Is gravity a pull?

[Strange]

Hi Royston

It's a side effect (further conclusion) of Hubbles expanding universe.

 

 

No it isn't.

[swansont]

Hi RoystonIt's a side effect (further conclusion) of Hubbles expanding universe.

What evidence do you have to support this?

[Capiert]

No it isn't.

Hubble didn't make that conclusion if that is what you mean (=want to hear).

What evidence do you have to support this?

Only what I can tell you.

You did give an invitation to develop a push model.

I'm not sure it was a compliment!

Good. Descartes said to doubt is the greatest thing you can do.

[swansont]

Only what I can tell you.

You did give an invitation to develop a push model.

 

Where is it? I don't see any development.

 

You're drawing conclusions from a nonexistent model - the invitation to develop was before you were doing that.

[Capiert]

The push theory has some interesting consequences (conclusions):

(1 is)

the free fall aceleration g depends on the (e.g. earth's) radius;

not the mass.

(That's assuming no rotation.)

 

E.g. 2 bodies positioned at rest (=same speed), apart (a small distance) in space,

deals with 2 radii (not their masses).

They (=their surfaces) will accelerate together.

 

Such a theory can be used to calculate the various planets' g,

& they are different from what is accepted,

so when you get a chance to measure (free_fall acceleration on other planets & moons, etc)

you will have evidence

to evaluate the theories.

Edited March 19, 2017 by Capiert

[swansont]

The push theory has some interesting consequences (conclusions):

(1 is)

the free fall aceleration g depends on the (e.g. earth's) radius;

not the mass.

(That's assuming no rotation.)

E.g. 2 bodies positioned at rest (=same speed), apart (a small distance) in space,

deals with 2 radii (not their masses).

They (=their surfaces) will accelerate together.

Such a theory can be used to calculate the various planets' g,

& they are different from what is accepted,

so when you get a chance to measure (free_fall acceleration on other planets & moons, etc)

you will have evidence

to evaluate the theories.

So look up planetary data and confirm your conjecture, or refute it.

[Capiert]

So look up planetary data and confirm your conjecture, or refute it.

I've seen them (=the data in a CRC handbook)

& they are not the same,

& can not be the same

because they are calculated

from 2 different formulas

Newtons & mine.

So I confirm

my conjecture

that the

the 2 theories

produce different data.

 

An(other) advantage

of the push theory

is that it does NOT need black holes,

so does not use them.

Some distant galaxies (only) appear

to be shrinking, instead.

Edited March 20, 2017 by Capiert

[John Cuthber]

I've seen them (=the data in a CRC handbook)

& they are not the same,

& can not be the same

because they are calculated

from 2 different formulas

Newtons & mine.

So I confirm

my conjecture

that the

the 2 theories

different data....

Bu the ones in the CRC book agree with the real observations.

Are you saying that the fact that it gives the wrong answer (as experimentally determined) is confirmation of your model.

 

Incidentally, I have seen this model before- in the back of a Dilbert cartoon book.

[Strange]

An(other) advantage

of the push theory

is that it does NOT need black holes,

so does not use them.

Some distant galaxies (only) appear

to be shrinking, instead.

 

 

No theory needs black holes. We just happen to observe them.

[swansont]

I've seen them (=the data in a CRC handbook)

& they are not the same,

& can not be the same

because they are calculated

from 2 different formulas

Newtons & mine.

So I confirm

my conjecture

that the

the 2 theories

produce different data.

 

 

Orbital periods and planetary radii are observational data. Let's see your calculations.

[Capiert]

The scaling equation for a (e.g. planet's) free_fall acceleration (caused by matter expanding, wrt absolute space) is (simply)

g=(ge/Re)*R

the earth's_free_fall aceleration ge=9.8 m/(s^2) (I use Pi^2 instead)

per the earth's_radius Re (ruffly 6.378 million meters)

ratio,

then multiplied by the (desired planet's) radius R.

 

That's only matter's expansion's results,

no rotational,

nor electromagnetic effects.

No theory needs black holes. We just happen to observe them.

I don't know how you can observe what you don't see. ?

But the ones in the CRC book agree with the real observations.

I'm not sure about that. They are too perfect wrt Newton's formula.

I once heard a professor say,

the 1st manned Apollo moon (landing) orbit had to be fine tuned

(recalculated on the fly)

because they didn't know it exactly. Theory didn't match, the numbers.

Are you saying that the fact that it (=?, your prounoun confuses me, please clarify) gives the wrong answer (as experimentally determined) is confirmation of your model.

No, because that statement doesn't sound logical.

I'm saying, if simply e.g. dropping a ball experiments,

give results different from Newton's

then here is an alternative (model, for comparison).

I can't trace the values in the books, because they are interdependent.

& you guys (& gals) must have better resources to checkup on experiments, than (little ol) me.

 

 

Incidentally, I have seen this model before- in the back of a Dilbert cartoon book.

I'm sorry, I don't know him.

But nice to here you have fun.

I don't believe this (theory) is (originally) my own

I'm just helping you all model it

because (I suspect) I understand it (a bit).

I can at least try to discribe it

the way I see it.

 

It's not difficult.

