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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?

ajorna
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    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 – Sal Feb 04 '23 at 21:41
  • Looking into the isotropic oscillator I read its trajectories will be elliptical, am I correct? – ajorna Feb 05 '23 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.

MathFont
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    "...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.) – Semiclassical Feb 04 '23 at 22:56
  • 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. – ajorna Feb 05 '23 at 20:38
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    @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. – David K Feb 07 '23 at 21:05
  • @DavidK My point is this is a way to let the inverse square law follow out of the force directly proportional to radius: that they are identical; and that the force proportional to radius is actually primary; the inverse square law (physically) needs it to exist for itself to come into action. It is only due to the shell theorem that we can pretend there can be an inverse square law without an object (that is more than pointlike). But in reality this is not the case – ajorna Feb 11 '23 at 21:59
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    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. – David K Feb 11 '23 at 22:10
  • @DavidK Newton goes into attraction at different shapes in one of his chapters indeed, I still need to read that part. I would be very interested in a proof of the proportional law by the inverse square law; it is not in the Principia. Would you know where I could find such a proof? – ajorna Feb 14 '23 at 10:31
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    @ajorna https://en.wikipedia.org/wiki/Shell_theorem – David K Feb 15 '23 at 01:51
  • @DavidK It makes sense. I know that the inverse square law and the proportional law are equal to each other on the surface of a massive sphere. And applying the shell theorem means the exterior shell nullifies in attraction when you go deeper inside the sphere, basically shrinking the effective size of the massive sphere for the attraction at your new point. You are now at the surface of basically a smaller sphere with a smaller mass, where again the inverse square law equals the proportional law, ad fundum linearly apparently until the tank is empty. Thanks, it clicks now! – ajorna Feb 15 '23 at 22:57