We first briefly recall method used for qualitative discussion of motion in one dimension. Many of these ideas will be used here for radial motion in three dimensions.

Simplest physical systems of interest, for example, Sun and Earth system, are two body systems interacting via a potential-dependent only the distance between the two bodies only. This problem can be reduced to an equivalent one-body problem in a spherically symmetric potential.

#### 2.1 Recall and Discuss — One dimension

**Equilibrium Points**

If a particle, moving in one dimension, is released from rest at some point, it will in general move towards lower potential energy. If it is released at a minimum or maximum of the potential, it will remain at rest It therefore follows that at these points \( \dot x = \ddot x = 0 \)

Let \( x_0 \) be a point where the particle in equilibrium.

If the point \( x_0 \) is a minimum of the potential and the particle is disturbed slightly, it will execute oscillations about the minimum. In this case we say that the point \( x_0 \) is a point of *stable equilibrium*.

If the equilibrium point is a maximum of the potential and even a slightest disturbance will make the particle move away from the equilibrium. In this case we say that the *equilibrium is unstable*.

**Turning points**

- A particle moving in a potential cannot go to regions where its energy is less than the potential energy. Its motion is confined to those values of \( x \) where

$$ V (x) \leq E $$ To see this note that we must have $$ E = \frac {p^2} {2m} + V (x) \hspace{5em} (1) $$ $$ \therefore E \geq V (x) \hspace{5em} (2) $$ $$ \because K.E. = \frac {p^2} {2m} > 0.$$ - The region, the set of values of \( x \) where (1) holds, is called
*classically accessible region*. - The points where \( E = V (x) \) are called
*turning points*. At a turning point the velocity becomes zero, \( \dot x = 0 \).

**Range of energies for bounded motion**

Assuming a continuous potential \( V (x) \), the potential will have a minimum or a maximum between two turning points. For a given energy, a particle will execute a bounded motion if the potential has two turning points such that it has a minimum between the turning points.

In general, whether the motion is bounded, or unbounded, depends on the initial conditions and energy of the particle.

For all possible initial conditions the motion is bounded only if the energy is less that the values of potential \( V (±∞) \) and if both \( V (±∞) \) are greater than the absolute minimum of the potential.

**2.2 Recall and discuss - Reduction of two body problem**

The Lagrangian of two body system $$ L = \frac 1 2 m_1 \left( \frac {d \vec x_1} {dt} \right)^2 + \frac 1 2 m_2 \left( \frac {d \vec x_2} {dt} \right)^2 - V \left( \vec x_1 - \vec x_2 \right) \hspace{5em} (3) $$

decomposes as sum of Lagrangian for the centre of mass and the for the relative motion:

\( \hspace{9em} L = L_{cm} + L_{rel} \hspace{5em} (4) \)

\( \hspace{9em} L_{cm} = \frac 1 2 M \left( \frac {d \vec X_{cm}} {dt} \right)^2, \hspace{5em} (5) \)

\( \hspace{9em} L_{rel} = \frac 1 2 \mu \left( \frac {d \vec x} {dt} \right)^2 - V \left( \vec r \right). \hspace{5em} (6) \)

Here \( M = m_1 + m_2 \) is the total mass, and \( \mu = \frac {m_1m_2} {m_1 + m_2} \) is called reduced mass of the two body system. The energy of the relative motion is given by $$ E = \frac 1 2 \mu \left( \frac {d \vec x} {dt} \right)^2 + V \left( \vec r \right). \hspace{5em} (7) $$