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On Wikiversity it states that for central forces: enter image description here Wouldn't $\ddot{\vec{r}}_1$ be the same as $\vec{g}$?

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No. For some reason this article is including two different forces: (1) an attractive or repulsive central force between the two masses, described by the potential energy function $V(r)$, and (2) a uniform external gravitational field acting on each mass, described by the acceleration $\vec{g}$.

The acceleration $\ddot{\vec{r}}_1$ of $m_1$ is due to both of these forces, amd will not be equal to $\vec{g}$.

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I'm a bit confused here, but as far I'm understanding that there are 2 particles whose Position is described as two position vectors, $\vec r_1$ and $\vec r_2$.

So, $$K_1 = \dfrac 12 m_1 \vec r_1^2$$ and silmilarly for $K_2$ of particle 2. $$K_2 = \dfrac 12 m_2 \vec r_2^2$$

We'll write Potential energy here too which are there dependent on two forces of interaction, first Gravitational force and other as Force between them.

So, $$V_1 = m_1\vec g.\vec r_1 $$ $$V_2 = m_2 \vec g. \vec r_2$$

Now, the last function which is same for particle 1 and 2 which will be added once (You may add it twice to both particles, but that depends on the type of Force and it's nature of definition!)

We get Lagrangian as:

$$L = \dfrac 1 2 (m_1\vec r_1^2 + m_2 \vec r_2^2) - V(\vec r) + m_1 \vec g . \vec r_1 + m_2 \vec g. \vec r_2 \blacksquare$$

Here $\vec g$ is opposite to the direction of gravitational forces

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  • $\begingroup$ Your kinetic energy should have a dot for the time derivative. Your $V_1$ and $V_2$ have the wrong sign. And you did not answer the OP’s question. $\endgroup$ – G. Smith Jun 16 at 17:46

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