Since Planetary motions plays a crucial role in our life, we would like to study about two body problems in a elaborate way.
Let us consider two masses $m_1$ and $m_2$ at a distance $\vec{r_1} $ and $\vec{r_2} $ from the origin of the coordinate system. Let say $\vec{r} $ is the relative position vector of $m_2$ from $m_1$. The centre of mass of these two masses are described by vector $ \vec{R}$ We would like to reduce this problem to one body problem, because it is easy to work with one body and its motion other than working two masses at same time. And we would like to solve this using Lagrangian principle.
Lagrangian for a conservative system is equal to the difference between Kinetic energy and Potential Energy i.e. L = T - V where T- total kinetic Energy and V - total potential Energy.
Kinetic energy of the system of two particles is, $$ T = \frac{1}{2} m_1 \dot{\vec{r_1}}^2 + \frac{1}{2} m_2 \dot{\vec{r_2}}^2 \,\,\,...eq.(1)$$
Using the centre of mass concept, $$ \dot{\vec{R}} = \frac{m_1\dot{\vec{r_1}} + m_2\dot{\vec{r_2}}}{ m_1+m_2} \,\,\,...eq.(2)$$
and $$ \dot{\vec{r_1}} + \dot{\vec{r}} = \dot{\vec{r_2}} \,\,\,...eq.(3)$$
With the help of eq.(3) and eq.(2),
$$\dot{\vec{r_1}} = \dot{\vec{R}} - \frac{m_2}{m_1+m_2} \dot{\vec{r}} \,\,\,...eq.(4)$$
and $$ \dot{\vec{r_2}} = \dot{\vec{R}} + \frac{m_1}{m_1+m_2} \dot{\vec{r}} \,\,\,..eq.(5) $$
Substituting eq.(4) and (5) in eq.(1) we get,
$$ T = \frac{1}{2} m_1 \left(\dot{\vec{R}} - \frac{m_2}{m_1+m_2} \dot{\vec{r}}\right)^2 + \frac{1}{2} m_2 \left(\dot{\vec{R}} + \frac{m_1}{m_1+m_2}\dot{\vec{r}}\right)^2 $$
Doing all simplifications, we will get, $$ T = \frac{1}{2} (m_1+m_2) \dot{\vec{R}}^2 + \frac{1}{2} \frac{m_1m_2}{m_1+m_2} \dot{\vec{r}}^2 \,\,\,...eq.(6)$$
The first term involving the position of centre of mass doesn't play any role in further discussion, if we are considering the potential that depend on the relative distance between the masses. For, example you can always choose your origin of the coordinate system at the centre of mass position, since it just moves at constant velocity or rest.
The reason is so simple that if you consider the two masses as a whole system, then there is no any external force acting on this system, and so the linear momentum of the system always remains constant i.e. the linear momentum of centre of mass.
Thus we finally get our Lagrangian for two particle system as,
$$ L = \frac{1}{2} \frac{m_1m_2}{m_1+m_2} \dot{\vec{r}}^2 - V(r) \,\,\,...eq.(7)$$
taking $ \frac{m_1m_2}{m_1+m_2} = \mu $ as reduced mass, the problem simplifies into simple lagrangian that looks like the lagrangian for a single particle, $$ L = \frac{1}{2} \mu\dot{\vec{r}}^2 - V(r) $$
Since central force acts only in the radial direction, the problem has spherical symmetry and so the total angular momentum is conserved. One of the coordinate can be chosen to by cyclic and the motion is confined to the plane perpendicular to angular momentum vector. All central force motions are confined to planar motion.
Due to the simplification of the problem, we just need two polar coordinates to completely describe the motion of the system.
Lagrangian again simplifies in the polar coordinates as,
$$ L = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2) - V(r)\,\,\,...eq.(8) $$ [we can use m for $\mu$]
$$ \frac{\partial{L}}{\partial{\theta}} = 0 $$ where $\theta$ is a cyclic coordinate.
