# Acceleration due to central forces in the Lagrangian

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

## 2 Answers

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 $$\vec{g}$$.

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

• 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. – G. Smith Jun 16 at 17:46