Eureka

Newton’s Mistake?

I want to keep this simple so I’ll get to the technical aspects later -
for now I just want to ask the question…
Did Newton make a mistake in his decision to use the inverse square rather than the inverse cube for the relationship between gravity and distance?

Kepler noted that the cube of the radius of the distance of a planet from the sun was directly proportional to the square of its OP.

Now since the magnitude of gravity is inversely proportional to the radius this suggests that the orbital Period is related to the square root of the CUBE of the acceleration due to gravity.

If you take the distance in metres that Pluto is from the sun and divide it by the distances from the sun of the other planets you find than the Earth ratio is (2Pi)^2 (39.47842) while Mercury, naturally, has the largest the largest value at almost 102.

If, next, you find the reciprocal of the magnitude of the acceleration due to the sun’s gravity for each planet and again use Pluto’s value to be divided by the value for the other planets Earth gets a value of 1558.556 and Mercury has 10401.095 while Pluto obviously has a value of unity.

If you have performed the previous tasks correctly you will find that the square of the relative gravity values will be equal to the relative distances… hence there is no doubt that gravity is proportional to the square of the distance but is it also proportional to the cube?

The interesting thing about the planetary orbits is the spread range - if you consider the range of values for the different aspects of orbits then there are four different spreads that become apparent.
1 The Orbital Velocity and values related to it have a spread of around 0 - 10
2 The various units of Radius and related values have a spread of around 0 - 100
3 The Orbital Period and values related to it have a spread of around 0 - 1000
4 Gravity has the highest spread of all with values covering a range of 0 - 10,000

In fact, if you calculate the spreads precisely it turns out that they are all powers of the same value…
Approximately 10.1 (10.0988, 101.987, 1029.948, 10401.3)
The values of g in m/s/s actually range from around 0.039577 for Mercury
through to approximately 0.000003805067 for Pluto.
While the range of the Orbital Velocity is from 47873.4 for Mercury to 4740.5 for Pluto
The range of the planets’ distance to the sun starts at around 57909175000 metres for Mercury
Reaching 5905910000000 metres for Pluto giving a range of around 101.98574

If you multiply the square of the Orbital Velocity by the distance of the planet from the sun (both in metres) then you get a constant value of 1.327E+20 metres.
To find the Orbital Velocity you simply need to divide that constant by the radius of the planet’s orbit and take the square root of the result…
since the Orbital Velocity does not seem to be greatly reliant on the acceleration due to gravity (as mentioned earlier the Orbital Velocity diminishes by a factor of 10 while g diminishes by a factor of 10,000!)
it would seem more likely that the Orbital Velocity is actually dependant on the planet’s distance from the sun because although its inertia of curving increases as it gets closer to the sun its orbital inertia more than compensates and so it velocity increases.

It seems to me that Newton might have overlooked the fact that it gets easier to curve with increasing radius and so allows a progressively higher Orbital Velocity relative to the acceleration due to gravity at that radius.

This would mean that the inverse square would work okay with orbits but not with the true gravitational attraction between two or more massive bodies!

If you think about Einstein’s solution to getting rid of ‘spooky’ action at a distance you realise that his solution was to make gravity 3 dimensional.

Whether or not you believe in Space-time it is easy to see that gravity can’t be treated like, say, light whose intensity is simply governed by the square of the distance from its source.

Why not? Because obviously the intensity of gravity at any radius depends on the locality in that, unlike light, it depends to some extent on what is around it.

Thinking in terms of Einstein’s curved space-time then obviously the gradient of the curve of the field at any point must be strongly effected by the curve of the space around it.

Having understood that it might make you wonder whether gravity will actually vary with the cube of the distance rather than the square.
So why does Newton think it is the square?

If you go back to the principle of a simple pendulum you find the Period of its swing is related to the square root of the radius of the swing divided by the acceleration due to gravity.

If you check the result of this you find that the Period is directly proportional to the square root of the radius - in other words it follows a similar relationship to the Orbital Period of a planet.

However, as I pointed out earlier, with planetary orbits it is limited to the square root of the CUBE of the radius!

Since you will find that the radius of a planet in metres (the distance from the sun) times the gravity in m/s/s gives the square of the Orbital Velocity i.e. R(metres)*g = OV^2 then we see that the OV = sqrt(R*g) m/s...
You can also divide the radius by the number of seconds in a year (31557600) and multiply the result by 2PI. Both methods return to using the relationship between the metre and the second as used by a pendulum.

