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The Engineer

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# Solar System Kinetic Energy Problem

11/22/2006 1:28 PM

What object in our solar system has the most kinetic energy. I'll limit the question to the reference frame for which the Sun is fixed, but allowed to rotate. For first estimates, please assume circular orbits for the planets, and assume the Sun and planets each are perfect spheres with uniform density. After that answer has been had, feel free to go crazy and see what you get without the above assumptions.

Kinetic Energy = 1/2 Iω2

Isolid sphere= 2/5 mr2

ω = Angular Velocity = v/r

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Guru

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#1

### Re: Solar System Kinetic Energy Problem

11/22/2006 2:47 PM

Maybe we don't need all of the math? Isn't the problem simpler if we look at the mass of each planet and multiply that by the planet's instantaneous orbital velocity, then divide by 2?

So, if we do that, we get the following Excel table:

Planet (Mass (Earth = 1) * Mean Orbital Velocity (Km/sec)) / 2 = KE

Mercury (0. 055 * 47.89 ) / 2 = 1.316975

Venus (0.815 * 35.03) / 2 = 14.28

Earth (1 * 29.79) / 2 = 14.90

Mars (0.107 * 24.13) / 2 = 1.29

Jupiter (318 * 13.06) / 2 = 2076.54

Saturn (95 * 9.64) / 2 = 457.90

Uranus (15 * 6.81) / 2 = 51.08

Neptune (17 * 5.43) / 2 = 46.16

Pluto (0.002 * 4.74) / 2 = 0.005

Looks like Jupiter wins the bout. Pluto is a non-contestant because its mass is too little and its rank has been pulled. ;-(

Another clue is that Jupiter perturbs the Sun the most of all planets.

So what am I doing wrong? Besides not being able to make a table in this weird forum!!!!

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#2
In reply to #1

### Re: Solar System Kinetic Energy Problem

11/22/2006 3:25 PM

Hero,

Actually Kinetic Energy = 1/2 Mass x Velocity2 = 1/2 mv2

The velocity is squared.

Also, planets are spinning as they rotate around the sun, that energy should be counted as well. Also, the Sun is spinning in this problem, and so that has to be considered.

I suspect it will come down between Jupiter and the Sun, but I'm not entirely sure and I'm interested to see if my guess is wrong.

I think if you use velocity squared in your equations, average velocity is a good way to calculate the kinetic energy for the orbital motion of each planet.

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#3
In reply to #2

### Re: Solar System Kinetic Energy Problem

11/22/2006 3:34 PM

You are right. I was too much hurried and forgot to square the velocity.

I did not think of spin as part of the question.

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#4
In reply to #2

### Re: Solar System Kinetic Energy Problem

11/22/2006 4:39 PM

One example of what I mean;

Mercury has average orbital velocity of 47.36 km/s and has mass of 3.3 x 1023 kg

So Orbital Kinetic Energy would be;

Orbital KE = 1/2(47360 m/s)2(3.3 x 1023 kg)
Orbital KE = 1/2(2.2 x 109 m2/s2)(3.3 x 1023 kg)
Orbital KE = 3.7 x 1032 kg-m2/s2

The radius of Mercury is 2439.7 km and the Rotational Velocity is 10.9 km/h;

So Rotational KE would be

Isolid sphere = 2/5 (3.3 x 1023 kg) * (2439700 m)2
Isolid sphere = 2/5 (3.3 x 1023 kg) * (6.0 x 1012 m2)
Isolid sphere = 7.9 x 1035 kg-m2

Rotational KE = 1/2(7.9 x 1035 kg-m2)*((3.0 m/s)/(2439700 m))2
Rotational KE = 1/2 (7.9 x 1035 kg-m2) * (1.5 x 10-12 s-2)
Rotational KE = 5.9 x 1023 kg-m2/s2

Total Kinetic Energy = Orbital KE + Rotational KE =
Total KE = 3.7 x 1032 kg-m2/s2 + 5.9 x 1023 kg-m2/s2

Total KE = 3.7 x 1032 kg-m2/s2

Hopefully I've calculated that correctly. So that would be the total KE for Mercury. It's notable that the rotational KE for Mercury isn't very important, but I don't know if that will always be the case.

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#5
In reply to #4

### Re: Solar System Kinetic Energy Problem

11/22/2006 5:29 PM

Ok, you've done all the first bit - except if we're to reply with an 'answer', we have to look up & plug in the data for the other planets, thumb our calculators & find the biggest result.

Are you suspecting that there may be bigger deviations than expected when non-circular orbits are allowed? Or maybe when a planet is given a non-uniform density?

How about totalling the KEs from the known asteroids - can that count?

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#6
In reply to #5

### Re: Solar System Kinetic Energy Problem

11/23/2006 12:15 AM

I'm honestly not sure. I think density may play a role and rotation too, I'm just not sure how big a role till I crunch some numbers.

I'm curious to see if there are big differences in the kinetic energy of the planets and the sun. If there is are big differences, why?

I just saw that Jupiter rotates at 12,600 m/s as opposed to the Sun that rotates at 1,992.8 m/s. Why should that be? What mechanism controls these rotational speeds and Energies. I was hoping seeing the energies for these planets might provide some insight.

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#7
In reply to #6

### Re: Solar System Kinetic Energy Problem

11/23/2006 4:21 AM

Hi Roger, you wrote: "I think density may play a role and rotation too". I think only inhomogeneous density plays a role, not density per se. Rotation definitely!

I'm curious as to why you want to compare kinetic energies and not total energies? This would include mc2, kinetic and potential energy.

I believe our Sun rotates far slower than what formation theories of the Solar System predict and nobody quite knows why! There is speculation about another star that may have collided with our Sun in the distance past.

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#13
In reply to #7

### Re: Solar System Kinetic Energy Problem

11/24/2006 3:28 PM

I agree with your correction, inhomgeneous density plays a role. As for why kinetic energies, I guess I just was curious about the kinetic energy. Total enegy (including mass energy) would be dominated by the sun just by the sheer mass of the sun. Potential energy could be interesting but that excludes the sun completely in this competition. Plus, this question comes from a conversation I was having where the person was assuming that Jupiter would have the most kinetic energy and I thought consdering rotation might not make that a slam dunk.

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#8
In reply to #6

### Re: Solar System Kinetic Energy Problem

11/23/2006 1:03 PM

Roger

I think you missed the 1/2 out in the orbital KE calc.

Also from mass and radius given, average density of Mercury works out about spot on same as Earth. Is this correct? I haven't had time to check but I thought Earth was rather denser at SG ~ 5.5. I know it's much denser than the big planets.

Also I remember reading somewhere that the angular momentum of the planets (in total) is much greater than the Sun's, but maybe not for energy due to the squared term.

Another thought is - does the sun have a unique angular velocity? Being fluid, perhaps it doesn't all rotate at same speed.

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#10
In reply to #8

### Re: Solar System Kinetic Energy Problem

11/24/2006 4:19 AM

Correction to #9 - if the angular momentum of the planets is greater than Sun's so is the planets' KE, by a bigger factor. Just as a bullet gets a lot more KE than the rifle, even though they have equal (and opposite) momentums.

According to my copy of CRC Handbook of Chemistry & Physics (old issue but I doubt it's changed much) SGs are -

Sun 1.410

Mercury 5.431, Venus 5.256, Earth 5.519, Mars 3.907, Jupiter 1.337, Saturn 0.688, Uranus 1.603, Neptune 2.272, Pluto 1.65.

So Mercury and Earth are about the same. Neptune surprisingly high. It doesn't give Sun's rotation speed.

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#15
In reply to #10

### Re: Solar System Kinetic Energy Problem

11/24/2006 7:02 PM

What does SG's stand for?

I'm happy to just calculate this myself. Just from observing that Jupiter rotates faster, ihas more mass, and has higher orbital velocity than the planets beyond it's orbit, I know I can eliminate them from this calculation. That leaves Earth, Mars, Venus, Mercury, and the Sun. The fastest planet of those is Mercury, but it's small. The sun is massive and so it's rotational kinetic energy may be larger than Jupiter's rotational + orbital kinetic energy combined. I'm going to calculate Jupiter to see (I've already done the Sun in this thread).

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#27
In reply to #15

### Re: Solar System Kinetic Energy Problem

11/27/2006 10:20 AM

Reply to #15 - SG = specific gravity.

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#12
In reply to #8

### Re: Solar System Kinetic Energy Problem

11/24/2006 3:23 PM

You Wrote: "I think you missed the 1/2 out in the orbital KE calc."

You're right! I fixed it in that comment, thanks for bringing it to my attention.

You Wrote: "Another thought is - does the sun have a unique angular velocity? Being fluid, perhaps it doesn't all rotate at same speed."

The sun does have an angular velocity, just like the Gas Giants. Jupiter seems to rotate very quickly for it's size which made me wonder if all of Jupiter rotates at that rate, or the whole planet does (including it's small rocky core) and how scientists can tell. So long story short, to your question, I'm not 100% sure.

You wrote: "Also I remember reading somewhere that the angular momentum of the planets (in total) is much greater than the Sun's, but maybe not for energy due to the squared term. "

I don't know, I'd like to find out!

