Thinking about a spaceship orbiting Earth as being a balance between the gravitational force on it and the inertial force resulting from the ship’s constant change of direction it occurred to me that all bodies in a stable, circular, orbit would be ‘weightless’ in respect to the thing they were orbiting.
From this it seems likely that as we are in orbit around the sun we are ‘weightless’ as regards its gravitational attraction, as in fact would all the planets be and hence they would feel their own gravity unaffected by that of the sun.
Pondering why the planets velocity around the sun did not decrease with the square of the distance to the sun it occurred to me that it is because orbits are a balance between the gravitational attraction of their sun and the amount of inertia caused by the path of the planet - in other words it is a matter of balancing curves. When the curve of the orbit matches the curve of gravity then the planet is in orbit.
Since the curve of gravity depends on the distance to the sun and the curve of the planet depends on its orbital speed then a body that is travelling too slow will fall and a body that is travelling too fast will ‘rise’.
In a way it is a battle between the straight line and the curve as the planet wants to move in a straight line but it is forced to curve by gravity.
The strength of the gravitational field simply results in varying scales, presuming that the inertia of curving remains constant, since a circle is a circle is a circle.
I suppose that I ought to explain why going faster takes the planet ‘higher’ (further from the sun) even though the orbital speed of planets drops as you go ‘higher’…
The crucial factor is that the strength of the gravitational pull must be strong enough to ‘fit’ the curve that the orbital body is describing with that of a circle at that distance from the sun.
In other words if its speed is too great for the pull of gravity (at that distance from the sun) to curve it enough then it will move outward and if its speed is too low then it will curve too much and start to move closer to the sun.
Naturally, this is only true for a circular orbit but for other orbits it is similar but with inertial dances of acceleration caused by variations in the balance between the inertia of nucleuses of the atoms and the direction of the inertial frame they compose as the orbit the sun...
As I’ve pointed out elsewhere the Wall Of Death (in which a motorbike circles a vertical wall) is much the same principle except the wall forces the motorbike to curve around the circle rather than gravity.
In both cases it is the inertia of curving that supplies the force to counteract gravity and yet it seems difficult to see how curved space might be involved in the Wall Of Death.
Certainly the fact that gravity pulls from a central point does give an effect that appears to be curved but it may well be that it has little to do with space beyond the fact that it diminishes with distance.
Kepler’s second law - The Law Of Areas - tells us that, due to the conservation of angular momentum, a line that connects a planet to the sun (essentially the planet’s varying orbital radius) sweeps out equal areas in equal times.
In other words that the relationship between distance from the sun and gravitational force is a result of the inverse square law of gravity.
So, in theory, at twice the distance the force should be quartered yet the orbital velocity remains higher than expected.
The Earth's mean orbital velocity is almost 30 Km/s yet Neptune which is over 30 times more distant from the sun still manages almost 5.5 Km/s despite the force of gravity being some 900 times weaker.
Checking using Kepler’s Third Law.
Since Neptune is approx 30 times as far from the sun as the Earth then it must traverse 30 times the distance and it turns out it travels at approx 5.5 times slower than the Earth so it takes approx 30x5.5 = 165 years.
The actual figure is about 164.8 so it is close enough for me. Yet I'm still staggered that Neptune can orbit the sun at over 1/6 of the velocity of the Earth on 1/900th of the gravitational force that the Earth gets!
Having checked Kepler’s laws if there is a mistake it must be in Isaac Newton’s assumption that gravity follows the inverse square law.
I'll give it some thought...
centripetal acceleration = v squared / radius
for the Earth 29.79 squared = 887.44/1 (AU)
for Neptune 5.43 squared = 29.48/30.6 (AU)
giving 887.44 to 0.963. close enough to 900 to 1...
(but interestingly multiply instead of dividing gives 887.44 to 902...
Ah, setting the Earth’s distance from the sun as unity (1 AU) has set the square of the Earth’s velocity as the value of the square of the orbital velocity times the distance from the sun for all the planets!
