Maximum force of gravity?

tvanflandern
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21 years 11 months ago #4430 by tvanflandern

Reply from Tom Van Flandern was created by tvanflandern

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>The downward force we experience on the surface of the earth is 1 g (by definition). What is the largest downward force that could exist on a planet of "infinite" size?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

The answer is still unknown. We have some numerical constraints on the graviton model in two papers in Pushing Gravity, the ones by Slabinski and by myself. However, the graviton flux by itself (as opposed to combinations of flux with other parameters) is not yet well-constrained.

BTW, it is not the size of a planet that matters, but its mass. For example, Jupiter has 300 times the mass of Earth. But its large size takes its surface farther from most of its mass, resulting in a much smaller surface gravity. Jupiter has only about 2.6 times the surface gravity of Earth. If Jupiter were larger still and of lower density, it could have less surface gravity than Earth. -|Tom|-

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glittle
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21 years 11 months ago #3370 by glittle

Replied by glittle on topic Reply from Glen Little

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>Jupiter has only about 2.6 times the surface gravity of Earth.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Is this known by observation or by theory? And if by theory, which one? <img src=icon_smile.gif border=0 align=middle> Since the density of material close to the "surface" is not great, maybe the local gravitational shielding from the planet is reduced, so the surface gravity is less.

Or... if it is indeed 2.6 g, is it also more "focused"?

I'm not sure how to explain my thought, but try this... Imagine a cone, the tip of which coincides with where I am standing on the surface of the earth at, for this example, the North pole. The sides of the cone correspond with a line indicating 0.8 (for example) of the downward force (grey area below). Maybe on earth, the angle at the tip where I am standing is about 75°. On a denser planet, it might be 90° while on a gaseous planet, more like 40°. (Red lines are trying to show direction of push by gravity. Straight down, the force is 1g. As the angle comes up, the force would reach 0g at about horizontal.)
<img src=" members.shaw.ca/gplittle/misc/earth-gravity2.gif " border=0>
Is this thinking correct? Would the angle depend on the density of the planet?

I wonder what that would do to our sense of balance? Would it be easier to tip over on a denser or lighter planet? Maybe we'll have to wait until we visit Jupiter more properly to find out!

Glen

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tvanflandern
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21 years 11 months ago #4766 by tvanflandern

Replied by tvanflandern on topic Reply from Tom Van Flandern

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>[tvf]: Jupiter has only about 2.6 times the surface gravity of Earth.

[glittle]: Is this known by observation or by theory?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Both. We have had orbiters at Jupiter, and sent a probe deep into the upper portions of the gas giant planet.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>And if by theory, which one? <img src=icon_smile.gif border=0 align=middle><hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

I don't understand the question. There is no alternative to Newtonian gravitation until you get out to about the eight decimal place, where relativity effects begin to matter.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>Since the density of material close to the "surface" is not great, maybe the local gravitational shielding from the planet is reduced, so the surface gravity is less.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

If gravitational shielding exists at all, it is too weak an effect to have been detected yet in any but the most ultra-high-precision experiments with satellites and lasers.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>Or... if it is indeed 2.6 g, is it also more "focused"?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

I don't understand the rest of your questions -- perhaps because they were based on some assumptions in conflict with the points I just outlined. -|Tom|-

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MarkVitrone
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21 years 11 months ago #4432 by MarkVitrone

Replied by MarkVitrone on topic Reply from Mark Vitrone

How does the surface area of a body in relation to its compostition change the force of gravity generated by the body. For example does a large sphere with low density generate less gravity due to lower shielding than a smaller body with higher density where shielding is increased? Does this relate to this discussion about the strength of gravity or the newtonian mechanics of known bodies? MV

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glittle
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21 years 11 months ago #3429 by glittle

Replied by glittle on topic Reply from Glen Little

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>
I don't understand the rest of your questions -- perhaps because they were based on some assumptions in conflict with the points I just outlined.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Fair enough...

Since reading your book many years ago, I've been trying to change my thinking about gravity from a force that pulls me toward the earth, to an external force that is pushing me toward the earth. Would you say that is the right way of thinking about it?

So when I think about gravitational sheilding, I'm thinking of the effect induced by the presence of a massive object (the planet) next to me - it is sheilding me from the impacts of the gravitation "particles" from that side. Since I'm being pushed by more gravitons coming from the other direction, the net effect is that I'm pushed toward the earth.

Is that a valid way of thinking about it? How would you rephrase that into more correct terms?

Now see if I can rephrase the question above...

Standing on the surface of the earth, I'm being pushed by x (zillion) gravitons from directly above my head (pressure exerted: 14 pounds per square inch?). From directly below me feet, the planet is blocking y (zillion) of the gravitons, so there is a net force downward exerted by (simplisticly) x-y gravitons. If I look toward the horizon, from that direction, I'm still being pushed by about x gravitons, since nothing is blocking their path. (From the opposite horizon, I'm pushed by the same amount, so no net force.)

When I look 10° below the horizon, at the dirt 100 feet away, from that direction I'm being pushed by x-z gravitons where z is the number blocked by the little bit of the planet they had to pass through. Let's move to 45° below the horizontal. From there, x-w gravitons are pushing me. The size of w (and z) will depend on the density of the material that the gravitons had to pass through. On earth, that might be 40% of y, but on Jupiter that might be 20% of y. So the amount and distribution of the gravitons reaching me from below, through the planet, will depend on the material in the planet in various locations.

