How does the Sun get orbital energy to start orbiting a barycenter?

[Robittybob1]

Here's an interesting thing. Imagine that the Earth is a hollow sphere, the mass of which adds up to the true earth mass. If you were outside of it, you would feel that attraction "down" towards the center of the earth. But! If you were *inside* of it, you would feel no gravitational attraction at all!

 

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

OK and I assume the same works for a ring, disc or a torus . In the middle of the ring, disc or torus there is no gravity either. So when the protosun was forming in the middle of the protoplanetary disc it was in a gravity free zone (only contending with its own collapse.) Everything outside of it depending on its own orbital energy to remain in the disc.

At what point is the inner mass going to start orbiting? It had angular momentum but no orbital energy.

[whiskers]

This proto area is very asymmetrical so the degree one applies the shell theorem is limited. My point there was the parts of the earth for instance can be circling the center of the earth without being attracted to it. Can you indicate what definition of 'orbital energy' you are referring to.

[J.C.MacSwell]

I have seen a number of links that claim that the planets orbit the SSB exactly and not the Sun. I think the logic is flawed. If Jupiter was set in stable orbit always further from the Sun, the SSB would move further from the Sun, and at the same time make Jupiter less of an influence on the orbits of the inner planets. If Jupiter was moved far enough, the SSB would have been moved entirely outside the orbit of Mercury. Unless one argued that the orbit of Mercury was hundreds of years rather than 88 days, it would be pretty clear that Mercury was in fact orbiting the Sun.

 

I think they are taking the results of a two body problem and assuming it holds when more bodies are involved.

 

OK and I assume the same works for a ring, disc or a torus . In the middle of the ring, disc or torus there is no gravity either. So when the protosun was forming in the middle of the protoplanetary disc it was in a gravity free zone (only contending with its own collapse.) Everything outside of it depending on its own orbital energy to remain in the disc.

At what point is the inner mass going to start orbiting? It had angular momentum but no orbital energy.

Reasonable guess, but that wouldn't be correct.

Edited February 14, 2015 by J.C.MacSwell

[Robittybob1]

Here's an interesting thing. Imagine that the Earth is a hollow sphere, the mass of which adds up to the true earth mass. If you were outside of it, you would feel that attraction "down" towards the center of the earth. But! If you were *inside* of it, you would feel no gravitational attraction at all!

 

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

 

Nasa and all these other places describe the bodies going around the barycenter, but this is the kind of language which is more like saying that the Sun is setting - it is a casual description which does the job, but it can lead to misinterpretations.

I'm starting to favour the idea that each planet orbits it own barycenter with the Sun but the Sun is moving for it is "orbiting" (not orbiting but moves toward) the SSB.

 

Demonstrated by removing each planet one at a time http://homepages.wmich.edu/~korista/solarsystem_barycenter.pdf

Edited February 14, 2015 by Robittybob1

[whiskers]

Nothing is moved by the SSB. It is a mathematical point, such as "the geographic center of the contiguous United States"

https://en.wikipedia.org/wiki/Geographic_center_of_the_contiguous_United_States

It exerts no influence on anything else.

 

If you are out away from 2 bodies - such as Pluto & its moon Charon, the center of their circling appears as the barycenter, because presumably you are in or close to an inertial frame. You understand that all motion is relative, right?

 

http://pluto.jhuapl.edu/News-Center/News-Article.php?page=20150212

[Robittybob1]

I have seen a number of links that claim that the planets orbit the SSB exactly and not the Sun. I think the logic is flawed. If Jupiter was set in stable orbit always further from the Sun, the SSB would move further from the Sun, and at the same time make Jupiter less of an influence on the orbits of the inner planets. If Jupiter was moved far enough, the SSB would have been moved entirely outside the orbit of Mercury. Unless one argued that the orbit of Mercury was hundreds of years rather than 88 days, it would be pretty clear that Mercury was in fact orbiting the Sun.

 

I think they are taking the results of a two body problem and assuming it holds when more bodies are involved.

 

Reasonable guess, but that wouldn't be correct.

That question about taking Jupiter further and further away means that the JSB moves as that happens but the SSB would be somewhat more stable I presume.

Nothing is moved by the SSB. It is a mathematical point, such as "the geographic center of the contiguous United States"

https://en.wikipedia.org/wiki/Geographic_center_of_the_contiguous_United_States

It exerts no influence on anything else.

