Gravity below the surface ...

[12345ask]

Hi! I think I understand this, but I am having trouble explaining it fully, which is making me think that I don't fully understand it. I've tried looking everywhere on the internet, but don't seem to be able to find what I'm looking for.

 

Gravitational field strength is inversely proportional to the square of the distance between the larger and smaller and objects at distances greater than the radius of the larger object. However, it is directly proportional to the distance from the centre of the larger mass for distances less than the radius of the larger massed object.

 

This is in reference to a solid spherical object of uniform density, e.g. a planet. My question is basically, why does field strength vary in this way? Why is g proportional to r below the surface of M?

[InfiniteNow]

I would guess it's because once you get below r, you are below the surface of the mass in question (in some sense, now part of it).

 

Once your distance from center is less than the radius, you must then also take into account the gravity above you (or, at a greater distance from center than you are), as this matter would now be pulling you in the opposite direction from center. I'm just guessing though... :hihi:

[cwes99_03]

I would guess it's because once you get below r, you are below the surface of the mass in question (in some sense, now part of it).

 

Once your distance from center is less than the radius, you must then also take into account the gravity above you (or, at a greater distance from center than you are), as this matter would now be pulling you in the opposite direction from center. I'm just guessing though... :hihi:

 

Precisely. When below the surface of a solid body like a planet you get a bit of gravitational pull in the outward radial direction (pointing away from the center). The same is true for all physicial forces. Check out a physics book and look into the electric field inside a spherical shell with a surface charge and then at a solid sphere with an equal distribution of charge throughout.

 

Complex equations must be used for putting satelites into orbit because of non-uniformities in the earth's gravitational field, but largely (when talking about planetary motion or motion of a car across the surface of the earth) we can ignore these things, or just use a more precise measure of g for the location of the experiment.

 

Cwes

[Erasmus00]

This is in reference to a solid spherical object of uniform density, e.g. a planet. My question is basically, why does field strength vary in this way?

 

Inside a spherical shell of radius R you feel no gravitational force. This has to do with the way the gravitational field varies, but you can think of it roughly as the sphere pulling you equally in all directions.

 

Outside a spherical shell, all the mass acts like its at the center. So, if you are inside the earth, the pull on you is proportional to the volume of the sphere beneath you, divided by r^2. Volume goes like r^3, so r^3/r^2 goes like r.

-Will

[infamous]

Precisely. When below the surface of a solid body like a planet you get a bit of gravitational pull in the outward radial direction (pointing away from the center).

Very logical answer cwes99_03; The closer to the center and more distant from the surface one becomes, the more balanced the gravitational effect. This is true because at the center of a body such as the earth, the gravitational effect will become the same in all directions. This brings to mind a question which I have also not been able to find the answer to. Suppose for the sake of argument we were able to dig a hole completely thru the earth, bisecting it's center exactly. Let's imagine that our hole is of uniform diameter thru-out. Now let's take a lead ball, say of ten pound weight and drop it into the hole. My question is this: Will the ball, because of accumulated velocity, pass thru the center only to reach a lesser height on the other side and return to dropping in the opposite direction back to the center in a pendulum motion until finally comming to rest at the center. Or, will it slow down on it's first trip down and come to rest at the center without passing thru it? The answer to this question may tell us much about the true character of gravity..........Infy

[Erasmus00]

Newtonian mechanics and gravity would predict that your ball would pass through the center and make it almost to the other side of the planet. If you neglect any air-resistance or damping processes, this oscillation will go forever. If you don't, then the ball will slow down in its oscillations and eventually come to a rest in the center of the planet,

-Will

[Qfwfq]

You are also supposing the hole to be from pole to pole else there'd be trouble due to Earth's rotation.

 

In saying exactly proportional to radius you are also supposing the sphere to be of uniform density, Earth's density is presumeably greater at the centre.

 

Neglecting any r-dependence of density, an amusing result is that the oscillation period of the ball would be exactly the same as the period for an orbit of the same radius.

[infamous]

Thank you both, Erasmus00 and Qfwfq for your considerate and intelligent responses. It is evident that both of you gentleman are highly educated in your respective fields and deserve recognization for such. The reason for my initial question being; The current controversy between the 'Standard Model' and 'The Final Theory' concerning the relevant character of Gravity. If this experiment could be done on a smaller scale, one within our current technological ability, it could settle the question once and for all: 'Is gravity a pulling force or is it a pushing force?'

 

Consider the expected results according to the Standard Model. As both you fine gentlemen have stated, an ocillation thru and past the center would be expected.

 

Consider the Final Theory, or with respect to it's real author; 'Gravity is the 4th dimension'. This theory suggests that gravity is only the effect of an expanding material existence. If this is the case, then the body dropped into the thru hole of the main body will travel to the center slowing down upon approach and stopping at the center on the first trip down. No ocillation would result, the body would find center and remain there undisturbed.

 

This experiment would, regrettably, be a mammoth undertaking and will most likely never be feasable or affordable. A body of enough mass, isolated from external gravitational influence could theoretically be constructed, so this experiment could in theory be conducted. The chances are however, next to nothing that this experiment will ever take place, this is why I term the situation, regrettable

 

 

Testable knowledge is such an expensive thing...................Infy

[TheBigDog]

Consider the Final Theory, or with respect to it's real author; 'Gravity is the 4th dimension'. This theory suggests that gravity is only the effect of an expanding material existence. If this is the case, then the body dropped into the thru hole of the main body will travel to the center slowing down upon approach and stopping at the center on the first trip down. No ocillation would result, the body would find center and remain there undisturbed.

