Two-body problem

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" Two-body problem" redirects here. For other uses, see Two-body problem (disambiguation).

In classical mechanics, the two-body problem is to determine the motion of two point particles that interact only with each other. Common examples include a satellite orbiting a planet, a planet orbiting a star, two stars orbiting each other (a binary star), and a classical electron orbiting an atomic nucleus.

The two-body problem can be re-formulated as two independent one-body problems, which involve solving for the motion of one particle in an external potential. Since many one-body problems can be solved exactly, the corresponding two-body problem can also be solved. By contrast, the three-body problem (and, more generally, the $n$ -body problem for $n\geq 3$ ) cannot be solved, except in special cases.

Two bodies with similar mass orbiting around a common barycenter with elliptic orbits.

Two bodies with a slight difference in mass orbiting around a common barycenter. The sizes, and this particular type of orbit are similar to the Pluto-Charon system.

1 Statement of problem
- 1.1 Center of mass motion (1st one-body problem)
- 1.2 Displacement vector motion (2nd one-body problem)
2 Two-body motion is planar
3 General solution for distance-dependent forces
- 3.1 Application to inverse-square force laws
4 See also
5 References

[edit] Statement of problem

Let $\mathbf{x}_{1}$ and $\mathbf{x}_{2}$ be the positions of the two bodies, and $m 1$ and $m 2$ be their masses. The goal is to determine the trajectories $\mathbf{x}_{1}(t)$ and $\mathbf{x}_{2}(t)$ for all times $t$ , given the initial positions

$\mathbf{x}_{1}(t=0)$ and $\mathbf{x}_{2}(t=0)$

and the initial velocities

$\mathbf{v}_{1}(t=0)$ and $\mathbf{v}_{2}(t=0)$ .

When applied to the two masses, Newton's second law states that

$\mathbf{F}_{12}(\mathbf{x}_{1},\mathbf{x}_{2}) = m_{1} \ddot\mathbf{x}_{1} \quad \quad \quad (Equation \ 1)$

$\mathbf{F}_{21}(\mathbf{x}_{1},\mathbf{x}_{2}) = m_{2} \ddot\mathbf{x}_{2} \quad \quad \quad (Equation \ 2)$

where

$\mathbf{F}_{12}$ is the force on mass 1 due to its interactions with mass 2, and

$\mathbf{F}_{21}$ is the force on mass 2 due to its interactions with mass 1.

Adding and subtracting these two equations decouples them into two one-body problems, which can be solved independently. Adding equations (1) and (2) results in an equation describing the center of mass (barycenter) motion. By contrast, subtracting equation (2) from equation (1) results in an equation that describes how the vector $\mathbf{r} \equiv \mathbf{x}_{1} - \mathbf{x}_{2}$ between the masses changes with time. The solutions of these independent one-body problems can be combined to obtain the solutions for the trajectories $\mathbf{x}_{1}(t)$ and $\mathbf{x}_{2}(t)$ .

[edit] Center of mass motion (1st one-body problem)

Addition of the force equations (1) and (2) yields

$m_{1}\ddot\mathbf{x}_{1} + m_{2}\ddot\mathbf{x}_{2} = (m_{1} + m_{2})\ddot\mathbf{x}_{cm} = \mathbf{F}_{12} + \mathbf{F}_{21} = 0$

where we have used Newton's third law $\mathbf{F}_{12} = -\mathbf{F}_{21}$ and where

$\mathbf{x}_{cm} \equiv \frac{m_{1}\mathbf{x}_{1} + m_{2}\mathbf{x}_{2}}{m_{1} + m_{2}}$

is the position of the center of mass (barycenter) of the system. The resulting equation

$\ddot\mathbf{x}_{cm} = 0$

shows that the velocity $\dot\mathbf{x}_{cm}$ of the center of mass is constant, from which follows that the total momentum $m_{1}\dot\mathbf{x}_{1} + m_{2}\dot\mathbf{x}_{2}$ is also constant (conservation of momentum). Hence, the position and velocity of the center of mass can be determined at all times from the initial positions and velocities.

[edit] Displacement vector motion (2nd one-body problem)

Dividing both force equations by the respective masses, subtracting the second equation from the first and rearranging gives the equation

$\ddot \mathbf{r} = \ddot\mathbf{x}_{1} - \ddot\mathbf{x}_{2} = \left( \frac{\mathbf{F}_{12}}{m_{1}} - \frac{\mathbf{F}_{21}}{m_{2}} \right) = \left(\frac{1}{m_{1}} + \frac{1}{m_{2}} \right)\mathbf{F}_{12}$

where we have again used Newton's third law $\mathbf{F}_{12} = -\mathbf{F}_{21}$ and where $\mathbf{r}$ (defined above) is the displacement vector from mass 2 to mass 1.

