Orbits and Kepler’s Third Law

An overview Kepler’s third law of planetary motion.

Kepler’s Third Law of Planetary Motion

GPT aided overview of Kepler’s third law of planetary motion. This equation was mentioned in the Moons, VSI.

In Newtonian gravitation, two bodies of mass $m_1$ and $m_2$ attract each other with a force given by $F=G{m}_1{m}_2/r^2$, where $r$ is the distance between them and $G$ is the gravitational constant. This force is mutual, equal in magnitude and opposite in direction, and causes both bodies to accelerate toward each other. In the absence of other forces, the two masses will orbit around their common center of mass, known as the barycenter. The location of the barycenter divides the distance between the two bodies in inverse proportion to their masses: that is, $m_1{r}_1=m_2{r}_2$, where $r_1$ and $r_2$ are the distances from each body to the barycenter, and $r=r_1+r_2$ is their total separation.

If the mutual orbit is circular, each mass moves in a circle centered on the barycenter. Although the radii of these circular paths differ, the angular velocity $\omega$ is the same for both masses, so they complete each orbit in the same time. The more massive body moves in a smaller circle, and the less massive body in a larger one. Importantly, the two bodies remain exactly opposite one another across the barycenter at all times—the line connecting them always passes through the barycenter, and their angular positions differ by precisely $\pi$ radians. This geometric configuration ensures that the gravitational force between them always points directly toward the center of each orbit, maintaining the circular motion.

The centripetal force required to keep a body of mass $m$ in a circle of radius $r$ with tangential speed $v$ is $F=m{v}^2/r$, Section 3. Equating this to the gravitational force, we can write $$\frac{G{m}_1{m}_2}{r^2}=\mu \frac{v^2}{r},$$ where $\mu ={\displaystyle \frac{m_1{m}_2}{m_1+m_2}}$ is the reduced mass. This equation arises from reducing the two-body problem to an equivalent one-body problem: treat one body of mass $\mu$ orbiting a fixed center at distance $r$, subject to the same gravitational force. Substituting $v=\omega r$, we get $$\frac{G{m}_1{m}_2}{r^2}=\mu {\omega }^{2}r.$$ Rewriting $\omega =2\pi /T$, we have $${\omega }^2={\left(\frac{2\pi }{T}\right)}^2=\frac{G(m_1+m_2)}{r^3},$$ since the $\mu$ terms cancel. Solving for $T^2$, we arrive at the general form of Kepler’s third law: $$T^2=\frac{4{\pi }^2{r}^3}{G(m_1+m_2)}.$$ This equation shows that the square of the orbital period is proportional to the cube of the distance between the two bodies, divided by the total mass. In the special case where one mass is much larger than the other (as with the Sun and a planet), the total mass can be approximated by the larger mass alone, recovering the classical form of Kepler’s third law: $T^2\propto r^3$. The mutual circular motion about a shared barycenter, with constant angular velocity and opposite positions, is not an approximation but a direct consequence of Newton’s laws of motion and gravitation.

Barycenters

The barycenters for the Sun and Earth and Earth and Moon both lie inside the more massive body, but the Moon–Earth one is noticeably offset.

Sun–Earth system

Mass of Sun: $M_{\odot }\approx 1.989\times {10}^{30}\text{kg}$

Mass of Earth: $M_{\oplus }\approx 5.972\times {10}^{24}\text{kg}$

Distance between them: $r\approx 1.496\times {10}^{11}\text{m}$

Barycenter distance from Sun’s center:

$$r_{\odot }=\frac{M_{\oplus }}{M_{\odot }+M_{\oplus }}\cdot r\approx \frac{5.97\times {10}^{24}}{1.99\times {10}^{30}}\cdot r\approx 449\text{km}$$

That’s well within the Sun’s radius ($\sim \mathrm{696,000}\text{km}$), so the barycenter is deep inside the Sun.

Earth–Moon system

Mass of Earth: $M_{\oplus }\approx 5.972\times {10}^{24}\text{kg}$

Mass of Moon: $M_{\text{Moon}}\approx 7.348\times {10}^{22}\text{kg}$

Separation: $r\approx \mathrm{384,400}\text{km}$

Barycenter distance from Earth’s center:

$$r_{\oplus }=\frac{M_{\text{Moon}}}{M_{\oplus }+M_{\text{Moon}}}\cdot r\approx \mathrm{4,670}\text{km}$$

Earth’s radius is about 6,371 km, so the barycenter lies inside the Earth, but off-center — roughly 73% of the way from the center to the surface.

Centripetal Force and Acceleration

The formula $F=M{v}^2/r$ gives centripetal force, the inwards force required for circular motion. It assumes the object moves in a circle of radius $r$ with constant speed $v$. The acceleration is centripetal, always directed toward the center, with magnitude $a=v^2/r$.

Suppose: $$\mathbf{p}(t)=r\left(\begin{array}{c}\cos \theta (t)\\ \sin \theta (t)\end{array}\right)$$ If angular velocity $\omega ={\displaystyle \frac{d\theta }{dt}}$ is constant: $$\mathbf{p}(t)=r\left(\begin{array}{c}\cos \omega t\\ \sin \omega t\end{array}\right)$$ Differentiate to obtain velocity: $$\mathbf{v}(t)=\frac{d\mathbf{p}}{dt}=r\omega \left(\begin{array}{c}-\sin \omega t\\ \cos \omega t\end{array}\right)\Rightarrow \parallel \mathbf{v}(t)\parallel =r\omega$$ and again for acceleration: $$\mathbf{a}(t)=\frac{d\mathbf{v}}{dt}=-r{\omega }^2\left(\begin{array}{c}\cos \omega t\\ \sin \omega t\end{array}\right)=-{\omega }^2\mathbf{p}(t)$$ So, acceleration points inward with magnitude $a={\omega }^{2}r=v^2/r$.

Thus, starting from position and differentiating twice gives you the centripetal acceleration. Tangential speed $v=r\omega$, and the force becomes: $$F=M\cdot a=M\cdot \frac{v^2}{r}.$$