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<!-- google_ad_section_start -->Identifying the Gravitational Constant<!-- google_ad_section_end -->
Identifying the Gravitational Constant
Gravity as a Force related to the Energy Density of the Vacuum
Published by davidgow77
05-13-2005
Introduction

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


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 F
V2 = AFM / r
Then F
R = FV1 x FV2
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  #1  
By Casey_deJong on 07-04-2005, 08:41 PM
Exclamation gravity as it is measured

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 :)
Last edited by Casey_deJong; 07-04-2005 at 10:16 PM..
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  #2  
By mkirkpatrick on 02-10-2006, 03:10 PM
Smile

Quote:
Originally Posted by 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 :)
Is not the moon one sixth of
the earths gravity,I am sure it is!

regards michael.
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  #3  
By G_burnett on 05-28-2008, 02:26 PM
Re: Identifying the Gravitational Constant

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

英国《应用物理学》封面文章报道理化所最新研究成果
 
--------------------------------------------------------------------------------

 作者:   发表时间:2008-02-29   摘自:中国科学院网




  在近期出版的英国《应用物理学》杂志Journal of Physics D:Applied Physics上,以中国科学院理化技术研究所研制的室温金属流体芯片散热器的实验系统图为封面,报道了直 接利用芯片产生的热量驱动的液态金属散热技术。

  近年来,随着高端计算机芯片、光电器件技术的飞速发展,热障问题日趋严重,成为制约芯片技术进一步发展 的瓶颈之一。理化所科研人员为此提出了一种突破性的技术观念,他们在计算机热管理领域首次引入了室温金属流 体散热工质,并获得该领域内首项液体金属芯片散热技术发明专利,此后进一步提出了旨在实现自然界导热率最高 的液体材料-纳米金属流体(Physics Letters A,2007),上述系列工作先后在国内外发表多篇论文并获得多项专利授权。在此次论文“Hea t driven liquid metal cooling device for the thermal management of computer chip”取得的进展中,刘静小组融合了温差发电技术与磁力泵技术,实现了无需外界电能供应、而仅利用计算 机芯片运行中所释放热量来驱动的液体金属散热器。由于无任何运动部件及风扇,这种装置彻底实现了无噪音运行 。温差驱动方法之所以有效,得益于液态金属冷却剂自身拥有的两个独特优点:一是热导率高,为传统水类换热介 质的数十倍,因而传热能力优异;二是作为导电性流体,易于采用无运动部件的磁力泵驱动。以往,这种泵需要在 大电流条件下工作,才能产生足以驱动液态金属循环的洛仑兹力,但如此一来会引起电路功耗及产热较高,从而严 重削弱技术的实用性。研究小组通过探索,成功地将驱动电流降至最低200毫安以下,这一进展使得利用温差发 电驱动流体散热成为可能。

  通常,工作中的计算机芯片表面具有较高温度,其与环境之间会形成自然的温差,因而利用这种温差,可借助 半导体发电片获得电能后,转而供应磁力泵并驱动循环通道内的金属冷却剂流动,从而完成热量的输运。由此发展 的散热器可实现微型化及低功耗。据此项研究的第一作者马坤全博士生介绍,目前不使用任何风扇及外加电流,已 能实现50瓦的散热量,已能满足普通计算机芯片的冷却降温需求,但要实现对更高功率密度芯片散热,则还需辅 以一定的外加电流。随着半导体技术的发展,其热电转换效率越来越高,因而由此发展的温差驱动散热技术预计会 在各类光电设备如笔记本电脑、台式机、投影仪等发挥作用。

  液体金属冷却计算机芯片技术近年来在国际上引起强烈反响,已成为先进散热管理领域内崭新的发展方向之一 。美国研究者在此方面随后开展的工作,相继得到多个国际著名科学媒体及产业界新闻杂志如New Scientists、Technology Review、The Wall Street Journal等的报道,并获得超过1130万美元的投资,凸显这类有别于传统技术观念的计算机热管理方法 的重大价值。理化所是这一领域最早提出、申请并获得知识产权的单位,其领先性工作为我国今后成功地主导该技 术的应用创造了条件。

  相关工作得到国家自然科学基金资助。


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