# 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?

• Consider the following: What would happen if $m \to M$ (or even larger)? In this case, it's not so obvious which mass you would argue is orbiting the other. – jim Sep 11 '17 at 21:44
• hello, thanks for your comment, it confuses me even more, haha. The acceleration a massive body has under the influence of gravity doesn't depend on its own mass, only on the mass of the other thing that is atracting it. So I don't understand how can the period depend on mass. The limit you consider shouldn't affect its trayectory, only the trayectory of the other particle. – P. C. Spaniel Sep 11 '17 at 21:50
• Well, when the smaller mass is negligible compared to the larger then the difference between results for T gets lost in the decimal places, if you compare M + m with simply M. Consider, the mass of the Earth compared to that of the Moon is about 100x, while compared to the Sun is about 1/5,000,000. – Mick Sep 12 '17 at 0:55
• What we mean is that when you start with the topic, you will solve question where one planet revolves around star. You will take star to be stationery. This is because magnitude of gravitation is same on both but the star has much greater mass than planet and hence much less acceleration. If you do so, then your expression here would be independent of m. When you want things to be more accurate, you have to account for the star's motion and then you would get the given eqn. Similarly, the relative velocity between a dropped ball and earth depends on the mass of earth if we do it accurately. – Red Floyd Sep 12 '17 at 3:24
• Perhaps this works as an explanation for why, on a theoretical level, the period of an orbiting object could depend on its mass. Although the net acceleration of an object in a gravitational field does not depend on its mass, the mass of the orbiting planet can move the larger object, which over time changes the force on the smaller object by altering the distance between them. – QuantumFool Sep 12 '17 at 6:38

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 $m$.

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.

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

$$\frac{T^2}{r_2^3} = \frac{4\,\pi^2}{G\,m_1}. \tag{7}$$

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.

• +1 Finally an answer that actually starts by addressing the issue ("because I believed that gravitational motion was independent of the mass of the orbiting planet"). – JiK Sep 12 '17 at 11:08