It might be interesting to set an imaginary planet with a distance from the sun as 1 light second - or maybe a light minute.
1 AU is about 500 light seconds so perhaps an imaginary planet at 8 light minutes might work well.
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I've played with an imaginary planet using C as the basic unit of length ie Light Seconds and it is starting to look as if Newton was almost right but he was trying to solve a three dimensional problem in two dimensions.
It seems to me that I was close with my concept of it being a battle between the curve and the straight line but like Newton I was thinking in two dimensions.
I'm not well at the moment but as soon as I feel better I will consider if could be a battle between the cube and the sphere and if gravity could be a consequence which might well explain the connection of curve doubling with the square root of eight.
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What really interest me is whether the inertia of curving is constant or not - does it require more force to produce identical curves (i.e. to turn through an identical angle while travelling an equal distance) when it is closer to the gravitational source than it does when it is further away?
Although it might seem simple to check this out I’m not sure how anything in space can control its rate of curving at two separate distances from the sun without changes in its acceleration - can you think of a means of doing it?
I think I must be suffering from the Davro effect* because of course the LHC is exactly what is needed to test if the force needed to ‘bend’ things around a curve is always proportional to the speed of the thing being bent ;O)
What the LHC does is force things to bend ever faster around the same curve.
So is there any discrepancies in a comparison between the results taken from a linear accelerator and that of the LHC?
If I’m right then the extra effort needed to curve would add an increasingly large increment to the force needed to attain the same linear speed.
Certainly it will make a huge difference to the results once the LHC swaps to using iron and I must say I have trepidations about what effect the LHC might have once it does - due to the dramatical increase in inertia.
Makes you wonder if Einstein was thinking about the inertia of curving when he said that as a thing approached the speed of light its mass approached infinity - perhaps he was calculating the increasing effect of inertia when accelerating through curved space.
It is worth bearing in mind that a straight line is a quite probably a hypothetical concept in a curved Universe or even in a Universe where the effects of gravity are ubiquitous.
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When I wrote earlier that the Earth should show little or no gravitational effect from the sun (obviously it supplies the gravitational attraction to keep the Earth in orbit) I had considered the Moon as simply that, a moon, but on checking I found that although the Earth/Moon centre of gravity lies within the Earth) that, to some extent, they dance like partners as they circle the sun - with the Earth, the heavy partner, swinging the dainty Moon around it but both revolving around their common centre.
Without doubt that must alter the effect the sun’s gravity has on the Earth.
It seems that the Sun is 391 times as far away from the Earth as the Moon but its force on the Earth is about 175 times as large and yet its actual tidal influence is less than 45% of that of the Moon.
This is explained by the fact that being nearer the moon's gravity has a wider effect on the earth - it is obviously presumed that because gravity, like light, follows the inverse square law that its ‘gravitational rays’ must also act like light rays.
Strangely, the same reasoning is followed for the Land Tides and yet it is even more obviously unlikely since you would expect a force that is 175 times larger yet concentrated into a smaller area to show a much larger and not smaller effect on solid land!
Because of the Moon's dance with the Earth affecting the earth's orbit I have no reason to argue with the belief that the sun’s gravitational effect on the Earth is less than half that of the moon but I can’t say I believe the commonly accepted reason for that.
* http://www.abctales.com/story/terrence-oblong/observation-many-people-bo...
Comments
Mangone | April 25, 2011 - 14:40
I probably should mention that I haven’t got around to calculating how much difference the elliptical elements of the planets orbits might effect the ‘balance’ of the ‘gravitational cancellation’ which relies on a ’constant acceleration’ (or no acceleration if you see it from a different perspective).
I haven’t had chance to give it much thought yet but it seems to me there is a possibility that there may be some slight gravitational anomalies when a planet is closest or farthest from the sun - and with the Earth it could be especially noticeable when the moon lines up with the sun during an eclipse.