If I understand the expansion model, the falling object would approach, but never reach the middle of the hole as the precise center of the hole would never expand up the the 'falling' object. This is similar to the idea... that thing where you can travel half the distance to someplace an infinite amount of times, therefore you can never get all the way there. I can't recall the name. Anyway, if that were true you would never reach the middle of the earth. Since it is not true, I would reason that the standard model is better.

 

Bill

[infamous]

Anyway, if that were true you would never reach the middle of the earth. Since it is not true, I would reason that the standard model is better.

 

Bill

You make a fair point Dog, thanks for your observation...............Infy

[cwes99_03]

Reminds me of the slingshot approach theorized in the most recent reincarnation of an old sci fi TV show whose name skips my mind at this moment.

 

I can't see how one would expect the object to slow down as it reaches the center. The reason being that gravitational effects are completely radial.

 

Assuming that the density of this massive object with a through whole is uniform throughout, the object would begin accelerating radially inward toward the center of mass. As it travels the effects of the mass surrounding it would be negated by the perfect symmetry of the massive object's density (i.e. all mass not in the radial direction of the center of the mass would be negated due to symmetry.) Now the actual gravitational pull would decrease, so that the rate of acceleration toward the center would decrease, but the objects velocity would have no reason to decrease without any force to cause it (say friction which i have wonderfully neglected).

 

As far as rotation of the "planet with the through hole" I believe the object would most likely not hit the walls, but let's investigate this since it has been some 5 years since I've done any real thought on the issue.

 

We drop the ball from a stationary position at the "planet's" surface. By stationary of course we mean traveling at the tangential velocity of the surface of the earth. I guess the only question then is whether that tangential velocity is greater or less than the tangential velocity of the wall of the whole at any point along the lead ball's path through the "planet", and whether the tangential velocity of the ball increases or decreases as it moves downward. My assumption would be that since we are neglecting friction (which would decrease the tangential velocity as well as the radial velocity) the initial tangential velocity of the ball would cause it to make contact at (r_hole)/(v_i-v_t) where v_t is the tangential velocity of the wall at time t when ball makes contact with the wall. Besides a little fenagling with the algebra to get t on one side of the equation does that sound accurate?

[Pyrotex]

...This is similar to the idea... that thing where you can travel half the distance to someplace an infinite amount of times, therefore you can never get all the way there. I can't recall the name.

Xeno's Paradox

[Pyrotex]

...We drop the ball from a stationary position at the "planet's" surface. By stationary of course we mean traveling at the tangential velocity of the surface of the earth. ...?

I remember that I was reading an old SF novel about the Earth being hollow, and a whole civilization lived on the inside surface of that hollow -- at the same time I was taking Freshman advanced physics.

 

We had the problem of what would be the gravitational force within just such a hollow Earth, with the hollow sphere concentric with the Earth itself. You pick any point on the surface of this hollow. You pick any tiny chunk in the body of the Earth, and calculate the gravity on you from that chunk.

 

Now you have to do Integral Calculus, in three dimensions. You integrate the "lattitude" angle until you have the gravitational force from a ring of Earth's material. You integrate the "longitude" angle until you have the force from a thin shell of material. You integrate over the radius of the Earth to get the total force. And it was freakin zero! No matter how big the hole inside the Earth. That ruined the SF novel for me. :hihi:

 

So, if the ball starts at the equator, it had an initial tangential velocity of about 1000 mph. So does the surface of the Earth at the equator. But as the ball drops, would it not KEEP its 1000 mph vector? Half way to the center of the planet, the walls of the well would only be rotating at 500 mph. So if the ball were still IN the well, it would have to have lost 500 mph of its tangential velocity in friction with the wall. Or am I missing something?

[infamous]

Or am I missing something?

Actually, Qfwfq pointed out earlier that the hole should go from pole to pole to eliminate this influence................Infy

[Qfwfq]

Unlike Earth, the moon doesn't seem to have a fluid interior yet its gravity isn't negligible. Care for a spot of diggin'? :lol:

 

That's the first time I've seen it spelt Xeno's paradox Pyrotex, sure it isn't Xeno's paradoz? :hihi:

 

Google gives 165,000 hit to 44,400.

 

I guess the only question then is whether that tangential velocity is greater or less than the tangential velocity of the wall of the whole at any point along the lead ball's path through the "planet", and whether the tangential velocity of the ball increases or decreases as it moves downward.

Ask Coriolis.

 

If you drop the ball without any force the two are initially equal. In inertial coordinates, avoid a wild goose chase by considering angular momentum conservation. The ball would follow an eliptical orbit if there were nothing for it to crash into but meanwhile the tunnel is turning around. To make it even simpler use Earth coordinates but without forgetting omega-cross-v.

[infamous]

Unlike Earth, the moon doesn't seem to have a fluid interior yet its gravity isn't negligible. Care for a spot of diggin'? :hihi:

 

If you have your spade in hand, might we also consider a large asteriod? One far removed from the local gravitational influence of the earth and moon, an asteroid might be an even better candidate.................Infy

[cwes99_03]

If you drop the ball without any force the two are initially equal. In inertial coordinates, avoid a wild goose chase by considering angular momentum conservation. The ball would follow an eliptical orbit if there were nothing for it to crash into but meanwhile the tunnel is turning around. To make it even simpler use Earth coordinates but without forgetting omega-cross-v.

 

See that's where i'd have to brush the dust off my old books to see if my above theory was right.

 

Pyro, I think you are on the same line of thought that I am, however, if the lead ball were to read the midway point(r/2) then the through hole must be awful wide.