_{\text {tot }}$ have extrema at the same values of $a_{i j}=a$. (In addition, the values of $a$ and $\omega$ should obey the relation written above.) It is easy to show that it is a maximum. Thus, the quantity $V_{t o t}$ has a maximum at $a_{i j}=a$.

This means that our three masses can remain in fixed distances only if these distances are equal to each other:

$$
a_{12}=a_{23}=a_{13}=a
$$

and if the relation

$$
\omega^{2} a^{3}=G M
$$

where $M$ the total mass of the system, holds. We have obtained the conditions (9) and (11) again.
As the system is isolated, its total energy, i.e. the sum of the kinetic and potential energies, is conserved. The total potential energy of the points $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ in the inertial system (i.e. when there are no inertial forces) is equal to the sum of the gravitational potential energies of all the pairs of points $\left(\mathrm{P}_{1}, \mathrm{P}_{2}\right),\left(\mathrm{P}_{2}, \mathrm{P}_{3}\right)$ and $\left(\mathrm{P}_{1}, \mathrm{P}_{3}\right)$. It depends only on the distances $a_{12}, a_{23}$ and $a_{23}$ which are constant in time. Thus, the total potential energy of the system is constant. As a consequence the kinetic energy of the system is constant too. The moment of inertia of the system with respect to the axis $\sigma$ depends only on the distances from the points $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ to the axis $\sigma$ which, for fixed $a_{12}, a_{23}$ and $a_{23}$ do not depend on time. This means that the moment of inertia $I$ is constant. Therefore, the angular velocity of the system must also be constant:

$$
\omega=\text { const. }
\tag{1}
$$

This is the first condition we had to find. The other conditions will be determined by using three methods described below. However, prior to performing calculations, it is desirable to specify a convenient coordinates system in which the calculations are expected to be simple.

Let the positions of the points $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ be given by the vectors $\mathbf{r}_{1}, \mathbf{r}_{2}$ and $\mathbf{r}_{3}$. For simplicity we assume that the origin of the coordinate system is localized at the center of mass of the points $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ and that all the vectors $\mathbf{r}_{1}, \mathbf{r}_{2}$ and $\mathbf{r}_{3}$ are in the same coordinate plane, e.g. in the plane $(x, y)$. Then the axis $\sigma$ is the axis $z$.

In this coordinate system, according to the definition of the center of mass, we have:

$$
m_{1} \mathbf{r}_{1}+m_{2} \mathbf{r}_{2}+m_{3} \mathbf{r}_{2}=0
\tag{2}
$$


Let us consider again the point $\mathrm{P}_{1}$ with the mass $m_{1}$ and the forces $\mathbf{F}_{21}$ and $\mathbf{F}_{31}$ given by the formulae (3) and (4). It follows from the text of the problem that the total moment (with respect to any fixed point or with respect to the mass center) of the forces acting on the point $\mathrm{P}_{1}$ must be equal to zero. Thus, we have:

$$
\mathbf{F}_{21} \times \mathbf{r}_{1}+\mathbf{F}_{31} \times \mathbf{r}_{1}=0
$$

where the symbol $\times$ denotes the vector product. Therefore



$$
G \frac{m_{1} m_{2}}{a_{12}^{3}}\left(\mathbf{r}_{2}-\mathbf{r}_{1}\right) \times \mathbf{r}_{1}+G \frac{m_{1} m_{3}}{a_{13}^{3}}\left(\mathbf{r}_{3}-\mathbf{r}_{1}\right) \times \mathbf{r}_{1}=0
$$

But

$$
\mathbf{r}_{1} \times \mathbf{r}_{1}=0
$$

Thus:

$$
\frac{m_{2}}{a_{12}^{3}} \mathbf{r}_{2} \times \mathbf{r}_{1}+\frac{m_{3}}{a_{13}^{3}} \mathbf{r}_{3} \times \mathbf{r}_{1}=0
$$

Using the formula (8), the last relation can be written as follows:

$$
\begin{gathered}
\frac{1}{a_{12}^{3}}\left(-m_{1} \mathbf{r}_{1}-m_{3} \mathbf{r}_{3}\right) \times \mathbf{r}_{1}+\frac{m_{3}}{a_{13}^{3}} \mathbf{r}_{3} \times \mathbf{r}_{1}=0, \\
-\frac{m_{3}}{a_{12}^{3}} \mathbf{r}_{3} \times \mathbf{r}_{1}+\frac{m_{3}}{a_{13}^{3}} \mathbf{r}_{3} \times \mathbf{r}_{1}=0, \\
\left(\frac{1}{a_{13}^{3}}-\frac{1}{a_{12}^{3}}\right) \mathbf{r}_{3} \times \mathbf{r}_{1}=0 .
\end{gathered}
$$

The vectors $\mathbf{r}_{1}$ and $\mathbf{r}_{3}$ are non-collinear (and different from 0 ). Therefore

$$
\mathbf{r}_{3} \times \mathbf{r}_{1} \neq 0
$$

and

$$
\frac{1}{a_{13}^{3}}-\frac{1}{a_{12}^{3}}=0
$$

hence,

$$
a_{12}=a_{13} .
$$

Similarly, one gets:

$$
a_{12}=a_{23}(=a) .
$$

We have re-derived the condition (11).

Taking into account that all the distances $a_{i j}$ have the same value $a$, from the equation (7) concerning the point $\mathrm{P}_{1}$, using the relation (2) we obtain:

$$
\begin{gathered}
G \frac{m_{1} m_{2}}{a^{3}}\left(\mathbf{r}_{2}-\mathbf{r}_{1}\right)+G \frac{m_{1} m_{3}}{a^{3}}\left(\mathbf{r}_{3}-\mathbf{r}_{1}\right)+m_{1} \omega^{2} \mathbf{r}_{1}=0 \\
-\left(G \frac{m_{1}}{a^{3}}+G \frac{m_{2}}{a^{3}} G \frac{m_{3}}{a^{3}}\right) m_{1} \mathbf{r}_{1}+m_{1} \omega^{2} \mathbf{r}_{1}=0 \\
\frac{G M}{a^{3}}=\omega^{2}
\end{gathered}
$$

This is the condition (9). The same condition is got in result of similar calculations for the points $\mathrm{P}_{2}$ and $\mathrm{P}_{3}$.

Final answer: the equation $a_{12}=a_{23}=a_{13}=a$ stands, and the angular velocity $\omega$ is constant and the relation $\omega^{2} a^{3}=G M$ holds.