Nolan | April 28, 2011 - 05:47
“As I’ve pointed out elsewhere the Wall Of Death (in which a motorbike circles a vertical wall) is much the same principle except the wall forces the motorbike to curve around the circle rather than gravity.”
“In a way it is a battle between the straight line and the curve as the planet wants to move in a straight line but it is forced to curve by gravity” “ in other words it is a matter of balancing curves”.
The geometry of flesh on the bone- A Matter of Curvature.
Mangone | April 28, 2011 - 07:53
Indeed Nolan - an altogether more fascinating geometry.
Nolan | April 28, 2011 - 17:05
And how.
Mangone | April 30, 2011 - 05:46
The wobble and the bounce the curve and the sway
The geometry in motion that takes your breath away ;O)
Nolan | April 30, 2011 - 18:35
That is actually non-linear dynamics- in the framework of classical physics. Much more challenging but well worthwhile.
oldpesky | May 2, 2011 - 15:48
Jings! Think I might have to phone my good friend Mr Hawking to help me understand what's going on here.
Mangone | May 3, 2011 - 08:47
I’ve just had a crazy thought. Realising that putting in Earth stats to Newton’s model gives us answers relative to Earth I wondered if the fact that all observations that Newton based his calculation on were also made from, and on, earth might have distorted the results, or it may be that some relationship which was thought to be constant was in fact relative.
Since, essentially it comes down to a cubic problem and the Earth’s gravity is roughly a third of the earth’s orbital velocity, when multiplied by 10 to the 3, it makes me wonder whether this is because it should also be taken as unity to fit with the other relative values… ;O)
Another crazy thought - is Pi truly constant or does it grow infinitesimally with the radius and that's why it is transcendental?... since there is a minimum finite length.
oldpesky | May 3, 2011 - 10:20
It's not that crazy a thought. I believe Pi now lives in Canada after sailing across the Pacific with Richard Parker.
Mangone | May 3, 2011 - 14:24
Thanks Pesky, that explains everything. I wonder if he misses his cat :O)
Nolan | May 3, 2011 - 18:08
Last I heard he was looking for his keys.
Mangone | May 4, 2011 - 05:19
Aren't we all, Nolan ;o)
Continuing the search for a possible way that Earth measurements might have crept into what was meant to be a Universal law I considered what was the most likely candidate.
Obviously weight/mass as weight is simply the measurement of mass in Earth's gravity.
Now, my main contention with Newton’s laws of Gravitation is that they don’t seem to take enough account of the increasing inertia of curving with reducing radius. In other words I wonder if Newton's law of centripetal force is accurate everywhere or just on Earth.
If that was wrong then it would mean that part of the effect of the inverse square orbital velocity could be attributed to curving and hence gravity wouldn’t weaken as quickly as Newton predicts.
I’m not sure how the centripetal force equation is tested but I have seen several explanations which don’t seem to take into account that the experiments are done in Earth’s gravity and so its force is constant - which it is obviously not when considering orbits.
It’s a starting point.
If that doesn’t pan out does anyone have any other suggestions why gravity might not weaken as quickly as predicted by Newton?
I mean do you really believe in Dark Matter and Dark Energy?
Nolan | May 5, 2011 - 11:39
Dark matter and black holes, black magic and dark energy
it's all rock & roll to me.
Mangone | May 5, 2011 - 12:36
Remembering back to when well over 97% of the Universes's mass was missing I laughed when someone came up with the idea that it was there as Dark Matter but we just couldn't see it - being dark and all.
When they came up with Dark Energy I wondered how on Earth energy could be dark - their answer was it couldn't, on Earth ;O)
Certainly the music of the spheres has got darker since Holst day, Nolan :O)
Not being well I’ve been doing things that aren’t too challenging for a swimming head.
I created a table of the planets with the Earth’s orbital velocity, distance from the sun and orbital period all set to unity so that it clarified the relationships.
Next I decided to create a table of relative orbital velocities from 1 to 15 times that of Earth…
It’s very easy because the square of the velocity x the distance is constant so as we’ve made the constant unity then the distance is the reciprocal of the square of the velocity.