[Strange]

I don't know how you can observe what you don't see. ?

 

 

This is symptomatic of most of what you post. You jump to implausible and supportable conclusions based largely on a lack of knowledge.

 

But you are correct, we haven't observed black holes directly (yet). We have only observed the effects of the presence of a black hole. These can really only be explained by the presence of black holes.

 

There is a project to directly view the black hole at the centre of our galaxy.

http://www.eventhorizontelescope.org

 

How can you see something black? The same way you can see shadows or eclipses.

[swansont]

The scaling equation for a (e.g. planet's) free_fall acceleration (caused by matter expanding, wrt absolute space) is (simply)

g=(ge/Re)*R

the earth's_free_fall aceleration ge=9.8 m/(s^2) (I use Pi^2 instead)

per the earth's_radius Re (ruffly 6.378 million meters)

ratio,

then multiplied by the (desired planet's) radius R.

 

 

 

So now see if this can be made to fit in with planetary orbits. We know that closed orbits only happen for 1/r^2 behavior. According to your model, is the earth's acceleration at the location of the moon the same as the moon's at the location of the earth, as it must?

[Capiert]

The 1/(r^2) rule is (electrostatic), Gauss's (surface_charge_density) law sigma=Q/A,

with the modification of using protons (positive charges)

instead of electrons.

The protons(' charge Q) are easily (ac)counted (by) using mass m (instead of charge, directly)

(e.g. AtWt also includes the neutrons' protons;

& electrons' mass is insignificant in comparison).

The substitution then looks like

sigma~mass/area

sigma~m/A, sphere's_Area A=4*Pi*(r^2)

 

sigma~m/(4*Pi*(r^2)).

 

It has nothing to do with (matter's) expansion, per se (as I know it (til) yet).

 

So Caution: Newton's G formula is a summary, or mixture,

of 2 formula F=m*a & sigma~m/A.

Edited March 21, 2017 by Capiert

[John Cuthber]

The 1/r^2 law is also the only power law that gives stable orbits.

Any other power law is impossible.

[swansont]

The 1/(r^2) rule is (electrostatic), Gauss's (surface_charge_density) law sigma=Q/A,

with the modification of using protons (positive charges)

instead of electrons.

The protons(' charge Q) are easily (ac)counted (by) using mass m (instead of charge, directly)

(e.g. AtWt also includes the neutrons' protons;

& electrons' mass is insignificant in comparison).

The substitution then looks like

sigma~mass/area

sigma~m/A, sphere's_Area A=4*Pi*(r^2)

 

sigma~m/(4*Pi*(r^2)).

 

It has nothing to do with (matter's) expansion, per se (as I know it (til) yet).

 

So Caution: Newton's G formula is a summary, or mixture,

of 2 formula F=m*a & sigma~m/A.

 

 

How does your model predict gravity at some distance away from a planet?

 

The moon's gravity is ~1/6 of that of the earth. (1.6 m/s^2) We've been there. The calculations that got us there used the same physics that tells us that.

 

The radius of the moon is ~1640 km, which is about 0.22 of the earth, predicting a gravity of 2.15 m/s^2. So there's one big problem.

 

We might also look at Mars. The radius ratios predict 4.5 m/s^2. The actual value is 3.7 m/s^2. Once again, we used the accepted physics to get multiple probes there. Oops.

 

 

 

Once you provide the formula I asked for, we can check against the orbital characteristics of e.g. satellites we've put in place, though that's not really necessary at this point. To almost nobody's surprise, you're wrong. The only real question now is whether you will accept that, or not.

[imatfaal]

!

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Hijack by JohnLesser hidden - please do not respond to questions or discussion with personal speculations.

 

To clarify - all responses to questions and participation in an ongoing discussion should be based in mainstream science. The only place for a new idea or hypothesis is in its own thread within the Speculations Forum.

 

 

[JohnLesser]

!

Moderator Note

 

Hijack by JohnLesser hidden - please do not respond to questions or discussion with personal speculations.

 

To clarify - all responses to questions and participation in an ongoing discussion should be based in mainstream science. The only place for a new idea or hypothesis is in its own thread within the Speculations Forum.

 

 

Hello you give me a warning point for putting mainstream,

 

 

Gravity is M1 attracted to M2 and M2 is attracted to M1 and the distance between M1 and M2 contracts like something falling to the ground.

 

Both M1 and M2 ''pull'' each other.

 

I think by ambiguity you misunderstood my post, Can you please remove the unjust warning point?

 

Pull to me means only one body is being forced to move.

Edited March 23, 2017 by JohnLesser

[studiot]

Hello you give me a warning point for putting mainstream,

 

 

Gravity is M1 attracted to M2 and M2 is attracted to M1 and the distance between M1 and M2 contracts like something falling to the ground.

 

Both M1 and M2 ''pull'' each other.

 

I think by ambiguity you misunderstood my post, Can you please remove the unjust warning point?

 

Pull to me means only one body is being forced to move.

 

It's best not to argue with the referee.

 

I understand your technical point and it's wrong.

 

Have you never played 'tug of war'?

 

There is no reason why a 'pull' can't be the resultant of two or more 'pulls'.