 

If you are out away from 2 bodies - such as Pluto & its moon Charon, the center of their circling appears as the barycenter, because presumably you are in or close to an inertial frame. You understand that all motion is relative, right?

 

http://pluto.jhuapl.edu/News-Center/News-Article.php?page=20150212

I suppose it is about the right time to answer the orbital energy question. Can it be shown that both Pluto and Charon have both got the right amount of orbital energy to orbit their center of mass?

http://en.wikipedia.org/wiki/Orbit#mediaviewer/File:Orbit2.gif

http://en.wikipedia.org/wiki/Specific_orbital_energycovers it pretty well.

 

Orbital energy = Negative sum of the standard gravitational parameters of the bodies/2 times the semimajor axis

[whiskers]

That question about taking Jupiter further and further away means that the JSB moves as that happens but the SSB would be somewhat more stable I presume.

I suppose it is about the right time to answer the orbital energy question. Can it be shown that both Pluto and Charon have both got the right amount of orbital energy to orbit their center of mass?

http://en.wikipedia.org/wiki/Orbit#mediaviewer/File:Orbit2.gif

http://en.wikipedia.org/wiki/Specific_orbital_energycovers it pretty well.

 

Orbital energy = Negative sum of the standard gravitational parameters of the bodies/2 times the semimajor axis

 

Once you get into this stuff you have to really have a solid grounding in motion being relative.

 

A lot of physics involves "accounting" quantities which will deceive you if you don't realize that the numbers come out differently in different frames of reference.

 

One spaceship coasts past another one in space. Each pilot feels himself not to be moving and the other one appears to be moving. Each therefore computes a zero kinetic energy and zero momentum for himself and a positive value for each of these quantities for the other guy. Ok?

I have seen a number of links that claim that the planets orbit the SSB exactly and not the Sun. I think the logic is flawed. If Jupiter was set in stable orbit always further from the Sun, the SSB would move further from the Sun, and at the same time make Jupiter less of an influence on the orbits of the inner planets. If Jupiter was moved far enough, the SSB would have been moved entirely outside the orbit of Mercury. Unless one argued that the orbit of Mercury was hundreds of years rather than 88 days, it would be pretty clear that Mercury was in fact orbiting the Sun.

 

I think they are taking the results of a two body problem and assuming it holds when more bodies are involved.

The Sun is very closely at one focus of the ellipse for each planet's orbit - due to gravitational attraction.

If you were coming in from outside the SS, looking at the bodies, you would find the SSB at the center of the orbiting bodies - due to inertial frames.

[Robittybob1]

 

... The Sun is very closely at one focus of the ellipse for each planet's orbit - due to gravitational attraction.

If you were coming in from outside the SS, looking at the bodies, you would find the SSB at the center of the orbiting bodies - due to inertial frames.

Please explain yourself thanks. "find the SSB at the center of the orbiting bodies - due to inertial frames." What do you mean by that?

....

Reasonable guess, but that wouldn't be correct.

If all the matter in the nebula was orbiting it would never have collapsed into a star.

 

Once you get into this stuff you have to really have a solid grounding in motion being relative.

 

A lot of physics involves "accounting" quantities which will deceive you if you don't realize that the numbers come out differently in different frames of reference.

 

Can you start me off then using pluto and Charon? How do we know they are orbiting a common barycenter as the .gif was showing?

http://en.wikipedia.org/wiki/Orbit#mediaviewer/File:Orbit2.gif

[swansont]

OK so what you have made me think about is what bearing has the barycenter got when approached from multiple angles. They often use the example of a hammer for it has a heavy head end and a lighter handle, but when anyone works out it CoM that is the same CoM no matter from which angle you look at the hammer, and I suppose that too includes the situation of an ant walking across the head. I'm struggling to accept this but is it wrong?

 

The SSB is the result of the interaction of many objects. If an ant is far from the hammer, the CoM of the hammer+ant system might be outside the hammer, but that doesn't move the CoM of the hammer itself.

 

Mercury feels a force from the sun directed toward the center of the sun, not the SSB.

 

 

Now when you ask "Why does simply applying gravitational attraction not account for the sun's motion?" That is exactly what I'm proposing but just had my doubts that those forces are enough to make the Sun orbit the SSB if you make the SSB the FoR.