I had to stop at 15 because I was just within the radius of the sun and so it was pointless to continue.
I noticed that the rotational period of the sun was about 1/15 of a year…
a little less at 24.33 days.
Now the sun’s radius is actually about 1/215 AU and so taking the square root of 215 we get about 14.66 for the orbital speed relative to Earth’s - which gives about 24.9 days…
Checking I found :
The regions of the Sun near its equator rotate once every 25 days.
http://www.windows2universe.org/sun/Solar_interior/Sun_layers/differenti...
Interesting coincidence. Almost as thought the sun is spinning and dragging everything around with it ;O)
As I said I'm not well so I could have made a glaring mistake in my calculations and not noticed - so if anyone notices one please tell me.
Nolan | May 6, 2011 - 16:51
They tell me this kind of stuff is hell. Attempts at reasoning are futile. Arguing must be a nightmare. One luck is when the student queries his mark of 1% he never asks where he got that one mark. Fortunately I'm a poet and not a glorified teacher.
Mangone | May 7, 2011 - 07:10
It certainly is hell when your head's spinning, faster than the planets, Nolan :O)
Converting the planetary data to relative measurements gives :
Mer 47.89 km/s @ 0.39 AU 0.24 years
Ven 35.03 km/s @ 0.72 AU 0.62 years
Ear 29.79 km/s @ 1.00 AU 1.00 years
Mar 24.13 km/s @ 1.52 AU 1.88 years
Jup 13.06 km/s @ 5.20 AU 11.86 years
Sat 09.64 km/s @ 9.54 AU 29.46 years
Ura 06.81 km/s @19.18 AU 84.01 years
Nep 05.43 km/s @ 30.60 AU 164.8 years
Going one step further and making the velocity relative too…
When the orbital velocity for Earth is also set to unity for Earth then V becomes Vr:
R = 1 / (Vr squared) = Vr x P
Vr = R / P
P = R / Vr
(Vr squared) x R = 1 (approx due to values being rounded - esp. R and P)
Where R is the distance from the Sun in AU’s
An AU (Astronomic Unit) is the mean distance of Earth from the Sun.
Vr is the orbital velocity at that distance (divided by Earth's orbital velocity of 29.79 Km/s)
P is the orbital period in years.
Giving -
Planet....…Vr…….....……R……….......P……….
Mercury 1.6076 @ 0.39 AU 0.24 years
Venus.. 1.1759 @ 0.72 AU 0.62 years
Earth.... 1.0000 @ 1.00 AU 1.00 years
Mars.... 0.8111 @ 1.52 AU 1.88 years
Jupiter.. 0.4384 @ 5.20 AU 11.86 years
Saturn.. 0.3236 @ 9.54 AU 29.46 years
Uranus.. 0.2286 @ 19.18 AU 84.01 years
Neptune 0.1823 @ 30.60 AU 164.8 years
You will find that the relationships given above will give slightly different values but I expect that is simply down to the fact that they are approximate values.
It could well be that P / Vr gives the reciprocal of the Centripetal force for a planet relative to its value for Earth.
Eg. 164.8 / 0.1823 gives Neptunes Ca as 904 times less than Earth's.
Mangone | May 7, 2011 - 11:23
The plan was to produce a table which was relative to the Earth's year so that the orbital velocity was always relative to R/P through Vr and of course - since the Mean Orbital velocity is always the circumference of R divided by the planet's year...
Then we get V = K * R/P where K = 2 Pi.
Or using the table where V = Vr * 29.79 Km/s.
Vr should equal R/P but may be slightly different due to approx values.
NB. 2Pi* R(AU) / P(years) is a perfectly valid velocity...
if you are happy to work in AU's and years.
For Km/s then - Earth it is 2*Pi* (1 AU / 1 year)
= (149,597,870 Km) / (31,557,600 seconds) * 2Pi
= 4.74 km/s multiplied by 2 * Pi which gives 29.79 Km/s.
