Gravity

[isaacknewtonnes]

Gravity is the attraction of two masses as discovered by Newton and we know that the “rate” of gravity is inversely proportional to the square of the distance of the two masses, however this does not include any third party masses eg atmosphere, other things of large mass.

 

So surely when measuring gravity at a fairly accurate rate not using the universal gravitation constant, we would never be able to accurately replicate a reading of g. This would be due to the interference of other sizeable enough planets causing some mass of the measured object to be “attracted” to their gravitational field. This alongside of the constant flux of the composition of the atmosphere, and space, which would interfere.

 

Anybody able to provide clarity on this ?

[Strange]

You can take into account as many other masses as you want. The calculations can get quite lengthy, but there is no reason why you can't work out exactly how much the Sun, the Moon, Mount Everest and the car outside your house add together to have an effect on you.

[isaacknewtonnes]

You can take into account as many other masses as you want. The calculations can get quite lengthy, but there is no reason why you can't work out exactly how much the Sun, the Moon, Mount Everest and the car outside your house add together to have an effect on you.

 

But surely, with the ever expanding universe and the constant increase of entropy, the rate of gravity on a fixed surface to a degree of accuracy, will never be the replicated in time ?

[Strange]

 

But surely, with the ever expanding universe and the constant increase of entropy, the rate of gravity on a fixed surface to a degree of accuracy, will never be the replicated in time ?

 

 

You can never calculate anything with perfect accuracy. But you can calculate it to whatever level of accuracy you need.

 

The expanding universe only affects things that are so far away that their gravity is completely insignificant on Earth.

[swansont]

The gravity map we have of the earth is precise to about a millionth of g. Compare that with the perturbing effect from other planets.

http://www.esa.int/Our_Activities/Observing_the_Earth/The_Living_Planet_Programme/Earth_Explorers/GOCE/ESA_s_gravity_mission_GOCE

[Timo Moilanen]

While needing mass of sun for some speculative calc. I found that masses because of earths mass and G uncertainity is quite uncertain and merely agreed upon. Look't into it an ended up calculatin earth mass and found out ,by my meaning ,that earth mass by GM/r2 is certainly not correct .The simple formula work only on homogenous density spheres and gravitation only on far objects . To my surprise I found out that g near perfect spheres behave not "linear" to 1/r2 but does a miniscule swank and a high before at surface diving to center zero. I woul be very happy to get knowlege if this is taken into ackount when measuring G . So far I think earth (not homogenous density) 0,15% too small G.* M (only flat density so far) and G is by my understandin less than exact. Coul someone tell me if the anomalies from straight GM/r2 compared to my calcs. for a 3-dimensional sphere are taken in some way in account when measuring G . It was not surpricing on the otherhand to see that same calculated values apply to any perfect sphere for F(1/r2) /r any density and dimension .Te distance R *x is given multiples of any r and the used body in this table is 0,1524 m 158kg Pleace see charts and table

Thank you for any interest It almost killed my 14 yo Dell summing up gravity of 36000 * 2pi (Guldin/Pappus) elements (could have used more than 200+ columns

19.6 g beräkn1.bmp

Cavendish1.bmp

Cavendish2.bmp

[swansont]

To my surprise I found out that g near perfect spheres behave not "linear" to 1/r2 but does a miniscule swank and a high before at surface diving to center zero. I woul be very happy to get knowlege if this is taken into ackount when measuring G .

 

 

Do you have a citation or link for this?

 

AFAIK, the most precise measurements of G use machined masses, so that they are quite uniform. The experiments do not depend on the local value of g. I saw one in Andrea De Marchi's lab in Turin some years ago. A pendulum being perturbed by two spherical masses that could be moved perpendicular to the oscillation plane, closer and further away, looking for the variation in oscillation frequency depending on that separation.

[Strange]

I would like to know what a "miniscule swank" (or even minuscule) is.

[Timo Moilanen]

Sorry I strugle with the language and . What I meant is that even the most perfect sphere can not be calculated as a point of mass . On the surface of the uniform density sphere g=mG/r2 is exact (not proven but assumed ) and by my calculations so close one get with a " computer" (15 digits ), but above the surface the simple equation for a mass point become inexact ( up to almost 0.3%) with a slightly stronger gravity field up to about 1,35 times the radius from center and then goes slightly under the simple value (18ppm at r*1.7).This "variation " is in the range of Cavendish type measurments for both equipment setup and measured differensies . I am sure this is not taken into account , and the mass of earth is said to be M= gr2/G and simple but laborious (computer so) math shows simply that for earth being "mixed density" it do not add up even on the surface .I could not find any precice details on done measurment , although there must be quite few (in last 100 years) . Could easily compare because the values ( I calculated ) depend only on the ratio R/r ( radius and distance) and G would be =F*r^2*k1*k2/(m1*m2)

Sorry that is r/R distance per radius

[swansont]

Sorry I strugle with the language and . What I meant is that even the most perfect sphere can not be calculated as a point of mass .

