How to prove that Gravity happens in one plane?

Question

How can we prove that an object which is orbiting another object by its gravity always moves in one plane in a full period? Not in different diffrent planes (two body system)

Answer 1

This is only true in a two body system. In many body systems like the Solar system the planets continually change their orbital plane (slightly) due to perturbations from other bodies.

In a two body system the orbital plane is constant because the Lagrangian is axially symmetric and that means angular momentum is conserved. This is a consequence of Noether's theorem. Since angular momentum is constant the orbital plane cannot change.

Answer 2

I had never heard of Noether's theorem and I still don't understand it, but here's an intuitive explanation...

Imagine that the initial velocity vectors of the two bodies lie in the same plane. The acceleration vector of each body will point toward the other body, and therefore will lie in the same plane as the velocity vectors.

Given that all velocity vectors and accelerations are within the plane, therefore future positions of the bodies must also be within the plane.

However, If the initial velocity vectors are not in a common plane, then the future positions will obviously not lie in a single plane. For example, a spacecraft orbiting a planet in a direction orthogonal to the planet's direction of motion would trace a helix through space.

Answer 3

It is not that gravity happens in a plane. It is that a two body system orbits in a plane. This can be seen fairly easily with Lagrangian mechanics. Consider the motion of a body in a plane. The position of the body in that plane has displacements $$ d\vec x~=~\hat rdr~+~\hat\theta rd\theta. $$ The unit vectors $\hat r$ and $\hat\theta$ are polar coordinates on a plane. The velocity is $\vec v~=~d\vec x/dt$ defines the square of velocity $v^2~=~\dot r^2~+~r^2\dot\theta^2$. For the problem of gravitation we have $$ {\cal L}~=~\frac{1}{2}m(\dot r^2~+~r^2\dot\theta^2)~+~\frac{GMm}{r} $$ There are two Euler-Lagrange equations for the two independent coordinates $$ \frac{d}{dt}\left(\frac{\partial{\cal L}}{\partial r}\right)~-~\frac{\partial{\cal L}}{\partial r}~=~0~=~m\ddot r~-~mr\dot\theta^2~+~\frac{GMm}{r^2} $$ and $$ \frac{d}{dt}\left(\frac{\partial{\cal L}}{\partial\theta}\right)~-~\frac{\partial{\cal L}}{\partial\theta}~=~0~=~mr\ddot\theta. $$ The last of these with $mr\ddot\theta~=~0$ gives the equation of constant motion $$ mr^2\dot\theta~=~L~=~constant. $$ This is the constancy of angular momentum. Conservation of angular momentum is what imposes two body orbits to lie on a plane. This also lets us write the Lagrangian as $$ {\cal L}~=~\frac{1}{2}m\dot r^2~+~\frac{L^2}{2mr^2}~+~\frac{GMm}{r}, $$ where angular momentum term gives an effective potential opposite the radial direction of gravity. It also means the variable $r$ and $\theta$ are not independent, which was observed to be the case by Kepler. The dynamical equation is then $$ \ddot r~-~\frac{L^2}{2mr^3}~+~\frac{GM}{r^2}~=~0 $$