I guess what I'm getting at is that the "shape" of the gravitional force that I feel on the surface will be dependant, not on the overall density of the planet, but on the distribution of that density. If Jupiter has a very dense core, but is very sparse near the surface, the "shape" will be different than on earth where the surface density is much more similar to the core density.

Is that any more understandable?

I don't know if there will be any practical use for this concept, but it may help clarify our thinking about how gravity works.

Glen

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tvanflandern
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21 years 11 months ago #3166 by tvanflandern

Replied by tvanflandern on topic Reply from Tom Van Flandern

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>[MV]: How does the surface area of a body in relation to its compostition change the force of gravity generated by the body. For example does a large sphere with low density generate less gravity due to lower shielding than a smaller body with higher density where shielding is increased? Does this relate to this discussion about the strength of gravity or the newtonian mechanics of known bodies? MV<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Shielding is not a factor in ordinary gravitation. If it exists at all, it is at the limits of our ability to detect. But when there is shielding, the greater the shielding, the less the resultant gravity. (Note distinction below between shielding and shadowing, which are opposite effects.)

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>[Glen]: Since reading your book many years ago, I've been trying to change my thinking about gravity from a force that pulls me toward the earth, to an external force that is pushing me toward the earth. Would you say that is the right way of thinking about it?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Your thinking is correct, but perhaps unnecessary. An important part of the Le Sage model is that gravity can be thought of as a push coming from the outside, or as a pull coming from the inside. The resultant force is identical either way. So if you are more comfortable, as most of us are, with thinking in terms of pulls, there is nothing wrong with that.

Think of it this way. A graviton heading toward the Sun might hit the Earth instead, giving Earth a push toward the Sun. But that push would not exist if the Sun had not blocked another graviton heading toward the Earth from the sunward side. So we can describe Earth's reaction as responding to a push from the outside, or equally well as responding to a negative-mass (missing) graviton on the inside that pulls instead of pushes.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>So when I think about gravitational sheilding, I'm thinking of the effect induced by the presence of a massive object (the planet) next to me - it is sheilding me from the impacts of the gravitation "particles" from that side. Since I'm being pushed by more gravitons coming from the other direction, the net effect is that I'm pushed toward the earth. Is that a valid way of thinking about it? How would you rephrase that into more correct terms?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

A point of terminology. We call the blockage of gravitons by masses "shadowing", not "shielding". Any two bodies in space shadow one another from some graviton impacts. Shielding occurs when bodies become so dense that gravitons can no longer get all the way through them, so some matter deep in the interior no longer blocks gravitons and therefore no longer contributes to the external gravitational force.

Here's another analogy to try out. Suppose a swarm of bees flies in front of the Sun. Ordinarily, that shadows us from some sunlight, and the amount of shadowing is proportional to the number of bees. But if the swarm were so dense that some bees were already in the shadow of other bees, then those shadowed bees would block no additional sunlight. They would fail to contribute. That's shielding.

Forget shielding for all but the most advanced purposes. Ordinary gravity works by shadowing.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>Standing on the surface of the earth, I'm being pushed by x (zillion) gravitons from directly above my head (pressure exerted: 14 pounds per square inch?).<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

The 14 psi is the weight of the atmosphere above us, not the pressure of the gravitons. The gravitons from above exceed the gravitons from below by 1 g (10 m/s^2 acceleration).

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>From directly below me feet, the planet is blocking y (zillion) of the gravitons, so there is a net force downward exerted by (simplisticly) x-y gravitons. If I look toward the horizon, from that direction, I'm still being pushed by about x gravitons, since nothing is blocking their path. (From the opposite horizon, I'm pushed by the same amount, so no net force.)<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

That looks right.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>When I look 10° below the horizon, at the dirt 100 feet away, from that direction I'm being pushed by x-z gravitons where z is the number blocked by the little bit of the planet they had to pass through. Let's move to 45° below the horizontal. From there, x-w gravitons are pushing me. The size of w (and z) will depend on the density of the material that the gravitons had to pass through. On earth, that might be 40% of y, but on Jupiter that might be 20% of y. So the amount and distribution of the gravitons reaching me from below, through the planet, will depend on the material in the planet in various locations.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

All true. But Newton long ago showed that all spherical shells of uniform density give rise to the same outside force as if the same mass was concentrated at the shell's center. Moreover, such shells give rise to zero internal force (everywhere interior, from shell to center).

Earth can be considered, to good approximation, as made up of many layers of uniform spherical shells. Density can change from shell to shell without changing the next effect. Local density variations are very minor compared to the force from the rest of the whole Earth or planet. But in fact, the direction of the vertical does change slightly, especially near mountains or seashores. Astronomers and geodicists have to take that into account in certain types of precise observations.

An extreme case is an asteroid with a highly deformed shape. Eros is such a case. There are places on Eros where the net gravitational force is sideways or even upward relative to the local surface.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>I guess what I'm getting at is that the "shape" of the gravitional force that I feel on the surface will be dependant, not on the overall density of the planet, but on the distribution of that density. If Jupiter has a very dense core, but is very sparse near the surface, the "shape" will be different than on earth where the surface density is much more similar to the core density.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

True in principle. But because of the uniform spherical shell axiom, there is not as much difference as you might think. However, Jupiter is a lot more flattened than Earth, largely in response to its faster spin.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>Is that any more understandable?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

It is indeed. I only

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