Ньютонівскька Проблема Двох Тіл

The Newtonian two-body problem is a problem in which one is given the masses, initial positions, and initial velocities of two objects and asked to calculate the positions of the objects at any future time. It is assumed that the two bodies interact only with each other.

Let $m_1$ and $m_2$ denote the two masses and $x_1(t)$ and $x_2(t)$ the positions of the objects at time $t$.

Let $v_1(t)$, $v_2(t)$, $a_1(t)$, and $a_2(t)$ denote the appropriate velocities and accelerations.

Let $F_{12}(t)$ denote the force that the first object exerts on the second at time $t$. Define $F_{21}(t)$ similarly.

Before we start, note that $x$, $v$, $a$, and $F$ are all vector quantities. Using the vector symbol for all of these will obscure the equations, which is why they have been omitted.

The general method to solve such a system is to decompose it into two one-body problems. The first body to solve for is the center of mass of the system. The second "body" is the difference in positions of the two objects, or $x_1-x_2$. Using some algebra, one can solve for $x_1$ and $x_2$ after obtaining the equations for these two.

Let $x_0$, $v_0$, and $a_0$ denote the position, velocity, and acceleration, respectively, of the center of mass. Let us first remember the definition of the center of mass: $$x_0=\frac{m_1x_1+m_2x_2}{m_1+m_2}$$

Differentiating this twice, we obtain the following equations: $$v_0=\frac{m_1v_1+m_2v_2}{m_1+m_2}$$ $$a_0=\frac{m_1a_1+m_2a_2}{m_1+m_2}$$ Using Newton's Second Law of Motion, $$a_1=\frac{F_{21}}{m_1}$$ $$a_2=\frac{F_{12}}{m_2}$$ Plugging this into our expression for $a_0$, we have: $$a_0=\frac{F_{21}+F_{12}}{m_1+m_2}$$ By Newton's Third Law of Motion, $$F_{21}=-F_{12}$$ Plugging this in, we find a very simple result: $$a_0=0$$ Thus, the center of mass undergoes no acceleration! This means that the center of mass is moving at a constant velocity.

An interesting sidenote: if we calculate the total linear momentum of the system, we find: $$p=m_1v_1+m_2v_2$$ $$p=(m_1+m_2)v_0$$ Since the two masses are constant, and since the center of mass has a constant velocity, the momentum of the system is constant.

Now, we may determine the position of the center of mass at all times: $$x_0(t)=x_0(0)+v_0t$$ $$x_0(t)=\frac{m_1[x_1(0)]+m_2[x_2(0)]}{m_1+m_2}+\frac{m_1v_1+m_2v_2}{m_1+m_2}t$$ Although the equation looks a little messy, the definition of the center of mass pops straight out at you in this equation, making this formula very easy and intuitive.

Because the center of mass has a constant velocity, it will be helpful to think in terms of the center of mass frame of reference. For the rest of this article, all of the position, velocity, and acceleration quantities will be redefined relative to the center of mass.

One thing we can do is to calculate the net torque caused by the forces about the center of mass. Let $\tau$ denote this net torque.

Calculating the net torque $\tau$, $$\tau = x_1\times F_{21}+x_2\times F_{12}$$ By Newton's Third Law of Motion, $$\tau = (x_1-x_2)\times F_{21}$$ Here, we will make an assumption that is usually true, but not always. Newton's Strong Third Law of Motion states that the forces that two objects exert on each other lie on the line connecting the two objects. This is true for the vast majority of cases, including the gravitational force and electrostatic forces.

Applying this assumption to this problem, we see that the vectors $F_{21}$ and $x_1-x_2$ are actually parallel. Thus, we may conclude: $$\tau = (x_1-x_2)\times F_{21}$$ $$\tau = 0$$ Thus, the net torque is zero. This means that the angular momentum about the center of mass is constant.

From the definition of angular momentum, denoted $L$, $$L = m_1(x_1\times v_1) + m_2(x_2\times v_2)$$ In the center of mass frame of reference, $v_1\parallel v_2$ and $x_1\parallel x_2$. Thus, we see that the plane perpendicular to $L$ contains $x_1$, $x_2$, $v_1$, and $v_2$.

This leads us to yet another important conclusion: in the center of mass frame of reference, all motion will occur in a plane.

Solving this system further would be futile, mainly because we are not given effective expressions for the forces.

However, in a specific problem, such as, perhaps, a problem about the Moon and the Earth, the expressions for the forces would be either given or solvable from other physical laws. Thus, the information presented above should be more than enough to solve the system for a specific problem.