Gas-Powered Planetary Expansion – Evidence And Calculations

From: Andrew Johnson

Date: 2015-04-14 21:49:32

Attachments : We’ve been working on this for some time (on and off for almost 1 year), so here it is, for those that are interested…   www.checktheevidence…     Gas-Powered Planetary Expansion – Evidence And Calculations (A Mechanism for Earth Expansion �C Part 2) © Peter Woodhead (peterchapel@googlema…) and Andrew Johnson (ad.johnson@ntlworld….) March/April 2015 Better Formatted PDF Version   Video Discussion www.youtube.com/watc…   Introduction   Following our previous article/posting “A Mechanism for Earth Expansion[1]”, it is the intention of this article to offer corrections and additional material in support of the theory of Gas Powered Planetary Expansion.     Serendipitously, on the same day as a related video was posted by us[2], new research was published in the New Scientist 12th June 2014 edition, in an article entitled “Massive ‘ocean’ discovered towards Earth’s core”[3]. This research received considerable exposure[4] in other publications[5], although research from approximately 7 years earlier[6] was not mentioned.     In our original article and video, a challenge was made to the seismologists regarding the real nature of the earth’s internal structure. This challenge was made because it is generally concluded, even before the appearance of the latest research about “underground oceans,” that at least part of the core is liquid iron[7]. We suggest that the core is actually gaseous �C mainly steam, and around this sits a layer of water, heated by one or more processes happening in the earth’s crust, which is above it.   An Ocean Towards the Earth’s Core?   One image which has appeared[8] following the recent postings about a “subterranean ocean” is shown below.   This model suggests much less water would be present than in the Gas-Powered Earth Expansion  (GPEE) model. However, a small amount of water does not seem to negate the GPEE model altogether – as has been previously stressed, steam expands to many times the volume of the water it is formed from[9], so it can be calculated that enough water will be still available based on the suggested depth and thickness of the transition zone shown above.   Ringwoodite and “Slush”   In the earlier article and video, we proposed that there was a layer of what we called “slush” �C some kind of material which was not quite water, but was not ice. We also calculated the volume of water in the inner earth, based on the figures we had suggested then. Now, with the new/additional information about the proposed Ringwoodite layer, we can improve these calculations.   In the new study, a type of rock called “Ringwoodite” (of which we were previously unaware) was mentioned. This is contained in the “water bearing” layer (Transition Zone) �C shown in dark blue �C in the diagram above.   Ringwoodite would be undergoing an expansion of its own, as the 5% or so of water is heated and released into the earth’s gas core.   The depth and thickness of the Transition Zone, as shown above, are 410km and 240km respectively. Radius of “bounding spheres” would then be 6370 – 410 = 5,960 and 6370-650 = 5,720km. Assuming a uniform thickness of this zone around the earth, the volume of this zone would be given by:     It is stated that the water content is between 1.5% and 2% by weight.[10] Since ringwoodite has a specific gravity of 3.9g cm3 then the water contained by volume will be somewhere between 3.9×1.5 = 5.85 and 3.9×2 = 7.8. These 2 figure are then averaged to 6.845% by volume. Hence the volume of water in this layer alone would be:   0.068 x 1.03 x 1011 = 7.02 x 109 km3   It has been estimated that our oceans currently hold a volume of 1,335,000,000 km[11]3 – the volume calculated above is approximately 5 times this figure. This therefore lends credence to the idea that the oceans could have come from underneath the present mantle.   One thought to consider regarding the origin and formation of the oceans is the age of the bed of the Mediterranean Sea (shown on the “rainbow” ocean floor map[12]). Could the first out-welling of basalt and water have occurred with “venting” from below the mantle in this area? Could that venting have been initiated by a meteoric or cometary impact? Perhaps further study of geological features may reveal more about this possibility.   An additional question, then, would ask why the ringwoodite contains the water in the first place �C how did it from down there below – all the hotter mantle material, if the accepted model of a hot/molten core is correct…? Surface Gravity and Centre of Mass Force   When the first article on this subject was published by us, it was inevitable that some estimates were made, including assumptions about the quantity of subterranean water and the depth to centre of gravity in the crust.  We have here attempted to make a better estimate for some of these figures, but we are still dealing with many variables and unknown quantities. However, we have also attempted to look at other related figures in an effort to improve and/or validate these estimates.   It became clearer that available evidence did not support the previously stated 300 km depth of the earth’s Centre of Gravity, or “Centre of Mass Force” as we have come to call it. The 300 km figure came about as a result of calculations that were over simplistic. Therefore, several attempts were made to formulate a means to calculate the surface gravity of a spherical shell, of uniform thickness. The method used was to consider the gas-filled earth being split into a number of segments, and consider the force of gravitational attraction on a mass at the surface. The following 2 diagrams are possible ways of visualising this:   Each segment exerts a force on masses on the surface. In the diagram above, the horizontal forces cancel each other out, but the vertical forces are additive. This is a kind of 2D-visualization. An enhanced 3D visualization is shown below.       