Or Vr * 29.79 Km/s = 29.79 Km/s
For Mercury -
then 149597870 * 0.39 gives 58343169
and 31557600 * 0.24 gives 7573824
58343169 / 7573824 = 7.7 Km/sec
and 7.7 * 2 * 3.14 = 48.356 Km/s, a little high.
Using our method 0.39/0.24 = 1.625 * 29.79 = 48.4
NB The value of Vr in the table (1.6013) was calculated using more accurate data and gives a better result.
Both of the others are well above the true value but since we calculated the first from the given values it shows that the fault lies with the accuracy of the approx values of the planet's distance from the sun and its year length.
My head is orbiting even faster than yesterday so please point out any serious flaws in my thinking and I'll try and fix them... or better still give me some thoughts on how to improve it.
Thanks.
Mangone | May 8, 2011 - 13:24
It seems to me that to get my table to work perfectly I need to find the ‘fulcrum’ of the Solar System.
I expect that will be when R/P = 1 and the orbital velocity = 2Pi.
Since then the orbital Period = 2Pi * R / 2Pi and hence Vr = 1(radian).
I'm expecting that this will be the distance from the sun at which the gravitational force will be balanced - but I'm not sure exactly what that will mean... should be interesting.
Once I've found that I can use more precise data and start to look for variations in the inertia of curving and the role that Mass plays in the dance.
My head is still 'swimming' and I can't concentrate so I thought I would look for something with an orbital velocity of 4.74 (the Earth's velocity / 2*Pi) and Pluto had exactly that and its distance from the sun is 39.44(AU).
Checking the square root of 39.44 unsurprisingly it was approx 2*Pi or 6.28.
So what's its orbital period - try (2*Pi) cubed... or 39.44 * 6.28.
Well the cube of 6.28 = 247.67 - good enough for me!
Also the square root of (4.74 squared * 39.44) = square root of 886 = 29.77... while the earth's Orbiting Velocity is 29.79.
Actually since the square root of (39.44) = 2 PI...
then if your calculator has a Pi button you get a better result by multiplying Pi by 2 and then by 4.74 for 29.783.
Mangone | May 9, 2011 - 05:57
A bit of thought brings the simple realisation that I have to divide the measurement of the distance from the sun (R) by 2*Pi to make it balance.
Obvious really since C = 2 Pi*R that the Orbital Period should reflect that. However, I think it would be easier to adjust the units of length which we use to measure the radius - if that doesn't work I can try redefining the time period from seconds to... ASecs :O)
As soon as I get time I'll see what happens when I adjust R to AU*2 Pi units (divide all the R's by 6.2832) and of course the Orbital Velocities and the Orbital Periods should be uneffected.
Since it is a PIAU, perhaps we could call them PiaUnits after Highhat :O)
Then Earth would be 29.79 km/s at R = 0.159155 and OP would remain at 1. So, now the orbital velocity * the Radius (distance from the Sun in new units) gives the value 4.74.
Checking Pluto we would get Orbital Velocity 4.74, R is 2 Pi and Orbital Period is 247.67 years and Orbital velocity * R gives 29.7823.
A bit more thought makes me wonder if I'm simply converting the Radius into the Circumference and if so then the original values would stand but then it would be the Orbital Circumference that was set as unity and not the Orbital Radius hence R would relatively become 1/2 Pi (C/R)... and give values as we have recalculated it for a Relative Table.
Mangone | May 9, 2011 - 08:13
I can't quite catch it yet as I'm constantly distracted by the missing planet between Mars and Jupiter and wondering if that was the 'fulcrum'.
I'm not considering this from a mass point of view as all the heavier planets would be on the same side... unless, of course, they're balancing the sun ;O)
The plan is to start again as the original table has been corrupted by my continous efforts to improve it and the relationships no longer work properly.
My aim is to produce a table which includes the Orbital Circumference as well as the Radius.
