Sign of gravitational law and potential energy Suppose we have two masses, $m1$ is on the origin and $m2$ on the positive $x$ axis.
We also define a unit vector from m1 to m2 so that the force from m2 to m1 is $$\vec{F}_{12} = -G\frac{m_1m_2}{r^3}\vec{r},$$  and the force from m1 to m2 is $$\vec{F}_{21} = G\frac{m_1m_2}{r^3}\vec{r}.$$
If we now define the potential energy of the gravitaional force on either of the two above forces we get one positive and one negative potential energy, namely $U = \pm Gm_1m_2\frac{1}{r}$. But in general the potential energy is defned with the negative sign.
What is the meaning of the positive potential energy which results from integrating the force $F_{21}$?
Thanks.
 A: The potential is the same for both particles:
$$U = -\frac{Gm_1 m_2}{||\vec r_1 - \vec r_2||}\,.$$
The equations of motion for two point particles of mass $m_1$ and $m_2$ located at $\vec r_1$ and $\vec r_2$ can be derived from the following Lagrangian
$$L = \frac12 m_1 \dot r_1^2 + \frac12 m_2 \dot r_2^2 + \frac{G m_1 m_2}{|| \vec r_1 - \vec r_2||}\,.$$
The physical trajectories $r_1(t)$ and $r_2(t)$ extremize the action $S = \int{\rm d}t L$, and thus satisfy the Euler Lagrange equations
$$\frac{d}{dt}\frac{\partial L}{\partial \dot r_i^j} = \frac{\partial L}{\partial r_i^j}$$
where $i$ labels the species $1$ or $2$ and $j$ labels the vector index $\hat x,\hat y,\hat z$. The resulting equations of motion are
$$0 = m_i\ddot r_i^j + \partial_{r_i^j} U\,.$$
We can rewrite this in vector notation
\begin{align}
0 &= m_1\ddot{\vec r_1} + \vec \nabla_1 U\,,\\
0 &= m_1\ddot{\vec r_2} + \vec \nabla_2 U\,,
\end{align}
where $\nabla_1$ is short hand for the vector $(\partial_{r_1^x},\partial_{r_1^y},\partial_{r_1^z})$ and similarly for $\nabla_2$. Importantly, because $U$ is a pure function of $||\vec r_1 - \vec r_2||,$ the gradient of $U$ with respect to $\vec r_1$ is negative the gradient of $U$ with respect to $\vec r_2$. In other words, we recover the fact that the force on $1$ must be equal and opposite to that on $2$,
\begin{align}
0 &= m_1\ddot{\vec r_1} + F_G\,,\\
0 &= m_1\ddot{\vec r_2} - F_G\,,
\end{align}
where
\begin{align}
F_G = \vec \nabla_1 U\,.
\end{align}