The FoR doesn't matter. The physics still works, regardless of the frame.

 

You keep making statements or asking questions that imply that you think there's something else required.

[Robittybob1]

 

Mercury feels a force from the sun directed toward the center of the sun, not the SSB.

Well then do you agree that the Sun is being wobbled at the same time as Mercury is orbiting the Sun? That to me, and I hope you agree, means that the Solar System mass is wobbling the orbit of Mercury too.

[whiskers]

Can you start me off then using pluto and Charon? How do we know they are orbiting a common barycenter as the .gif was showing?

http://en.wikipedia.org/wiki/Orbit#mediaviewer/File:Orbit2.gif

We "say" they are orbiting a common barycenter. Also, as per the gif, taken by a satellite, we "see" them orbiting a common barycenter. I don't think you understand that motion is relative, and that an orbit is built up (integrated) from instantaneous attractions. The orbit isn't "magic" and the center of the orbit isn't magic.

[Robittybob1]

We "say" they are orbiting a common barycenter. Also, as per the gif, taken by a satellite, we "see" them orbiting a common barycenter. I don't think you understand that motion is relative, and that an orbit is built up (integrated) from instantaneous attractions. The orbit isn't "magic" and the center of the orbit isn't magic.

I have always thought of an orbit a combination of tangential motion combined with falling toward a mass, where there is sufficient tangential velocity that the mass that is falling misses. If it was just instantaneous attractions it might be more like an asteroid passing by or worse still striking.

 

http://en.wikipedia.org/wiki/Orbit#Understanding_orbits

 

 

 

In physics, an orbit is the gravitationally curved path of an object around a point in space, for example the orbit of a planet around the center of a star system, such as the Solar System.[1][2] Orbits of planets are typically elliptical. But unlike the ellipse followed by a pendulum or an object attached to a spring, the central sun is at a focal point of the ellipse and not at its centre.

Current understanding of the mechanics of orbital motion is based on Albert Einstein's general theory of relativity, which accounts for gravity as due to curvature of space-time, with orbits following geodesics. For ease of calculation, relativity is commonly approximated by the force-based theory of universal gravitation based on Kepler's laws of planetary motion.

Understanding orbits[edit]

There are a few common ways of understanding orbits:

As the object moves sideways, it falls toward the central body. However, it moves so quickly that the central body will curve away beneath it.

A force, such as gravity, pulls the object into a curved path as it attempts to fly off in a straight line.

As the object moves sideways (tangentially), it falls toward the central body. However, it has enough tangential velocity to miss the orbited object, and will continue falling indefinitely. This understanding is particularly useful for mathematical analysis, because the object's motion can be described as the sum of the three one-dimensional coordinates oscillating around a gravitational center.

So it sounds about right.

Edited February 14, 2015 by Robittybob1

[Robittybob1]

For two bodies orbiting a common barycenter the centrifugal forces would have to balance the centripetal force caused by the gravitational attraction between the masses. So we would need to know where the barycenter is (defined by the mass of each body and their separation) and their speeds (orbital period and orbital radius)

Edited February 14, 2015 by Robittybob1

[astromark]

~ Great swaths of numbers are required, but obtainable..We know that to understand this a 'Graph' could be drawn to help with the understandings..

The planet Venus is a great deal smaller ( Mass, and gravity well.) than the Gas giant Jupiter., But the actuall 'felt effect' is close to the same. from the Suns point of view.

There is a formular for calculating mass effect. Involving distance and gravity mass.. and that if you add orbital velocities the calculations are part of orbital machanics.. It's a level of mathmatics I have zero experties at.. So can not contribute in that regard.. I hope my understanding helps..

[Robittybob1]

~ Great swaths of numbers are required, but obtainable..We know that to understand this a 'Graph' could be drawn to help with the understandings..

The planet Venus is a great deal smaller ( Mass, and gravity well.) than the Gas giant Jupiter., But the actual 'felt effect' is close to the same. from the Suns point of view.

 

There is a formula for calculating mass effect. Involving distance and gravity mass.. and that if you add orbital velocities the calculations are part of orbital mechanics.. It's a level of mathematics I have zero expertise at.. So can not contribute in that regard.. I hope my understanding helps..

Tell me about it! Someone else can do the maths I just need to understand the physics of it. I will be looking at the the Pluto-Charon combination since a spacecraft is honing onto it and it will really give us a perfect example of planets and moons orbiting a barycenter.

