Cutting a hole in the moon (hyptothetical physics)

Factual Questions
1

Hey everyone. Longtime lurker, first-time poster. I apologize for the length in advance, but I wanted to be as clear as possible.

A friend and I have been debating a hypothetical gravity issue and we’ve come to a standstill. Suppose that we cut a hole 10 feet in diameter directly through the center of our moon (or any other hypothetical moon or planet that you choose). If I jumped into this hole, I think everyone can agree that I would eventually come to rest in the middle of the moon. Where my friend and I differ is the reason why I would come to rest in that spot. Let me get the assumptions of the hypothetical out of the way, then I’ll explain the different opinions.

Assumptions:
The moon is perfectly round.
The hole is perfectly round and goes directly through the center of the moon.
The walls of the hole are built with indestructible super titanium that can’t possibly collapse.
The matter that was previously in hole is shot out into deep space and irrelevant.
I have a special suit so breathing and pressure aren’t an issue.

I figure that the center of the gravity is in the same spot (directly in the center of the moon), regardless of whether or not the hole exists. Therefore, when I jump in, I will fly right by the center, and almost come out the other side. The process will repeat and slowly degrade until I eventually float in the middle. What happens after that, I’m not sure (implode?), but it’s also irrelevant because it’s not the point we were debating.

My friend claims that by creating the hole, the center of gravity of the moon is no longer in the center of the moon. It’s hard for me to explain his opinion in text, but think of it this way:

( * | | * )

This is a cutaway side view of his model. Parenthesis are the edges of the moon, the asterisks are the new centers of gravity (which would be in that position uniformly around the whole sphere), and the parallel lines are the hole going down the middle.

My view would look like this:

( |*| )

Basically, his opinion is that I am at rest because all the gravity is now equally pulling me away in all directions. My opinion is the exact opposite; I am at rest because all gravity is equally pushing me in from all directions, because I’m right in the center.

I suppose that this question can be answered in a simpler manner. Can the center of gravity of an object be an empty space, or must there be mass in that space? Is the center of gravity in this moon unchanged when the hole is created?

Bonus question: What happens to me once I reach the middle. Do I implode?

2

Well, if this is the heart of the matter, you’re wrong: gravity is a purely attractive force, so the mass of the moon around you couldn’t possibly be “pushing” you into the center. Gravity always pulls, never pushes. So in that sense, your friend is correct: you can be at rest in the center of the moon hole because the pulling forces of all the mass around you sum to zero.

However, you’re right about the center of gravity, and your friend is totally wrong. The center of gravity is always defined as a single point; talking about a center of gravity that is a ring through the topography of your Moon donut doesn’t really make any sense. There is definitely no requirement that the center of gravity of some object be located at any point within its physical extent. All it is is the point where, for some purposes, you can treat the object as though it were a point mass; for example, if you replaced the moon with a tiny concentrated dot with the same mass, but located at the moon’s present center of gravity, the path of a satellite that was orbiting the moon would be unchanged. The same would hold true for your moon-with-a-hole-in-it, even though that center is not currently occupied by moon matter.

3

Ok, that definitely clears up one aspect that I wasn’t sure about. Thanks!

Now, what I meant when I said that the gravity would be equally pushing me in, is that if I’m falling down (although I understand that ‘down’ is arbitrary), once I pass the middle and start coming out the other side of the hole, I was thinking of that in terms of being pushed back toward the middle.

In other words, if you draw out a diagram showing which direction the forces of gravity are pointing towards, they are all pointing from the outer surface of the moon towards the center of the moon, correct?

Would they not all be pointing directly at me when I’m floating in the middle? To me, this looks like all of the force in the diagram is pushed directly at me from all directions. Am I seeing this wrong?