Since the Orbital period is the time it takes to travel around the circumference it should be possible to set both to 1 for a relative table which will get rid of the 2*Pi problem.
When I've done that I should be in a better position to place Planet X which I suspect was around 3.4 AU.
I'm also hoping to understand the relationship between Earth and Pluto better once the new Relative Table is completed.
Mangone | May 9, 2011 - 13:58
Planet ……R (Lm)….. ……OC (Lm) …..OP (weeks)… ..OV (Lm/Week)
Mercury … 3.22 …. … ……20.232 ….... …12.52… … … …1.616
Venus ……. 6.01 ……. ……37.76.……. ….. 32.35 …… … …1.167
Earth ………8.32 ……. ….. 52.276.….. ….. 52.18 … …… …1.0
Mars ………12.70 …… ..… 79.80 ……. ….. 98.09 …… ……. 0.8134
Planet X
Jupiter …… 43.30 …… ….. 272.0 …… ……618.84 …. ……. 0.4395
Saturn … … 79.20 …… … 497.6 …… …….1537.2 …. ……. 0.3237
Uranus … …159.6 ……. …1002.8 ….. …. 4383.64.….. ……0.2287
Neptune.. …249.6 ………..1568.28 …….…8599.26 …. …. 0.1824
Pluto ………. 327.8 ……….. 2060.0 ……….12925.0.….. ……0.1592
Mangone | June 4, 2011 - 08:10
I’ve been thinking about pendulums again today.
They interest me for several reasons but my main concern at the moment is how well they mimic orbits despite the fact that the acceleration of gravity is constant for them. What I mean is they obey the inverse square rule without the need for a steadily declining gravitational force.
I’m not convinced that the sun’s gravity diminishes as quickly as it is predicted it should and it would be very interesting to discover that gravity is not actually the main governing factor in planetary orbits.
Playing with the Hyper physics simple pendulum simulator I discovered that if first you set the force of gravity (g) to 39.4784 (about 4 times Earth's gravity)...
Well to be more precise it is (2*Pi) squared which is exactly the gravity needed for L = 1 metre when T = 1 second in the pendulum sim!
Once you have set the gravity you can then enter the distance of any planet from the sun (in AU’s) in the second (the metres) L box and then click on 'pendulum period' -
and it will supply you with quite a good approximation of the planet’s orbital period in Earth years divided by its distance from the sun.
That is the planet's orbital period divided by the orbital radius...
In other words if you multiply the ‘pendulum period’ by the number you entered in the L box it will give you the planet’s year length in Earth years…
Here is a list of the necessary data Mercury 0.39AU 0.24 years.
Venus 0.72AU 0.62 years. Earth 1AU 1 year. Mars 1.52AU 1.88 years.
Jupiter 5.2AU 11.86 years Saturn 9.54 AU 29.46 years.
Uranus 19.18AU 84.01 years. Neptune 30.6AU 164.8 years.
Now the results are not perfect but they are certainly good enough to demonstrate that the results do follow the actual measurements remarkably well.
At least this gives a simple orbital year calculator.
Now all I need is some hint of how long a year the missing planet might have had and I can try different distances until I find something close :O)
If you don’t have a calculator most versions of windows have one that will suffice once you change the ‘view’ from ‘Standard’ to ‘Scientific’.
The simple pendulum simulator is here - http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html
Nolan | June 9, 2011 - 19:04
Not bad Mangone. Imagination is more important than knowledge Einstein said so himself. Do you know what this “string theory” is all about? “Knot theory”? What is that?
Mangone | June 10, 2011 - 04:32
Thanks Nolan.
My brain is already tied in knots trying to figure out why Pluto is 4*(Pi squared) or 39.4784 times further from the sun than the Earth...
can't be coincidence!
If you enter 39.4784 into the pendulum sim as the length of the pendulum (essentially the radius) and set g = 1 then the period T = 39.4784
set g = Pi then the period T = 7 x Pi
Nolan | June 10, 2011 - 18:56
If you think Pluto is way-out try Mercury.