m1 is a mass on the surface, which is attracted by the mass in all of the mass of the earth m2 is a ring/shell. The volume of the shell is given by the area x thickness of ring and this is therefore proportional to the mass. The vertical force is proportional to sin θ where θ is the angle of the line of force measured perpendicular to the equatorial diameter, D above.   The Shell Theorem   Some people will immediately point out that “the gravitational force will still act as if it is at the centre of the earth �C even if the earth is hollow or gas-filled!” They will point to examples like this one[13]:   Such a shell can be envisioned as a stack of rings. Field Outside a Massive Spherical Shell   It is hardly surprising then that, in researching a method for calculating the CMF more accurately, we became aware of Newton’s Shell Theorem. Which is stated thus[14]:   Newton’s Shell Theorem states essentially two things, and has a very important consequence. First of all, it says that the gravitational field outside a spherical shell having total mass M is the same as if the entire mass M is concentrated at its center (center of mass). Secondly, it says that for the same sphere the gravitational field inside the spherical shell is identically 0.   This is explained quite well in a YouTube video entitled “Universal Gravitation — Shell Theorem”[15]. We would argue that calculations based on the “hollow shell” idea are only accurate when measurements are made at a sufficient distance away from the earth. If it is true that most of the mass of the earth is concentrated in a shell, and the core is actually gas-filled, then the centre of mass force (CMF) will not be in the core of the earth.   We would agree with questions asked comments made in a Maths/Physics forum in May 2009 by a poster named “geistkiesel”[16]   Where, and how, does the Newton Shell theorem (NST) place the CMF at the sphere COM? It doesn’t. The question of locating the CMF is not discussed! The NST model begins with a ring on the sphere of differential mass dM oriented perpendicular to m and centered on r. By summing the force for each dM on each ring then integrating over the surface of the sphere, the total force, F = GmM/r^2 results, which says nothing, absolutely nothing, regarding the location of the CMF.   Another poster (unsurprisingly) responded:   the result of the integration is:     which is exactly the same equation you would get if all the mass were at the center. how can you possibly not understand that?     Further discussion ensued, but posters did not seem to accept the validity of the question. For example, if “the force acts from the centre,” then would it not be the case that there should be no gravitational anomalies as measured from the surface?   Intuitively, when you look at a diagram showing an object in contact at the surface with the shell, it seems unlikely that the mass of the far hemisphere can exert the same force on the object as the near hemisphere. To establish where this CMF might be for the expanding earth, a spreadsheet was developed. We realise that to produce a more accurate figure for the CMF distance, we need a revised version of the calculations which are shown in the “Field Outside a Massive Spherical Shell” example above. We need to use integration, but that level of maths is beyond us at the moment.     With the above equation, it is assumed that the centre of mass is at the centre of the earth. R is the radius of the earth �C which is the same as the distance to the assumed centre of mass.   With a hollow or gas-filled shell, the centre of mass attracting a mass on the surface, would not be at the centre of the earth �C the mass of shell would be closer to the mass on the surface. Hence, the effective R would be smaller than the radius of the earth (Rc < Re).   It is therefore possible that the mass of the earth is actually much smaller than the accepted figure, so when divided by a smaller r2, we get the same overall resulting force at the surface...   CMF Calculations   A more accurate result could be determined by considering the earth's different layers (which, according the GPEE model would be lighter material as you descend). The actual distance to the CMF is proposed to be at around 1993 km - determined by using the spreadsheet calculations.   A similar spreadsheet created using pre-expansion/solid earth statistics puts the CMF at a depth of 3764 km �C close to pre-expansion earth’s centre.  Therefore,  gravity at the pre-expansion earth’s surface would be roughly 23% of that today if we use depth to CMF 1993 km (gravity measurement derived) and 36% using the a CMF depth of 2250km. This would therefore allow us to explain why the large flora and fauna existed on a pre-expansion earth with a surface gravity 25% of what it is today.   Gravity Anomalies at Altitude   In considering the force of gravity when on a high mountain, one might consider it can be affected by 2 factors.   a)      The force of gravity would be slightly reduced by being further away from the centre of the earth b)      The force of gravity would be slightly increased by having “dense hard rock” beneath you.   