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#9
In reply to #4

### Re: Solar System Kinetic Energy Problem

11/23/2006 10:51 PM

Roger,

I am interested as to why you are using Kgm2S-2 as the units for energy in post #5 rather than Joules?

Also do you have the remaining relevant info on the planets rotational velocities and diameters at hand so I can plug it into my spreadsheet and see what come out or do I need to look them up?

Finally if some sort of pattern emerges here how could we apply it to the asteroids in the asteroid belt and what could we deduce from it? For example could we calculate the total mass of all the asteroids or their average rotational velocity?

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#11
In reply to #9

### Re: Solar System Kinetic Energy Problem

11/24/2006 3:15 PM

You wrote: "I am interested as to why you are using Kgm2S-2 as the units for energy in post #5 rather than Joules?"

1 Kgm2/s2 = 1 Joule; So what I wrote is equivalent (which you probably already know). I kept it the way I did so that I didn't confuse anyone with the switch, though I suppose I could have just added a line that said the conversion.

You wrote: "Also do you have the remaining relevant info on the planets rotational velocities and diameters at hand so I can plug it into my spreadsheet and see what come out or do I need to look them up? "

Wikipedia has all the relevant data. I just searched each planet in wikipedia to get the data.

You wrote: "Finally if some sort of pattern emerges here how could we apply it to the asteroids in the asteroid belt and what could we deduce from it? For example could we calculate the total mass of all the asteroids or their average rotational velocity?"

To be honest, I don't know what to expect. The way this came about was in a discussion where the person I spoke to assumed it would be Jupiter. I wasn't convinced, I thought rotation could play a role.

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#14

### Re: Solar System Kinetic Energy Problem

11/24/2006 4:01 PM

Here's a calculation for the rotational kinetic energy of the Sun.

Sun's mass is 1.99 x 1030 kg and rotational speed of 1992.78 m/s and a radius of 6.96 x 108 m.

I=2/5 mr2 = 2/5(1.99 x 1030 kg) x (6.96 x 108 m)2
I=(7.96 x 1029 kg) x (4.84 x 1017 m2)
I= 3.85 x 1047 kg-m2

KE = 1/22 = 1/2 I(v/r)2
KE = 1/2 (3.85 x 1047 kg-m2)((1.99 x 103 m/s)/(6.96 x 108 m))2
KE = (1.93 x 1047 kg-m2) x (2.86 x 10-6 s-1)2
KE = (1.93 x 1047 kg-m2) x (8.18 x 10-12 s-2)
KE = 1.58 x 1036 kg-m2/s2

So the Rotational Kinetic Energy of the Sun is 1.58 x 1036 joules. I'd like to see how this compares to the total Kinetic Energy of Jupiter (orbital + rotational).

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#16

### Re: Solar System Kinetic Energy Problem

11/24/2006 7:53 PM

Here are the kinetic energy calculations for Jupiter.

Jupiter has mass of 1.9 x 1027 kg, radius of 71.5 x 106 m, avg orbital velocity of 13.1 km/s, and avg rotational velocity of 12.6 km/s

Orbital Kinetic Energy = 1/2 mv2 = 1/2 (1.9 x 1027 kg) x (13.1 x 103 m/s)2

KEorbital = (9.5 x 1026 kg) x (1.72 x 108 m2/s2)
KEorbital = 1.63 x 1035 kg-m2/s2

Rotational Kinetic Energy = 1/22

I = 2/5 mr2 = 2/5 (1.9 x 1027 kg) x (71.5 x 106 m)2
I = (7.6 x 1026 kg) x (5.1 x 1015 m2)
I = 3.88 x 1042 kg-m2

KErotational = 1/2 (3.88 x 1042 kg-m2) x ((12.6 x 103 m/s)/(71.5 x 106 m))2
KErotational = (1.94 x 1042 kg-m2) x (1.76 x 10-4 s-1)2
KErotational = (1.94 x 1042 kg-m2) x (3.1 x 10-8 s-2)
KErotational = 6.0 x 1034 kg-m2/s2

Total Kinetic Energy = KEorbital + KErotational
KETotal = (1.63 x 1035 kg-m2/s2) + (6.0 x 1034 kg-m2/s2)
KETotal = (2.2 x 1035 kg-m2/s2)

KETotal = 2.2 x1035 joules

So the total Kinetic Energy for Jupiter is 2.2 x 1035 joules , which is larger than the rotational kinetic energy of the Sun 1.58 x 1035 joules. It certainly appears that Jupiter probably holds the crown, but not by much.

As an aside, I noticed while doing these Jupiter calculations that the orbital kinetic energy for Jupiter and the rotational kinetic energy of the Sun are within 3% of each other. That seems way too close to me to be coincidence. I'd like to know why that is. It makes sense to me that as a solar system forms, the gas cloud that condenses is rotating and eventually forms the spinning Sun and the planets in their orbits. It makes sense to me in this light that orbital velocity of planets and the spinning of the sun should be close. Yet I didn't see that closeness in my Mercury orbital KE calculation and since I know the other gas giants are both slower and smaller than Jupiter, I won't see this effect in them either. Maybe it scales to mass? or distance? I don't know, but it's an interesting result worth some thought.

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#17

### Re: Solar System Kinetic Energy Problem

11/25/2006 3:13 AM

I got all the figures from Wikipedia and here is what I came up with for the total kinetic energy of the planets and dwarf planets.

Planet

Orbital Energy

Rotational Energy

Total Energy

Mercury

3.703E+32 Joules Orbital KE

1.015E+26 Joules Rotational KE

3.703E+32 Joules Total KE

Venus

2.985E+33 Joules Orbital KE

5.364E+26 Joules Rotational KE

2.985E+33 Joules Total KE

Earth

2.642E+33 Joules Orbital KE

4.341E+31 Joules Rotational KE

2.685E+33 Joules Total KE

Mars

1.860E+32 Joules Orbital KE

1.254E+30 Joules Rotational KE

1.873E+32 Joules Total KE

Jupiter

1.619E+35 Joules Orbital KE

1.010E+37 Joules Rotational KE

1.026E+37 Joules Total KE

Saturn

2.670E+34 Joules Orbital KE

1.857E+36 Joules Rotational KE

1.884E+36 Joules Total KE

Uranus

2.005E+33 Joules Orbital KE

1.955E+34 Joules Rotational KE

2.155E+34 Joules Total KE

Neptune

1.511E+33 Joules Orbital KE

2.478E+34 Joules Rotational KE

2.629E+34 Joules Total KE

Pluto

1.421E+29 Joules Orbital KE

7.529E+25 Joules Rotational KE

1.421E+29 Joules Total KE

Ceres

1.512E+29 Joules Orbital KE

3.217E+26 Joules Rotational KE

1.516E+29 Joules Total KE

Sun

2.653E+38 Joules Rotational KE

2.653E+38 Joules Total KE

Now the figure for the sun is a bit high because the whole thing is very fluid and the rotation at the equator once every 25 days but away from the equator it takes up to 35 days so the amount of rotational kinetic energy will be less than that stated. Since the sun is like this it is quiet possible that all the gas giants have the same problem so I don't know what we can read into this.

An interesting points to note is that for the solid planets the major component is their orbital velocity while for the gaseous planets it is the rotational component that is the dominant factor.

PS I apologize for the format but the CR4 editor dismantled my nice table when I posted it.

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#18
In reply to #17

### Re: Solar System Kinetic Energy Problem

11/25/2006 2:45 PM

Masu,

First of all, thank you for doing so much work to help me in this. Can you check your rotational KE calculations, they seem off to me. You did this in excel right? That was good thinking. I will do the same and see what I get.

Roger

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#19

### Re: Solar System Kinetic Energy Problem

11/26/2006 9:14 AM

As Roger pointed out the rotational energies in my earlier post were a little high. The problem was that the data I had listed the speeds in Km/h and dopey me multiplied this by 3.6 instead of dividing when I converted to ms-1. Here is an updated list of the energies:

Planet…...…Orbital Energy…… Rotational Energy…… Total Energy…..….Ratio

Mercury…. 3.703E+32 Joules…. 6.045E+23 Joules…. 3.703E+32 Joules….. 6.126E+08

Venus…… 2.985E+33 Joules…. 3.194E+24 Joules…. 2.985E+33 Joules….. 9.347E+08

Earth…...... 2.650E+33 Joules…. 2.585E+29 Joules…. 2.650E+33 Joules….. 1.025E+04

Ceres.……. 1.512E+29 Joules…. 1.914E+24 Joules…. 1.513E+29 Joules….. 7.901E+04

Mars…..….. 1.860E+32 Joules…. 7.467E+27 Joules…. 1.860E+32 Joules….. 2.492E+04

Jupiter….... 1.619E+35 Joules…. 6.014E+34 Joules…. 2.220E+35 Joules….. 2.691E+00

Saturn……. 2.670E+34 Joules…. 1.106E+34 Joules…. 3.776E+34 Joules….. 2.415E+00

Uranus….... 2.005E+33 Joules…. 1.164E+32 Joules…. 2.121E+33 Joules….. 1.722E+01

Neptune….. 1.511E+33 Joules…. 1.475E+32 Joules…. 1.659E+33 Joules….. 1.024E+01

Pluto….….. 1.421E+29 Joules…. 4.483E+23 Joules…. 1.421E+29 Joules….. 3.169E+05

The final figure in the table is the ratio of orbital KE to rotational KE. Note that for the rockey planets the figures are reasonably large numbers meaning the orbital KEs are considerably greater than rotational KEs. When we get to the gaseous planets though the number is considerably smaller which translates to the rotational KEs being a much greater part of the KE equation.