4

No, and this is why “center of gravity” is probably not a useful concept for talking about this specific scenario. It seems like you’re imagining gravitation as static lines of force that all point towards the Moon’s center of gravity, but this is not the case. Gravitation is simply an attractive force between all matter; pick any two particles, and there is a force pulling them both towards each other. When you are inside the Moon tunnel, you are surrounded by matter, and all of it is “pulling” on you. The force vectors on you would all be pointed out, away from the center of the Moon.

In that scenario, the net gravitational force on you is zero. When you’re standing on the surface, there is a net gravitational force on you that pulls you towards the Moon’s center of gravity, but in reality this is just the sum of the gravitational forces on you from every infinitesimal bit of matter in the Moon itself.

5

At the center of the tube through the Moon, you’re in an unstable equilibrium. If you moved off the tube axis a bit, the Moon’s gravity would pull you (very gently) towards the wall of the tube.

Your friend is right, if by “center of gravity” he is really talking about the location of lowest gravitational potential. Without the tube, this is obviously at the Moon’s center, but with the tube, the lowest potential would be a ring around the tube. Calling it the “center of gravity” doesn’t make any sense, though.

6

I’m not sure why you think you would implode, either. At the centre you would be effectively weightless - gravity would have very little net effect on you.

7

It can definitely be in empty space. Think of a ring (like a hula hoop) - the CofG is certainly not within the solid part of the ring.

Correct - its location is still the center of the sphere.

Why would you implode? The net gravitational force on you is zero.

Note that it would take a very long time for you to come to rest. Based on your description, the only friction damping your oscillations would be due to the relatively few molecules of gas you’d encounter in the vacuum found near the moon. You’d probably die of old age while still in motion.

8

I don’t believe this is correct. If you move away from the center you are closer to the mass on one side, but more mass is on the far side. These two factors balance out, giving no net force.

9

Interesting. Where is the location of lowest gravitational potential? I assume it’s going to be at some diameter larger then that of the hole (and thus within the solid mass of the planet), yes?

Does that imply that the point of lowest gravitational potential will always be within a solid portion of a mass (except for the mathematically perfect hollow sphere)?

10

For a very large hole, what’s left of the moon would look like a doughnut, more or less. As the hole gets smaller, the donut-ness decreases, but presumably the instability remains for even small holes (because otherwise there would be some finite sized hole above which the instability exists and below which it does not, which is hard to imagine).

11

As others have pointed out, the direction of the pull of gravity varies from place to place. Unless your object has special symmetry (like a sphere), the direction of the pull of gravity is NOT necessarily to the Center of Gravity of the object. As an example, if you had a magical giant cube of solid material that didn’t collapse under its own gravity into a sphere, the direction of the gravitational pull on itss surface almost never points in the direction of the center of the sphere. It does at the center of each side, and of each edge, and at the corners, but I don’t believe it does elsewhere. And if you drew all those gravity vectors, they wouldn’t all pass through the same point, or even a simple figure like a ring or a cube. You have to calculate it anew for each point.
In your example, the pull of gravity as you travel on a line through the center does indeed always point to the center, and it will do so effectively for a hole of diameter much less than that of the moon. When the hole starts to become an appreciable fracytion of the moon’s diameter, though, things will get more complicated. The gravity inside a spherical shell DOES cancel out – this is a classic physics problem. But it doesn’t inside a hollow tube within a solid sphere. You’ll have to work it out, but you will get pulled more towards the side you’re closer to than the other as you move away from the centerline. But, again, you’re not going to be pulled toward some ring through the solid remaining portion of the sphere. Exactly where the pull appears to originate from will depend upon your position within the hole and your distance from the centerline.

12

I don’t think you’d implode, but the force of gravity would be pulling your feet (which are past the center) upward while your head (on the opposite side) would be still be pulled downwards, right? Might feel a bit weird.

On the other hand, why would you come to rest at all? There’s no friction! :smack:

13

Gravity is not an issue. It does not get stronger in the middle.

