Swap Mercury for Jupiter

[Zarkov]

An interesting assumption Persol made in the Earth/ Moon thread. That was that orbits depend upon acceleration ONLY. Mass is not a player in the mathematical formulations.

So here is a theoretical exercise.

I propose to place the planet Jupiter in the position of Mercury in the Solar System.

I calculate the the orbital parameters at Mercury's distance will not change. A body there must (by theory ) travel at Mercury's orbital velocity (47.89 km/sec).

Therefore Jupiter if placed in that orbit, would travel at 47.89 km/sec.

Now this would give by calculation a rv^2 value the same as that of Mercury.
= 2.2 X 10^4 km^3 / sec^2 = Gcentral spin-Mercruy, this value now would be the value calculated for the new orbit of Jupiter....

BUT from NewtonianTheory, [ Gcentral spin = G X Mass ], can be shown to be true.
( The results obtained from this formula give results in total agreement with published data)

Interestingly, the new calculated value for "Jupiter's" mass would be equal to the small value of Mercury.

[ Mass Mercury 0.33 X 10^24 kg, Mass Jupiter 1900 X 10^24 kg ]

In other words, according to Newtonian theory, all planets MUST be where they are otherwise G is violated.

This is a curious situation because even though the density of the planets decrease up to Saturn, the density then increases, from Saturn to Uranus to Neptune, so it would appear that the positions of the planets are not ordered by gravity.

Any comments???

[James R]

Zarkov:

Your conclusion is incorrect, as usual.

A planet's orbital speed around the sun for a given orbital distance depends only on G and the mass of the Sun, not the mass of the planet. Therefore, if you put a Jupiter-size planet in Mercury's orbit, the period of the orbit would be the same as for Mercury itself.

There is no violation of Newton's laws here.

[Pete]

BUT from NewtonianTheory, [ Gcentral spin = G X Mass ], can be shown to be true.
Please define the term "GCentral spin", and specify what mass you mean.

[Zarkov]

>> Please define the term "GCentral spin", and specify what mass you mean.

Gcentral spin = distance from central body X the tangential velocity^2
or rv^2

Mass is that value calculated and called mass of orbiting body.

>> A planet's orbital speed around the sun for a given orbital distance depends only on G and the mass of the Sun, not the mass of the planet. Therefore, if you put a Jupiter-size planet in Mercury's orbit, the period of the orbit would be the same as for Mercury itself.

By theory, that is correct, except it leads to some strange conclusions.

[Zarkov]

>> Please define the term "GCentral spin", and specify what mass you mean.

I was misleading there, Gcentral spin is rv^2 but there is an interlocking rv^2 relationship.
With Mercury the value I gave was inferred given its displayed parameters. For Earth, a better example, Gcentral spin-earth = distance from centre X velocity^2 of a satellite at that distance.
Interestingly each of the observed values, one for each planet is diferent [ but constant within the satellite system ], the Sun system is a constant for all planets, and a planet's satellite system is also constant [ a different constant to the Sun's value ]

Each satellite system points back to the source of central spin and each is interlocked.


Maybe Mercury, because it has no satellites is a bad example.. sorry

So maybe Earth, it has natural and artificial satellites, and Jupiter would be a better example.

If Jupiter was placed in Earth's orbit ??

The calculated mass would still be inconsistent

[Pete]

Is the following consistent with your understanding of Newtonian gravity?

r = radius of circular orbit
v = tangential velocity of orbiting body
M_o = Mass of orbiting body
M_c = Mass of central body

When M_o is much less than M_c:
rv² = G.M_c

[Zarkov]

Yes you are correct. I was confused for a moment. I forgot the mass of the Sun would not change, ie the rv^2 at that position would stay constant.

I am satisfied I have made an error.

[Zarkov]

Some food for thought in the units of both sides.
Accelerated volume equivalent to mass X a constant
The constant G has units, of km^3 / sec^2 / kg of geomagnetic matter,
OR an accelerated volume to mass ratio

mmmh

[Zarkov]

However I fail to understand how

if I put Mercury on the end of a rope, Mercury's radius long and swing it, the Cf = 1 say

If I put Jupiter on the same length of rope and swing it at the same speed, the CF would be much greater, directly proportional to mass

So yes the theory is consistent ? but ? you are going to say that Jupiter would pull back, proportional to mass/ r^2, cancelling the needed force ?
I can't see it happening.

so I am confused.

