Questions regarding an Olympiad problem for perturbational solutions of the three-body problem This theoretical problem on the International Physics Olympiad talks about the three-body problem with respect to LISA. In one of their solutions to the problem:


Consider the case $M=m$. If $\mu$ is now given a small radial perturbation (along $\rm O\ \mu$), what is the angular frequency of oscillation of $\mu$ about the unperturbed position in terms of $\omega_0$? Assume that the angular momentum of $\mu$ is conserved.

In the solution, they give the total energy of $\mu $ as:
$$
E_T
=
-\frac{GM\mu}{r_1}
-\frac{Gm\mu}{r_2}
+\frac{1}{2}\mu\left(\frac{d\rho}{dt}\right)^2
+\frac{1}{2}\mu(\omega\rho)^2
$$
Where $\rho$ stands for the distance between $O$ and $\mu$. I understand the terms of the above expression apart from the $\frac{1}{2}\mu(\frac{d\rho}{dt})^2$ part. Where does this expression come from, I do get $\frac{d\rho}{dt}$ stands for velocity but why does it have to exist ?
(From what I know in circular motion, the velocity is tangential and that component can be found as $\omega\rho$, hence the kinetic energy expression $\frac{1}{2}\mu(\omega\rho)^2$ ).
Questions:

*

*Where does this expression $\frac{1}{2}\mu\left(\frac{d\rho}{dt}\right)^2$ originate from?


*What does it mean when it says a small perturbation is given in the radial direction? Also how does this fact apply in this question?


*What is the implication of angular momentum being conserved in this problem?


*

$$
E
=
-\frac{GM\mu}{r_1}
-\frac{Gm\mu}{r_2}
+\frac{1}{2}\mu\left(\left(\frac{d\rho}{dt}\right)^2
+\rho^2\omega^2\right)
\tag{15}
$$
Since the perturbation is in the radial direction, angular momentum is conserved ($r_1=r_2=\mathfrak R$ and $m=M$),
$$
E
=
-\frac{2GM\mu}{\mathfrak R}
+\frac{1}{2}\mu\left(\left(\frac{d\rho}{dt}\right)^2
+\frac{\rho_0^4\omega^2}{\rho^2}\right)
\tag{16}
$$

In this above quote why does $\rho$ suddenly become $\frac{\rho^4_0}{\rho^2}$ and $\omega^2=\omega^2_0$?
I appreciate the clarification.

EDIT:
There is one more portion I do not quite understand which is to the solution of problem 1.1 which is given by:

Two gravitating masses $M$ and $m$ are moving in circular orbits of radii $R$ and $r$, respectively, about their common centre of mass. Find the angular velocity $\omega_0$ of the line joining $M$ and $m$ in terms of $R$, $r$, $M$, $m$ and the universal gravitational constant $G$.
Let O be their centre of mass. Hence
\begin{align}
MR-mr & = 0 \\
m\omega_0^2 r & = \frac{GMm}{(GMm)^2} \\
M\omega_0^2 R & = \frac{GMm}{(GMm)^2} 
\end{align}
From Eq. (2), or using reduced mass, $\omega_0^2 = \frac{G(M+m)}{(R+r)^3}$. Hence, $$\omega_0^2 = \frac{GM}{r(R+r)^e} = \frac{Gm}{R(R+r)^2}.$$

I do not really understand what they mean by reduced mass and how those three expressions are equivalent. For instance I do understand how they got $\frac{GM}{r(R+r)^2}$ and $\frac{Gm}{R(R+r)^2}$ but I don't get how they got $\frac{G(M+m)}{(R+r)^3}$.
 A: angular momentum L is conserved thus
$$\omega\,\rho^2=L$$
with
$$L=\omega_0\,\rho_0^2$$
you obtain
$$\omega^2\,\rho^2\mapsto \frac{\rho_0^4\,\omega_0^2}{\rho^2}$$


your Edit
you obtain those 3 equations
$$ -m{\omega}^{2}r+{\frac {MmG}{ \left( R+r \right) ^{2}}}=0\tag 1$$
$$-M{\omega}^{2}R+{\frac {MmG}{ \left( R+r \right) ^{2}}}=0\tag 2$$
and the center of mass equation
$${\frac {mr-MR}{M+m}}=0\tag 3$$
solve equations (1) and (2) for $~\omega^2~,r~$ you obtain
$$\omega^2= {\frac {{m}^{3}G}{{R}^{3} \left( M+m \right) ^{2}}}\tag 4$$
solve equation (3) for  M and substitute to equation (4)
$$\omega^2=\frac{m\,G}{R\,(R+r)^2}$$
A: *

*The problem states that the mass is given a small radial perturbation. In other words, the mass is displaced from its equilibrium position in a direction parallel to $O$ and $\mu$. This means that $\rho$ will change as a function of $t$, leading it to start oscillating in the radial direction. The problem asks for the frequency of oscillations that are a result of this process. Additionally, there will be small addition of translation kinetic energy from oscillations which is given by the $\frac{1}{2}\mu (\text{d}\rho/\text{d}t)^2$ term.


*Angular momentum is conserved when there is no external torque. Therefore, when we displace the mass in the radial direction, the moment arm remains zero, leading to conservation. This is useful because we can can find $\omega (t)$ and $\rho (t)$ at any time which change because of the radial perturbation. As @Eli points out, this can be calculated via
$$L = \omega_0 \rho_0^2 = \omega \rho^2\implies \omega = \omega_0 \frac{\rho_0^2}{\rho^2}.$$
Hence, the $\frac{1}{2}\mu (\omega \rho)^2$ term becomes $\frac{1}{2}\mu \frac{\rho_0^4 \omega_0^2}{\rho^2}.$