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Well, it has been an interesting journey for me.
I started out with a virtual pendulum and ended up with a simple method of not only predicting circular orbits but also of demonstrating that wherever a body is orbiting it is in a perfect balance that it shares with all other orbiting bodies in its system.

That is to say that, for any gravitational system, there is always a constant the describes the product of the magnitude of the important variables.
This constant varies from gravitational system to system, but is easy to find, and, with it, all you need is the radius to be able to calculate all the other variables.

So, what do pendulums and orbits have in common?
Well, they work on essentially the same rules but since orbits demand 'Balance’ they share a constant that relies on the radius.

So orbits are essentially a function of the radius of any gravitational system which is constant - i.e. where the curve of gravity can be seen as having a single major source - and so become a subset of the possible pendulum swings.

However, the key point is that the both rely on inertia to control their timing.
It is the predictable ‘resistance’ that allows pendulums and orbits to have a definite time period.

If you think about Einstein’s theories of Relativity it tells us that orbits are based on the curve of space-time which leads to some very strange conclusions. Perhaps the most important one is the prediction that inertia increases with gravity, that clocks slow down, which I have never been keen on believing.

However, while doing some research for data to check my satellite orbit predictions I read that geostationary satellites are fitted with special clocks which run slow at ground level so that they will run at the correct speed when orbiting at some 42 million metres (radius, from the centre of the planet).

This set me thinking that if this is true then pendulums would have to run faster in lower gravity too…
Which, of course they do NOT - quite the reverse!
So, what would make a pendulum run faster even though the gravity was reduced?

What if inertia was relative too?
I mean what if the Solar System revolves and whatever it is that causes inertia revolves with it?

What difference would it make?
Well, it wouldn’t make any obvious difference to us because everything we know already circles the sun.
Yet, by the same principle, geosynchronous orbits would be seen to be satellites moving with the Earth’s revolving inertia rather than as having velocities of some 3075 meters per second.

Planets too would be revolving with the sun’s rotating inertial field and it would only be bodies that had a velocity (or maybe only an acceleration) RELATIVE to the rotating inertial field that felt the resistive effects.

I haven’t given it too much thought but it seemed possible that gravity could be a flow of inertia.
Certainly, there is no reason to think that Newton’s idea that, left to themselves, things go in a straight line would actually hold true in a spinning inertia field - they would more likely follow the curve.

So how would that explain pendulums running faster in lower gravity?
It wouldn’t :O) Unless the decreasing inertia of curving became progressively more important than the decreasing gravity…
thoughts for another time ;O)

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A Simple To Use Relative Table
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If you use my 2Pi principle to balance the units of measurement and so make all the variables directly proportional then you can see that, be it pendulum swing or orbit, the relationship between time and distance is simply a function of the curve that the body is following -
which is why it can be expressed as a function of radius.

I have adapted the original Xeters system which relies on an imaginary planet X (which is described later) to a simple relative table that retains much of the power of the original system while being much easier to use as the variables are all relative to Earth and not Planet X.

However, if you use it to calculate g (the acceleration due to gravity) by dividing the OV by the OP... it will give you g divided by 2PI.

Yet, the value of g seems to work just as well as the values according to Newton ...
My point being - how can we know which values are correct? ;O)

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In the following table to get the Curve Index then just divide the radius in AU's (essentially the distance between the chosen planet and the centre of the sun relative to the Earth's distance from the sun) by the OP (the time it takes the planet to orbit the sun expressed in Earth years).

The other radius in Xeters is important for the balance of the system and to simplify finding all the other values. 1 Xeter = 5022505 metres.
Ci = OV/29785.5 : OP = Square Root (R(AU)^3)
OV = R(Xeter)/OP : R(Xeter) = R(Au)*29785.5

Planet ……R(AU)….. ……OP (Yrs) ……R(Xeters)..… ..OV (m/s)
Mercury … 0.3871 … ……0.2408423..….11529.939……47873.4
Venus ……0.72333 …… ..0.615186. ….. 21544.813 … .35021.62
Earth …….1.……….....……1.……….......……29785.513……29785.513
Mars …… 1.52366… ..… 1.8807576.……45383.058 …..24130.2
Jupiter……5.20336 ……..11.869334...……15985.01……13057.6
Saturn ……9.5371………...29.452649 ……284067.41…. 9644.885
Uranus … 19.191242….…84.072625……571621.13...…6799.135
Neptune…0.06893……...164.88344......895618.81…....5431.83
Pluto .…….39.47858…….. 248.05175…….1175889.3……4740.5

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A Description of the Original System.