So what dose this mean? Could there be a reason for this apparent discrepancy in the figures.

If you took the gaseous planets and stripped off their outer gaseous sections leaving a hard rockey core then you would see ratios like that for the existing planets. Could this mean that originally all the planets were gas giants and some how the gaseous parts were stripped off leaving only the cores? If this were the case it would throw current thinking on planet formation out the window.

Has anybody got any other theories or observations?

PS I apologies for the layout again. I tried to do it a different way this time and it looked like a table till I posted it but the editor bit me again.

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#20
In reply to #19

### Re: Solar System Kinetic Energy Problem

11/26/2006 7:07 PM

Masu,

That's excellent, thank you for putting that info together. I won't be able to look at it in detail till tomorrow, then I'll give you my impressions and opinions.

Roger

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#21
In reply to #19

### Re: Solar System Kinetic Energy Problem

11/27/2006 1:42 AM

Hi Masu, you asked: "Has anybody got any other theories or observations?"

I have one, which I asked to Roger before, but I did not see a reply (maybe, I missed it?) Anyway, I'm still not sure what this particular analysis means! To view orbital kinetic and rotational energies of orbiting bodies alone makes no sense to me.

In post #7 I suggested that to get a complete picture, one must include potential energy and even mass/thermal/pressure energies. One can leave out the final trio for simplicity, but you cannot leave out potential energy!

Orbits in their simplest form work on a conservation of potential plus kinetic energy principle. As the original disk of matter contracted and lumped out into proto planets, the inner parts had less potential energy and more kinetic and rotational (spin) energy.

So what you can learn by leaving out one main component, I'm not too sure! Maybe Roger must give us some pointers.

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#22
In reply to #21

### Re: Solar System Kinetic Energy Problem

11/27/2006 4:03 AM

I agree Jorrie and I originally intended to include the PE as well but forgot, sorry. I only remembered some hours later. As soon as I can figure out how to fit all the info on one line I will post it.

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#23
In reply to #22

### Re: Solar System Kinetic Energy Problem

11/27/2006 4:27 AM

Masu, maybe a hint for your posting problem. I copy&paste an Exel table to a graphics package, save it as a JPEG and then import it into CR4 by the "insert image camera" of the editor. Roundabout, but it works for me!

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#24
In reply to #23

### Re: Solar System Kinetic Energy Problem

11/27/2006 7:30 AM

Jorrie I'm glad you're here as I have a question. I am trying to calculate the potential energy for each planet and to calculate the force of gravity between the each of the planets I was using the formula

g = (U x Mp x Ms)/S2

U = universal gravitational constant

Mp = Mass of planet

Ps = Mass of the Sun

S = Average distance between planet and Sun

But I am getting ludicrous figures like 3.092 x 1068 ms-2 for Mercury which I find somewhat hard to believe.So what am I doing wrong?

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#26
In reply to #24

### Re: Solar System Kinetic Energy Problem

11/27/2006 10:16 AM

Hello Masu (and apologies Jorrie for jumping in)

Not sure what you're doing. You say calc is force of gravity and the RHS of your equation gives this. But g is usually an acceleration and you give the (unexpected) result in m.s-2. I don't know which planet you're on (if you'll excuse the expression, it's not meant to be offensive!) but if I divide by U instead of multiplying I get somewhere near the 1068.

Also I worked out the orbital KEs and got very similar figures to yours. Total ~ 2*1035 J. But I'm puzzled by your figure for Sun's rotational, 2.653*1038 J. I found a rotational period 26 days on the web and using that calculate 2.56*1036 J, and that's assuming same speed and same density all through, both tending to give a high figure.

Also I need to correct a correction in post #10 re momentum. This applies to linear momentum but not angular.

I make Sun's angular momentum just under 1042 (SI units of course), total for planets just under 1045. Although planets' total mass is much lower than Sun's, total moment of inertia is higher so lower KE figures. I calculate MoI for planets 4.8*1051, for Sun 3.3*1047 kg.m2.

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#29
In reply to #26

### Re: Solar System Kinetic Energy Problem

11/27/2006 12:29 PM

Roger Your calculation for the rotational KE for the sun is out by a factor of 10. The mistake is in your calculation of I. You calculated as follows

I=2/5 mr2 = 2/5(1.99 x 1030 kg) x (6.96 x 108 m)2
I=(7.96 x 1029 kg) x (4.84 x 1017 m2)
I= 3.85 x 1046 kg-m2

which is incorrect, the answer is actually 38.56 x 1046 and that makes the rotational KE for the Sun 1.579 x 1036 which is greater than Jupiter's total KE so the Sun wins after all and your initial hunch was correct.

By the way its easier to work out if you substitute prior to doing the calculations as follows

Kinetic Energy = 1/2 Iω2

Isolid sphere= 2/5 mr2

ω = Angular Velocity = v/r

therefore

KE = Â½ 2/5 mr2 v2/r2

KE = 1/5 mr2for a solid sphere

Codemaster here is an example for the acceleration between Mercury and the Sun

GMS = U (MM MS)/S2 = 6.6742 x 10-11 x 6.419 x 1023 x 1.988 x 1030 /(2.28 x 108)2

GMS = 85.169 x 1042 / 8.294 x 1016

GMS = 10.2687 x 1026 = 1.02687 x 1027

If you use the equation for an object on the surface of earth you get 9.817ms-2 so where is everything going wrong.

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#31
In reply to #29

### Re: Solar System Kinetic Energy Problem

11/27/2006 1:01 PM

Reply to #29 (latter part)

Masu - I don't understand your figure for S, 2.28*108 m, presumably you mean distance from Mercury to the sun? Number I have is 4.6*1010 m. With the right distance calc gives the force (in newtons, naturally) between Mercury and the sun. If you leave out the mass of Mercury it gives the acceleration caused by the sun at Mercury's distance. This should of course = Mercury's centripetal acceleration r.ω2.

To get g = 9.817 m.s-2 (ie gravity at Earth's surface, I assume you're referring to) you need to use mass and radius of Earth.

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#32
In reply to #29

### Re: Solar System Kinetic Energy Problem

11/27/2006 1:17 PM

Hi Masu, I think I have answered your: "...so where is everything going wrong?" to some extent in my previous post. The average gravitational acceleration of Mercury relative to the Sun is, in my calculations: 3.96 x 10-2 ms-2, which is incredibly, only ~4 Milli-g!

I get this from the following values: G = 6.67 x 10-11 m3 Kg-1 s-2, Msun = 1.99 x 1030 Kg, s = 5.79 x 1010 m and a = GMsuns-2 ms-2. Divide this by ~9.8 ms-2 to get "g".

Does this make any more sense? Correct me if I goofed!

Jorrie

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#34
In reply to #29

### Re: Solar System Kinetic Energy Problem

11/28/2006 4:34 AM

Reply again to #29

Hello Masu - re Sun's KE, I agree pretty much with your 1.579 x 1036 J. (for the KE in my post #26 I'd forgotten to divide by 2 - ironic in light of my post #8 !) But where does this leave the figure 2.65*1038 J quoted from Wikipedia in your post #17? Looks as if they got it wrong.

If I use my rule-of-thumb value for sun's radius 6.4*108 m (= 100 x Earth's) and forget the 1/2 it comes very close to 2.65*1036. Just possible they did that and the 1038 is a typo.

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#28
In reply to #24

### Re: Solar System Kinetic Energy Problem

11/27/2006 12:23 PM

Hi Masu, you wrote: "... I am getting ludicrous figures like 3.092 x 1068 ms-2 for Mercury, which I find somewhat hard to believe. So what am I doing wrong?"

I spot two immediate problems with your equation:

1: U is normally used for the "gravitational constant of the Sun", meaning it represents GMs, so you cannot have another Ms inside the brackets.

2: If you are after "g", it is acceleration (as is clear from your units), the mass of the planet does not belong in the equation (all masses fall at the same acceleration).

So the proper equation for acceleration is simply:

g = U/S2 ms-2.

For reasons of errors like the above, I despise the use of U = GMs, but rather use Newton's gravitational constant G in all equations.

I would rather write: a = GMs/S2 ms-2 and reserve g for 9.8 ms-2, meaning "one Earth gravity".

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#25
In reply to #21

### Re: Solar System Kinetic Energy Problem

11/27/2006 9:55 AM

Jorrie,

This whole exercise came from a discussion in which the question popped up "what object in the solar system has the most kinetic energy?" The first answer given was Jupiter, because of it's mass and orbit, but kinetic energy includes rotation too, and I was sure that the spinning of the sun should give Jupiter orbit + rotation a run for its money. Unfortunately (for my Sun theory) we've found Jupiter won out. (So to answer your question as to why just kinetic energy, no good reason except curiousity as to how big the rotational component was as compared to the orbital component).

