Sorry Cecil but gravity is not greatest at the center of the earth. It is greatest at the surface. 2 things in support of my position. 1-Smaller planets have less gravity. 2- If you dig a hole to the center of the earth your feet will be pointing to the center. Once you reach the center, all directions would be up. Rendering you weightless. The density however would be greatest do to relativity.
If you’re referring to “What if you fell into a tube through the earth?”, then you have Cecil wrong, since he said:
Where does he say it’s greatest?
(And what does density have to do with relativity, unless you are equating gravity with relativity?)
I suspect Fauxguy is referring to http://www.straightdope.com/classics/a4_183.html where Cecil says
The reason gravity decreases as you go from the surface of a massive sphere downward toward the center is that the effect of the mass above your level cancels out, some of it pulling you toward the center, some of it pulling you back toward the surface. What Cecil was talking about is moving along the surface of the non-spherical earth from a point relatively close to the center (at the poles) to a point that’s relatively far from the center (at the equator). In this context, gravity does decrease the farther you get from the center, as Cecil says.
You are correct, mostly by accident.
1 Smaller planets generally have less gravity because they generally have less mass
2 Close. As already pointed out, the effect of mass above you will cancel out the effect of mass below you. As you move towards the surface, however, those forces will be out of balance and you will be drawn towards the center of gravity again. If you mathmatically modeled the earth as a point in space with the same math (kind of like a black hole singularity), the force of gravity would indeed be highest at the point. In fact, within about a centimeter of the center (Earth’s Schwarzschild radius) the gravity (and Schwarz) would be so strong that nothing would escape (exactly like a black hole).
3 Density has nothing to do with it
4 Unless you plan on traveling at or near the speed of light, relativity has nothing to do with it
The problem in the already-linked column is this:
Sorry, but there’s no other interpretation for that other than “boneheaded mistake.”
Yes and no. For purposes of gravity, as long as you are outside a sphere, all the mass can be treated as though it were concentrated at the central point. It is only when you actually go below the surface that this ceases to be true. Since the discussion is limited to observers at the surface, the center is the apparent point of maximum gravity.
This is somewhat complicated by the fact that the Earth is not a sphere. I haven’t the resources to say by how much.
All true, but it’s sufficiently incorrect (to state that the center of the earth is the point of maximum gravity) that the first reading set off my alarms. A moment’s reflection resolved the issue, but I still think the explanation would have been better served by referring to the “center of gravity” without the parenthesis.
The math is rather easy for spherical symetry. At any point outside of a spherically symetric object, it’s gravity is the same as a point-mass at its center. Going inside the object is just as easy. The gravity at any point inside of the object is the same as a point-mass at the center with the mass of all the material at a depth greater than the observation point. The mass above the observation point has no effect.
It may seem funny, but go ahead and do the math. The gravity anywhere inside of a hollow sphere, not just the center, is zero, i.e. the pull is equal in all directions. So when inside of a non-hollow sphere, you only feel the pull of what is below you.
But… That does not mean that gravity is always lower. A previous poster was wrong. Density does have something to do with it. For all of my previous statements to hold, you only need spherical symetry. That is, the density of the sphere must only be a function of the radius (distance from the center). The force of gravity at any depth will be the mass below times the gravitational constant, divided by the radius squared where the mass the integral of the density, as a function of radius, from the center, to the given depth (r^2 dr). If the mass decreases at a rate less than 1/r^2, gravity will increase with depth.
As a trivial example, a sphere with two layers. The inner sphere has a density of p1 and radius of one unit distance. The outer shell has a density of p2 and radius of two units. Gravity at the surface of the outer sphere is the total mass of the two spheres (volume times density for each) times the gravitaitional constant divided by radius squared,
G((4 pi/3)p1 + ((4 pi/3)(2^3) - (4 pi/3))p2)/4
=G((4 pi/3)p1 + (28/3 pi)p2)/4
=G (pi/3)(p1 + 7p2)
Gravity at the edge of the inner sphere is,
G(4 pi/3)p1
A simple comparison shows that if p1 > (7/3)p2, the gravity at the edge inner sphere is greater than at the surface. (And if p1 = p2 the gravity on the outer sphere is twice that of the inner, which we expect.)
So density does matter. The gravity at the surface may not be the maximum. It depends on the density gradient. We really all know this already without even realizing it. Here at the Earth’s surface, we’re still swimming around in this ocean of gases called the atmosphere. The Earth’s atmosphere definitely exerts gravitational force like the solid (and liquid) at and below the surface, but we don’t think twice about the fact that as you rise up in the atmosphere, the gravitational pull decreases even though the net mass below us, the solid-liquid planet plus all of the air below us, is increasing.
