Does Jupiter orbit the Jupiter-Sun barycenter or not?

[Robittybob1]

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.)

Edited February 21, 2015 by Robittybob1

[whiskers]

In the 2 body problem, if you just set the bodies down, they'll be attracted toward another and then collide. If either one has some velocity component perpendicular to the line between them, there will be one or another type of relationship depending on the eccentricity http://en.wikipedia.org/wiki/Orbital_eccentricity

It might be Parabolic, http://en.wikipedia.org/wiki/Parabolic_trajectory

Hyperbolic, http://en.wikipedia.org/wiki/Hyperbolic_trajectory

or an elliptical orbit, one special type of which is a perfect circle

this page is nice http://en.wikipedia.org/wiki/Two-body_problem

Do Jupiter and Sun orbit their common barycenter? Technically no because of the perturbations of other bodies in the SS, but they would if they were alone.

[Robittybob1]

In the 2 body problem, if you just set the bodies down, they'll be attracted toward another and then collide. If either one has some velocity component perpendicular to the line between them, there will be one or another type of relationship depending on the eccentricity http://en.wikipedia.org/wiki/Orbital_eccentricity

It might be Parabolic, http://en.wikipedia.org/wiki/Parabolic_trajectory

Hyperbolic, http://en.wikipedia.org/wiki/Hyperbolic_trajectory

or an elliptical orbit, one special type of which is a perfect circle

this page is nice http://en.wikipedia.org/wiki/Two-body_problem

 

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?

[Strange]

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.

 

Isn't that the wrong way round?

 

I would have expected, "Can you put that wordy explanation into an equation so I can enter it into Excel".

[Robittybob1]

...

Do Jupiter and Sun orbit their common barycenter? Technically no because of the perturbations of other bodies in the SS, but they would if they were alone.

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?

 

Isn't that the wrong way round?

 

I would have expected, "Can you put that wordy explanation into an equation so I can enter it into Excel".

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

Edited February 22, 2015 by Robittybob1

[Robittybob1]

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.

-1/2G*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.)

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.

Edited February 22, 2015 by Robittybob1

[Robittybob1]

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.

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.

Edited February 22, 2015 by Robittybob1

[whiskers]

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

If they are point masses and have elastic collision, the video of what happens after collision should be same as reversing the film on what happens before..

http://en.wikipedia.org/wiki/Elastic_collision

Edited February 22, 2015 by whiskers

[Robittybob1]

If they are point masses and have elastic collision, the video of what happens after collision should be same as reversing the film on what happens before..

http://en.wikipedia.org/wiki/Elastic_collision

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.

Edited February 22, 2015 by Robittybob1

[whiskers]

Do you understand that all motion is relative? And that all physics works the same if a constant velocity is added to all particles?

[swansont]

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.

That the PE. Regardless of whether it's in orbit.

 

 

and Ek = 1/2 G*m*M/r (kinetic Energy)

Kinetic energy is 1/2 mv²

It will be 1/2 the magnitude of the PE in a circular orbit.

[Robittybob1]

That the PE. Regardless of whether it's in orbit.

 

 

 

Kinetic energy is 1/2 mv²

It will be 1/2 the magnitude of the PE in a circular orbit.

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.

Do you understand that all motion is relative? And that all physics works the same if a constant velocity is added to all particles?

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.

Edited February 22, 2015 by Robittybob1

[swansont]

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.

 

You aren't free to choose any frame, i.e. you can't arbitrarily "split the motion between the bodies". Orbits do not represent inertial frames. The construction of the equations chooses the reference point for you.

 

Equations such as the KE being half of the PE are derived under certain assumptions. You can't violate those assumptions when using them and expect to get the right answer.

 

So if there was 1 car going 16.666 ... m/sec (60 km/hr) relative to another 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.

Kinetic energy is not an invariant. It does not stay the same when you go from one frame to another. Two objects moving 8 m/s will (in general) not have the same KE as either of them moving at 16 m/s.

 

[Robittybob1]

 

You aren't free to choose any frame, i.e. you can't arbitrarily "split the motion between the bodies". Orbits do not represent inertial frames. The construction of the equations chooses the reference point for you.

 

Equations such as the KE being half of the PE are derived under certain assumptions. You can't violate those assumptions when using them and expect to get the right answer.

 

 

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".

Edited February 22, 2015 by Robittybob1

[swansont]

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".