[astromark]

Tell me about it! Someone else can do the maths I just need to understand the physics of it. I will be looking at the the Pluto-Charon combination since a spacecraft is honing onto it and it will really give us a perfect example of planets and moons orbiting a barycenter.

On this I will use Alpha Centarous A and B.. The brighter of the pointers to the Southern Cross.

~ The double. The distance between these two stars is about the same as from Our Sun right on out to Uranus. Orbiting about a common berycenter that would be just a little eskew of halway between them.. A 88 year period is observed.. Of these two stars we observe that a 10% larger and 10% smaller than Sol..Our sun..

That this barycenter still allows for a near to star B, a planet. Orbiting the near to it star. Not the barycenter.. No obvious ecentricity is observed

and ( sorry for my spelling.) used a older lap top and do not see a spell check box.. and so slow..

 

On this subject of barycenteres.. Earths Moon sets up the neatest study of this for us.. as its tidal locked and does have a orbit that is not purfect.. Mostly inside of earth but some distance from the core.. A wiki search will show you more than I can on this old laptop.. Mark.

Edited February 16, 2015 by astromark

[Robittybob1]

On this I will use Alpha Centarous A and B.. The brighter of the pointers to the Southern Cross.

~ The double. The distance between these two stars is about the same as from Our Sun right on out to Uranus. Orbiting about a common berycenter that would be just a little eskew of halway between them.. A 88 year period is observed.. Of these two stars we observe that a 10% larger and 10% smaller than Sol..Our sun..

That this barycenter still allows for a near to star B, a planet. Orbiting the near to it star. Not the barycenter.. No obvious ecentricity is observed

and ( sorry for my spelling.) used a older lap top and do not see a spell check box.. and so slow..

 

On this subject of barycenteres.. Earths Moon sets up the neatest study of this for us.. as its tidal locked and does have a orbit that is not purfect.. Mostly inside of earth but some distance from the core.. A wiki search will show you more than I can on this old laptop.. Mark.

Some interesting facts about that planet:

 

 

Alpha Centauri Bb[edit]

Further information: Alpha Centauri Bb

On 16 October 2012, researchers, mainly from the Observatory of Geneva and from the Centre for Astrophysics of the University of Porto, announced that an Earth-mass (M⊕) planet had been detected in orbit around Alpha Centauri B using the radial velocity technique.[99][100] Over three years of observations had been needed for the difficult analysis.[12] The planet has a minimum mass of 1.13 M⊕.[13] It is not in the habitable zone, orbiting very close to the host star at just 0.04 AU and completing one orbit every 3.236 days.[13] Its surface temperature is estimated to be 1200 °C (about 1500 K),[101][102] far too hot for liquid water and also above the melting temperatures of many silicate magmas. For comparison, the surface temperature of Venus, the hottest planet in the Solar System, is 462 °C (735 K).

No life there Mark!

[astromark]

Yes, no life.. but the barycenter was the subject.. Wibbly wobbly and hot as hell..
You mentioned Pluto.. have a look at the 'SETI inst.,' site.. they have images of the wobble of pluto.. Charion is 1/8 of pluto.mass. and close..

[Robittybob1]

Yes, no life.. but the barycenter was the subject.. Wibbly wobbly and hot as hell..

You mentioned Pluto.. have a look at the 'SETI inst.,' site.. they have images of the wobble of pluto.. Charon is 1/8 of Pluto.mass. and close..

Have you seen the other thread? http://www.scienceforums.net/topic/87761-pluto-charon-combination-how-two-astronomical-bodies-orbit-barycenter/

 

Between now and July when the New Horizons probe gets there! Very very exciting.

 

On July 14, 2015, the Pluto system is due to be visited by spacecraft for the first time. The New Horizons probe will perform a flyby during which it will attempt to take detailed measurements and images of the Plutoid and its moons.

[Robittybob1]

I would like to try the same analysis we did to Pluto-Charon to the Sun-Jupiter combination.

http://www.scienceforums.net/topic/87761-pluto-charon-combination-how-two-astronomical-bodies-orbit-barycenter/#entry852932

 

But the Sun has a much more complex pattern of orbit than Pluto does. Where would we start? OK get some orbital distances and masses and accurate periods and plug them into the formula(s).

