Two-body problem

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

As described below, Newton's laws of motion allow us to reduce the two-body problem to an equivalent one-body problem, i.e., to solving for the motion of one particle in an external potential. Since one-body problems can usually 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.

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. Newton's second law states that

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

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

Our mission 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 $\mathbf{v}_{2}(t=0)$ (12 constants altogether).

The key trick to solving the two-body problem is to add and subtract these two equations, which decouples them into two one-body problems. Adding them results in an equation describing the center of mass motion, whereas subtracting them results in an equation that describes how the vector between the masses changes with time. When combined, the solutions to these one-body problems provide the solutions for the trajectories $\mathbf{x}_{1}(t)$ and $\mathbf{x}_{2}(t)$ .

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

Addition of the two equations 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.

Displacement vector motion (2nd one-body problem)

Subtracting the two force equations and rearranging gives the equation

\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 $\mathbf{F}_{12} = -\mathbf{F}_{21}$ .

We introduce a new vector $\mathbf{r}$

\mathbf{r} \equiv \mathbf{x}_{1} - \mathbf{x}_{2}

that is the displacement vector from mass 2 to mass 1. Importantly, the force between the two objects should only be a function of this displacement vector $\mathbf{r}$ 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. Therefore, the subtracted equation can be written

\mu \ddot\mathbf{r} = \mathbf{F}(\mathbf{r})

where $\mu$ 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)

Two-body motion is planar

Remarkably, the motion of two bodies always lies in a plane. 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, i.e., $\mathbf{F} \propto \mathbf{r}$ . 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 $\mathbf{L}$ .

Trajectory solution

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 $\mu \ddot\mathbf{r} = \mathbf{F}(r) \equiv F(r)$ 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 $L = \mu r^{2}\omega$ is conserved. The conservation of angular momentum allows us to solve for the trajectory $r(\theta)$ by making a change of independent variable from $t$ to $\theta$

\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)

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 $\alpha$ , 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 $A>0$ and $\theta_{0}$ are constants. This solution 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 $-\frac{\alpha \mu}{L^{2}}$ .

Newtonian Gravity

Applying the gravitational formula we get that the position of the first body with respect to the second is governed by the same differential equation as the position of a very small body orbiting a body with a mass equal to the sum of the two masses, because

m_1 m_2/\mu=m_1+m_2

Assume:

the vector r is the position of one body relative to the other (above called x)
r, v, the semi-major axis a, and the specific relative angular momentum h are defined accordingly (hence r is the distance)
$\mu=G(m_1+m_2)$ , the standard gravitational parameter (the sum of those for each mass)

where:

$m_1$ and $m_2$ are the masses of the two bodies.

Then:

the orbit equation applies; recalling that the positions of the bodies are $m_2/(m_1+m_2)$ and $-m1/(m1+m2)$ times r, respectively, we see that the two bodies' orbits are similar conic sections; the same ratios apply for the velocities, and, without the minus, for the angular momentum with respect to the barycenter and for the kinetic energies
for circular orbits $rv^2 = r^3 \omega^2 = 4 \pi^2 r^3/T^2 = \mu$
for elliptic orbits: $4 \pi^2 a^3/T^2 = \mu$ (with a expressed in AU and T in years, and with M the total mass relative to that of the Sun, we get $a^3/T^2 = M$ )
for parabolic trajectories $r v^2$ is constant and equal to $2 \mu$
h is the total angular momentum divided by the reduced mass
the specific orbital energy formulas apply, with specific potential and kinetic energy and their sum taken as the totals for the system, divided by the reduced mass; the kinetic energy of the smaller body is larger; the potential energy of the whole system is equal to the potential energy of one body with respect to the other, i.e. minus the energy needed to escape the other if the other is kept in a fixed position; this should not be confused with the smaller amount of energy one body needs to escape, if the other body moves away also, in the opposite direction: in that case the total energy the two need to escape each other is the same as the aforementioned amount; the conservation of energy for each mass means that an increase of kinetic energy is accompanied by a decrease of potential energy, which is for each mass the inner product of the force and the change in position relative to the barycenter, not relative to the other mass
for elliptic and hyperbolic orbits $\mu$ is twice the semi-major axis times the absolute value of the specific orbital energy

For example, consider two bodies like the Sun orbiting each other:

the reduced mass is one half of the mass of one Sun (one quarter of the total mass)
at a distance of 1 AU: the orbital period is ${1\over 2} \sqrt{2}$ year, the same as the orbital period of the Earth would be if the Sun would have twice its actual mass; the total energy per kg reduced mass (90 MJ/kg) is twice that of the Earth-Sun system (45 MJ/kg); the total energy per kg total mass (22.5 MJ/kg) is one half of the total energy per kg Earth mass in the Earth-Sun system (45 MJ/kg)
at a distance of 2 AU (each following an orbit like that of the Earth around the Sun): the orbital period is 2 years, the same as the orbital period of the Earth would be if the Sun would have one quarter of its actual mass
at a distance of $\sqrt$ ^*{2} \approx 1.26 AU: the orbital period is 1 year, the same as the orbital period of the Earth around the Sun

Similarly, a second Earth at a distance from the Earth equal to $\sqrt$ ^*{2} times the usual distance of geosynchronous orbits would be geosynchronous.

General Relativistic Gravity

In the general theory of relativity gravity behaves somewhat differently, but, to a first approximation for weak fields, the effect is to slightly strengthen the gravity force at small separations. Kepler's First Law is modified so that the orbit is a precessing ellipse, its major and minor axes rotating slowly in the same sense as the oribital motion. The law of conservation of angular momentum still applies (Kepler's Second Law). Kepler's Third Law would in principle be altered slightly, but in practice, the only way to measure the sum of the masses is by applying that Law as it stands, so there is effectively no change. These results were first obtained approximately by Einstein, and the rigorous two body problem was later solved by Howard Percy Robertson.

Examples

a binary star, e.g. Alpha Centauri (approx. the same mass)
a double planet, e.g. Pluto with its moon Charon (mass ratio 0.147)
a binary asteroid, e.g. 90 Antiope (approx. the same mass)

Other uses

The phrase two body problem is also used jokingly by scientists to refer to the difficulty of married graduate students or postdocs finding jobs at the same university.

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