A little question about $F_g = F_c$ It is already some time ago that I was fluent in classical mechanics.
I do not understand a certain detail in the formula $F_{grav.}=F_{centr.}$.
Probably I overlook something simple.
Suppose mass $m$ circles around mass $M$ at distance $d$ with constant speed $v = 2\pi d/T$.
Then $$GmM/d^2 = mv^2/d.$$
Now suppose I consider this circular movement with mass $m$ in the centre. Distance, speed and period are the same, so:
$$GmM/d^2 = Mv^2/d.$$
What is my error?
Both $m$ and $M$ rotate uniformly in a circle around the center of mass. This is the preferred point of view that is easy to understand.
Still one can assume a reference frame on $m$ in rest for an observer on $m$ in rest. $M$ definetely moves with the same speed, period and distance. Ergo sum the centripetal force in this reference frame can no longer be formulated as $Mv^2/d$.
But how does the observer on $m$ get to know that?
Could anybody elaborate on whether and how this particular situation may be calculated?
As a matter of fact it seems that in this reference frame besides gravity there must appear to be a "fictious" repelling force that equals $(m-M)v^2/d$.
 A: Neither formula is exact (assuming that Newtonian gravity is the only force affecting the two masses): a reference frame in which $m$ moves circularly while $M$ is stationary is not inertial, since in an inertial frame $M$ will move by the force exerted by $m$.
The same is true for the reverse.
The first formula holds in the approximation that $M \gg m$: then the force exerted by $m$ causes a negligible acceleration of the mass $M$, which we can then approximate as stationary.
We are considering $m$ to be a so-called "test mass".
The reverse, on the other hand, would be a really bad approximation, since the force on $m$ is definitely not negligible - it is what causes the motion, after all!
A: If mass $M$ is fixed in place and mass $m$ (which is only under the influence of Newtonian gravity) is in a circular orbit of radius $R$ and speed $v$, then
$$G\frac{Mm}{R^2} = \frac{mv^2}{R}$$
$$\implies v^2 R = GM$$
This is a constraint between $v$ and $R$.  In other words, circular gravitationally-bound orbits around $M$ cannot have arbitrarily chosen speed and radius; given some speed, there is only one radius at which such an orbit is possible, and vice-versa.
If instead we fix $m$ in place, then circular orbits around $m$ obey a similar constraint, but now we have that $v^2 R = Gm$.  If $m\neq M$, then circular orbits of radius $R$ around $M$ will have a different speed than circular orbits of radius $R$ around $m$.

Jacopo also makes a good point in his answer.  If you assume that both masses are only under the influence of Newtonian gravity, then neither one orbits the other; instead, they both orbit their mutual center of mass.  If one mass is much larger than the other, then the larger mass will be approximately stationary and we might say that the smaller mass is approximately in orbit around it, but that's not exactly true.
A: From a purely kinematic perspective, M is orbiting m and m is orbiting M with the same angular speed.
The Newtonian gravity supposes a force between the masses, that we can imagine as a tension in a giant massless rod: $$T = \frac{GMm}{r^2}$$
The system is rotating as a rigid body around its COM. The centripetal force is the tension in the rod:
$T = m\omega^2 r_m = M\omega^2 r_M$
As $\omega$ is the same:
$$m r_m = M r_M \implies m r_m = M (r-r_m) \implies r_m = \frac{Mr}{M+m}$$
$$\frac{GMm}{r^2} = m\omega^2 r_m = M\omega^2 r_M \implies \omega = \sqrt{\frac{G(M+m)}{r^3}}$$
As $\omega$, $r$ and $G$ can be determined, we can say that the masses are the output of the theory.
A: In continuation of the answer of Claudio Saspinksi:
Choosing M as the inertial frame of reference gives
$$\frac{GMm}{r^2} = m\omega^2 r \cdot \frac{M}{M+m}= m\omega^2 r 
\cdot (1 - \frac{m}{M+m})$$ where $ - \frac{m^2\omega^2 r}{M+m}$ is a (pseudo) centrifugal force experienced by m.
The observer on M cannot observe it.
Choosing m as the inertial frame of reference gives
$$\frac{GMm}{r^2} = M\omega^2 r \cdot \frac{m}{M+m} = M\omega^2 r \cdot (1 - \frac{M}{M+m})$$ where $ - \frac{M^2\omega^2 r}{M+m}$ is a (pseudo) centrifugal force experienced by M.
The observer on m cannot observe it.
The lesson I learn is that by not choosing the center of mass as the frame of reference, you will have to take (pseudo) centrifugal forces into account.
I don't think many people are aware of this...
Also this shows that you don't need a rotating frame of reference for pseudo forces to emerge. Also you don't need a non-inertial frame of reference. In both cases we have a non-rotating inertial frame of reference.