Jupiter mass - 1.898E+27 kg (317.8 Earth mass)

Jupiter period around Sun (Jupiter orbits the Sun every 11.86 Earth years (or 4,332 days) * seconds in a day (86400) = 374,284,800 seconds.

Distance to Sun Jupiter barycenter. barycenter with the Sun lies above the Sun's surface at 1.068 solar radii * 6.955E+8 m or 778,500,000,000 - 1.068*6.955E+8 = 777,757,206,000

 

Average total Distance from Sun: 778,500,000,000 m

 

Mass of Sun 1.9891E+30 kilograms

Period should be the same???

Distance Of Sun center to Sun Jupiter barycenter? 1.068*6.955E+8 = 742794000 m

 

Find out the force of gravitational attraction. G = 6.67E-11 Fg =G*m1*m2/(r^2) (equal and opposite) = 4.1573E+23 N

Find out their centrifugal forces: Fc=mV^2/distance to barycenter (Velocity = 2 * pi * r / T or in words circumference / time period

 

Problem here is that Wikipedia has Jupiter's orbital speed based on it orbiting the Sun and not the Sun-Jupiter barycenter.

So we don't agree on orbital speed.

13056.35 m/sec around baycenter
13068.82 m/sec if orbit centered on the Sun (assumed reason for difference) figure given was 13070 m/sec from http://en.wikipedia.org/wiki/Jupiter

 

Sun:

Jupiter: force = mV^2/r

4.16E+23 Which means the centrifugal force was greater than the calculated Gravitational force which must mean the Sun is effectively drawn toward Jupiter so to increase the Gravitational force (which then will shift the barycenter so the whole thing will require finer adjustments) but supports my original hypothesis that the Sun is displaced from the center of the Solar System, and from raw calculations that looks to be in the order of 254,704,106 meters or 0.37 Solar radii.

Compare them.

 

I'll edit this post as the data is found and converted to SI units kg, m and sec.

Edited February 17, 2015 by Robittybob1

[swansont]

 

Problem here is that Wikipedia has Jupiter's orbital speed based on it orbiting the Sun and not the Sun-Jupiter barycenter.

So we don't agree on orbital speed.

13056.35 m/sec around baycenter

13068.82 m/sec if orbit centered on the Sun (assumed reason for difference) figure given was 13070 m/sec from http://en.wikipedia.org/wiki/Jupiter

 

 

It's entirely possible that, given the available precision of the data,13068.82 m/sec = 13070 m/sec

[Robittybob1]

 

It's entirely possible that, given the available precision of the data,13068.82 m/sec = 13070 m/sec

Yes those two figures (13068.82 m/sec = 13070 m/sec) were considered the same. I was comparing 13056.35 with 13070 (as given by Wiki or 13068.82 as calculated by me using Jupiter-Sun distance), for when I use the average Sun-Jupiter distance their orbital velocity comes out as their calculated value. Well let's see what Janus has to say. I don't know how they got their figure at this stage.

I would like to try the same analysis we did to Pluto-Charon to the Sun-Jupiter combination.

http://www.scienceforums.net/topic/87761-pluto-charon-combination-how-two-astronomical-bodies-orbit-barycenter/#entry852932

 

But the Sun has a much more complex pattern of orbit than Pluto does. Where would we start? OK get some orbital distances and masses and accurate periods and plug them into the formula(s).

Jupiter mass - 1.898E+27 kg (317.8 Earth mass)

Jupiter period around Sun (Jupiter orbits the Sun every 11.86 Earth years (or 4,332 days) * seconds in a day (86400) = 374,284,800 seconds.

Distance to Sun Jupiter barycenter. barycenter with the Sun lies above the Sun's surface at 1.068 solar radii * 6.955E+8 m or 778,500,000,000 - 1.068*6.955E+8 = 777,757,206,000

 

Average total Distance from Sun: 778,500,000,000 m

 

Mass of Sun 1.9891E+30 kilograms

Period should be the same???

Distance Of Sun center to Sun Jupiter barycenter? 1.068*6.955E+8 = 742794000 m

 

Find out the force of gravitational attraction. G = 6.67E-11 Fg =G*m1*m2/(r^2) (equal and opposite) = 4.1573E+23 N

Find out their centrifugal forces: Fc=mV^2/distance to barycenter (Velocity = 2 * pi * r / T or in words circumference / time period

 

Problem here is that Wikipedia has Jupiter's orbital speed based on it orbiting the Sun and not the Sun-Jupiter barycenter.

