# If a stationary ball of neutrinos were released at the equator, what orbit would it take?

Students are traditionally asked how long it would take an object to fall through the "center" of the earth and reach the other side. In the demonstrations which I have seen, the initial velocity of the object due to the rotation of the earth is not taken into consideration.

Suppose one had a solid ball of "neutrino-ium" (or perhaps dark matter) stationary at the earth's surface at the equator which one releases and allows to fall through the earth.

Assuming a spherical earth of constant density, into what sort of orbit would the ball settle?

Because of the variation in gravity during the trajectory, I doubt that standard equations of orbital mechanics will be sufficient to determine the orbit, so I am at a loss.

• The equations are sufficient assuming you know the mass distribution inside the Earth; if you don't, you can make various approximations. But in any case the neutrinos will oscillate around the center of the Earth (assuming spherical symmetry). – Javier Oct 11 '16 at 21:18
• Related to 126077; also see this, crudely assuming a uniform earth and ignoring frame-dragging effects of the earth's rotation, of course, as a zeroth order approximation... – Cosmas Zachos Oct 11 '16 at 21:23
• How massive is the ball?? – user108787 Oct 11 '16 at 22:19
• The ball has an initial horizontal velocity because of the earth's rotation. The traditional exercise, at least as I have seen it presented, ignores this initial velocity and assumes the ball is falling down a perfectly vertical shaft which passes through the center of the earth. Such a ball would bounce from one wall of the shaft to the other if one considers its initial velocity perpendicular to the gravitational force. – John Wayland Bales Oct 14 '16 at 18:20
• My apologies, I misread the tangential velocity spec. Since the gravity force is central, angular momentum is conserved, so take it to be $l\equiv L/m=\dot{\theta} r^2$ contributing a centrifugal barrier term $m l^2/(2r^2)$ to the gravity potential, thus opposing the plain oscillator force. – Cosmas Zachos Oct 14 '16 at 22:46

Since the force is central, angular momentum is conserved, and so the angular motion can be solved in terms of the radial motion. Using the conventions of the spinless earth starting point you mention, that is $\omega^2=g/R$, for $R$ the radius thereof, we see the earth's motion is a small perturbation, as the 24hrs period is so much larger than the canonical 2π/ω=84.5 mins. In fact, the centrifugal barrier will eject the neutrino ball so the effective period is more like half the spinless case. It will come up ahead of where it was released---it will have effectively spun faster than the earth.

The conserved angular momentum is $L=m r^2 \dot{\theta}$, which it pays to re-write in terms of a dimensionless variable a. The vanishing a limit is the spinless earth, while a=2 corresponds to the notional earth spinning so fast as to give the neutrino ball the escape velocity. The particle's mass m is a canard that factors out. $$L\equiv \sqrt{a} m \omega R^2 .$$ Thus, given $r(t)$, integrating $\dot{\theta}=L/(mr^2)= \sqrt{a} \omega R^2/r(t)^2$ completely specifies the angular motion. For the "real" earth, $\dot{\theta}_R/\omega=\sqrt{a}\sim 0.06$, small.

In these variables, the energy of the ball for $r\leq R$ (only!) is $$E=\frac{m\dot{r}^2}{2}+\frac{mg}{2R} r^2+\frac{m a\omega^2R^4}{2r^2} .$$

The equation of motion then, in terms of the dimensionless x = r/R is the nonlinear Ermakov equation $$\ddot{x} = -\omega^2 x + \frac{a \omega^2}{x^3},$$ whose first integral is $~~\dot{x}^2/\omega^2+x^2+a/x^2=2E/(\omega^2 mR^2)=1+a$.

Consider the solution
$$x=\sqrt{\sin^2(\omega t +\pi/2) + a/(1-a) } ~~ \sqrt{1-a} .$$ It has a suitable $a\to 0$ limit, the spinless oscillation, $r=R\sin (\omega t +\pi/2)$, and x(t=0)=1. Nevertheless, it will have exactly the same 84.5 min period as the spinless case, since a and ω are independent. It should not be a surprise the periodic motion amounts to a closed elliptic orbit, of course: it is just an orbit in a harmonic potential, until surfacing (and splicing into a Kepler potential...).

It is also apparent that the neutrino ball will never get to the center, as the minimum r will be $\sim R \sqrt{a}$ for small a, the effective centrifugal barrier. For the "real" earth, this perigee would be ~0.06 R.

Perhaps somebody with graphics juices and time to lavish could plot the integrated r - θ orbits viewed from the North pole. Note the enormous eccentricity: $e\sim 1-2 \sqrt{a}$ !