 

No, you can't treat an accelerating object as if it was stationary using the same exact equations you would use in an inertial frame.

[Robittybob1]

 

No, you can't treat an accelerating object as if it was stationary using the same exact equations you would use in an inertial frame.

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

[swansont]

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

 

 

The fictitious Coriolis force has to be added if you want to use equations on the rotating earth, because the earth is in a rotating frame, rather than an inertial one. The centrifugal force is another fictitious force.

 

In other cases, the choice of FoR is not arbitrary. mv²/r only works in one frame.

[Robittybob1]

 

 

The fictitious Coriolis force has to be added if you want to use equations on the rotating earth, because the earth is in a rotating frame, rather than an inertial one. The centrifugal force is another fictitious force.

 

In other cases, the choice of FoR is not arbitrary. mv²/r only works in one frame.

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.

[swansont]

Even though they are called fictitious it doesn't mean they aren't real. It is an alternative meaning of fictitious.

 

They are fictitious in the sense that they are not present if you analyze the problem in an inertial frame of reference, i.e. where Newton's first law applies. You have to make them up and add them when in an accelerating frame, in order to make Newton's second law work properly.

[whiskers]

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

If I am on an accelerating rocket, I would measure all other local bodies to be accelerating in the opposite direction, with no force to explain their acceleration. If I am merely on a coasting rocket, I would measure all other local bodies to be coasting in the opposite direction, which needs no force to explain.

[Robittybob1]

If I am on an accelerating rocket, I would measure all other local bodies to be accelerating in the opposite direction, with no force to explain their acceleration. ....

To me that seems like a silly way of viewing the world.

[Greg H.]

To me that seems like a silly way of viewing the world.

While it may seem silly, it is mathemtaically accurate.

 

This is what makes physics deceptively easy at first blush. A lot of it makes good sense and jives with what we experience everyday. And a lot of it seems to make absolutely no sense whatsoever, and is just as correct. This is why I often say that doing physics by common sense is the worst way to go about it. You have to actually get a grasp of what the equations are telling you.

 

For example, in the case given by whiskers, if you had no other reference - only your rocket and a single planet - could you really PROVE which one of them was accelerating moving, if all you had to go one was the view out the window? According to the math, either case is viable - either you and your rocket are moving away from the planet, or the planet is moving away from you and your rocket. The equations make no distinction between the two cases, and will yield valid answers regardless of which case you assume. So while it may seem silly to you that a planet is wandering off away from your rocket, to a Peirson's puppeteer, that's just a Thursday afternoon.

 

And not even a very interesting one.

Edited February 23, 2015 by Greg H.

[imatfaal]

Hmm not sure I agree with the above two posts - acceleration is detectable and is not relative.

 

If I am in a sealed lift-car I can detect if I am in a gravitational field or accelerating (although I cannot distinguish locally between the two - only by tidal forces) - that is to say that the laws of physics in my little sealed box can tell the difference between me accelerating and somethings I can see outside accelerating in the opposite direction. Conversely if I am in a sealed lift-car I cannot tell if I am in constant velocity motion or the space outside that I can see is in constant motion in the opposite direction - ie the laws of physics are identical whether I am or not.

Edited February 23, 2015 by imatfaal
cleaning up

[Greg H.]

Hmm not sure I agree with the above two posts - acceleration is detectable and is not relative.

 

If I am in a sealed lift-car I can detect if I am in a gravitational field or accelerating (although I cannot distinguish locally between the two - only by tidal forces) - that is to say that the laws of physics in my little sealed box can tell the difference between me accelerating and somethings I can see outside accelerating in the opposite direction. Conversely if I am in a sealed lift-car I cannot tell if I am in constant velocity motion or the space outside that I can see is in constant motion in the opposite direction - ie the laws of physics are identical whether I am or not.

In retrospect you may be right - it's constant motion that's undetectable. I'm guessing that's what I meant, and I got caught up in the accelerating bit from the previous post. In any case, I should have known better. Thanks for the correction.

 

Edit

I corrected my post above.

Edited February 23, 2015 by Greg H.

[Robittybob1]

In retrospect you may be right - it's constant motion that's undetectable. I'm guessing that's what I meant, and I got caught up in the accelerating bit from the previous post. In any case, I should have known better. Thanks for the correction.

 

Edit

I corrected my post above.

It has happened to me. I'm sitting in a train waiting to move off, and then I notice the station moving, "silly me its the train moving instead", but for that moment in time your brain is fooled.