How does gravity work?

Question

In my physics class, the professor said that (theoretically) if we could drill through the center of the Earth and drop a rock down the hole, the acceleration due to gravity would decrease linearly as the rock descended to the center. Then in my Geophysics class I learned about some theorem (forget what it was exaclty) that said you could treat the gravity of an object as if all the mass of the object were concentrated at its center. If the latter were the case, the rock would continue to accelerate at 9.8 m/s^2 all the way through it's fall to the center.

Answer 1

The Law of gravity says that the force between two bodies is:

F = G M m / r^2

Where M is the mass of object 1 and m is the mass of object 2 and G is the universal gravitation constant.

When you talk about g=9.8 m/s^2 that is a valid aproximation near the surface of the earth. Where we write F = g m and
g = G M / R ^2 where M is the mass of the earth and R is the radius of the earth. It doesn't matter whether you are next to the Dead Sea or at the top of Mt. Everest g=9.8 m/s^2 is good aproximation. It is not addequate however for a satalite in geo-synchronous orbit, the hubble space telescope, or computing the orbit of the Moon around the Earth or the planets around the Sun.

One of the very important facts is that Gravity is an inverse square law, the force goes as 1/r^2 (r squared!) There are some very important properties that are only true if the force is an inverse square law. If the law becomes 1/r^(2+0.00001) that is enough to break these rules.

The therom you are refering to is valid because gravity is an inverse square law. However ... for this problem of a hole all the way through the middle of the earth its not exactly the best way to think about the question.

Consider the rock falling through the hole, to be part way between the top and the center of the earth. We need to draw a picture where you have two "shells" THe rock is at some position r above the center. You have one "shell" which is the sphere of radius r and has all the mass of the earth inside that sphere. For that sphere you can consider all the mass concentrated on the center.

You also have a hollow shell with all the mass of the earth between r and R ( the radius of the whole earth). Some of the mass in this shell is pulling the rock up away from the center, and some of the mass is pulling the rock towards the center.

It turns out for a spherical shell of uniform density, the gravitational force on an object inside the hollow part of that shell ( anywhere in the hollow space) always works out to be zero.
I tried to remember how to to the integral to prove that, an I'm afraid I can't remember how. (An appology is due for a hand waving arguement here, but now you know why I don't have a Ph.D. in either physics or applied mathematics.)

So the only mass that is pulling on the rock when it is at r from the center is only the mass of the earth inside the sphere of radius R.

Consider, the Volume of a shpere is 4/3 pi r^3 where r is the radius of the sphere. The mass of the sphere is M = rho * 4/3 * pi * r^3 where rho is the density of the material in the sphere.

So when the rock is at r it feels a force of F= G M m / r^2 and
the M is rho * 4/3 * pi * r^3 and r^3 / 3^2 is simply r
so we end up with

F = G 4/3 pi r m for the force on the rock when it is at height r above the center of the earth. So the acceleration is G 4/3 pi r
and does decrease lineraly as the rock decends to the center.

The key point is how to handel the spherical shell of matter outside r because this shell does apply a force to rock when it is at r. It turns out when you integrate this force it works out to zero.

Looking at this more closely you could use the theorm about considering all the mass concentrated at its center if you draw a plane tangent to the shpere at r, and consider the part of the shell above the plane as pulling the rock up and the part of the shell bellow the plane as pulling the rock down.

Have fun, Enjoy physics. With questions like this you are really on the right track to mastering the subject.

Answer 2

You have received a number of answers that are too theoretical for what you're trying to understand. In particular, the response involving general relativity isn't relevant to your needs here.

First, just to clarify what your physics professor said, the ACCELERATION of the falling rock will decrease as it falls through the earth, but its VELOCITY will continue to increase until it reaches the center. As long as there is even a little bit of acceleration downward, the downward velocity will continue to increase. (Of course "down" and "up" might reverse their meanings as the falling rock passes the center of the earth, but you see what I'm saying.)

What you were told in geophysics (that you can treat the body as if all its mass were concentrated at its center) is a reasonable approximation to how gravity works, and this approximation applies to the body's pull on objects OUTSIDE the body itself. For objects INSIDE the body (or inside a hole in the body), it should be clear that not all of the body's mass is pulling the object toward its center; some of the mass is trying to pull the object back up toward the surface.

Incidentally, the last sentence in your question is not correct. If you could treat the mass as if it were concentrated at the center of the body, then the acceleration (or gravitational force) would INCREASE as the rock fell toward the center. The closer it came to the center, the greater the attraction would be (based on the inverse square law).

If you have had enough calculus to understand the following concept, I think that this will give you a much more meaningful view of the issue you asked about:

The force of a body's gravity on any object is equal to the sum of the gravitational forces exerted on that object by each of the body's parts. To arrive at that overall force (i.e., to determine its magnitude and direction), we can INTEGRATE the "differential" force (i.e., the infinitesimal pull) exerted on the object by each point within the body. This would involve evaluating a triple integral (over the 3 linear dimensions) of the body's density (mass per unit volume at any particular point) divided by the square of the distance (from that particular point to the object), multiplied by the object's mass (a constant) and multiplied by the gravitational constant.

That sounds like a complex formula, but the point is that the body's gravity is simply the sum of the gravitational force of each element of mass within the body. Yes, we can assume that all of the mass is at the center of the body, but that produces only a reasonable approximation to the exact value of the gravitational force exerted by all of the units of mass that make up the body. It is a better approximation for objects that are far away from the body.

For objects that are ON THE SURFACE of the body (as we are on the surface of the earth), the effect is a bit different. We can see that the parts of the body near the surface and in the vicinity of the object (say, a rock that lies just below the earth's surface 100 yards away from you) will create more of a LATERAL pull on the object than a downward force.

And for objects that are BELOW THE SURFACE of the body, it is clear that some parts of the body are pulling the object UPWARD, not attracting it toward the center of the body.

Hope that helps you.

Answer 3

That would be true if there was no normal force.

An object sitting on the ground does not go toward the earth because the earth pushes upward on the object.

So your question would then be, why would the earth push up if its acceleration is down.

The answer is, you can only view the forces acting on the object itself. The forces on the object is gravitational force and normal force that have a net gain of zero.

The earth itself is another object; it pushes up because the dirt underneath is pushing up on the earth on the surface.

This explanation can go on and on until the dirt we speak of is the center of the earth (or magma). Simply put, something in the way of a falling object will provide a counter force to either stop it or slow it down.

Answer 4

It's a little more subtle than that. The theorem states that the gravitational field at distance R from a *spherically symmetric* body (one whose density depends only radius r) would be the same as if all the body's mass interior to r=R were concentrated at the center. Corollary to this is that the mass exterior to r=R has no net contribution.

Answer 5

General relativity says gravity is a curvature in the space-time contiuum caused by the mass in the area (space-time would curve around it, kind of like a ball on a rubber sheet).
Quantum mechanics says gravity is caused by a sub-atomic particle called the graviton which has yet to be discovered.
Because of this, no one really knows, and a theory uniting relativity and quantum mechanics would answer this.

Answer 6

The same thing that causes gravity everywhere else, mass. For gravity on Earth to change the mass of the planet would have to change or the radius of the planet would have to change.

Answer 7

No one really knows how gravity works.It's one of many basic theories that physicists are still trying to reconcile with all the other basic theories of how our universe works into what's know as a "Grand Unified Field" theory.

Answer 8

it works by pushing force down to a surface.

Answer 9

I don't know but with the world spinning so fast i'm surprised we don't fly out.....