However, in the process, I noticed that the kinetic energy of the rotating Sun and the orbital kinetic energy of Jupiter are almost identical. Now this seems way too convenient to be coincidence and that made me start to wonder why. Do you know? I'm not aware of a reason of why they should match up like they do, but I'm not aware of a lot of things. I would appreciate your insight in this, even if its to say "its coincidence".

My initial thought is that the Sun has synchronised with Jupiter somehow (like Earth and moon) or that the rotation rate of a star caps the orbital kinetic energy of a planet. That could explain why gas giants are all farther away from the sun in our solar system rather than closer as has been seen in far away stars. Anyway, this could all be completely wrong, I just am enjoying the exercise.

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#30

### Re: Solar System Kinetic Energy Problem

11/27/2006 12:54 PM

Thanks Jorrie, I had a thought that was where I was going wrong because if you used earths mass and radius and plugged it into the equation it give you 9.817ms-2. Here is an updated table containing the orbital kinetic, potential and total energy and the acceleration due to the Suns gravity. Do you think these are reasonable values for the Sun's gravity?

Finally here is an updated table of the kinetic energies as Roger first asked.

As you can se the one with the greatest total KE as Roger asked is the Sun and not Jupiter as we first thought.

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#33
In reply to #30

### Re: Solar System Kinetic Energy Problem

11/27/2006 1:55 PM

Hi Masu, I have just checked the right hand column of your first table and noted that we disagree with a precise factor of 106. You seem to have for Mercury: 3.96E+04 m/s2, while I have: 3.96E-02 m/s2. What's a factor of a million among friends? Except...

I checked your value for Earth and it is pretty obviously wrong: an acceleration of 5.9E+03 m/s2 towards the Sun would have ripped everything off the surface of Earth and put it in orbit around the Sun! Probably, it would have broken Earth in a million pieces and 'swallowed' all that rubble! Check it - it's 590g - simply not on!

So I suggest you check for the factor 106 and I recheck my (pretty low, I must say) accelerations again!

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#35

### Re: Solar System Kinetic Energy Problem

11/28/2006 6:13 AM

Sorry, sorry, sorry, I apologize for the erroneous data.

I found where I was going wrong with the gravity calculations and the problem was with the base units, I was using kilometers instead of meters. The confusion came about as some units were in meters and scientific notation while others were in kilometers with and scientific notation. Please forgive me, even NASA make mistakes like this.

I have decided to go back to square 1 and do it all again but this time using SI units and engineering notation which will hopefully eliminate the problem and make it easier for us engineering type people to understand. Pleas hang in there and I will post the new tables in a couple of hours.

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#36
In reply to #35

### Re: Solar System Kinetic Energy Problem

11/28/2006 6:20 AM

Masu - do you have Mathcad? It's brilliant, it looks after all the units for you, so it's much easier to focus on the physics of a problem without getting distracted by various constants, etc.

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#37
In reply to #35

### Re: Solar System Kinetic Energy Problem

11/28/2006 6:50 AM

No problem! Fortunately, we have lots of chances to get it right (so had NASA with that mix of SI and imperial units that lost them a Mars orbiter once - they found out a tad too late!)

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#38
In reply to #35

### Re: Solar System Kinetic Energy Problem

11/28/2006 8:48 AM

Again, thanks for doing this. I'm trying to get involved but I keep getting distracted. Your total for the Sun's rotational energy looks off by a factor of ten, but that may be because my calculation was off by a factor of ten ;) When I have time later this morning, I'll check it.

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#39
In reply to #35

### Re: Solar System Kinetic Energy Problem

11/28/2006 11:22 AM

Masu,

I've checked my calculation for the rotational kinetic energy of the sun and I was wrong, your calculated value was correct (Man, is there anything more dangerous than actually doing the math? I'm making mistakes left and right!)

So the Sun would have the highest kinetic energy, I guess that makes me happy, but that means that Jupiter's orbital energy is 1/10th the size of the Sun's rotational energy. So much for the Sun and Jupiter Synchonizing. Oh well, this is why we do the math.

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#40
In reply to #35

### Re: Solar System Kinetic Energy Problem

11/28/2006 11:47 AM

Hi Masu, since my 'forte' is gravitation theory and not quite mechanical engineering, I decided to provide a check for your gravity calcs...

I used these primary values: G = 6.67E-11 m3kg-1s-2; M_sun = 1.99E+30 kg.

The mean orbital velocities I calculated from the primary values and distances as a check on reality.

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#41
In reply to #40

### Re: Solar System Kinetic Energy Problem

11/28/2006 12:25 PM

It will be interesting, once we get our numbers straight, if the Virial theorem holds. If it does, the potential energy should be twice the kinetic energy.

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#42
In reply to #41

### Re: Solar System Kinetic Energy Problem

11/28/2006 12:49 PM

Yep Roger, it would be very interesting!

I've always been scared of the Virial theorem, because it sounds a bit viral to me!

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#43
In reply to #42

### Re: Solar System Kinetic Energy Problem

11/28/2006 12:53 PM

It's funny, I've got a cold right now and I was thinking the same thing as I was typing it. Notice if Masu's potential energy calculations are off by 103 then the virial theorem will be correct.

Also, now that Masu caught my mistake on the rotating kinetic energy of the Sun, that if you added up all the kinetic energy of the planets the total would still be less the kinetic energy of the rotating Sun. Wow.

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#48
In reply to #41

### Re: Solar System Kinetic Energy Problem

11/28/2006 1:38 PM

Roger, you said: "... if the Virial theorem holds. If it does, the potential energy should be twice the kinetic energy."

This is in fact a property of circular Newtonian orbits: if you ignore the energy of own rotation (angular momentum-energy), the potential energy will be exactly twice the kinetic energy.

If M is the mass of the primary and m the mass of the orbiting body, then:

P.E. = GmM/r

K.E. = 0.5mV2

V2 = GM/r

Thus:

Kinetic Energy. = 0.5GmM/r = half Potential Energy.

Conclusion: including the rotational energy into the kinetic energy 'violates' the Virial theorem! What does this tell us? I don't know, so let's think about it!

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#51
In reply to #48

### Re: Solar System Kinetic Energy Problem

11/28/2006 2:07 PM

I wrote in my previous post: "Conclusion: including the rotational energy into the kinetic energy 'violates' the Virial theorem! What does this tell us? I don't know, so let's think about it!"

I only thought up this (before pouring a sundowner and retiring to "slow-living"):

The Virial Theorem is valid for particles. Actually, I think, for point-particles. Can such things have angular momentum due to spin? Unlikely.

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#52
In reply to #51

### Re: Solar System Kinetic Energy Problem

11/28/2006 2:20 PM

Jorrie,

I agree. I think the Virial theorem applies only to orbital (kinetic and potential) energy and it checks out (which is reassuring).

I'm a bit surprised how strong the rotational kinetic energy is for many of the planets, I probably would have expected less. It probably is less once we start figuring in the density variations. Take the Sun for instance. I treated the mass as if it was distributed evenly throughout the sphere of the sun. More likely the Sun is denser near it's center, which would result in a downward correction to its rotational kinetic energy. Anyway, I've learned a bit. There's something about getting your hands dirty (and Masu's) with actual numbers that makes the theories you read about sink in.

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#59
In reply to #48

### Re: Solar System Kinetic Energy Problem

11/29/2006 4:06 AM

Jorrie you showed that

Kinetic Energy. = 0.5GmM/r = half Potential Energy

So I thought I would plot the results so far and this is the result

As you can see the data tends to indicate that the relationship between the orbital KE and PE is

PE = 2.2413 x KE0.9986

Which is close to being linear and complying with your analysis but not exactly. Why the discrepancy? It seems to be too good a fit on the graph to be just error in the calculations. It seems to indicate that the KE is less than expected and the loss is more significant for the smaller bodies. Have you any idea what is going on?

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#60
In reply to #59

### Re: Solar System Kinetic Energy Problem

11/29/2006 4:28 AM

masu, your "PE = 2.2413 x KE0.9986" cannot be right, because the calculated ratio must be 2:1 exactly. You have not perhaps included the rotational energy into the kinetic energy?

If you did, you must also make a correction on the potential energy (caused by the rotational energy), which will bring the ratio back to 2:1, I think.

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#61
In reply to #60

### Re: Solar System Kinetic Energy Problem

11/29/2006 6:53 AM

Hello Jorrie

I can see that by calculation the KE must be 1/2 the (negative) PE, because it comes out of the algebra, but I'm not sure what significance the PE has. I believe it's the PE lost in falling from infinite distance to current orbit radius from sun. But presumably it was never at infinite distance, and if it was, where has the other 1/2 of the PE gone? Can you enlighten me please?

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#85
In reply to #61

### Re: Solar System Kinetic Energy Problem

11/30/2006 1:07 PM

Hi Codemaster, I'm sorry, in this very busy thread, I missed this one for a while! Your "... but I'm not sure what significance the PE has. I believe it's the PE lost in falling from infinite distance to current orbit radius from sun." is in fact pretty close to correct.