You’d still be under the effects of moon gravity (1/6 earth) and the center of gravity would not change by removing the matter there. As the rest of the moon remains a contiguous mass, it has only one center of gravity. (The earth actually has two. Center Mass and the point between the center of earth’s mass and the moons. - the latter one is moving.) Pluto has one and there is a place between pluto and its moon that the gravity is balanced (the center of their combined masses)

Your assessment is most accurate.

14

I think its probably more accurate to treat a moon-sized object with a tube of 5ft radius through its center as sphere-like than donut-like.

From the staff report (by Chronos) of your link:

15

That works for a spherically symmetric body, but it isn’t true in general. The gravitational force for the Moon with a tube removed is the same as the force for the full Moon (:)), minus the force for the tube of matter that gets removed.

At a point inside the Moon a distance R from the center, the force due to the full Moon is just the force for the portion of the Moon within radius R. This increases linearly with R, assuming uniform density.

Assuming R and R[sub]tube[/sub] << R[sub]moon[/sub], the force due to the tube of matter of radius R[sub]tube[/sub] and length R[sub]moon[/sub] can probably be well approximated by the force due to an infinitely long cylinder of radius R[sub]tube[/sub]. For R > R[sub]tube[/sub], this will decrease like 1/R.

At R = R[sub]tube[/sub], the force due to the tube is larger than the force due to just the sphere, so at some point, those two forces are equal, and you’re at the minimum of the gravitational potential. I’d WAG that this occurs at something like two or three or pi times R[sub]tube[/sub], but I really don’t know.

Yes, that would be true. In electrostatics, for a distribution of charge (of all one sign), you’d say it was true by Gauss’s law. And of course, the equations are essentially the same.

16

You’re missing my point.

For a truly toroidal planet, like that described in the Staff Report, there’s no net force on a body in the exact center of the torus, but any body slightly off-center will have a net gravitational force that pulls away from the center–an unstable equilibrium, as ZenBeam describes. That means someone standing on the inside diameter of the torus would actually be standing there, attracted more by the mass under his feet than by the mass on the other side of the hole.

If that’s true for a torus, it’s still going to be true if you add mass to the torus shape to make it look like a hollow cylinder, right? And it’s still going to be true if the inside diameter of the hollow cylinder is 1/2 of the outside diameter, or 1/10 of the outside diameter, or 1/100 or 1/1000 or 1/10000000000. There’s no magical ratio where it stops being true, and it’s still true if the inside has a 5 ft radius and the outside is the same size as the moon.

What changes as the hole size decreases is that the magnitude of the force gets smaller and smaller. So the gravitational force on the inside wall of the moon-tube is, I suppose, practically zero, but it’s not exactly zero.

17

And this is why Ringworld needs attitude thrusters.

18

In tube-through-a-planet problems, we usually make the approximation that the diameter of the tube is negligible compared to the size of the planet, and thereafter forget about it. But if you do insist on considering it, then if you’re even a little bit off-center in the tube, you will be pulled a little more toward the nearest wall.

Quoth Stealth Potato:

Not quite true. Satellite orbits around the moon-with-a-hole would in fact be slightly different than they would be around intact-moon or moon-compressed-to-a-point. The famous result that the gravitational field is the same if you treat all of the mass as being at the center of mass is true only if you start with a spherically-symmetric mass distribution, and our moon-with-a-tube isn’t quite spherically-symmetric any more. The definition of center of mass is mathematical, but the physical significance is that it’s the motion of the center of mass that determines the momentum of the system, and a force pointing through the center of mass of a system will not exert any torque on the system.

19

I see the point, and agree.

20

:smack: Using Gauss’s law, it’s easy to show that the force at the surface of an infinite cylinder of a given radius is 3/2 the force at the surface of a sphere of the same radius. Using the variation of the force with radius for the two cases, the forces cancel at sqrt(3/2) * R[sub]tube[/sub]. So for a ten foot diameter tube, for example, the minimum gravitational potential would be along a ring of diameter 12.25 feet.

From the above, it would be half the strength you’d feel at the surface of a sphere of the same radius (yeah, pretty small).