Gravity Graphs

ihwip · Jul 18, 2007

I was thinking about the 'shell effect' of gravity cancellation that occurs when inside a sphere. This is where the gravity of the shell of the sphere above you cancels itself out. I know there is a technical term but still... Wouldn't this mean that on the traditional gravitational graph the Earth would appear more as a 'ring' where the very center of the Earth would spike back up to the same level as the space outside of its gravitational field? I am not sure how to explain what I am visualizing through text so please bare with me. My main question would be then...what height would the center of the Earth be? The same as geosynchronous orbit perhaps? Or would it be the same as a LaGrange point? (My guess)

billslugg · Jul 18, 2007

IHWIP I do not know what you mean by "traditional gravitational graph", "appearing as a ring", or "spike back up." I can tell you that as you dig down into the Earth, the gravity increases until you meet the iron core. At that boundary it is about 5% higher than on the surface of the Earth. As you dig further into the core, the gravity decreases until it reaches zero at the center of the Earth. At the distance of the geosynchronous orbit, the gravity would be somewhat less than the surface of the Earth, maybe about 50%? At any LaGrange point, the gravitational forces add to zero by definition. <div class="Discussion_UserSignature"> </div>

nexium · Jul 18, 2007

Bill slugg may be correct, but old calculatons yield a non-linear decrease all the way to the center, because most of the mass above you subtracts from the gravity of the mass below you. We have only been able to test to about 8 miles below sea level, and the gravity change is typically smaller than the local variations due to nearby ore masses and voids. We do have accurate data for GEO altitude and it is about 3% of average sealevel gravity. Neil

ihwip · Jul 19, 2007

The graph I am speaking of is the blackhole graph used to describe the gravity well 4-dimensionally. Basically what I am saying is that due to the loss of gravity as you dive closer to the gravitational center, the graph would start going back up. Thus instead of being a dimple it would be more similar to a crater with the center spike.

vandivx · Jul 19, 2007

it doesn't 'spike up' for Earth and I don't think why it should for BH [donning sci-fi hat] IF and I say IF BH did have singularity then you would be either outside of it and so would suffer 'outside pull' of it, that is full dose of its gravitation and as you fell into the singularity and became part of it, then ... then I just don't know, I go splat and my brains fry and all fuses burn and all that [/donning sci-fi hat] vanDivX <div class="Discussion_UserSignature"> </div>

billslugg · Jul 19, 2007

There is a graph here of the gravity inside the Earth. <div class="Discussion_UserSignature"> </div>

emperor_of_localgroup · Jul 20, 2007

The graph is very strange. We learned in our physics class 'g' goes down as we travel to the center because mass decreases within the sphere at the point of location. Now the graph shows 'g' actually going up in the Mantle compared to 'g' on the surface. Isn't gravity mysterious? I'm sure the scientists have an 'explanation' of this apparent breakdown of mass-gravity relation. <div class="Discussion_UserSignature"> Earth is Boring </div>

heyscottie · Jul 20, 2007

I'd say so. The 'explanation' is that the crust and mantle is less dense than the core. With a uniformly dense object, g would decrease as you remember. With a uniform object, as we move inward, radius decreases linearly, and mass within my radius decreases cubicly. With a non-uniform object where the middle is denser, radius still decreases linearly, but mass decreases less than cubicly. This holds up until we reach the core, where density is greater and, from there on in, more uniform. g then decreases to 0. If you want, you can think about yet another "discontinuity". The atmosphere of the planet contributes to the planet's mass as well. If we consider the new "surface" of the planet to be the nominal edge of the atmosphere, we calculate a starting g less than 9.8 m/s/s. Then, as we move to the center of the earth, g rises quickly (up to 9.8 m/s/s) until we reach the outer edge of the crust, increases less quickly until we reach the outer edge of the core, and then decreases until we reach the center.

emperor_of_localgroup · Jul 20, 2007

The decrease in 'g' above the earth surface is of course due to the distance, and mass of atmosphere is negligible compared to mass of the earth. But when you enter the earth, it's not distance but mass will control the value of 'g'. If the graph is correct, it is telling us mass of the crust and mantle is insignificant compared to mass of the inner and outer cores. In other words, mass of the earth is solely determined by mass of the outer and inner cores. Interesting... <div class="Discussion_UserSignature"> Earth is Boring </div>

MeteorWayne · Jul 20, 2007

Not solely, but the Iron-Nickel; core IS much denser than the lithosphere. <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

heyscottie · Jul 23, 2007

Exactly! Or, at least mostly exactly. It has everything to do with density. You don't look at mass or distance decoupled from each other, but rather tied together -- which is density. Density of atmosphere << Density of crust << Density of core

Saiph · Jul 29, 2007

IHWIP: You're essentially correct. The "gravity" graph typically used to illustrate black holes, where stronger gravity is a "deeper" depression will indeed show a spike in the center for planets and stars. For black holes...it gets a little weird with the singularity an all. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

ihwip · Jul 31, 2007

OK, that sounds about right but how 'high' will the spike be compared to the fabric of space outside the Earth's surface? I am thinking it would be on the same level as the LaGrange points.

Saiph · Jul 31, 2007

at the very center....flush with zero potential level, i.e. as if there was no gravity due to the object itself. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

emperor_of_localgroup · Aug 1, 2007

If that is the explanation for earth, then how would you explain gravity graph of gas planets? For those planets the 'g' will keep increasing and increasing until one gets to the small solid core of the planets??? The sentence, 'Jupiter's g is 2.54 times higher than earth's g' doesn't make much sense. Does it? Because which point would you consider surface of the Jupiter? I guess we have to settle for a different explanation for gas planets. <div class="Discussion_UserSignature"> Earth is Boring </div>

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