[Pete]

There are only three relationships you need to understand to resolve this puzzle:

Force = G x M₁.M₂ / r²

Acceleration = Force / Mass

Centripetal Acceleration (circular orbit) = v²/r

[Zarkov]

That is certainly correct.

rv^2 = MG which I call Gcentral spin.

[Zarkov]

From the Physics/maths stream,,, re G Gravitational constant

>> G is the ratio of the mass to the charge of the subatomic particles (using 4pi*q as the charge).

It seems that this is a correct interpretation of the G M 'fudge factor'.

Makes one wonder what exactly is "mass".

[Pete]

So yes the theory is consistent ? but ? you are going to say that Jupiter would pull back, proportional to mass/ r^2, cancelling the needed force ?
I can't see it happening.

so I am confused.
Are you still confused?

From the Physics/maths stream,,, re G Gravitational constant

>> G is the ratio of the mass to the charge of the subatomic particles (using 4pi*q as the charge).

It seems that this is a correct interpretation of the G M 'fudge factor'.

I don't think it seems that way at all. G has different units to a mass/charge ratio, so no equality is possible even if the numbers are the same.

[Zarkov]

re this

>> Are you still confused?

Yes I am,,, could you enlighten me, re the centripetal force of Mercury v Jupiter in the context of this discussion.
:)

[Pete]

Equations and VariablesForce = G x M₁.M₂ / r²
Acceleration = Force / Mass
Centripetal Acceleration (circular orbit) = v²/r

G = 6.7x10^-11Nm²/kg²
r = 5.8x10¹⁰m
M_S = Mass of Sun = 2.0x10³⁰kg
M_J = Mass of Jupiter = 1.9x10²⁷kg
M_M = Mass of Mercury = 3.3x10²³kg


Force between Mercury and the Sun (F_M)F_M = G.M_SM_M/r²
F_M= 6.7x10^-11Nm²/kg² x 2.0x10³⁰kg x 3.3x10²³kg / (5.8x10¹⁰m)²
F_M= 1.3x10²² N


Acceleration of Mercury (a_M)a_M = F_M / M_M
a_M= (G.M_SM_M/r²) / M_M
a_M= G.M_S/r²
a_M= 6.7x10^-11Nm²/kg² x 2.0x10³⁰kg x / (5.8x10¹⁰m)²
a_M = 0.04 m/s²
Note that the Acceleration of Mercury does not depend on the Mass of Mercury. This means we should exactly the same result if we give Mercury a much bigger mass, or replace Mercury with Jupiter:


Force between Jupiter and the Sun, with Jupiter at Mercury's orbital radius (F_J)F_J = G.M_SM_J/r²
F_J = 6.7x10^-11Nm²/kg² x 2.0x10³⁰kg x 1.9x10²⁷kg / (5.8x10¹⁰m)²
F_J = 7.6x10²⁵ N


Acceleration of Jupiter, with Jupiter at Mercury's orbital radius (a_J)a_J = F_J / M_J
a_J = (G.M_SM_J/r²) / M_J
a_J = G.M_S/r²
a_J = 6.7x10^-11Nm²/kg² x 2.0x10³⁰kg x / (5.8x10¹⁰m)²
a_J = 0.04 m/s²
Which is the same as a_M, as expected.
This indicates that the acceleration of the orbiting body depends only on the mass of the central body and the radius of orbit. The mass of the body is irrelevant (as long as it isn't large enough to seriously perturb the central body).


Orbital Velocity

The velocity required to maintain a circular orbit directly depends only on acceleration and radius of orbit:
a = v²/r
v = √(r.a)

In this case:v_J = √(r x a_J)
v_J = √(5.8x10¹⁰m x 0.04 m/s²)
v_J = 48000 m/s

I hope this helps!

