Third Kepler law and mass dependance The third Kepler law states that:
\begin{equation}
\frac{T^2}{R^3}=\frac{4\pi^2}{G(M+m)}
\end{equation}
Where $T$ is the period of the orbital movement, $R$ is the semimajor axis, $M$ is the mass of the sun and $m$ is the mass of the planet.
This is counterintuitive to me because I believed that gravitational motion was independent of the mass of the orbiting planet, since the mass $m$ cancels out from the beginning when one states newton's law. Furthermore, I thought that this had to do with some fundamental things asociated with gravity being a geometrical theory that doesn't depend on your mass but just on the geometry of your trajectory.
Why does the period depend on the mass of the planet then?
 A: The $M+m$ in third Kepler's law is a vestige of the reduced mass associated to the two body problem. Roughly speaking we map a coupled and complicated system of two interacting particles into an equivalent problem of decoupled differential equations, one of them describing the motion of a particle of reduced mass $\mu$ under a central potential corresponding to gravitational interaction.
By integrating Kepler's second law, $dA/dt=L/2\mu$, over a complete orbit we obtain
$$\frac AT=\frac{L}{2\mu},$$
where $A$ is the area of the orbit and $L$ the angular momentum of the particle of mass $\mu$. For simplicity let us consider a circular orbit or radius $R$. Then
$$T^2=\frac{4\pi^2\mu^2R^4}{L^2}.\tag1$$
In the circular orbit, the centrifugal force matches gravity, thus
$$\frac{GMm}{R^2}=\mu\omega^2R=\mu R\frac{L^2}{\mu^2R^4},$$
since $L=\mu R^2\omega$. Solving for $\mu^2 R^4/L^2$ and plugging back into (1) we obtain
$$T^2=\frac{4\pi^2R^3}{G(M+m)}.$$
Note that for the solar system we normally have $M\gg m$ so we normally neglect $m$.
A: The statement that the mass of the smaller body can be ignored is a useful approximation in many real-world situations.  The difference between $M$ and $M+m$ for the earth-moon system is only about 1%.  For earth and satellites or the sun and planets, this amount can be ignored unless you're getting into several decimal points of precision.
Often these simplified formula come with the stated assumption that $M \gg m$.
The reason the mass of the smaller body matters is that the gravitational field it moves through is not static, but depends on the other body. The smaller body is able to accelerate the larger one, so that the field changes over time, and this change reduces the period.
$m$ only cancels out in the limit that the gravitational field is static.
A: One has to consider the meaning of $R$, since normally the semi-major axis can refer to half the distance between periapsis and apoapsis from the perspective of a the center of mass of the two celestial bodies (assuming only a two body problem). However in this case it is actually referring to the distances between the two bodies.
For a circular orbit this can be shown to be true, but it should also hold for elliptical orbits. The body of mass $m_1$ will be orbiting at a distance of $r_1$ from the center of mass and The body of mass $m_2$ will be orbiting at a distance of $r_2$ from the center of mass. As stated before $R = r_1 + r_2$ and from the center of mass it follows that $m_1\,r_1 = m_2\,r_2$. Solving for $r_1$ and $r_2$ gives $r_1 = m_2\,R\,(m_1 + m_2)^{-1}$ and $r_2 = m_1\,R\,(m_1 + m_2)^{-1}$. The total force between the two bodies follows from Newton's law of gravitation
$$
F_g = \frac{G\,m_1\,m_2}{R^2}. \tag{1}
$$
But each body will be moving along their own circular path around the common center of mass. Namely if a body is subjected to only a force of constant magnitude perpendicular to its velocity then it will move along a circular trajectory at a constant rate according to
$$
F_{\perp} = \omega^2\, r_i\, m_i, \tag{2}
$$
with $\omega = 2\,\pi\,T^{-1}$ the angular velocity. The only force acting on each body is gravity, so $F_{\perp} = F_g$. Equating the right hand sides of  equations $(1)$ and $(2)$ and either substituting in $1$ or $2$ for $i$ both gives
$$
\frac{G\,m_1\,m_2}{R^2} = \frac{4\,\pi^2\, m_1\,m_2\,R}{T^2 (m_1 + m_2)}. \tag{3}
$$
Simplifying this expressing indeed gives
$$
\frac{T^2}{R^3} = \frac{4\,\pi^2}{G (m_1 + m_2)}. \tag{4}
$$
However if you would prefer to define the semi-major axis as half the distance between periapsis and apoapsis from the perspective of a the center of mass, then equation $(4)$ can be rewritten to
$$
\frac{T^2}{r_1^3} = \frac{4\,\pi^2 (m_1 + m_2)^2}{G\,m_2^3}, \tag{5}
$$
$$
\frac{T^2}{r_2^3} = \frac{4\,\pi^2 (m_1 + m_2)^2}{G\,m_1^3}. \tag{6}
$$
However when $m_1 \gg m_2$ then equation $(6)$ still simplifies to 
$$
\frac{T^2}{r_2^3} = \frac{4\,\pi^2}{G\,m_1}. \tag{7}
$$
A: From the perspective of trying to understand how the mass of the orbiting planet could impact its period, consider that although the acceleration of an object in a gravitational field does not depend on the mass of the object ($ a = \frac{GM}{r^2} $), the motion of the larger mass is impacted by the mass of the smaller object. The motion of the larger object induced by the presence of the smaller object over time can make the distance between the two objects dependent on the mass of the smaller one, and the acceleration felt by the smaller mass is directly related to the distance between the two objects. 