That said, I have no idea what the density gradient of the Earth exactly looks like. I know it is much more dense at its nickel-iron core than at the surface, but is it enough for the maximum gravity to be below the surface?
If we interpret “max gravity” to be shorthand for “deepest point in the gravity well”, then “max gravity” is at the center of the Earth, and (which I think is the point of 7), you will age most slowly at the center of the Earth, and progressively more quickly as you climb out of the well.
An excellent point yheffen - and one I only realised a few years back when someone corrected me about it in a thread on this board.
Apparently, though there are large uncertainties involved, gravity actually remains roughly constant in the upper layers of the Earth’s interior. According to the (rather badly labelled) graph in this undergraduate course exercise, gravity g rises slightly down to the core-mantle boundary and only starts falling once you’re inside the core. But that’s only in one model and the details will be model dependent.
The gravity you feel standing at the edge of the hole (1G, vector towards the Earth’s center) is the sum of the gravitational vectors of all the Earth’s atoms, and after you’re pushed in (just assumption, I sure wouldn’t jump in) the gravity changes due to more and more of the atom’s vectors moving behind you as you fall deeper into the Earth. The gravity does not decrease (did the Earth suddenly lose mass?) only the vectors change. After all the oscillations damp out and you are suspended motionless at the Earth’s center you are not feeling 0 G, you are feeling about 1/2 a G pulling radially outwards in all directions from the center of your body.
No, you feel 0g (note capitalization, by the way; capital G has a very other meaning in connection with gravity). If gravity worked the way you think it does, the Apollo astronauts would have had very different experiences.
If the gravity just cancels itself out at the center of the sphere and becomes zero doesn’t that mean it would be perfectly safe to observe two 10 solar mass black holes orbiting each other 100 yards apart from dead center between them on the 50 yard line? I think the fact that your navel is in a comfortable 0g (thanks for the G/g tip) is little comfort to your head and feet as they accelerate towards their respective black holes with more force than the tensile strength of your body.
Well, first of all, your black holes would be well inside each other’s event horizon, so they will soon be merging into one black hole of 20 solar masses.
Second, you are talking about a Lagrange Point. If you have two black holes of 10 solar masses each in stable orbit around each other, there is no reason you could not rest comfortably between them.
The only reason your head and feet would be pulled in opposite directions would be if the change in gravity increased at such a rate that the forces pulling your feet were significantly higher than those pulling at your head. Kind of like if you fell into the black hole itself. But basically, as long as you stay out of the event horizon, a black hole is no more dangerous and does not pull any harder than any other object with the same mass.
You’re confusing tidal forces - the change in gravitational attraction over a set distance - with gravity itself.
A large mass concentrated in a tiny point, as in a black hole, will have such a steep gradient that the head and feet of a human would have different enough stresses on it to pull it apart at close range. (At a distance, as said, it would behave exactly like any other object with similar mass.)
There’s no such gradient at the center of the earth because the mass is spread out over a comparatively huge area.
That is to say, as comfortable as one can be in the frozen vacuum of space.
This isn’t correct (or it reads strangely). Gravitational fields behave according to superposition, so the two clauses above are equivalent (gravity decreasing, or adding vectors in opposing directions). So your body at an ideal point in the center) isn’t feeling the stress of being tugged in a million different directions outward–the vectors sum to nothing so there is, equivalently, no gravity at that point.
Yes, I’m talking about gravitational gradients, and didn’t we just prove there is one when the top of the hole is at 1g and the center is not? Not that it’s black hole strong, but reducing the scale of the experiment (as long as we stay above quantum levels) should only reduce the magnitude of the force, not eliminate it. msmith537 brought up Lagrange points, but a Lagrange point is just that, a geometric point. Any object with size, even if it’s center of mass perfectly coincides with, say, one of the Earth/Moon Lagrange points, has part of it’s mass closer to Earth and part closer to the Moon. This should create a tensile force (weakly) stretching the object between the Earth and the Moon. Can’t we pair up every atom in the Earth (nearly every atom, as the Earth is not perfectly round or smooth) with the atom that is the same distance away from the center in the opposite vector and treat the center of the Earth as the coincident Lagrange points of several quintillion pairs of atoms?
There is no gravity inside a hollow sphere from the hollow sphere. Now you’ve introduced a whole new situation that has nothing to do with the original question.
No, the Lagrange solution only works if m[sub]a[/sub]>>m[sub]b[/sub]>>m[sub]c[/sub]. There is no Lagrange point when m[sub]a[/sub]=m[sub]b[/sub].