I am interested in the position of the point of zero gravity within the Earth as a function of the gravity of the moon. Take the example of the moonless Earth. The position of zero gravity would be approximately at the center of the Earth (the center of mass of the Earth). However, when a moon-like mass is added at the appropriate distance, the moon acts by its gravity on the Earth and changes the Earth's zero gravity position in the direction of the moon.

I guess that logic is fine, but I’m not entirely sure. I would certainly be interested in some link that describes this problem mathematically.


1 Answer 1


For a single spherical mass with uniform density and finite radius, there is one point in space with zero gravitational field: the center of the mass. This was first proved by Newton and is usually called the “shell theorem.” Specifically, the gravitational force on a particle with mass $m$ towards the center of a sphere with mass $M$ and radius $R$ is

$$ F=\frac{Gm M^\text{enclosed}}{r^2} $$

If the sphere has uniform density $\rho=M\cdot\left( \frac{4\pi}3R^3 \right)^{-1}$, then you can find $r^{-3}M^\text{enclosed}=R^{-3}M$ and get $F\propto r$ inside the sphere, and $F=0$ at the center. But the Earth has a denser metal core and a less-dense rocky mantle, so the simple linear result only applies in the uniform-density region near the core.

For two bodies without rotation, there are three such zero-force points: one near the center of each body, and one somewhere on the line connecting them. If we call the line between the centers of the masses the $x$-axis, and say the mass $M_1$ is at $x_1$ and $M_2$ is at $x_2$, the balance point $x$ obeys

$$ \frac{GmM_1}{(x-x_1)^2}=\frac{GmM_2}{(x-x_2)^2} $$

This unstable gravitational equilibrium point is closer to the smaller body — unlike the barycenter, which is closer to the larger body.

To find the stable zero-force point near the center of $M_1$, you can solve the force-balancing equation using $M_1^\text{enclosed}$ as a function of $r=x-x_1$. But now you have to know the density profile. Also, $M_1$ will itself fall towards this point, so its location will move towards $M_1$’s surface until the two masses crash into each other.

With rotation, we say that the bodies are “orbiting,” and it is convenient to analyze the problem in a rotating reference frame and include a centrifugal pseudo-force. In this frame, there are seven points where the total pseudo-force vanishes. There are two stable points at (not near) the center of each mass. The other five are the Lagrange points. The unstable L1 is near the point where the gravitational forces balance — but not exactly at the gravitational balancing point, because of the centrifugal pseudo-force. The unstable L2 and L3 lie “beyond” the two masses. And the points L4 and L5, lying “ahead” and “behind” the orbit of the smaller body around the barycenter, become stable equilibria in the limit where the “moon” is much less massive than the “planet.”


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