[James R]

The force on, say, Jupiter at a particular distance r from the Sun is:

F = GM_JM_S / r²

where M_J and M^S are the masses of Jupiter and the Sun, respectively. Notice that the force is directly proportional to M_J. If we use Mercury instead, we replace M_J with M_M, the mass of Mercury, and end up with a significantly smaller force at the same distance.

The acceleration of Jupiter at this distance, on the other hand, is given by:

a = F/M_J = GM_S/r²

Notice that the acceleration does not depend on the mass of Jupiter. So, if we replace Jupiter by Mercury, the acceleration is unchanged.

[Pete]

I'm not sure why you're confused, since your GCentral Spin relationship is easily derived from the equations I used, and produces the same result:

M_SunG = rv² = constant for all circular Solar orbits.

This means that orbital velocity depends only on the Mass of the Sun, the value of G, and the orbital radius:
v = √(M_SunG/r)

This means you can figure out Mercury's orbital velocity without knowing its mass:
v = √(2x10³⁰kg x 6.7x10^-11Nm²kg^-2 / 5.8x10¹⁰m)
v = 48000 m/s

[Zarkov]

>> I'm not sure why you're confused, since your GCentral Spin relationship is easily derived from the equations I used, and produces the same result:


correct Pete...

>> This indicates that the acceleration of the orbiting body depends only on the mass of the central body and the radius of orbit.

mmmh, that is what I concluded as well.... but I don't believe this... I have seen mass cancel itself out in my equations..... seems strange

Please criticise this...

"if I put Mercury on the end of a rope, Mercury's radius long and swing it, the Cf = 1 say

If I put Jupiter on the same length of rope and swing it at the same speed, the CF would be much greater, directly proportional to mass "

You must admit the momentum of each situation is very different.

I have a real problem with massive mass equating to minute mass at an orbital velocity.

[James R]

Zarkov:

In terms of common sense, there is no reason the mass should cancel out. The reason for that is that there are two kinds of mass in the equations for the orbit. There are the masses in the gravitational force equation, which are gravitational masses, which determine the strength of an object's gravity. Then there is the mass which appears in F=ma; that is inertial mass, which is concerned with an object's resistance to being accelerated by a force.

From observation, it seems that inertial mass and gravitational mass are exactly equal to each other. Careful experiments have been done, and this is known to hold to at least a factor of around 10^-18. People like Newton just accepted the fact without much question.

However, Einstein's theory of general relativity actually explains why the two types of mass must be exactly equal; it follows from the Principle of Equivalence.

It is because of the equivalence that the mass cancels out in the equations given above.

[Pete]

if I put Mercury on the end of a rope, Mercury's radius long and swing it, the Cf = 1 say

If I put Jupiter on the same length of rope and swing it at the same speed, the CF would be much greater, directly proportional to mass

Yes, that's exactly correct.
Why?
Because the gravitational force on the planet is directly proportional to the planet's mass!

This is exactly the same reason that the rate of fall of an object is independent of its mass.

[Pete]

mmmh, that is what I concluded as well.... but I don't believe this... I have seen mass cancel itself out in my equations..... seems strange

It does seem strange, but it's borne out by careful observations... I wonder what that means?

[Zarkov]

Thanks guys, yep that is all correct.
There certainly is an uncanny link between gravitational mass and the field induced.

Sometimes I think I am in a closed shop with no windows, but I am sure I am missing something.

I get the feeling we are all dealing with charge..... charge is independant from mass,

and the equations I have drawn from theory certainly look that way.

Mass keeps popping in and getting anhilated.... I still have a way to go
:)

[Zarkov]

>> It does seem strange, but it's borne out by careful observations... I wonder what that means?

The problem I have is that in the case in point,
Jupiter/Mercury would be moving.... not stationary.... The Sun just sits there relatively (yes I know it is moving, but to a different drum beat)

So we have a dynamic mass and a static mass, and it seems that the amount of dynamic mass is not material at all (the dynamic mass is viewed as static or worse, just non existent).

I really expect that the motion of charge creates a magnetism that holds the system together, much like an electron in a magnetic field that gets drawn into an orbital path.

