Robittybob1

Even though they are called fictitious it doesn't mean they aren't real. It is an alternative meaning of fictitious. I've learned a lot about barycenters and how to calculate them. Tonight I have been listening to lectures online about gravitational and orbital energy. I still wonder about how the nebula collapsed as rapid as it did. If 99% of the mass ends up in the Sun how does the 1% remainder get it to orbit a barycenter, I'm not sure if it can. It is time for a topic change really. Thanks for your assistance in understanding inertial frames. The next big one was understanding the switching mechanism for the solar magnetic field. Coriolis force effect could have a part to play there. I just don't know. G-nite.

What example could you give me to illustrate your point please?

We had looked at those two masses (above) doing a perfect binary orbit around each other (and their barycenter) and we calculated their velocities and they were inversely related to their masses. OK so who's frame of reference am I looking at that situation from? Could you say a "distant observer" (perpendicular to the plane of the orbit and the barycenter)? So I'm not splitting the motions up but if they were measured quantities and found to be different by this "distant observer" would the calculations still be valid? OK on paper it may seem as if I'm splitting them up arbitrarily but each division represents a situation that could be measured (so I thought). I think I get it now, the escape velocity away from a mass is always from that masses FoR. You can't say the other mass will need to have a different Ek simply because the primary mass has some basic motion. You would always treat it as "stationary".

They have been slowed down, and that is acceleration.

So potential energy is negative and kinetic is positive and orbital energy is the addition of the two, so it is equal in magnitude to the kinetic energy but negative. So in a binary situation eg the Earth-Moon do you look at the orbital energies separately or just from a geocentric point of view? For I was getting lower total energy out of situations where I split the motion between the bodies. So if there was two cars going 16.666 ... m/sec (60 km/hr) relative to each other, if you measure the Ek it is different if you were to split that motion up to 8.333 ...m/sec each. So their gravitational interaction on each other will be different I'd say. OK if they were in flat space they may not feel their freefall towards each other but stars that were previously aligned up ahead have now veered off to the side and the change of direction will depend on their motion. I haven't come to any conclusion on this yet, but in the back of my mind it seems interesting. Maybe I'm not that sure about that. For I often think of that situation when I'm diving along in my car and I think whether it will take the same amount of energy to throw a ball forward as backwards (the same speed relative to me as a driver, so I can't do the experiment as I'm driving). Have you got proof of this? For if the ball went out the side window has it got the same energy as a ball traveling as fast on the outside? I know it seems simple but I have yet to be convinced.

Yes I had started going through that page and got sidetracked, into lectures on the analysis of elastic collisions (YT) but I now see lower down the words "Elastic collision of masses in a system with a moving frame of reference. The velocity of the center of mass does not change by the collision:..." I always think of how the Solar System formed. As the nebula collapsed material had to be brought in from such vast distances it virtually had to free-fall for 150,000 years and after that time of falling it had gained a lot of speed in the central direction but very little motion in the tangential direction for if there had been too much angular momentum the particles in nebula would still be just a gas cloud. So it presumably there were less of elastic type collisions and more of the inelastic collisions so a portion of the kinetic energy was repeated lost in the form of heat. Fortunately heat is able to radiate out through the infalling dust. Once you have the dust colliding at even 99% elastic collisions with only 1% of the impact energy being lost as heat, and the resultant masses still gravitationally bound the same "particles" are capable of repeated serial collisions. See the animations below the words "As can be expected, the solution is invariant under adding a constant to all velocities, which is like using a frame of reference with constant translational velocity." Imagine if you were to now add the the extra factor of gravitational attraction between the masses, so they end up with this sort of elastic band between them allowing them to oscillate in the moving frame of reference they now are incapable of bouncing away from each other.

Interesting results for to conserve momentum and energy how many different speeds can they bounce off each other? Only 1 result conserves both energy and momentum.

So what is the escape velocity for the 1 kg mass from a 2 kg mass with the density of 3 kg/1000 cm^3? From that we should be able to calculate the radii of the masses. What has become evident is that in orbiting situations the orbital velocity is inversely related to the mass of the bodies. In the above example where the situation is perfect circle perfect binary orbits, the mass ratio is 1:2 and the speed ratio is 2:1 in favour of the lighter mass, so I am planning to test out the escape velocity formula varying the speeds of each mass but keeping the relative velocity the same at the start. So when these two masses bump into each other and they conserve momentum they will then separate varying speeds inversely proportionally to their mass.

