Identifying the Gravitational Constant

[davidgow77]

This short paper is designed to address the dimensional problem of the value of G being given in Nm2 kg2. It aims to show that the gravitational constant is actually a constant describing the action of two very weak forces, namely the force exerted by the spatial vacuum upon one mass, and the force that the same vacuum exerts upon a second mass.

Gravity as a Force related to the Energy Density of the Vacuum

davidgow77@aol.com

Tuesday 1st March, 2005

Summary

This short paper is designed to address the dimensional problem of the value of G being given in Nm^2 kg^2. It aims to show that the gravitational constant is actually a constant describing the action of two very weak forces, namely the force exerted by the spatial vacuum upon one mass, and the force that the same vacuum exerts upon a second mass. This paper also introduces the constant of accommodation (A), which describes the maximum amount of energy that can fit into 1m^3 of space, and states that the vacuum energy density (AV) is -8.168-06J/m^3. It also gives a constant of proportionality for the force that the vacuum exerts upon any given mass (AF), measured in N(1/2) m kg.

Identifying The Gravitational Constant

We know that the force of gravity is proportional to the mass of the object in question, and inversely proportional to the square of the distance between the center of the mass-carrying objects, but the problem with gravitational force is that, in absence of a second mass carrying body, the force does not exist except as a potential force. An analysis of G needs to take into account the context within which the gravitational constant is used, so we will first look at the Newtonian equation used to calculate the gravitational force between two mass carrying objects:

F = GM1 M2 / r^2

My first observation here is that the gravitational constant is only used once; it does not need to be used a second time in conjunction with the second mass (M2). This would seem to imply that the gravitational constant, G, must be a proportional constant that describes the properties of both gravitational fields (of both masses) simultaneously, since when we add the second mass into the equation the constant does not change or need adjusting. What I would like to suggest here is that we substitute G and replace it with a constant that purports only to explain the properties of one gravitational field.

Given that the units of G are Nm^2 kg^2, it may be better to represent the equation by stating that F = (sqrt(G))^2 M1 M2 / r^2 . This gives a value of -8.168-06 N(1/2) m kg for the square root of G (sqrt(G)). This also allows us to split the equation into two parts and allows us to show exactly how the gravitational constant should be related to each mass. For the ease of notation, we shall say that AF = sqrt(G) (thus AF = -8.168-06 N(1/2) m kg), where AF is the constant of proportionality of the force of the vacuum (FV) exerted upon any mass, and can be accurately represented by stating that the force of the vacuum (FV) is proportional to the mass of the object, and inversely proportional to the distance from the center of the mass:

FV = AFM / r

To show how this forms the basis of Newton’s equation for the force of gravity, we take the FV for two masses, and represent them as:

F = (AFM1 / r) x (AFM2 / r)

or

F = (AFM1) x (AFM2)
r^2

If we substitute the values into the above equation and use two familiar masses for M1 and M2, then we should get exactly the same result as when using G. We shall say M1 is the Earth, and M2 is a 1kg object at the earth’s surface. Thus:

(AFM1) = (8.168-06 N(1/2) m kg x 5.98^24 kg) = 4.884464^19 N(1/2) m

(AFM2) = (8.168-06 N(1/2) m kg x 1kg) = 8.168-06 N(1/2) m

The force of gravity is essentially the product of the force exerted by the vacuum by each of the two masses. In this respect, the force exerted upon one mass by the vacuum is multiplied by the force exerted upon the second mass. This results in an attractive force (gravity) between the two masses, which is strong enough to give rise to motion where one of the masses is substantially large enough.