The force between the two objects should only be a function of $\mathbf{r}$ and not of their absolute positions $\mathbf{x}_{1}$ and $\mathbf{x}_{2}$ ; otherwise, physics would not have translational symmetry, i.e., the laws of physics would change from place to place, in this case the two bodies. Therefore, the subtracted equation can be written

$\mu \ddot\mathbf{r} = \mathbf{F}_{12}(\mathbf{x}_{1},\mathbf{x}_{2}) = \mathbf{F}(\mathbf{r})$

where $μ$ is the reduced mass

$\mu = \frac{1}{\frac{1}{m_{1}} + \frac{1}{m_{2}}} = \frac{m_{1}m_{2}}{m_{1} + m_{2}}$

Once we have solved for $\mathbf{x}_{cm}(t)$ and $\mathbf{r}(t)$ , the original trajectories may be obtained from the equations

$\mathbf{x}_{1}(t) = \mathbf{x}_{cm}(t) + \frac{m_{2}}{m_{1} + m_{2}} \mathbf{r}(t)$

$\mathbf{x}_{2}(t) = \mathbf{x}_{cm}(t) - \frac{m_{1}}{m_{1} + m_{2}} \mathbf{r}(t)$

as may be verified by substituting into the defining equations for $\mathbf{x}_{cm}(t)$ and $\mathbf{r}(t)$ .

[edit] Two-body motion is planar

The motion of two bodies with respect to each other always lies in a plane (in the center of mass frame). Let us define the linear momentum $\mathbf{p} = \mu \dot\mathbf{r}$ and the angular momentum

$\mathbf{L} = \mathbf{r} \times \mathbf{p}$

The rate of change of the angular momentum equals the net torque $\mathbf{N}$

$\frac{d\mathbf{L}}{dt} = \dot\mathbf{r} \times \mu\dot\mathbf{r} + \mathbf{r} \times \mu\ddot\mathbf{r} = \mathbf{r} \times \mathbf{F} = \mathbf{N}$

However, Newton's strong third law of motion holds for most physical forces, and says that the force between two particles acts along the line between their positions. Therefore, $\mathbf{r} \times \mathbf{F} = 0$ and angular momentum is conserved. Therefore, the displacement vector $\mathbf{r}$ and its velocity $\dot\mathbf{r}$ are always in the plane perpendicular to the constant vector $\mathbf{L}$ .

[edit] General solution for distance-dependent forces

It is often useful to switch to polar coordinates, since the motion is planar and, for many physical problems, the force $\mathbf{F}(\mathbf{r})$ is a function only of the radius $r$ (a central force). Since the r-component of acceleration is $\ddot{r} - r \dot{\theta}^{2}$ , the r-component of the displacement vector equation $\mu \ddot\mathbf{r} = \mathbf{F}(r)$ can be written

$\mu\frac{d^{2}r}{dt^{2}} - \mu r \omega^{2} = \mu\frac{d^{2}r}{dt^{2}} - \frac{L^{2}}{\mu r^{3}} = F(r)$

where $\omega \equiv \dot\theta$ and the angular momentum $L = μ r 2 ω$ is conserved. The conservation of angular momentum allows us to solve for the trajectory $r (θ)$ by making a change of independent variable from $t$ to $θ$

$\frac{d}{dt} = \frac{L}{\mu r^{2}} \frac{d}{d\theta}$

giving the new equation of motion

$\frac{L}{r^{2}} \frac{d}{d\theta} \left( \frac{L}{\mu r^{2}} \frac{dr}{d\theta} \right)- \frac{L^{2}}{\mu r^{3}} = F(r)$

This equation becomes quasilinear on making the change of variables $u \equiv \frac{1}{r}$ and multiplying both sides by $\frac{\mu r^{2}}{L^{2}} = \frac{\mu}{L^{2} u^{2}}$

$\frac{d^{2}u}{d\theta^{2}} + u = -\frac{\mu}{L^{2}u^{2}} F(1/u)$

[edit] Application to inverse-square force laws

If $F$ is an inverse-square law central force such as gravity or electrostatics in classical physics

$F = \frac{\alpha}{r^{2}} = \alpha u^{2}$

for some constant $α$ (negative for an attractive force, positive for a repulsive one), the trajectory equation becomes linear

$\frac{d^{2}u}{d\theta^{2}} + u = -\frac{\alpha \mu}{L^{2}}$

The solution of this equation is

$u(\theta) \equiv \frac{1}{r(\theta)} = -\frac{\alpha \mu}{L^{2}} + A \cos(\theta - \theta_{0})$

where $A$ and $θ 0$ are constants, $A\ge0$ and for $α > 0$ (repulsive force) the additional requirement $A\ge\frac{\alpha \mu}{L^{2}}$ . This solution, applicable for values of $θ$ for which u > 0, shows that the orbit is a conic section, i.e., an ellipse, a hyperbola or parabola, depending on whether $A$ is less than, greater than, or equal to $-\frac{\alpha \mu}{L^{2}}$ , and a straight line for $α = 0$ .