IMO, the easiest way to look at it is: suppose the Sun was the sole star in the universe and we take a particle with minuscule mass and shoot it away from the Sun at exactly escape velocity, Ve = √[2GM/s], for the starting distance s.

This particle would travel away from the Sun forever, but asymptotically approach zero velocity as distance gets extremely large. Now reverse this situation: let the particle have near zero velocity towards the Sun, at extremely large distance. Newton's gravity theory assigns a total energy of zero to this particle (zero potential energy and zero kinetic energy).

Now as this particle falls towards the Sun, its total energy must remain zero, while its kinetic energy obviously steadily increases (KE = 0.5mv2). This means that its potential energy must decrease from zero, i.e., become negative (PE = -GMm/s).

If we sum the two energies, we get: E = 0.5mv2-GMm/s, which must equal zero, as per Newtonian assumption. Solve this for E=0 and we get the escape velocity:

Ve = Â±√[2GM/s], which is known to be true from experiment.

This only holds for s > 0 in the Newtonian case and in the relativistic case, only for s > 2GM, because the velocity cannot exceed the speed of light.

Think this over and ask if still not clear.

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#62
In reply to #60

### Re: Solar System Kinetic Energy Problem

11/29/2006 7:37 AM

Jorrie

The data for the graph are taken directly from the table for kinetic and potential energy that I posted earlier. The calculation for the KE and PE are

KE = Â½ MassPlanet x Velocity2Orbit

PE = MassPlanet x OrbitRadius x GravitySun

Here is the line of best fit for the known planets

And this is the exponential equation that seems to fit exactly

So the question is where are we going wrong. The kinetic energy calculation isn't really open for debate so that would mean the error must be with the potential energy calculation.. Are we calculating the gravity of the sun correctly? Should we be using the suns gravity at the orbital distance?

If you think about it the potential energy is the energy you would expend if you lifted a object from the center of the Sun to the height of its orbit. But gravity isn't constant over these distances so I think we need to rework the equation for potential energy to take this into account.

Boy this just keeps getting more and more complicated all the time. Fascinating isn't it?

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#63
In reply to #62

### Re: Solar System Kinetic Energy Problem

11/29/2006 9:16 AM

Hey guys,

I think the 2xKE=PE is for circular orbits, whereas these orbits are perfectly circular. That could be a source of error.

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#65
In reply to #62

### Re: Solar System Kinetic Energy Problem

11/29/2006 9:22 AM

Masu - if we do the algebra, we can show that for any planet the orbital KE = G*Msun*Mplanet/(2*Rplanet), and PE = G*Msun*Mplanet/Rplanet, so ratio must be 2:1, as Jorrie said. But there are several ways to calculate these things, using different inputs. Values for these inputs are derived from observations and have some uncertainty, so unlikely to lead to exactly 2:1. Also if you add rotational KE to orbital, ratio obviously changes.

Also PE is not from centre of sun to Rplanet, it's from Rplanet to infinity (see post #60)

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#66
In reply to #65

### Re: Solar System Kinetic Energy Problem

11/29/2006 10:57 AM

Let's see… If I type "(m_sun) (m_earth) G / 1 AU" into the google search line, I get:

(m_sun * mass of Earth * G) / (1 AU) = 5.30020186 Ã— 1033 joules.

If I then type in "0.5 G (m_sun) (m_earth) / AU" I get:

(0.5 * G * m_sun * mass of Earth) / AU = 2.65010093 Ã— 1033 joules

If I then type in "5.30020186/2.65010093" I get:

5.30020186 / 2.65010093 = 2

That seems close enough to a 2:1 ratio to suit me.

This could be repeated for each planet (with the awkward inconvenience that the non-earth orbital radii are unknown to the Google calculator) and in each case, the ratio will have to be 2:1, precisely, assuming we use the same source for distance and mass for each planet. Even if there is disagreement between sources on mass, the radii must disagree in a precisely compensatory way (it's not as if we measure the radii with a tape measure and mass with a scale – they have to correspond perfectly in any individual's thought system: in other words, you can't have one source saying that the mass is a little higher but the orbit radius is the same, given equal periods.)

I am impressed with Masu's industry – and impressed by the time all of you have put into this, but I fear that Masu has made a mistake in his spreadsheet. It is impossible for the ratio not to equal precisely 2, without saying that the masses, distances, or periods for the KE vs PE calculations are different. Perhaps Masu has added in the individual rotational kinetic energies for each planet? That would be OK, I suppose, if we were working in a different frame of reference (as would be adding in caloric effects, rest energy, etc.) but for our purpose – which is to decide how hard we'll have to whack to knock a given planet out of the system, then we don't want to add in those other energies.

(A golf ball will fall with the same acceleration through a vacuum, regardless of whether you give it a spin or not.)

Incidentally, I think it was Masu who correctly stated that the PE is best thought of as the energy required to lift the planet out to its orbit, against that planet's sun/planet gravitational attraction. To view it the other way around, from infinity in, doesn't make a lot of sense, at least to me, given that, by convention, we do things the other way around (e.g. a golf ball on a meter high counter has some PE relative to the floor, rather than some PE relative to the edge of the universe.)

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#68
In reply to #66

### Re: Solar System Kinetic Energy Problem

11/29/2006 12:50 PM

Ken - only bit I'd comment on is PE definition. Formula G*M1*M2/R is from integrating force G*M1*M2/r2 from ∞ to R, = G*M1*M2*(1/ - 1/R) so has to be that way round. Potential from centre of sun to R is something else. Need to split the integration into 2 parts, from orbit to sun's surface Rsun, and from sun's surface to centre, because the force/distance relationship changes. I make it 1.5*G*Msun*Mearth/Rsun

= 7.67*1035 J compared to your 5.30020186 Ã— 1033. I did that fairly quick so if anybody can find a mistake.....

I'll post the workings if you want.

Cheers

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#71
In reply to #68

### Re: Solar System Kinetic Energy Problem

11/29/2006 1:19 PM

Correction to #66

Good start - I've found a mistake already. I integrated from ∞ (instead of from Rearth) to centre of sun, so PE 7.67*1035 J includes Ken's 5.30020186 Ã— 1033 joules. Latter almost gets lost in the total, but for the record.

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#76
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### Re: Solar System Kinetic Energy Problem

11/30/2006 12:19 AM

Codemaster:

Using your PE of 7.67 * 1035 J and the agreed KE for earth of 2.65 x 1033 J, then our ratio would be about 289:1, substantially different than 2:1. Could this be?

If anyone is feeling energetic, perhaps finding specific orbital energies (and then multiplying by masses) would give us numbers for comparison.

I notice, in the Wikipedia example re the space station, that the PE:KE ratio is 2:1, lending credence to what we have been saying.

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#70
In reply to #62

### Re: Solar System Kinetic Energy Problem

11/29/2006 1:03 PM

Hi Masu, you wrote: "The data for the graph are taken directly from the table for kinetic and potential energy that I posted earlier."

I think the difference lurks in the value taken for velocity, which is usually given in the tables as the mean orbital velocity of an elliptical orbit. I'm not sure how exactly this is defined.

As Roger has pointed out in his reply, the 2:1 probably only holds for circular orbits, with a constant velocity of SQRT(GM/s). The mean velocity may be slightly different, perhaps explaining the discrepancy?

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#72
In reply to #70

### Re: Solar System Kinetic Energy Problem

11/29/2006 2:55 PM

But wait!! Take a look at Masu's chart (the second one) in post 45. His own values always indicate a 2:1 relationship, not a 2.2413:1 one. It is only his curve fitting or his transportation of data that went wrong. I'd blame the problem on his obsessively focusing too much on pretty pictures!! (OK, only kidding – I like all the stuff he does.) Clearly all those 2:1 ratios cannot fall on a line that suggests a 2.2413:1 (almost) ratio.

Actually, if you do the math, I think you'll find that the ratio holds perfectly for elliptical orbits as well (with the gain in speed at a point compensated by the reduction in radius at that point -- equal swept area, etc). I think it also holds for falling golf balls, because the average speed is Â½ the speed at which it hits the floor, if that makes any sense.

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#44

### Re: Solar System Kinetic Energy Problem

11/28/2006 12:55 PM

Fun. I arrived here at the perfect time: all the number crunching's been done. So just how large would a golf club need to be to knock Jupiter into Earth's orbit? Where would you you stand to make your swing? How fast would Jupiter then orbit? No ... please... only kidding... don't fire up your spreadsheets.

I know where Roger is going with this. He remains upset by the demotion of Pluto, and is planning to blast Jupiter out of its orbit in spite! He wanted to know the KE so he'd know how much rocket fuel to buy.

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#46
In reply to #44

### Re: Solar System Kinetic Energy Problem

11/28/2006 1:27 PM

You Wrote:"He remains upset by the demotion of Pluto"

How can I be upset about something that hasn't happened. I mean, think about it, some organization you've never heard off tells you Pluto is no longer a planet, then gives you a reason that doesn't make any sense? Why would you believe them? I don't believe them. I guarantee they'll change their mind the second they realized nobody's listening to their nonsense.

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#49
In reply to #46

### Re: Solar System Kinetic Energy Problem

11/28/2006 1:46 PM

I agree. Pluto is clearly a planet, as far as I can tell.