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I've been playing with this fabulous implementation - done by an old friend - of my ideas regarding choosing a distance from the sun as a fulcrum or 'balance point' and it is absolutely amazing!

It gives a wonderful insight into the forces at work in planetary orbits with changing radii.

What’s amazing about this system is that it demonstrates that a body in orbit is always in balance - that what it gains on the one hand it loses on the other… although I think it has at least three hands :O)

As I suggested in an earlier post my friend has found a way to make the forces equal (kind of) by using 2PI for the time Period for a complete orbit of a circle.

I hadn’t imagined that it would then show that changing the radius simply alters the ratios between all the main values (gravity, velocity and distance) while maintaining a constant, overall value.

It demonstrates that all bodies orbiting the sun can have a value of about K = approx 4205638288402 (cube root approx 16141.5) and this constant can be used to find all the values just from the radius.

To get K simply multiply all four values together (g*OV*OP*R(X))
NB Don't include R(PXU)(the ! is to remind you not to) and always use R(Xeters) as the radius when dividing K.

Actually, further research has revealed that everything can be calculated from 16141.5 for the planets in our Solar System...
and checked by its square 260548022

The radius R(PXU)is very useful for finding the OP (Orbital Period - as you can do with AU's on an Earth centred system) but this measurement of the radius is involved in many other equations and very useful if you want to check values or find some of the dozens of relationship you could never have imagined before - it is approx R(Xeters/16141.5)... R(X) is also OV squared/g.

In practice the system’s precision is now good enough to keep the check numbers perfectly constant, and using it you soon start to see the way that gravity is balanced by the inertia of curving as expressed by the Orbital Period.

First convert the distance in metres to Xeters by dividing by 31557600 to get the radius in Xeters.
Now find the OP by dividing the radius in Xeters by 16141.5 to get the Radius(PXU).
Take the square root of the cube of Radius(PXu)
to get the OP (orbital Period).
Also R(PXU) * OV^2 is a good check and should always be exactly 16141.5^2 = 260548022

Dividing the radius in Xeters by the OP gives the Orbital Velocity, and dividing that by the OP gives the acceleration due to gravity.

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I’m not sure if the Xeter value is related to the spin of the Solar System but there is a huge amount of power stored in that spin and I think that any complete theory of gravity would need to take this spin into account.

If you think about it, it is this spin which makes sure that most bodies end up orbiting the sun rather than diving into it.

It also explains how bodies end up orbiting each other... if you consider that they would almost certainly start out at different distances from the sun and so any force that would draw them together would also cause them to curve differently due to their spin differential.

Once you start to think about the inertia of curving you realise it works whether you are approaching the sun or moving away from it and tends to bend the body toward an orbit.

Although it is the orbital velocity that 'chooses' the orbital radius it is the inertia of curving that having 'bent' the bodies path into a orbit keeps it in check.

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The top 'table' is simply for comparison...
The tables don't display well and can't contain enough decimal places to display the true precision that the system is capable of and so I have abandoned it and urge you to calculate your own.

I suggest you use Windows calculator set to 'Scientific' and copy the tables into a spread sheet or perhaps a word processor so that you can then cut and paste values between the calculator and the table...

Actually, it is so difficult to line things up with a word processor that it can drive you to distraction but on ABCtales, you can't use spaces to seperate your data and the result isn't worth the effort so I've not tried to update the tables.