If the magnetic field is constant then the vortex ( as rv^2 shows for any cosmic spinning system) formed really only reacts to the charge (magnetic-electric) of any dynamic body. All observations would be interlocking and consistent and logical.

But as it is ????????

[Zarkov]

I know the current model is incorrect, but it is hard to prove at this stage.

[Persol]

I know the current model is incorrect, but it is hard to prove at this stage.Old saying: "If you can't explain it, then you don't know it"

[Zarkov]

>> Old saying: "If you can't explain it, then you don't know it"

Yep "know" is a bit strong... gdamn

Logically what we see is, and the explanations are consistent.

It is still curious

I expect the explanation is contained in the sheer massive difference in inertia between the Sun and the orbiting body.

Otherwise, yep, I don't know it
:)

[Pete]

Zarkov,
You must remember that the situation is an approximation because one body is so much smaller than the other. In point of fact, the mass is indeed material. Jupiter at Mercury's position would have a slightly different orbit because it would induce a small wobble in the Sun.

That's how a number of extra-solar planets have been discovered.

[Zarkov]

>> In point of fact, the mass is indeed material

well all cosmic matter is weightless, except when it is not inertial. So at times, especially evident in Mercury's orbit, apparent mass changes, ie non inertial mass changes, is indeed material in affecting the orbital parameters.
Mercury's mass at perihelion is over 3.5 times the minimum mass at aphelion.

[Pete]

Mercury's mass at perihelion is over 3.5 times the minimum mass at aphelion.

I think you're making that up!

[TruthSeeker]

Huuumm... I don't undertnd. Whyis the mass ofthe planet irrelevant? Is it because it is too small compared to the mass and gravity of the sun? :confused: :confused: :confused:

[TruthSeeker]

I don't understand. Why is the mass of the planet irrelevant? Is it because the mass of the planet is too small compared to to the mass and gravity of the sun? Otherwise they would orbit a common centre of gravity just like binary stars. Is that right? :confused: :confused: :confused:

[Zarkov]

>> > I think you're making that up!


LOL, never would contrary to some opinion.

The mass is calculated mass, ie according to an inertial situation, given the displayed parameters at perihelion and aphelion, that is what the mass should be for a body to display the parameters observed. The gravity on the surface also more than doubles in a complete orbit.

>> I don't undertnd. Whyis the mass ofthe planet irrelevant? Is it because it is too small compared to the mass and gravity of the sun?

According to theory, if I place a mass Z at the end of a string and spin it, then the force on the string is dependant upon, length, spin and mass.
Now length and spin are kept constant.... so double the mass the force is doubled (direct proportion)

Now the gravity exchange, which must balance the outward inertial force above is

F = G m1 m2 /r^2

The mass of the Sun remains the same and eg m2 is doubled
so we have

f = G m1 2m2 / r^2 = 2 ( G m1 m2 /r^2)
so this force is doubled.

It appears that the relation of attraction and inertial escape both rely on a direct proportion to mass of the orbiting body.

very convienent isn't it, but it is what we see,,,, a large satellite or a small one both get in orbit at a set distance with the same set velocity.
:)



All well and good and as it is supposed to be.... after all an ejected planet comes in all sizes and must then slowly spiral away from the Sun taking it's intrinsic parameters with it..

BUT what a trick !!!!!!!!!!!!!!

[James R]

Truthseeker,

I answered your question earlier in the thread. Please go back and read the rest of the thread.

[Pete]

(Zarkov)>> Mercury's mass at perihelion is over 3.5 times the minimum mass at aphelion.
(Pete)> I think you're making that up!

LOL, never would contrary to some opinion.

The mass is calculated mass, ie according to an inertial situation, given the displayed parameters at perihelion and aphelion, that is what the mass should be for a body to display the parameters observed.

I suggest that your calculations are flawed. Can you reproduce them for us?

The gravity on the surface also more than doubles in a complete orbit.
Again, I think your either making that up, or basing it on flawed calculations.
Can you explain why you think the surface gravity changes?

>> I don't undertnd. Whyis the mass ofthe planet irrelevant? Is it because it is too small compared to the mass and gravity of the sun?