So we don't agree on orbital speed.

13056.35 m/sec around baycenter

13068.82 m/sec if orbit centered on the Sun (assumed reason for difference) figure given was 13070 m/sec from http://en.wikipedia.org/wiki/Jupiter

 

Sun:

Jupiter: force = mV^2/r

4.16E+23 Which means the centrifugal force was greater than the calculated Gravitational force which must mean the Sun is effectively drawn toward Jupiter so to increase the Gravitational force (which then will shift the barycenter so the whole thing will require finer adjustments) but supports my original hypothesis that the Sun is displaced from the center of the Solar System, and from raw calculations that looks to be in the order of 254,704,106 meters or 0.37 Solar radii.

Compare them.

 

I'll edit this post as the data is found and converted to SI units kg, m and sec.

What can we say about the Sun's centrifugal forces compared to just Jupiter's gravitational force? I''ll be back later to discuss this.

Edited February 17, 2015 by Robittybob1

[Robittybob1]

Thinking this through is quite tricky, lets assume the above conclusion is correct, most of that distance comes off the Jupiter side of the barycenter.

So let the Sun orbit with that radius 742794000 m.

Now we should recalculate the Gravitation attraction, for we are saying they are closer because the Sun is being drawn toward Jupiter. I call this distance the Effective Gravitational Distance (EGD) and from the centrifugal force calculations I estimate that to be 7.78245E+11 m.

If my logic is right that should give a gravitational force equal to the centrifugal force of Jupiter. (Logic correct). There still could be a problem for I am trying to calculate centrifugal force from a modified radius. But ignore that for the moment and we'll come back to it.

 

Centrifugal force of the Sun = mV^2/r, r being 742794000 and v is the circumference/period, the period being that of Jupiter, period being 742794000 seconds. Velocity of the Sun going around its wobble 12.47 m/sec (12.46941459)

So the Centrifugal force = 4.16371E+23 - a way more than it should be.

 

All I can say for sure is the Sun does not orbit the barycenter but how to sort it out is beyond me at the moment.

The radius was 599,361,605 longer that needs to account for the G force. That is nearly as much as the solar radius 6.96E+08

Edited February 18, 2015 by Robittybob1

[Mordred]

What your doing is quite complex, I must admit your going at it with diligents. An aid to help with the visualization grab some graph paper as well as a compass from your geometry set. Set units of force according to length on the graph, use the compass for the directional force component of the force sums.

 

It's more a help keep track aid. Than accuracy, similar to a flow chart. Thus far your doing good keep it up

The graphing will help sort out the vectoral components of the force of gravity. Ideally you will want to apply sine and cos relations to the vector component of force.

For example you have three different mass objects. The Sun, Jupiter and let's say Venus. Each planet influences the CoM in the multibody problem. Force being vectoral will also influence the barycenter with both a scalar as well as a directional influence.

To start pick simpler direction relations, all in a straight line. Larger mass planet (Jupiter) one side, Venus opposite,

 

 

Lol keep in mind more than two body calculations is extremely tricky. Google three or more body problem.

P.S. we measure the mass of planets and stars according to their influences upon other bodies. If you think about that statement. You will realize the complexity and error margins in assigning a mass value

[Robittybob1]

What your doing is quite complex, I must admit your going at it with diligents. An aid to help with the visualization grab some graph paper as well as a compass from your geometry set. Set units of force according to length on the graph, use the compass for the directional force component of the force sums.

 

It's more a help keep track aid. Than accuracy, similar to a flow chart. Thus far your doing good keep it up

The graphing will help sort out the vectoral components of the force of gravity. Ideally you will want to apply sine and cos relations to the vector component of force.

For example you have three different mass objects. The Sun, Jupiter and let's say Venus. Each planet influences the CoM in the multibody problem. Force being vectoral will also influence the barycenter with both a scalar as well as a directional influence.

To start pick simpler direction relations, all in a straight line. Larger mass planet (Jupiter) one side, Venus opposite,

 

 

Lol keep in mind more than two body calculations is extremely tricky. Google three or more body problem.

It is tricky even in the situation of Jupiter and the Sun. You would think they would both orbit their common barycenter end of story, maybe I have messed up somewhere.