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#50
In reply to #49

### Re: Solar System Kinetic Energy Problem

11/28/2006 1:53 PM

Excellent. Hopefully you've looked through a telesope once, and I owned one when I was 7, so we are both clearly amateur astonomers. We'll issue a press release from our "CR4 Amateur Astronomer Union" saying after long deliberations and finally a vote we have reestablished Pluto as a planet. Furthermore, we have also voted Uranus "Planet we'd least like to visit" and Mercury "Fastest Planet". Earth of course wins "best planet" again, sorry Mars.

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#53
In reply to #50

### Re: Solar System Kinetic Energy Problem

11/28/2006 3:25 PM

Perfect. Let's have our people draft the release. Actually, as the only person here (I wager) who really is from Mars*, I'm a little sad to see that it missed out (again!) as "Best planet". I was happy to see that Uranus took both the coveted "Planet we'd least like to visit" award and the "Most unfortunate name" award.

* A small town in Pennsylvania, where I lived for 10 years, or so.

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#54
In reply to #53

### Re: Solar System Kinetic Energy Problem

11/28/2006 3:33 PM

I'll tell you what, we'll run our press release and we'll give Mars serious consideration next year.

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#57
In reply to #54

### Re: Solar System Kinetic Energy Problem

11/28/2006 11:49 PM

If it looks, sounds, smells and tastes like a planet then chances are it's a planet.

I like my definition,

If you can stand on it then it's a planet.

because you can PLANT your foot on IT (PlanIt)!!!!

This would let Pluto and Ceres plus all the moons and a few of the bigger asteroids in but kick out the gas giants Jupiter, Saturn, Uranus and Neptune because they have no surface and you can't stand on them. What dose that make the count of planits now?

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#69
In reply to #57

### Re: Solar System Kinetic Energy Problem

11/29/2006 12:58 PM

"What does that make the count of planits now?"

Well for me, 9, but only because I'm stubborn. Otherwise, somewhere in the hundreds (at least).

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#45

### Re: Solar System Kinetic Energy Problem

11/28/2006 1:06 PM

OK folks I have completely re-done the spreadsheet and here are the answers to Rogers initial question

This is somewhat different to my original post but I believe this is now correct. The original errors came form using mixed units in the complete spreadsheet so this time I have converted all the raw data to engineering format. I am not sure what can be drawn fro this except that the total KE of the Sun's KE is about six times as great as the KE of all the planets combined and Rogers initial gut feeling was correct. The ratio also shows that the rotation KE for the rockey planets is small compared to it orbital KE but with the gas giants the rotational KE is relatively much greater. I have absolutely no idea what this could mean it's just an observation

No for the more useful table containing KE and PE.

You must admit that its more understandable when I use engineering notation rather than scientific notation. I will leave it up to others to see what we can deduce from these tables. I should have done what I preach and converted everything to standard units in the first place, it would have made life a whole lot simpler.

By the way I now have an Excel spreadsheet that has the following information for each of the planets:

Distance to Sun, Orbital Path Length, Eccentricity, Perihelion, Aphelion, Orbit Period, Average Velocity, Maximum Velocity, Minimum Velocity, Orbit Inclination, Equatorial Diameter, Polar Diameter, Average Diameter, Volume, Mass, Density, Gravity, Axial Tilt, Rotational Period, Rotational Velocity, Escape Velocity and Min, Max and Mean Surface Temperatures.

in both scientific and engineering notation. If anybody would like a copy please ask and I will email a copy to you.

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#47
In reply to #45

### Re: Solar System Kinetic Energy Problem

11/28/2006 1:32 PM

Masu,

I'd like a copy. Also, well done. I think it's cool that the kinetic energy worked out to be half the potential energy, just as expected.

So the Sun has the most kinetic energy.

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#55
In reply to #45

### Re: Solar System Kinetic Energy Problem

11/28/2006 5:51 PM

Something else occurred to me looking at these numbers. The distribution of mass in our solar system follow a exponential fairly well (see graph). The highest is Jupiter on the right, the smallest is Pluto on the left. I'm going to look at the moons of Jupiter to see if a similiar mass distribution exists.

My instinct tells me that if we included the Sun and all the tiny objects in the solar system (KBO, comets, asteroids, etc.) the distribution would actually be a power law distribution. Power Law distributions are interesting because of a number of features. If so, one might be able to anticipate the number of certain size planets in a system based only on the Stars mass and it's largest planets mass.

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#56
In reply to #55

### Re: Solar System Kinetic Energy Problem

11/28/2006 11:05 PM

Hi Roger, you wrote: "Something else occurred to me looking at these numbers."

During my "slow-living" phase last night, something occurred to me as well! Suddenly, my 'relativistic' side got the better of the engineer in me and I took a new look at the figures...

The proper 'accounting' of energies in any system includes rest-mass and also, we 'erroneously' added potential energy as positive, while it should be added negatively, i.e.,

E ≈ mc2 + 0.5 mv2 - GMm/s (Newtonian approximation).

In all planets, the rest energy dwarfs the potential/kinetic components by factors of 108 or more! For Earth it is: mc2 = 5.22E+41 J, KE = 2.57E+33 J, PE = -5.15+E33 J

So, you are right in that we should look at the masses - they are the most important parts!

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#67
In reply to #56

### Re: Solar System Kinetic Energy Problem

11/29/2006 10:57 AM

You wrote "The proper 'accounting' of energies in any system includes rest-mass and also, we 'erroneously' added potential energy as positive, while it should be added negatively, i.e."

Good Point, the gravitational force is attactive so the potential is negative. The kinetic energy is given by the equation;

So you can see your absolutely right about the negative sign.

As for rest mass, that is definitely the largest energy by far. Still, I found the distribution of mass surprising. For some reason, I was under the impression Saturn and Jupiter were closer in size than they are. Furthermore, I was surprised that Uranus and Neptune were smaller than both by a lot. It seems a misnomer to call them all "gas giants" when one is almost 20 times bigger than the other (Jupiter to Uranus).

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#58
In reply to #55

### Re: Solar System Kinetic Energy Problem

11/29/2006 1:47 AM

Hi Roger, on your graph, I can see that you used Earth's mass as unity on the vertical scale, but what does your horizontal scale mean?

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#64
In reply to #58

### Re: Solar System Kinetic Energy Problem

11/29/2006 9:21 AM

I graphed it as a distribution. So there is only one set of numbers, the masses of the planets. I used a log scale axis for which an exponential function would appear as a straight line (lnex=x) to demonstrate how well the exponential fit is (I think its easier to see with a straight line) for the mass distribution.

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#73

### Re: Solar System Kinetic Energy Problem

11/29/2006 9:47 PM

I think the idea that the error is caused by the elliptical nature of the orbits and the fact that I used the average orbital velocity and average orbital radius is what is causing the problem. When I was gathering the data for the original table I had this funny feeling that I should collate as much info as possible. Since my table has values for aphelion and perihelion I will generate another table with KE and PE for both and see if it applies then. Pleas hang in there here as at CR4 we always get to the bottom of the problem.

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#74
In reply to #73

### Re: Solar System Kinetic Energy Problem

11/29/2006 10:36 PM

Hi masu, I have rechecked my velocity calcs and I do find very minor differences (probably round-off errors) between mine an Mitton 1991's tables. Here is the table I posted before, but with Mitton's values for mean velocity:

You can compare your source's data with these if you wish.

The only possible significant difference between the calculated speed and Mitton's table is for Pluto, with its very elliptical orbit.

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#75
In reply to #73

### Re: Solar System Kinetic Energy Problem

11/29/2006 11:12 PM

But... but... but what error? Your figures in the second chart on your post #45 are just fine: 2:1 ratio in every case. I think you just need to come up north so you're not hanging by your feet all the time. Then you will feel better, and realize that your numbers are OK, its just the graphing that went haywire.

If you cruise around Wikipedia, touching on specific orbital energy (vis-viva) etc. I think you'll find little to support eccentricity as a source of a 10% discrepancy in the 2:1 ratio -- which I believe should remain precisely 2:1 regardless of e.

Relax... your numbers are fine -- something just got goofed up in the graphing or curve fitting.

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#77
In reply to #75

### Re: Solar System Kinetic Energy Problem

11/30/2006 1:17 AM

Hi Ken and Masu, pardon my jumping in here.

Ken, there are small differences in Masu's table 2 of #45 for kinetic and potential energies from the 2:1 ratio, e.g., Mercury and Pluto are quite off the ideal and they are the two very eccentric orbits.

This probably skews the curve fitting algorithm.

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#84
In reply to #77

### Re: Solar System Kinetic Energy Problem

11/30/2006 12:16 PM

Hi Jorie:

It's true, there are small differences. (And, in fact, I have to admit having done only a (not very random) sample of the total, doing the ratios in my head, and jumped to the incorrect conclusion that they all were effectively 2.0:1, allowing for rounding errors and the possibility of different sources for one piece of data versus another.

