# Why is an elliptical orbit with central mass is not allowed in two-body problem?

I'm aware that 2-body problem has an analytical solution. However I was wondering what if we guessed for an answer and then checked whether it is a valid one. I know the equations and expressions governing the 2-body problem are the following:

\begin{align} E &= \frac{1}{2} m_* \dot{r}^2 + \frac{1}{2} \frac{l^2}{m_* r^2} + U(r) = \frac{1}{2}m_*\dot{r}^2 + U_{\rm eff}(r), \\ U_{\rm eff}(r) &= \frac{1}{2} \frac{l^2}{m_* r^2} + U(r). \\ \frac{{\rm d}\theta}{{\rm d} t} &= l / m_* r^2;~~~l \equiv |\mathbf{L}|, \\ m_* &= \frac{m_1 m_2}{m_1 + m_2}, \\ \rho(\theta) &= \frac{1}{1 + e \cos \theta} \end{align}

I'm aware that these equations are from the reference frame of the reduced mass and I'm assuming one mass is much larger than the other one such that only one mass orbits around the other approximately. I tested the validity of values such as when $${\rm d}\theta/{\rm d}t$$ when $${\rm d}r/{\rm d}t$$ at extremities when either is $$0$$ but that told me nothing. I was wondering if there's a way to validate what laws are being violated if the guess was such that one mass is at the center of the ellipse and other mass orbiting (this is assuming this ellipse has different values for semi-major and semi-minor axis) with values of $${\rm d}\theta/{\rm d}t$$ being max and $${\rm d}r/{\rm d}t = 0$$ at $$r_{\rm max}$$ (semi-major axis) and $${\rm d}\theta/{\rm d}t$$ is min and $${\rm d}r/{\rm d}t = 0$$ at $$r_{\rm min}$$ (semi-minor axis).

This seems like a very basic question but I could not figure out how it is possible to rule out certain solutions without analytically solving the question.

• I have converted your image equations to LaTeX because Phys.SE supports MathJaX. Please check that I've transcribed them correctly. Aug 31, 2020 at 18:43

In general, if two particles are attracted to each other by a central force (not necessarily a gravitational force), and you know the trajectory $$r(\theta)$$ and the angular momentum of the system $$L$$ about its center of mass, then it turns out that there's only one possible force law that can create that trajectory. This is because from Newton's Second Law written in terms of the separation and the reduced mass $$\mu$$, $$\mu \ddot{\vec{r}} = F(r) \hat{r},$$ and from the conservation of the angular momentum $$\vec{L} = \mu \vec{r} \cdot \dot{\vec{r}}$$, we can derive the Binet equation: $$F\left( r \right) = - \frac{L^2}{\mu r^2} \left( \frac{d^2 (1/r)}{d\theta^2} + \frac{1}{r} \right).$$
For example, for an ellipse with a focus at the origin, we have in polar coordinates $$\frac{1}{r} = A + B \cos \theta$$ for some constants $$A$$ and $$B$$; it is not hard to see in this case that $$F(r) = -L^2 A/\mu r^2$$, i.e., the particles interact via a force obeying an inverse square law.
On the other hand, for an ellipse with its center at the origin, we have $$\frac{1}{r} = \sqrt{ A \sin^2 \theta + B \cos^2 \theta},$$ and one can show (after much algebra) that the Binet equation reduces to $$F(r) = - \frac{L}{\mu r^2} \left[ \frac{2 A B}{(A \sin^2 \theta + B \cos^2 \theta)^{3/2}} \right]= - \frac{L A B}{\mu} r,$$ i.e., this is the trajectory taken by a particle in a 2-D (or 3-D) harmonic potential.