You’ll end up at rest in the center iff the only source of friction is something that scales with speed, like air resistance. If the walls of the tube are of some merely mortal material, you’ll probably get stuck before that. If there’s no friction at all, then not only will you oscillate indefinitely, but you’ll do so with the same period regardless of how off-center the tube is, which is also the same as the orbital period for a low-Earth-orbit satellite.
Thanks again to everyone for posting, I didn’t expect to generate so many interesting sidetracks. With hindsight I should have omitted any reference to a tunnel, it was only meant as a means of transport to get Dave to the centre of the planet, but turned out to be somewhat misleading.
It seems the solution has been given by hdc_bst - thank-you, I am now confused at a much higher level. I’m now on a mission to properly understand Shell Theory - time to dust off the old calculus books.
You can do it much more easily with Gauss’ Law. Of course, that just shuffles off the calculus burden onto proving Gauss’ Law, but it’s at least intuitive.
Mohawk?
Fridgemagnet said:
You can understand the concept without resorting to calculus. Simple vector math and a few cases can be illuminating.
Consider you floating in space nearby a large object - say a pointmass planet. You will be pulled toward that planet, no?
Okay, let us substitute three smaller spherical orbs of uniform density in place of that planet, so they form a line one on top of the other in front of you.
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Note that each object will try to pull you toward its own center - acting like three point masses. Now consider that they are the same size and mass. Break the vector forces into an X-Y grid horizontal and vertical, and you will see that the middle orb pulls you in the negative X, and that the upper and lower orbs pull you somewhat in the negative X and somewhat Y and -Y respectively. But the Y and -Y components are equal and opposite, exactly canceling, so you only get the combined pull in the -X.
Now look at a second case, where you move the upper and lower orbs to the far side so you are at the center of an equilateral triangle. Similar vector math will show that the force from each orb is offset by the vector parts of the other orbs.
Calculus does the same example, but places you in a sphere of infinitely small point masses infinitely spaced around you and then sums the infinite points. Same answer - all the pieces vector cancel each other out.
That’s simple enough when you’re at the center of the sphere, but the force is also zero everywhere else inside the spherical shell, too. That’s a little harder to prove, or to see intuitively.
Thanks also to Chronos and Irishman for discouraging me from attempting calculus in my advancing years - you’re quite right, there are still plenty of chores around the house that need doing…
You’re right I’m sure, but sadly I was wondering if these two scenarios are not the same in all respects. I have no problems grasping the concept when it comes to electromagnetic fields, so I’m not sure why I would imagine gravity would work in the same way as string. Not 10/11/26-etc-dimensional superstring string, just string, tugging away. The vector analysis would be the same, and if the object being tugged on by various bits of string is stationary then the resultant force would be zero, but the object would still be subject to very real tensions.
I shall write down a hundred times: “Gravity isn’t string”.
For the record HMHW, Dave’s hair is a mess, what with the mud and the floating about and that.
The difference between gravity and string is that gravity pulls on every bit of you at once. If you attach one string to my left ear and another to my right ear and pull them in opposite directions, my head feels no net force, but my ears will be pulled off my head. If you attach a left-pulling string at every point where you have a right-pulling one, though, they will completely cancel out, and no bits of me have to be pulled off.