Therefore, $$ \frac{d}{dt}\left(\frac{\partial{L}}{\partial\dot{\theta}}\right) = 0$$ or $$\frac{\partial{L}}{\partial{\dot{\theta}}} = const. = l \,\,\,...eq.(9)$$
$$\rightarrow\,\,\,\,\, l = m r^2 \dot{\theta} $$
From, this we can prove that the Areal velocity is constant in central force motion, which is exactly the Kepler's second law.
For radial component, Euler lagrange equation gives,
$$ \frac{\partial{L}}{\partial{r}} = \frac{d}{dt} \left(\frac{\partial{L}}{\partial{\dot{r}}}\right) $$ or $$ \frac{d}{dt} (m\dot{r}) = mr\dot{\theta}^2 - \frac{\partial{V}}{\partial{r}} $$
Applying for $\dot{\theta}$
$$ m\ddot{r} = \frac{l^2}{mr^3} - \frac{\partial{V}}{\partial{r}} \,\,\,...eq.(10)$$ multiplying on both sides by $\dot{r}$ we get.
$$ m\ddot{r}\dot{r} = \left(\frac{l^2}{mr^3} - \frac{\partial{V}}{\partial{r}}\right) \frac{dr}{dt} $$
or $$ \frac{d}{dt} \left(\frac{1}{2}m\dot{r}^2\right) = \frac{\partial}{\partial{r}} \left( \frac{-l^2}{2mr^2} - V\right) \frac{dr}{dt} = 0 $$Taking out the differential with repect to time,
$$\frac{d}{dt} \left(\frac{1}{2}m\dot{r}^2 + \frac{l^2}{2mr^2} + V(r)\right) = 0 $$
$$ \rightarrow\,\,\,\,\, \frac{1}{2}m\dot{r}^2 + \frac{l^2}{2mr^2} + V(r) = const. = E \,\,\,\,...eq.(11)$$
Where this const. E is the total Energy of the system.
Eq.(11) is our desired differential equation. From this equation we can derive all the result that we need to understand the system completely.
Let us consider two masses $m_1$ and $m_2$ at a distance $\vec{r_1} $ and $\vec{r_2} $ from the origin of the coordinate system. Let say $\vec{r} $ is the relative position vector of $m_2$ from $m_1$. The centre of mass of these two masses are described by vector $ \vec{R}$ We would like to reduce this problem to one body problem, because it is easy to work with one body and its motion other than working two masses at same time. And we would like to solve this using Lagrangian principle.
Lagrangian for a conservative system is equal to the difference between Kinetic energy and Potential Energy i.e. L = T - V where T- total kinetic Energy and V - total potential Energy.
Kinetic energy of the system of two particles is, $$ T = \frac{1}{2} m_1 \dot{\vec{r_1}}^2 + \frac{1}{2} m_2 \dot{\vec{r_2}}^2 \,\,\,...eq.(1)$$
Using the centre of mass concept, $$ \dot{\vec{R}} = \frac{m_1\dot{\vec{r_1}} + m_2\dot{\vec{r_2}}}{ m_1+m_2} \,\,\,...eq.(2)$$
and $$ \dot{\vec{r_1}} + \dot{\vec{r}} = \dot{\vec{r_2}} \,\,\,...eq.(3)$$
With the help of eq.(3) and eq.(2),
$$\dot{\vec{r_1}} = \dot{\vec{R}} - \frac{m_2}{m_1+m_2} \dot{\vec{r}} \,\,\,...eq.(4)$$
and $$ \dot{\vec{r_2}} = \dot{\vec{R}} + \frac{m_1}{m_1+m_2} \dot{\vec{r}} \,\,\,..eq.(5) $$
Substituting eq.(4) and (5) in eq.(1) we get,
$$ T = \frac{1}{2} m_1 \left(\dot{\vec{R}} - \frac{m_2}{m_1+m_2} \dot{\vec{r}}\right)^2 + \frac{1}{2} m_2 \left(\dot{\vec{R}} + \frac{m_1}{m_1+m_2}\dot{\vec{r}}\right)^2 $$
Doing all simplifications, we will get, $$ T = \frac{1}{2} (m_1+m_2) \dot{\vec{R}}^2 + \frac{1}{2} \frac{m_1m_2}{m_1+m_2} \dot{\vec{r}}^2 \,\,\,...eq.(6)$$
The first term involving the position of centre of mass doesn't play any role in further discussion, if we are considering the potential that depend on the relative distance between the masses. For, example you can always choose your origin of the coordinate system at the centre of mass position, since it just moves at constant velocity or rest.