Several web pages allow the force of gravity at altitude to be calculated[17] (all other factors being equal). Using the summit of Everest at 9,000 metres as an example,  the page calculates the force of gravity will be reduced by 0.28%. This figure is essentially calculated as follows:  Explanations for gravity anomalies vary �C for example, a study called “Geophysical anomalies along strike of the Southern Appalachian Piedmont”[18] suggests   A regional trend in the Bouguer field is determined for an observed eastward decrease in crustal thickness based on seismic refraction measurements. A comparison of the calculated regional with surface boundaries shows that in places the crustal thinning occurs more than 50 kilometers west of the Inner Piedmont-Charlotte belt transition.   Could this “crustal thinning” be a result of the gas-filled structure of the earth? The differences in crust thickness would, arguably, have a much more significant effect on the local force of gravity, due to the reduced distance from the surface to the centre of mass force.   In August 2013, New Scientist published an article entitled “Gravity map reveals Earth’s extremes”[19]. The article notes some extremes of gravitational acceleration. For example a reduction of about 0.5% (based on the accepted average acceleration due to gravity of 9.81 ms-2) has been measured at the summit of mount Nevado Huascaran Peru in the Andes at 7,000 m �C i.e[20].   The Nevado Huascarán summit (Peru) with an estimated acceleration of 9.76392 ms-2   Using the afore-mentioned webpage[21], we would expect to find the acceleration due to gravity to be, 9.7814 ms-2 -a reduction of 0.32%   If we were to suggest this was the result of the CMF being at 1993km depth, we can calculate the expected change as follows.   This study identifies the places with the highest and lowest gravity acceleration:   A candidate location for Earth’s maximum gravity acceleration was identified – outside the SRTM area, based on GGE-only �C in the Arctic sea with an estimated 9.83366 m s-2. This suggests a variation range (peak-to-peak variation) for gravity accelerations on Earth of about ~0.07 m s-2, or 0.7 %, which is about 40 % larger than the variation range of 0.5 % implied by standard models based on a rotating mass-ellipsoid (gravity accelerations are 9.7803 m s-2 (equator) 9.8322 m s-2 (poles) on the mass-ellipsoid, cf. Moritz [2000]). So far such a simplified model is also used by the Committee on Data for Science and Technology (CODATA) to estimate the variation range in free-fall acceleration on Earth [Mohr and Taylor, 2005]. However, due to the inhomogeneous structure of Earth, presence of topographic masses, and decay of gravity with height the actual variations in free-fall accelerations are ~40% larger at the Earth’s surface   Could some of these anomalies be explicable by the CMF being in a position other than the core of the earth? If we take the highest and lowest stated figures for g, we can explain the total variation measured fits well with the calculation made above i.e.:   9.83 (Arctic Ocean) – 9.76 (Nevado) = 0.07 ms-2 variation   0.07/9.81 = 0.0071 = 0.7%   In a paper entitled “Specific Gravity Field and Deep Crustal Structure of the ‘Himalayas East Structural Knot’”[22], a number of Gravity readings are shown in Figs 5, 6 and 7 (Fig 7 is shown below). When first discovered, the this survey of the Tibetan plateaux was very exciting, it seemed to point to a 0.1% reduction in g for an increase in altitude of 1,000 metres. That would have supported a depth to CMF of around the 1993 km mark giving a value for g atop Everest of 99.1% of that at sea level. However, the stated gravity figures were not actual measurements but were “calculated”. Since we do not know how they were calculated we could not determine what the actual measurements were. So, we did not have any measurements to corroborate the proposed 0.1% drop in g per 1,000 metres, it was then realised that the figures actually pointed to a drop of just 0.01 mGals per 1,000 metres, that would give a drop in g atop Everest of just 0.09%, less than a third of the 0.28% reduction calculated from the accepted “solid” earth model. Until we gain some actual “weight” readings at altitude this will remain a puzzle.      These and similar studies tend to explain these anomalies by inferring that beneath the mountain areas were areas of “less dense substrate”, also that under the low altitude areas lay “much denser substrate”. We would argue that the reductions are more significant because the CMF is closer to the surface than is accepted/used in the calculations.   We would like to appeal to anyone who would be able to take weight measurements at high altitude �C e.g. anyone who lives in the Rockies in the USA or anywhere in a mountain range and is able to weigh a 100g or 200g weight at sea level and then go to as high an altitude as possible with a sensitive weighing scale and weigh the same weight and compare the results. (This was tried on a holiday flight at approx 10,000m but the aircraft was simply not stable enough to give a usable reading.)   Angular Momentum   Since part 1, additional (simple) calculations were done to consider the earth’s angular momentum and the current day length. From this, the approximate length of the earth’s day in the time of the dinosaurs can be calculated. Similarly, the equatorial centripetal force can be calculated, but this is essentially negligible. These figures are shown below. So this does not really “factor in” as an explanation for the gigantism of flora and fauna.   MYA Day Length (Hours) Equatorial Centripetal Force (N/kg) 300 8.38 0.16 250 9.09 0.14 200 9.86 0.13 150 10.95 0.11 100 (Message over 64 KB, truncated)

Related articles...

Comments are closed.