Sun’s gravity...Orbital Velocity...Orb Period......Radius........Radius
@ planet..........m/s/s........m/s........years........metres.........AU
Mercury = 0.0395753567~47872.5~0.2408467~57909175000~0.3871
Venus = 0.01133454012~35021.4~0.61519726~108208930000~0.7233
Earth = 0.00593056385~29785.9~~~~~1~~~~ 149597890000~~~1
Mars = 0.0025546598~24130.9~~1.8808476~ 227936640000~~1.5237
Planet X = 0.000511490~16141.5~~6.2832~~509384094857~~3.40502
Jupiter = 0.00021944282~13069.7~~11.862615~~778412610000~5.2034
Saturn = 0.00006557346~9672.4~29.4475~1426725400000~9.5371
Uranus = 0.00001627322~6835.2~84.016846~2870972200000~19.1912
Neptune = 0.000006566755~5477.8~164.79132~4498252900000~30.069
Pluto = 0.000000381518~4740.57~247.6~5905910167200~39.478
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© J M KNIGHT July 7th 2011. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Sun’s gravity~~~~Orb Velocity~~Orbit Period~~~ Radius~~!~R(PXU)
@ Planet~~m/s/yE~~m/s~~~Planet X years~~~R (Xeters)~!
Mercury = 1248979~47872.5~0.038332~~~~1835.05~~!~0.113686
Venus = 357691~~~35021.4~~~~0.09791~~~3429~~!~0.21243
Earth = 187155~~~29785.9~~~~0.15915~~~~4740.5~~~!0.29368
Mars = 80614.9~~~~24130.5~~~0.29933~~~7223~~~!0.44747
Planet X = 16141.5~~~16141.5~~~~~1~~~~~~16141.5~~~~!~1
Jupiter = 6925~~~~13069.7~~~~~~1.888~~~~24666.7~~~!1.5282
Saturn = 2070~~~~9672.4~~~~4.6865~~~~ 45195.8~~~~!~~~2.8
Uranus = 513.78~~~~6835.2~~~~13.371~~~~ 90933~~~!~~~5.6335
Neptune = 207.23~~~~5477.8~~~~26.242~~~~142541~!~~~8.83077
Pluto = 120.08~~~~4740.5~~~~39.478~~~~ 187147~~~!~~~11.5942
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Actually it looks as though it might be that the cube root of K = 2569 * 2PI = approx 16141.503
Since this is a slightly different value and K is its cube it means that everything needs recalculating but I can't be bothered since it is such a pain to try and get things to line up in the table.
When all the values are finalised then I will re-edit them but until then the answers will be rough approximations - but still good enough to allow the innumerable relationships to be found and checked.

So, if we call 16141.503 Kcr (the cubic root of K) then -
R(X)/R(PXU) = Kcr and R(X)= K /(OV squared)
while R(PXU) = (Kcr squared) / (OV squared)

OV = g*OP (PXyears) : OP = sqrt (R(PXU) cubed): K/R(X) = OV squared.
16141.503 = OV*(OP/R(PXU)) = R(X)/R(PXU) = (OV squared/(g*R(PXU))

1 Xeter = approx 31557600 metres : Light travels at approx 9.5 Xeters/second = 299792458 m/s.
I PX year = 2 PI Earth years hence 1 Earth year = 1/2PI PX years (0.159155)
1AU = 0.29368 PXU or I PXU = 3.40506674 AU
16141.503 / 4740.5 = Cube root of 4(PI squared) = 3.40502

Just a change in radius from 1 Xeter to 9.481 Xeters = about a 90 fold increase in gravity but only about a 3 fold increase in Orbital Velocity… where is all that energy going; into curving?
It would be great to get rid of gravity altogether but I can’t think of how to maintain the balance.

Gravity~~~~~~~~OV~~~~OP(PX years)~Radius R(X)~ Radius (PXU)
4205638288402 , 2050813 , 0.0000004876134 , R = 1Xe (31557.3 kilometres) ! 0.0000619513
(133275.7 m/s/s) Result = 4205654689105 ~ check by dividing by K =1.0000038997, okay.

46788824703 666037 0.000014235 R = 9.481 Xeters ! 0.00058736
(1482.7 m/s/s) NB gravity m/s/y = m/s/s/ 31557274.
Result K = 4205833832573.5 check by dividing by K = 1.000003870, okay.

It’s easy to start a new orbit if you covert the radius -into PXU (Planet X Units).
R Xeters are simply 16141.5 times a PXU.
Divide K by R(Xeter) and take the square root for OV.
Take the PXU and cube it then take the square root for the Orbital Period in Planet X years.

For gravity: divide K by the square of R(Xeter) and it should give you g in m/s/y.

NB Although using K should be fairly accurate you can check by multiply all four values together (g*OV*OP*R(X)) it is highly unlikely the result will be exactly K but it should be close.