Therefore, the force of gravity between the two masses must be:

F = (4.884464^19 N(1/2) m) x (8.168-06 N(1/2) m kg)
r^2

or

F = (4.884464^19 N(1/2) m / r) x (8.168-06 N(1/2) m / r)

The distance, r, between the center of the earth and the 1kg mass (essentially the radius of the earth) is 6376500m, so if we input this final information we should get:

F = 3.98977395214 Nm^2
6376500m^2

F = 9.81 N

Certainly using the value AF demonstrates why the dimensions of the gravitational constant G are Nm^2 kg^2, but its primary advantage is that it shows us from a classical mechanics point of view how gravity is propagated, and indeed what it is; gravity is a resultant force (FR) of the product of the two partial forces exerted on two mass carrying bodies by the vacuum, thus:

Where FV1 = AFM1 / r
And FV2 = AFM / r
Then FR = FV1 x FV2

[BREAK]
The question now is, how does a vacuum exert a force on a mass carrying body? What I would like to propose here is that the vacuum of space is a “negative energy plenum”, where each cubic meter of space can accommodate a maximum amount of energy of 8.168-06J, termed the Constant of Accommodation (A). This essentially means that the vacuum of space has a vacuum energy density (AV) of -8.168-06J/m^3, and that space is inclined to be accommodated with the energy held within the matter in order to cancel out these energies and give a zero value. This inclination manifests itself as a partial force (measured in N(1/2)) determined by the constant of proportionality AF.

The important point to note here is that this approach requires that mass carrying bodies do not occupy space, but rather exist within their own spatially extended dimensions. This is something that Einstein thought significant when he stated in Relativity:

“Physical objects are not in space, but these objects are spatially extended. In this way, the concept of empty space loses its meaning” [1]

In response to this I would say that empty space is the volume of space that was previously occupied by energy but which, due to the propagation of matter, has now been vacated. From this, we should deduce that there is a corresponding amount of energy (A) in the observable universe for every cubic meter of space (or conversely, every cubic meter of space can only accommodate 8.168-06J of energy (A), which in turn suggests a vacuum energy density (AV) of -8.168-06J/m^3. If this is true, then we should be able to calculate a rough estimate of the Critical Mass Density (CMD) simply by using the constant A (derived form the gravitational constant), and the approximate volume of the universe.

According to WMAP[2], the radius of the observable universe is 13.7 billion light years, giving an approximate radius of 1.296120075^26 m. Given that the universe is uniform in all directions, an accurate way to calculate its volume would be to calculate the volume of a sphere with radius Ru:

Vu = (4/3p)x Ru^3

V u = 4.188790205 x Ru^3 (where Ru = 1.296120075^26 m)

V u = 4.188790205 x (1.296120075^26m)^3 = 9.120619146^78 m^3

From here, two equations are needed to find the mass of the universe. The first shows the equation for finding the energy in the universe (Eu), while the second converts the units of measurement from joules to kilograms, and thus from energy into matter.

Eu = Vu x A

E u = 9.120619146^78 m^3 x 8.168843247-06J/m^3

E u = 7.450490812^73J

The solution to this equation states that the amount of energy in the universe should be equal to 7.450490812^73J. All that is left to do now is to convert this into mass to give a rough approximation of the amount of matter in the observable universe (Mu):

Mu = Eu /C2

Mu = 7.450490812^73J / 8.987551787^16 m/s

Mu = 8.28978902^56 kg

This calculation assumes that all energy within the universe exists in mass form, which is clearly not the case, but the value 8.28978902^56 kg does give a good indication of the absolute maximum amount of matter in our 13.7 billion year old observable universe. From this information, it should also be possible to ascertain the critical density of the universe. If we accept that the constant of accommodation, A, sets the energy density of the universe at 8.168843247-06J/m^3 then the calculation for the critical density of the universe should be:

A/C^2 = 8.168843247-06J/m^3 / 8.987551787^16 m/s = 9.089063897-23 kg/m^3

or

9.086063897-28g/cm^3

This value is slightly lower that the 1 x 10-29 g/cm^3 that seems to be the general consensus for the value of the critical mass density (CMD) of the universe. The most important point to take from this rudimentary calculation is that G, A, AV and the CMD of the universe are inextricably linked. The value given for the CMD of 9.086063897-28g/cm3 should also be viewed as a maximum value, given the fact that it is based upon the notion of all the energy in the universe existing in matter form.

