# What shape would the orbit of a free falling object inside a 'massive' planet be according to Newton?

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?

• 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 4 at 21:41
• Looking into the isotropic oscillator I read its trajectories will be elliptical, am I correct? Feb 5 at 20:32

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.