I agree but do you agree that it would take a long time for the Solar System to settle down? Imagine if all the planets were finally amalgamated into this one big planet called "Super Jupiter" it wouldn't instantly find its ideal position would it? That depends how you work it. If I know roughly what it is in words I can create a formula in Excel, but if it is just symbols as on that page I can't comprehend it so easy. An attempt at an excel equation could help, but in Excel if I copied a formulated cell it would be meaningless to anyone else. e.g Velocity = 2 * pi * r / T becomes =2*PI()*C3/A18 because in "C3" I have my value for radius r and in "A18" I have the value for period T. so I prefer it just in words Velocity = 2 * pi * radius / Period

On that two body problem page http://en.wikipedia.org/wiki/Two-body_problem#Laws_of_Conservation_of_Energy_for_each_of_two_bodies_for_arbitrary_potentials Can you say those expressions (equations of conservation of energy) in words so I can begin to understand how I could put them into my Excel calculator sheet please. For I want to see if the energy can be transferred from one body to another through motion and gravity as I expressed in http://www.scienceforums.net/topic/87803-does-jupiter-orbit-the-jupiter-sun-barycenter-or-not/page-3#entry853878. Was I close to being right?

I am frustrated that I still don't know how to calculate orbital energy properly. Eg = -G*m*M/r (gravitational potential energy) NOT in orbit, it will just fall right back down. G*m*M/r orbital energy and Ek = 1/2 G*m*M/r (kinetic Energy) Total energy = Eg + Ek Binding energy = - Energy total Escape velocity, the Kinetic energy required is equal to the binding energy = 1/2mv^2 = G*m*M/r = Sqrt(2GM/r) Orbital energy = Eg + Ek = -1/2G*m*M/r (r = orbital radius not the height.)

If you can afford one of these, you buy the petrol station as well.

It was just my bad maths.

I was asking what xD meant. I hadn't come across it before. I'm backward in time.

I'm not that dumb!

I'm just a dummy ... what does xD mean?

Did you have infinity if you can add 100 to it? You can add a hundred more 9s as in 999999999 ...... to a recurring series.

73 - like Stephen Hawking

Ultimate McLaren revealed ahead of Geneva debut http://www.msn.com/en-nz/news/other/ultimate-mclaren-revealed-ahead-of-geneva-debut/ar-BBhIhOE?ocid=UP97DHP It has speed written all over it. Great design!

It is not as if the two masses have to approach each other at the right speed to result in a perfect circular orbit, but it is obvious it wouldn't take too much extra velocity to get to escape velocity. But my thoughts are that as a mass passes by at a higher than required speed ("speed" is hardly the right word since they are only moving in the order of 30 microns/sec!) but less than the escape velocity the mass being overtaken will be moved along with that mass for the gravitational force will keep on acting over enormous distances. and it will curve the path of the passing mass. So when the overtaking mass finally runs out of kinetic energy it will fall back down to the now moving primary mass but the primary mass will keep gaining kinetic energy and the relative speed as they pass for the second time will be lower. Now it is possible this dampening down could take billions of years to circularise the orbits, the lighter mass will appear to be orbiting the primary but they have not yet established their true binary orbit. Gradually more and more energy is being transferred to the primary as it gains orbital energy at the expense of the secondary's kinetic energy. In these situations the orbital energy formulas will not work accurately for they are not true orbiting bodies until the kinetic energy has been "equalised". Could that account for the difficulty of making Jupiter's orbital parameters fit the orbital formulas? That the Sun is not in a true barycentric orbit with Jupiter as yet.

Orbital period is 444.9665645 years Mass 1 kg velocity = 0.00000029830275 m/sec Mass 2 kg velocity = 0.00000014915138 m/sec In opposing directions and orthogonal to the gravity between them. Could that have been worked out if I hadn't used reactive centrifugal forces around the barycenter point? So what would happen if mass 1 (1 kg) approached the mass 2 (2 kg) at 1000 meters when mass 2 has no motion, would mass 1 go into an orbit? Would there be a way for mass 1 getting mass 2 to start orbiting the common barycenter?

I'm quite good at farting so will you go and have a look?

OK

I searched for the origin of the "1.068 solar radii" expression and came across this paper http://www.researchgate.net/publication/200710983_On_the_estimated_precession_of_Mercury%27s_orbitthat really question the precision of these facts. For if we can't tell how the Sun orbits its barycenter with Jupiter, and Mercury is orbiting the Sun how precisely did they calculate Mercury's precession? That paper just looks at the maths but didn't give values of the S-J distances. I agree the barycenter is the center of mass of the two bodies but how did you define it as "the center of their mutual orbit"? If they were binary maybe. Let's look at a contrived situation where a 2 kg mass is gravitationally orbiting a 1 kg mass in free space. What is their barycenter point and how fast do they orbit? Make the orbits perfectly circular. In my way of thinking they will only be able to do this where the reactive centrifugal forces balanced their gravitational force. There are obviously multiple answers to this and it would depend on their relative speed, so can it happen with a distance between them of 1000 m? So the barycenter point is 1/3 and 2/3 position long the distance between them. The heavier object will orbit the barycenter at a distance 1/3*1000 meters and the 1 kg mass will be 2/3*1000 m from the barycenter but they will be 1000 m apart, so what will their G force be? Newtonian G force = G* m1*M2/r^2 = G * 1*2/10^6 = 2G*10^-6 = 6.67384*10^-11*2.0*10^-6 = 13.34768 *10^-17 N So what speed do the masses have to travel to stop them just from being gravitationally brought together?

The ones who did die at the time of dying in their dreams aren't here to tell us about it.