In practice, the only way to measure the values of A, AF, AV and the critical density is with more accurate measurements of G, but with the knowledge of what G actually is (G is the square of the constant of proportionality of the force of the vacuum, AF), the accommodation constants may provide, within a classical framework, a deeper understanding when it comes to grand unified theories and theories of quantum gravity.

[Casey_deJong]

Unfortunatly a 1 kg item here is much less of a burden say on the moon.
If we wish to know the true effect of gravity then we must no look at size or weight but density of mass. Then again i might just be saying something because i'm starved for attention :)

[mkirkpatrick]

Unfortunatly a 1 kg item here is much less of a burden say on the moon.
If we wish to know the true effect of gravity then we must no look at size or weight but density of mass. Then again i might just be saying something because i'm starved for attention :)

Is not the moon one sixth of
the earths gravity,I am sure it is!

regards michael.

[G_burnett]

There are some interesting developments along these lines in this research.

　Author: published: 2008-02-29 is an excerpt from: Chinese Academy of Sciences network in the United Kingdom recently published in the journal of applied physics Journal of physics d: applied physics, Chinese Academy of Sciences study of the Technical Institute of physics at room temperature metal fluid chip radiater experimental system for the photo on the cover, as reported by directly using chip heat driven liquid metal heat dissipation technology.
　　In recent years, as the high-end computer chips, photovoltaic technology, increasing thermal barrier and become constraints on the further development of chip technology. Repetitedly physichemical the puot for this breakthrough technology concepts, they are in the computer for the first time in the field of thermal management introduces a room-temperature fluid cooling of metal working fluids and endorsed by the first item in the field of liquid metal chip technology patent 1978and further put forward aimed at reaching the highest thermal conductivity of the natural world of liquid materials-nanometal velocity (Physics Letters a, 2007), the above series of work published in papers at home and abroad, and to obtain a number of patent licensing. In this thesis “ t t-driven liquid ª÷äý cooling device for the thermal management of computer chip ” progress had been made, a difference in temperature between exposure to the Group fusion power generation technology and magnetic pump technology that you do not need external power supply, but only by means of computer chips in the operation of the release of heat to drive the liquid metal radiators. Because there are no moving parts and a fan, a quiet running. Effective temperature difference driving is, to benefit from the liquid metal coolant itself has two distinct advantages: first, the high thermal conductivity, to traditional water heat transfer media much, and excellent heat transfer capacity; (2) as a--(ⅰ) with no fluid, easy-to-moving part of magnetic pump drive. In the past, this type of pump needs in the high-current conditions in order to produce sufficient to drive the liquid metal loop Lorentz force, but this can be caused by circuit power consumption and higher heat, thus seriously weakened the practicality of the technology. Exploratory research group, succeeded in driving current as little as possible the following 200 milliamps, that makes use of thermoelectric cooling fluid drive.
　　Typically, the work of computer chip surface has a higher temperatures, and environment form the difference in nature, and thus take advantage of this temperature difference, you can use a power semiconductor power generation film, turn to the supply of magnetic pump and driving cycle channel with metal coolant flow to heat transport. From this radiater can be achieved in the development of miniaturation and low power consumption. According to the study's author Ma, luculent, at present, all doctoral describes any fan and have been impressed current, onlyfailing heatflux 50-watt, satisfies the General computer chips of cooling demand, but to achieve a higher power density on the chip, it will be supplemented by a certain amount of impressed current. As semiconductor technology, its heat and power conversion efficiency higher and higher, and the difference in temperature by this development driven cooling technology is expected to be all kinds of opto-electronic devices such as laptops, desktops, projectors, and so on.
　　Liquid metal cooled computer chip technology in the international arena in recent years caused backlash has become an advanced thermal management within the territory of one of the new direction. United States researchers in this regard the work then was to get a number of international manufacturers-the eminent scientific media and news such as New Scientists, Technology Review, The Wall Street Journal reports, and more than $ 45 million in 1130, give prominence to this class as opposed to a traditional techniques of the concept of computer thermal management methods of significant value. Physichemical is the first in this area put forward, obtaining intellectual property rights, the leading work to serve China's future successfully led the application of the technology.
　　Related work was nsfc'funding.

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