But even in the most extreme case, the ratio goes to 2.06. We can reasonably say that the .9986 exponent on Masu's curve means that the curve is a straight line. But if all but 2 values on that line essentially reflect a 2.0:1 ratio, then how can the slope be 2.2413? If we drew a line of best fit by hand, with a really sharp pencil, that line would be indistinguishable from one with a slope of roughly 2.01, wouldn't it? (I thought perhaps the two anomalous values were close together in their energy values, tilting the line unrealistically. But that is not the case, with Mercury being near the middle of the plot.)

I think Masu's original calculations used mean distances and velocities, so all the values should be in the ratio 2:1 precisely, I think. But even with the anomalies, the 2.413 figure seems out of whack even for a cranky curve fitting algorithm.

I wonder if the curve fitting algo is just plain wrong? Maybe if I can find the time, I'll plug the values into my son's fancy-schmancy calculator, and see what it comes up with.

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#86
In reply to #84

### Re: Solar System Kinetic Energy Problem

11/30/2006 1:35 PM

Hi Ken, you wrote: "I think Masu's original calculations used mean distances and velocities, so all the values should be in the ratio 2:1 precisely, I think. But even with the anomalies, the 2.413 figure seems out of whack even for a cranky curve fitting algorithm."

I must agree, something looks wacky there! But Masu has a tenacity that will bring him (and us) to the crux of the issue, I'm sure!

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#87
In reply to #86

### Re: Solar System Kinetic Energy Problem

11/30/2006 2:13 PM

Yes, I second that opinion: Masu is remarkably industrious!

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#78

### Re: Solar System Kinetic Energy Problem

11/30/2006 2:22 AM

OK fellow CR-4 astro-physicists I have the data for potential and kinetic energies at perihelion aphelion and I don't think you're going to like it. Here is a table of the energy calculated from the figures I obtained from Wikipedia.

And this is a plot of the data.

The Blue points are the aphelion data set and the yellow points are for perihelion. The green line represents the equation PE = 2 x KE while the two dashed red lines represent PE error of Â± 25% so you can get some idea of the errors involved. The dots that represent each of the planets are also sized to represent an error of roughly Â± 25%.

So the question is do we accept an error of Â± 25% for out calculations of energy or is there something else going wrong. I have also included the eccentricity of the planets orbit in the table and there are some interesting things showing up. For the Perihelion the ratio of energies seems to be less than the expected 2 to 1 by a factor that appears to in some way be related to the eccentricity of the orbit. For the aphelion the ratio it is greater than the expected 2 to and again the value also appears to be related to the eccentricity. Venus is however the one glaring exception to this with both less than the expected value of 2. If it weren't for Venus I would say it was the orbits being elliptical was the cause of the errors but currently I don't know. Has anybody got any ideas?

To keep it interesting the photograph that is the background is from NASA. It is a time exposure of the pre dawn launch of the space shuttle Atlantis and shows its path from lift off to orbit.

PS Something I just noticed after I posted this is that the total energy for each of the planets at perihelion and aphelion is not the same. This there is something wrong with the calculations somewhere, the question is what and where?

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#79
In reply to #78

### Re: Solar System Kinetic Energy Problem

11/30/2006 5:22 AM

Hi Masu, your "PS Something I just noticed after I posted this is that the total energy for each of the planets at perihelion and aphelion is not the same. This there is something wrong with the calculations somewhere, the question is what and where?" is to be expected. It is your 'punishment' for taking little notice of what Roger and myself have been saying above (my #56 and Roger's #67).

Potential energy is negative! Add them that way and it stays constant.

Anyway, I'm glad you made this mistake, because it hammers home the common error of taking PE as positive!

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#80
In reply to #78

### Re: Solar System Kinetic Energy Problem

11/30/2006 8:44 AM

Masu, you wrote: "So the question is do we accept an error of Â± 25% for out calculations of energy or is there something else going wrong?"

I think there is no "error"! What's going on is that PE = -2 KE only holds for circular orbits or when calculated for the mean distances and velocities of non-circulars.

When you add the (negative) potential energy and the kinetic energy due to the transverse (or tangential) orbital velocity component, you get a parameter called "effective potential" in the literature. This parameter is only constant for circular orbits and changes along any elliptical orbit, so it is expected that the PE/KE ratio will also change along the orbit.

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#81
In reply to #78

### Re: Solar System Kinetic Energy Problem

11/30/2006 10:33 AM

Masu,

As far as I can see, your tables look right. It's a good thing that your seeing the difference in KE to PE ratio between Perihelion and Aphelion. If you didn't, we'd have serious problems. Here's some facts on Elliptical orbits that might help you make sense of what you calculated;

1. Perihelion is the closest point in an elliptical orbit to the sun, aphelion is the furtherst point in an elliptical orbit to the sun. (This one I think you knew)

2. The speed of an object in orbit varies as it's distance from the sun changes. At aphelion a planet reaches its slowest orbital velocity. At perihelion a planet reaches its fastest orbital velocity.

3. A circular orbit has aphelion = perihelion (the distances are the same) and the orbital velocity is constant for the entire orbit. The more eccentric the orbit, the larger the difference in the orbital velocity between aphelion and perihelion will be.

So think about it, the kinetic energy is directly related to orbital velocity (the mass of the planet is constant), So the kinetic energy of any planet in an elliptical orbit (all of them) will be smaller at aphelion as compared to perihelion. Now potential energy increases (becomes less negative) the further you get from the sun, so the potential energy is lower (in magnitude) at aphelion as comapared to perihelion. The point though is that the energies don't change the same amount, The kinetic energy drops off slightly faster at aphelion as compared to perihelion than potential energy does, so the ratio doesn't hold for non-circular orbits.

So when I saw your data I was reassured by the pattern that the ration was slightly off at perihelion and apheion. Also, notice the difference is largest for Pluto, the plane with the largest eccentricity to its orbit. You did a lot of good work here Masu and I appreciate it because I really feel I've learned a bit. Thanks. (That goes for everyone contributing to this conversation as well)

Roger

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#82
In reply to #81

### Re: Solar System Kinetic Energy Problem

11/30/2006 11:28 AM

Roger can you send me an email at masu@internode.on.net so I can send you a copy of the spreadsheet you asked for.

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#83
In reply to #82

### Re: Solar System Kinetic Energy Problem

11/30/2006 11:38 AM

Masu,

I've sent an email to you again. I originally sent an email to your other address but I guess you didn't get it. I check my private CR4 messages, (I know you can't send attachments), so if you want to provide me with your email address in the future, send it by cr4 mail (like you did earlier). That way your address isn't exposed on the web for spammers to pick up and use.

Roger

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#88

### Re: Solar System Kinetic Energy Problem

12/01/2006 5:19 AM

Serendipitously this article rocked up in my in-tray today. If you havn't already read it, it discusses the problems of trying to establish stable lunar orbits and how because of Earth the only way is to have highly elliptical orbits.

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#89

### Re: Solar System Kinetic Energy Problem

12/01/2006 11:03 AM

Ok fellow CR-4 astro-physicists, many thanks for the confidence in my skills. I have, I hope, the definitive answer and this time the news is good. So we don't need to rebuild the universe after all. Here is the table.

As you can see the potential energy is subtracted from the kinetic energy to get the total energy and an error in the original data for Venus that was used to do the calculations has been corrected.

The new table at the bottom shows the median energy for each of the planetary orbits calculated from the aphelion and perihelion energy rather from the mean orbital elements. It also has a column labeled (Total Energy Deviation) that shows the mean deviation of the total energy between the three values expressed as a percentage. I am happy to say that things are finally starting to look like they should with only minor variations in the total energy. If I understand things correctly the negative value for the total energy is due to the fact that for these tables we havn't included the mc2 component of the total energy.

Finally the columns labeled Energy Ratio show the ratio of the mean potential to the mean kinetic energy. The error is the amount the calculated value differs from the expected value of 2 expressed as a percentage.

Lets go back now and look at aphelion and perihelion tables and see how much the ratio of potential to kinetic energy varies from the predicted value of 2. You can see that there appears to be a correlation between this value and the eccentricity of the orbit.

The table above shows amount the PE/KE ratio varies from 2 verses the eccentricity of the orbit. The yellow dots and line represent the aphelion values while the blue line and dots represent the perihelion values. I don't know about everybody else but it certainly looks to me like the change in the ratio is indeed related to the eccentricities of the orbits.

For the moment if we leave Pluto and Mercury out, the values we have do indeed fit the mathematics extremely closely. As for Pluto and Mercury, they are the smallest planets and are at the extremes of the solar system so one would expect any errors would be exaggerated so I believe we can say that the we have safely explained the errors and they are indeed due to the eccentricity of the orbits.

So folks what do you think? Can we bury this now or should we dig deeper?

PS The photograph in the background of the graph is courtesy of NASA and shows a phenomenon I myself have been lucky to have both see and on one lucky time predicted. Under certain still conditions, if you look at the sun as it sets over water, just before it dips below the horizon you can see part of it turn a brilliant emerald green. The photograph is good but unfortunately doesn't do it justice. It's extremely rare and I have seen only a handful of times but it can be seen. It apparently occurs within a few degrees of 40Âº north and south of the equator usually during autumn (fall in America) looking over water with absolutely no wind. So if you ever find yourself in those circumstances grab your camera and see if you can catch it.

