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I first posted this at the physics stack but was suggested to go here for real answers.

Imagine a hypothetical spherical planet with a massive core but which is somehow internally traversable without friction (f.i. due to a tunnel or a superfluid core). What shape would the orbit of a free falling object inside it be according to Newton, traversing such tunnel that happens to be identical to its trajectory, or through the frictionless superfluid? We know outside of the planet a moon would have an elliptical orbit according to Kepler, but what shape would the orbit of a moon inside this hypothetical planet be?

Imagine the thought experiment of Newtons cannonball , but then shot from somewhere midway the planet instead of on the surface of the planet. And without escape velocity for this matter.

The reason I ask for such a mental stretch is because in the Principia, Newton gives an example to explain the brachistochrone curve being a cycloid using the attractive force inside a hollow planet (prop. 49 theorem 17). In prop. 52 cor. 1 he uses the case of a Tusi couple (special case of the cycloid) inside the hollow planet, assuming a linear attractive force instead of the inverse square law. However in reality (for as far as I understand) it turns out the linear attractive force is only the case inside a massive planet.

Hence the case of my current hypothetical planet. The Tusi couple resembles a harmonic oscillator (which it should with a linear attractive force) when it equals half the diameter of the planet. However if you extend its reach by making it a trochoid, it will create elliptical orbits surrounding the planet. My quest is to discover if a smaller trochoid will resemble the orbit of a free falling object inside the planet as well, or not.

Thus, will that orbit under the influence of the linear attractive force inside the massive planet be elliptical, circular, or something else?

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    $\begingroup$ If the planet has uniform density then the enclosed mass at a radius $r$ is proportional to $r^3$ so the force ends up proportional to $r$. The equation of motion in the plane is $r''=r^{-3}-r$ where constants are omitted, which is the isotropic oscillator equation $\endgroup$
    – Sal
    Feb 4 at 21:41
  • $\begingroup$ Looking into the isotropic oscillator I read its trajectories will be elliptical, am I correct? $\endgroup$
    – ajorna
    Feb 5 at 20:32

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Expanding on a previous comment, in the case of an isotropic solid ball, the potential energy within its interior is a quadratic function of distance from the center (see Wikipedia's discussion of gravitational potential energy.) It is a linear function of $r^2= x^2+y^2$. Consider an object constrained to travel without friction within a straight-line tunnel that is a chord of that ball, say defined by $y=a$. The conservation of total energy then provides a relation of the form $$\dot x^2 + x^2=c$$ (after suitable adjustment of the units of measurement and other constants). Here $x$ is the rectilinear distance measured along that chord. This conservation equation is exactly that of the classic harmonic oscillator; and the trajectories are the familiar ones of an oscillating spring.

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    $\begingroup$ "...and the trajectories are the familiar ones of an oscillating spring." Which is to say, generically they're ellipses which are concentric with the ball. (By contrast, the usual inverse-square law force gives ellipses with the attracting body at one focus. So while it may look similar it's rather different.) $\endgroup$
    – Semiclassical
    Feb 4 at 22:56
  • $\begingroup$ I find it interesting though! As I hear physicists say the inverse square law seemed to be rather random by Newton, I guess the proof is wonky. That's why this Tusi couple got me intrigued. Thanks for the clarification about the trajectory for the oscillator. $\endgroup$
    – ajorna
    Feb 5 at 20:38
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    $\begingroup$ @ajorna My recollection from studying orbits in physics is that in the classical model the only power laws that give you closed, repeating orbits in a central force field are the inverse square law and the law with force directly proportional to radius. It takes a bit of work to prove this but it isn't particularly "random", just special. I suppose you could call it an interesting coincidence that the actual forces between celestial bodies happen to follow one of the only two power laws that have this property. $\endgroup$
    – David K
    Feb 7 at 21:05
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    $\begingroup$ I would say it is more correct to say that the proportional law inside the ball needs the inverse square law, since we can use the inverse square law to prove it. Moreover, the proportional law only works in the special case of a spherical ball of uniform density--not for the Earth, for example--whereas you can compute the gravitational field of any body (any shape, any distribution of density) by integration of the inverse square law over all parts of the body. I prefer a law that works universally to one that works only in a single fictional case. $\endgroup$
    – David K
    Feb 11 at 22:10
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    $\begingroup$ @ajorna en.wikipedia.org/wiki/Shell_theorem $\endgroup$
    – David K
    Feb 15 at 1:51

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