 

Sure you can prove this, for a 1/r^2 law. It's Gauss's law. It's more often applied to Electrostatics, but since Coulomb's law and Newton's law have the same form, it applies to both

 

 

https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity

[Markus Hanke]

What I meant is that even the most perfect sphere can not be calculated as a point of mass .

 

 

That's not correct - in fact, if the body is spherically symmetric and the field is irrotational, then, in three dimensions, you are automatically dealing with an inverse-square law. Since this implies that the field must have vanishing divergence outside the source, a perfect sphere is exactly equivalent to a point source, so far as the form of the force law is concerned. As swansont has correctly pointed out, this is precisely Gauss's theorem, which is in turn a special case of Stoke's theorem, which follows from elementary topological considerations.

[Timo Moilanen]

The above mentioned lavs are sure accurate , but now we are adding up only the horizontal component , the down force (about integer cos(a/2) of total field ) a being 0 to 2pi around a continous cylinder "cable" .At surface distance being 67% of total in center 0 and far away about equal .The sphere further more have different mass points for crosection area and mass center point for each element . Simpler calculations with say 4-5 number accuracy can suggest the sqrt theory right as is it for the all over field , and a few section calculation can lead anywhere (inexact). Furthermore the difference is larger than 10^-4 only up to 1/10 of the radius 1,1 that .The smaller sphere in a Cavendish experiment is near enough and calculating earth mass (heterogen dencity) I came up with the r= 6369067 is equal point for inversce sqr and drops the mass of earth a stunning 0,12% (the mean radius 3671000 have little meaning in my calcs. but I'we not changed it)

[Markus Hanke]

The above mentioned lavs are sure accurate , but now we are adding up only the horizontal component , the down force (about integer cos(a/2) of total field ) a being 0 to 2pi around a continous cylinder "cable" .At surface distance being 67% of total in center 0 and far away about equal .The sphere further more have different mass points for crosection area and mass center point for each element . Simpler calculations with say 4-5 number accuracy can suggest the sqrt theory right as is it for the all over field , and a few section calculation can lead anywhere (inexact). Furthermore the difference is larger than 10^-4 only up to 1/10 of the radius 1,1 that .The smaller sphere in a Cavendish experiment is near enough and calculating earth mass (heterogen dencity) I came up with the r= 6369067 is equal point for inversce sqr and drops the mass of earth a stunning 0,12% (the mean radius 3671000 have little meaning in my calcs. but I'we not changed it)

 

 

To be honest, I have no idea what you are talking about

Gauss's theorem is applied in three dimensions here, not just "horizontally", or else we aren't dealing with a sphere at all.

[Timo Moilanen]

Sorry again , meant vertical because we feel only the force downward not those pulling equally strongly to each side . Of the 13,5 m/s2 field only 9,8 is downward . And this means that in center the field is strongest (not absolutely true however rational) but have no direction or all directions are equally stong

[swansont]

 

AFAIK, the most precise measurements of G use machined masses, so that they are quite uniform. The experiments do not depend on the local value of g. I saw on in Andrea De Marchi's lab in Turin some years ago. A pendulum being perturbed by two spherical masses that could be moved perpendicular to the oscillation plane, closer and further away, looking for the variation in oscillation frequency depending on that separation.

 

Haven't read it, but this was just made available

 

A Frequency Metrology approach to Newtonian constant G determination using a pair of extremely high Q simple pendulums in free decay

http://iopscience.iop.org/article/10.1088/1742-6596/723/1/012046

[imatfaal]

 

Haven't read it, but this was just made available

 

A Frequency Metrology approach to Newtonian constant G determination using a pair of extremely high Q simple pendulums in free decay

http://iopscience.iop.org/article/10.1088/1742-6596/723/1/012046

 

What a great paper - his enthusiasm for his subject shines through "A great deal of beautiful error analysis" etc. One thing he said the experiment was designed to avoid was the noise caused by 0.2 hertz ocean waves - couldn't they just rent a lab somewhere midcontinental USA, Asia, Africa or would that open up more difficult cans of worms

[swansont]

 

What a great paper - his enthusiasm for his subject shines through "A great deal of beautiful error analysis" etc. One thing he said the experiment was designed to avoid was the noise caused by 0.2 hertz ocean waves - couldn't they just rent a lab somewhere midcontinental USA, Asia, Africa or would that open up more difficult cans of worms

 

 

They're in Italy, so the commute would be harsh. Those locations may be quieter, but there's vibrational noise everywhere.