The reason is so simple that if you consider the two masses as a whole system, then there is no any external force acting on this system, and so the linear momentum of the system always remains constant i.e. the linear momentum of centre of mass.
Thus we finally get our Lagrangian for two particle system as,
$$ L = \frac{1}{2} \frac{m_1m_2}{m_1+m_2} \dot{\vec{r}}^2 - V(r) \,\,\,...eq.(7)$$
taking $ \frac{m_1m_2}{m_1+m_2} = \mu $ as reduced mass, the problem simplifies into simple lagrangian that looks like the lagrangian for a single particle, $$ L = \frac{1}{2} \mu\dot{\vec{r}}^2 - V(r) $$
Since central force acts only in the radial direction, the problem has spherical symmetry and so the total angular momentum is conserved. One of the coordinate can be chosen to by cyclic and the motion is confined to the plane perpendicular to angular momentum vector. All central force motions are confined to planar motion.
Due to the simplification of the problem, we just need two polar coordinates to completely describe the motion of the system.
Lagrangian again simplifies in the polar coordinates as,
$$ L = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2) - V(r)\,\,\,...eq.(8) $$ [we can use m for $\mu$]
$$ \frac{\partial{L}}{\partial{\theta}} = 0 $$ where $\theta$ is a cyclic coordinate.
Therefore, $$ \frac{d}{dt}\left(\frac{\partial{L}}{\partial\dot{\theta}}\right) = 0$$ or $$\frac{\partial{L}}{\partial{\dot{\theta}}} = const. = l \,\,\,...eq.(9)$$
$$\rightarrow\,\,\,\,\, l = m r^2 \dot{\theta} $$
From, this we can prove that the Areal velocity is constant in central force motion, which is exactly the Kepler's second law.
For radial component, Euler lagrange equation gives,
$$ \frac{\partial{L}}{\partial{r}} = \frac{d}{dt} \left(\frac{\partial{L}}{\partial{\dot{r}}}\right) $$ or $$ \frac{d}{dt} (m\dot{r}) = mr\dot{\theta}^2 - \frac{\partial{V}}{\partial{r}} $$
Applying for $\dot{\theta}$
$$ m\ddot{r} = \frac{l^2}{mr^3} - \frac{\partial{V}}{\partial{r}} \,\,\,...eq.(10)$$ multiplying on both sides by $\dot{r}$ we get.
$$ m\ddot{r}\dot{r} = \left(\frac{l^2}{mr^3} - \frac{\partial{V}}{\partial{r}}\right) \frac{dr}{dt} $$
or $$ \frac{d}{dt} \left(\frac{1}{2}m\dot{r}^2\right) = \frac{\partial}{\partial{r}} \left( \frac{-l^2}{2mr^2} - V\right) \frac{dr}{dt} = 0 $$Taking out the differential with repect to time,
$$\frac{d}{dt} \left(\frac{1}{2}m\dot{r}^2 + \frac{l^2}{2mr^2} + V(r)\right) = 0 $$
$$ \rightarrow\,\,\,\,\, \frac{1}{2}m\dot{r}^2 + \frac{l^2}{2mr^2} + V(r) = const. = E \,\,\,\,...eq.(11)$$
Where this const. E is the total Energy of the system.
Eq.(11) is our desired differential equation. From this equation we can derive all the result that we need to understand the system completely.
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