Two body problem
Newtonian mechanics
The two body problem is denoted by a set of two point-particles $\vec{x}_i$, $i \in \left\{1, 2 \right\}$, with their mutual gravitational force :
\begin{equation} m_i \ddot{\vec{x}}_i(t) = \sum_{i \neq j} G \frac{m_i m_j}{\| \vec{x}_i - \vec{x}_j \|^2} \end{equation}Giving the system of equations
\begin{eqnarray} \ddot{\vec{x}}_1(t) &=& G \frac{m_2}{\| \vec{x}_1 - \vec{x}_2 \|^2}\\ \ddot{\vec{x}}_2(t) &=& G \frac{m_1}{\| \vec{x}_1 - \vec{x}_2 \|^2} \end{eqnarray}In a lot of interesting cases, we have the situation in which $m_1 \ll m_2$, in which case we can assume negligible acceleration on the first mass. Let's consider both this and the general case and compare them later.
Central body method
If we allow the assumption that $m_1 \ll m_2$, let's say calling $m_1 = M$ and $m_2 = m$, we can compare the two accelerations by
\begin{eqnarray} \frac{\ddot{\vec{x}}_1(t)}{\ddot{\vec{x}}_2(t)} = \frac{m}{M} \approx 0 \end{eqnarray}(the acceleration is never zero from what we can see of the equation)
Therefore, we have that
\begin{eqnarray} \ddot{\vec{x}}_1(t) = \frac{m}{M} \ddot{\vec{x}}_2(t) \approx 0 \end{eqnarray}The acceleration of the larger mass is very small compared to the smaller one. If we assume it to be zero, then $\vec{x}_1(t) = x_0 + v_0 t$, which can be set to $\vec{x}_1(t) = 0$ after a Galilean transform. If we rename $x_2$ as simply $x$ in this case, our system reduces to
\begin{eqnarray} \ddot{\vec{x}}(t) &=& G \frac{M}{\| \vec{x} \|^2}\\ \end{eqnarray}To simplify our problem, it is best to express this vector in terms of spherical coordinates.
\begin{eqnarray} r &=& \sqrt{x^2 + y^2 + z^2}\\ \theta &=& \arccos(\frac{z}{\sqrt{x^2 + y^2 + z^2}})\\ \varphi &=& \mathrm{atan2}(y,x) \end{eqnarray} [...]In those coordinates, the position vector can be expressed as
\begin{eqnarray} \vec{x}(t) &=& r(t) \vec{e}_r(t)\\ \dot{x}(t) &=& \dot{r}(t) \vec{e}_r(t) + r (\dot{\theta} \vec{e}_\theta(t) + r(t) \dot{\varphi} \sin(\theta) \vec{e}_{\varphi}(t)) \end{eqnarray}Full solution
Lagrangian mechanics
\begin{eqnarray} \frac{m}{2} \dot{\vec{x}}(t) \cdot \dot{\vec{x}}(t) + \end{eqnarray}Hamiltonian mechanics
Differential geometric resolution
One aspect of the standard classroom resolution of the two body problem is its rather awkward handling of time-dependent basis vectors, which can more elegantly be done using differential geometry. If we consider the Newton-Cartan theory,
The $\mathrm{SO}(4)$ method
One specific aspect of central forces is a hidden $\mathrm{SO}(4)$ symmetry. If we look at the symmetries of the Lagrangian,