According to theory, if I place a mass Z at the end of a string and spin it, then the force on the string is dependant upon, length, spin and mass.
Now length and spin are kept constant.... so double the mass the force is doubled (direct proportion)

Now the gravity exchange, which must balance the outward inertial force above is

F = G m1 m2 /r^2

The mass of the Sun remains the same and eg m2 is doubled
so we have

f = G m1 2m2 / r^2 = 2 ( G m1 m2 /r^2)
so this force is doubled.

It appears that the relation of attraction and inertial escape both rely on a direct proportion to mass of the orbiting body.

very convienent isn't it, but it is what we see,,,, a large satellite or a small one both get in orbit at a set distance with the same set velocity.
:)

Lovely explanation! I am impressed :cool:

Note that this only works when the mass is small compared to the Sun.
If the mass is large enough the Sun would also move, meaning that the radius of the planet's circular path would not be the same as the gravitational radius.

[Zarkov]

I still don't quite believe it.

Theoretically even the moon system of Jupiter should remain intact.

I havn't done the calcs, maybe later,... but the differential field spin from the Sun's spin field should equal that predicted by Gcentral spin-jupiter (defined by the moon system) for this new position.

I will need to recalculate L1 and L2 for the above.

I can't see this happening... the outer moon of Jupiter is quite a way out...

I will get back on this one... I do not have my notes with me at the mo..

[Blindman]

Where does the extra inertia come from? Do you add it when you exchange the planets?

[Zarkov]

OK, Elvara (jupiter moon) is about 11 X 10^6 km from Jupiter
Mercury is about 60 X 10^6 km from Sun.
So my concern was unfounded... yep Jupiter and moons would fit where Mercury is now..
except this would be a most unnatural state as I am sure it took many millions of years for (a) jupiter to get where it is now and (b) for it to eject that many moons and have them spiral out to their distance.

But, it would work.

>> Where does the extra inertia come from? Do you add it when you exchange the planets?

Hi Blindman.... inertia was magically added when we moved Jupiter :)
For Jupiter to exist in a stable orbit at Mercury's distance we had to speed it up somewhat from 13.06 km/sec to 47.89 km/sec.... just a bit of slight of hand, so quite a lot of invisable inertia had to be added.

[RawThinkTank]

Jupiter will pull sun towards itself more than mercury, hence to maintain same distance from sun that of mercury Jupiter will have to complete orbit around sun faster than mercury in order to generate centrifugal force for compensating gravitational pull.

By the way, humans wont understand this.

[Zarkov]

>> By the way, humans wont understand this.

Lucky I am not human...
You are incorrect IMO.

[Janus58]

Jupiter will pull sun towards itself more than mercury, hence to maintain same distance from sun that of mercury Jupiter will have to complete orbit around sun faster than mercury in order to generate centrifugal force for compensating gravitational pull.

By the way, humans wont understand this.

What happens is that both the Sun and Mercury orbit around their mutual CoG. This CoG lies on the line joining the Sun and Mercury, much closer to the Sun's center than Mercury's. This CoG forms the focus of Mercury's orbit, and it is the distance from Mercury to this CoG that represents the radius of Mercury's orbit.

If you place Jupiter in place of Mercury, you shift this CoG further from the Sun's center and closer to Jupiter's. This has the effect of decreasing the radius of Jupiter's orbit when compared to Mercury's (even though the distance between Sun and planet hasn't changed).

The formula for centripetal force (The force needed to hold a moving obect in a circular path) is F = v²m/r

Note that for Jupiter m becomes larger and r smaller.

m isn't a factor, since the increase is centripetal force due to its increase is exactly canceled by the increase in gravitational force caused by its increase.

So r is the only factor for determing the velocity needed to maintain the orbit. Since r is smaller for Jupiter than Mercury it takes less velocity for Jupiter to maintain is orbit than Mercury when both are placed the same distance from the Sun.(47.969 km/sec for Mercury and 47.946 km/sec for Jupiter.)

Of course, Jupiter will also have a shorter distance to travel per orbit and, even with its slower orbital velcocity, it will complete its orbit sooner than Mercury. (by One hour)