The two-body problem (Lagrangian approach) to derive the orbital period of Earth: Happy New sidereal Year!

It is nearly 2026 and I cannot think about anything except celestial mechanics. The idea that a full cycle -which datum has been established by our species- has been completed: a new sidereal year, the completion of a full orbit.

So, to celebrate with you, here, I will use the principle of least action (the Lagrangian approach) to show a beautiful model that, at our scale, approximates nicely the sidereal year (without perturbations), i.e.: T -> 365.2564 days. I will avoid cumbersome algebraic calculations since I am aiming for a brief text.

Let’s then use the elegant Lagrangian approach to solve the two-body problem where the system is made of two entities: our Sun and our Earth. With the equations of motion, Kepler’s laws can be deduced along with the Earth’s orbital period.

I will use planar polar coordinates since conservation of angular momentum allows it. The coordinates (r,θ) will describe, respectively, radial position and angular position.

The equations of motion

Let M be the mass of the Sun and let m be the mass of Earth such that M>>m. This will allow us to treat the Sun as a fixed body though I will keep the general reduced mass (rigor), i.e.: μ= (Mm)/(M+m)→m.

Consider this two-body system in an inertial frame of reference such that r is the relative position vector whose magnitude r is the distance bewteen the bodies. Now we look for the Lagrangian ℒ = T−V, where T is the kinetic energy of the system and V is the potential (gravitational) energy. Taking G as the gravitational constant:

By Euler-Lagrange, for angular motion:

It is worth mentioning that this confirms Kepler’s second law about constant areal velocities:

For radial motion

Therefore,

Is the radial equation of motion.

The elliptical shape of the orbit

Now, let’s solve for the orbit’s shape (elliptical). For that, we shall use Binet transformation. Let ϕ=1/r :

After some cumbersome algebra, we get this linear, inhomogeneous differential equation:

Solving it as simple harmonic motion with constant shift ψ:

Thus, taking the perihelion at θ = 0, ψ = 0 and defining eccentricity:

Letting

And, finally, substituting ϕ, we obtain the elliptical orbit (first Kepler’s law):

The orbital period

Finally, we derive the orbital period (one Earth year cycle). Let T be the orbital period, i.e.: the time for θ to achieve 2π radians.

Takin a as one AU, we get, as expected: T -> 365.2564 days.

Happy New sidereal Year!