On Validity of the formula for gravitational potential energy The formula for gravitational potential energy,
$$-G\frac{m_1 m_2}{R},$$
is found by using the fact that the change in potential energy is equal to negative of the work done ( by conservative forces). One of the assumptions is that the 2nd larger mass remains stationary relative to each other, and thus only the work done on the much smaller has to be taken into account.
This is obviously true for something like A satellite and the earth, but what about the case when the masses are similar?
Is this formula still when the two gravitating masses are of similar mass?
I tried to derive the formula without the assumption by adding a pseudo force on $m_2$ to take $m_1$ at rest. I arrive at the formula
$$U(R) = -G \frac{m_2 (m_1 + m_2)}{R},$$
which does actually reduce to the usual formula for $m_1\gg m_2$, but is clearly wrong because of the asymmetry about the 2 masses. Moreover, I cant find any source on any such formula. What am is the error in my reasoning here?
 A: The answer is yes the formula
$$U(r) = -G\frac{m_1 m_2}{r}$$
is general where $r$ is the distance between the two masses. In Newtonian gravity the axiom is that the gravitational force felt by body 1 from body 2 is
$${\bf F}_{12} = G \frac{m_1 m_2\, {\bf r}}{r^3} = G \frac{m_1 m_2\, ({\bf r_2 - r_1})}{|{\bf r_2}-{\bf r_1}|^3}$$
from that and the definition of potential,
$$U({\bf r}_0) = \int_\infty^{{\bf r}_0} {\bf F}\cdot {\rm d}{\bf r},$$
the formula is derived (easiest way is to put one of the masses at the origin, then this integral is trivial). You should think about the potential energy as being assigned to a certain configuration of masses. You can build the configuration by steps. Step 1 there is no mass in the region under consideration so you can bring $m_1$ from infinity to the region of interest without spending any energy, then in step 2 keeping $m_1$ where it, is you bring $m_2$ to its final distance $r$. You should think about this system as being frozen in time, to which we assign the given gravitational potential, so it is a label essentially to a instantaneous configuration, this being a consequence of Newton's version of gravity acting immediately in all of space, but this point will carry us off-topic.
Another question and a completely different matter is, how does this system evolve. If you "release" this system you must now study its dynamics and start making use of Newton's laws and so on to determine equations of motion, trajectories and so on.
A: Suppose the (spherically symmetric) bodies are identical and initially separated by distance $R$. We will now take both bodies to infinity synchronously (that is keeping the midpoint in one place – relative to the fixed stars!) If we measure distance $r$ from the centre of mass of the system, that is midway between the centres of the bodies, then the work done on each body taking it from $r=\frac R2$ to infinity is
$$\int_{R/2}^\infty\frac{GM_1M_2}{(2r)^2}dr=\frac 14\int_{R/2}^\infty\frac{GM_1M_2}{r^2}dr=\frac 12 \frac{GM_1M_2}{R} $$
So the total work done is
$$\frac{GM_1M_2}{R}$$
This is the gain in PE of the system, so its PE in the original configuration was, relative to the PE at infinite separation,
$$-\frac{GM_1M_2}{R}$$
With a little more thought we can adapt this treatment to bodies of unequal, but not necessarily hugely unequal, mass.
