Wonderful!

Glad it helped!

The time it takes to go through the Earth is the same as the time for a satellite to go around the Earth once. (A one to one relationship).

The matter outside the sphere defined by its position, does not apply a force on the vehicle, only the mass inside the shrinking sphere as it approaches the center of the Earth.

Michel: Please explain, why wouldn't it? If a small object with mass m was touching the center of a flat surface of a huge spherical cap with mass M floating somewhere in the empty space, would you argue that Newton's Law of Gravity (F =GmM/x^2) does not apply if, at the same time, this small object was attracted by a nearby sphere? I would argue that it does. The center of gravity for the cap would be the same as its geometrical center, i.e. z = (3/4)(2R-h)^2/(3R-h) where h = cap height and z = centerline distance from flat base. Note that in this case x = r + z where r is the distance of the small object from the sphere's center.

Think about it this way. If you are at the center of the Earth. All the mass around you pulls on you but all the forces will cancel and thus there is no force of gravity at the center. This same principle holds no matter where inside the sphere you are.

​@@MichelvanBiezen Thank you for your explanation, however, why would it be wrong to assume that the gravity force balance acting on the vehicle falling to the center of the Earth is determined by an expanding spherical cap above the vehicle (F upward) and by a shrinking truncated sphere below the vehicle (F downward)? The motion would still be harmonic, although math-wise this model is complex and probably requires a numerical solution. Sorry, but I'm stubborn and have spare time while in "house arrest" during this pandemic.....

I like you tenacity. Typically I encourage my students not so spend too much time trying to figure what they did wrong, but rather learning how to do it correctly. As it turns out we have an entire playlist of videos that shows how to calculate the gravity caused by mass distributions which will show you why it works that way: PHYSICS 18.1 GRAVITY WITH MASS DISTRIBUTION

Yes, you can use a function approximating the density, but then the problem would not be a pure oscillator

@@MichelvanBiezen 👍

Glad it was helpful. 🙂

@@MichelvanBiezen a lot..

A "restoring" force is defined as a force that will push the object to the equilibrium point, which in this case would be to the center of the Earth.

Were does gravity start?

The rotation is just a mathematical analogy, but in reality the "angular" frequency does not require rotation in the real physical world. Simple harmonic motion with a mass oscillating on the end of a spring is such an example.

When we are dealing with a spring, the k represents the spring constant. k = (force on spring) / (elongation of the spring) Here we need to find the "equivalent" k for the Earth system.

yes, this is a very interesting problem.

Have a look at Gauss's law

As long as there is a force that acts like F = - kx there will be simple harmonic motion. That force does not need to come from a spring.

Interesting analogy! 🙂

Yeah@@MichelvanBiezenman.. Binary Beefburgers are the way forward when it comes to Space Travel.. B=MC Squared... Burgers = Mac Sauce and Chocolate Shake to the Crack Pipe Power of Two Baby.!! 0️⃣🤜🧢🍔💥😜1️⃣