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#90
In reply to #89

### Re: Solar System Kinetic Energy Problem

12/01/2006 12:21 PM

Hi Masu, yep, I think you have now stitched it together pretty well! Your excellent analysis clearly shows that the 2:1 "law" only holds precisely for circular orbits, which are rare things in nature.

If you will pardon me for nitpicking, I still do not like the notion of "subtracting PE from KE" - I prefer to add energies together, with PE having a negative value, as I elaborated upon in post #85.

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#91
In reply to #89

### Re: Solar System Kinetic Energy Problem

12/01/2006 4:22 PM

Hmmm… I'd like to agree, because you have put so much work (or fun) into this.

But I can't quite agree that eccentricity causes a change in the 2:1 ratio. That ratio can only make sense when considered for the orbit as a whole: it obviously cannot hold at the extremes of an eccentric orbit because of the large differences in speed. Once you consider the orbit as a whole then the maths treat it as a circle, whether it is or not. (Did you notice how I tried to speak Australian there, with "maths".)

First, consider this:

If the 2:1 ratio changed with eccentricity, then the deviation from that value for Pluto should be noticeably larger than it is for Mercury. But your total energy chart shows the opposite: 1.89 for Pluto, 1.92 for Mercury.

Then consider this:

At first, I thought that perhaps averaging the energies (peri vs aph) rather than working with the semimajor axes (as representative of mean distance) might make a difference, but it does not appear to. I calculated values for mercury, based on a semimajor axis of .387 AU and a mass of .33 x 10^24 and got values of 7.57 x 10^32 and 3.78 x 10^32. Similarly, got a values of 2.91 x 10^29 and 1.45 x 10^29 for Pluto.

If unconvinced, consider this:

We can all agree that the total energy is the same at every point along an orbit, circular or eccentric. We have to also agree that calculated energies for eccentric orbits (based on semimajor axis) have to be the in the same ratio, because there is no term in the calculation that varies with eccentricity: whether the obit is circular or not, our formulas treat it as circular. I played with velocities from the chart to verify that v^2 really does equal GM/r. It does. So that means that the ratio must be exactly 2:1 for each planet, per Jorries post #48. I think your chart shows that, except for Mercury and Pluto.

My guess is that for Mercury and Pluto you have some incorrect values. The chart I used is this one, (along with it's links to additional data on Mercury) which matches Google's values well.

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#92
In reply to #91

### Re: Solar System Kinetic Energy Problem

12/02/2006 12:24 AM

Roger you stated

"I played with velocities from the chart to verify that v2 really does equal GM/r. It does."

I disagree. The following table shows the values of v2 and GM/r for each of the planets at aphelion and perihelion and the difference between the two normalized to GM/r.

and the values are not exactly equal. If you now take the variation from this table and plot it against the eccentricity for that particular orbit you get a chart that looks like this.

The blue are the perihelion values and the Yellow the aphelion. As you can see the relationship between the value that v2 differs from GM/r to the eccentricity of the orbit is about as linear as you could possibly expect even for Pluto and Mercury.

I could very well be missing something but the variations in the value of V2 to GM/r seem to be related directly to the eccentricity of the orbit. What do you think?

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#93
In reply to #92

### Re: Solar System Kinetic Energy Problem

12/02/2006 5:57 AM

Masu, your "I disagree. The following table shows the values of v2 and GM/r for each of the planets at aphelion and perihelion and the difference between the two normalized to GM/r." seems to be a misunderstanding between Roger and yourself.

That v2 = GM/r holds strictly only for circular orbits, or when using mean distance and mean velocity. I figured out that the mean parameters actually mean: the equivalent circular orbit with the same total energy or angular momentum as the elliptical orbit.

The mean distance is actually equal to the semi-major axis (a), as Ken has stated in post #91. Therefore, the mean orbital velocity is v2 = GM/a.

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#94
In reply to #92

### Re: Solar System Kinetic Energy Problem

12/02/2006 2:43 PM

Masu,

Oh boy, now you've done it! Roger will be justifiably enraged by your having attributed my words to him. Roger actually knows his math(s), whereas I start to get confused when it gets more complicated than anything I can jot down on a cocktail napkin with a Magic Marker. (Which, I'd guess, but don't know, is a brand that carries over to Australia, like Xerox)

(I mention the following only as proof of my eminent distractibility: I think "maths" makes more sense than "math" as shorthand for "mathematics". In a sentence, you could say: "The mathematics show that …" or "The maths show that …." Here we'd say "The mathematics show that …" but "The math shows that …" I assume in Australia you say "Ford are bringing out a new car." Whereas here we say "Ford is bringing out a new car." You say "Chicago is in Illinois". We say (at least something like 70% of our high school graduates) "Chicago… ummm, give me a hint -- is that a country or a state?")

(More evidence of distractibility: People (especially tourists) gather quite often at the southernmost point of the US, in the town of Key West, to see the "Green Flash." I've been there several times, and think I saw it once, but it was quite subtle. Here, I think Key West is "the" place to see it.)

Actually, re velocities, "the chart" I used was not yours but the one I referenced later in the post, which quotes mean velocities. I think Jorrie's post #93 clears up the math pretty well. Also, your graph does, in the sense that if you draw the average between blue and yellow, you'd find a straight line (which logically, chromatographically, and artistically, should be green) at zero variation.

I feel confident that we have the situation under control, and can now make plans for selecting a planet to remove from orbit: I don't want to simply blast it to oblivion – I think we should strive for something more educational. For instance, some amount of thrust, with its directional vector continuously changed, could render the chosen planet "stationary" relative to the averaged plane of orbits at some instant. If that chosen planet is then released from our tampering, it would fall straight toward the sun, were it not for the effects of the other planets. If we did the math, I think we'd find that the gravitational effects of the other planets (depending, of course, on their positions) would be so insignificant as to hardly alter a straight line path into the sun. Perhaps we'd also find that the effect on the sun's location would also be insignificant – although a big planet could become one heck of a bullet.

How long would it take from the stop point to oblivion? Suppose the international community convinced us of the folly of our experiment after having started it, and we wanted to fire up the thrusters at a point half way to oblivion for the planet, and to put it into a different, highly elliptical orbit. The educational benefit of the highly elliptical orbit would be that we could then occasionally get a very close look at a planet that otherwise would always have been very far away. We could repeat this process to study each planet in turn. Let's bring the planets to us, rather than traveling to them!!!

Perhaps, one of the people here less lazy than me can make a coherent question out of this. And maybe those less lazy than I can do the math… hopefully leaving me to arrive late to the party, only to make my usual nit-picky comments. Of course, we'd want to start a new thread.

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#95
In reply to #94

### Re: Solar System Kinetic Energy Problem

12/03/2006 2:05 AM

My apologies Ken and Roger for mixing your names up. I thought I had corrected that oversight but I was obviously wrong.

Interesting concept of crashing a planet into the sun though I would suggest we use either Mercury or Venus so it didn't take earth out on the way through. Only problem is that if you try and slow a planet down it actually speeds up as the orbit decays.

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#96
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### Re: Solar System Kinetic Energy Problem

12/03/2006 12:13 PM

Mercury or Venus??!! Where's your spirit of adventure? Having the planet fly past Earth is part of the thrill! Re the orbit decay, that's the reason for continuously adjusting the thrust vector. At the beginning, the thrust would be in the direction that slows the orbital speed. As the orbit then attempts to decay, the thrust would angle progressively further "outward", away from the sun, to cancel the tendency for the planet to be pulled into the sun prematurely. Once the orbital speed was 0, the thruster would be pointed straight outward, holding the planet away from the sun, until we released it.

Our thruster would best be aimed out from the planet's center all the time, which would require a huge gimbal system encircling the planet. That entire system could be mounted on a planet sized ball bearing, to permit the planet to continue spinning around its own axis. Alternatively, we could use a pair of thrusters mounted on opposite sides, tangent to the planet's surface, to first stop the rotation -- but then that seems like less of an engineering challenge. (Handling the gaseous planets would be a real trick.)

I'm going to the local hardware store to start buying some of the parts. Seems a small black hole might be useful.

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#97
In reply to #96

### Re: Solar System Kinetic Energy Problem

12/04/2006 1:57 AM

I have a better idea. It would be much easier to play about with Mars and stick it at the LaGrange point between the Sun and Jupiter. Then we could play around with it and do all sorts of nasty things without worrying about keeping it in the same place.

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#98
In reply to #97

### Re: Solar System Kinetic Energy Problem

12/04/2006 8:46 AM

Oooo! I like that idea! BTW, I was unable to find a small black hole at my local hardware store, but I'll keep looking around. It seems that we might want to keep one or two in our back pockets, so if anything goes wrong with our experiment, we can create a new universe, so we have somewhere to live.

According to Brian Greene, we need only compress about 5 KG into a volume of about 10-25 cc to create such a micro black hole from which a universe could spring. My son and I have been working on this, starting with a pumpkin left over from Halloween. We whacked it many times with a large sledge hammer, leaving our kitchen covered everywhere with a slippery orange slime. Unfortunately, before we were able to see if we had succeeded in creating a very small black hole somewhere in that mess, my wife cleaned it all up, possibly flushing evidence of the black hole creation down the drain. A sad moment. So, my inclination is to simply buy one.

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