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How do I demonstrate gravitational influence mathematically?


Butch

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Given 2 gravitational points in a close oscillatory relationship and a 3rd at a relativly greater distance, how would I demonstrate mathematically that the influence of the first 2 upon the 3rd decreases as the distance between the first 2 increases?

Assume the oscillatory relationship to be circular, or an average.

Edited by Butch
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I am only looking for a vector statement to show that the gravitational influence of the 2 point system on the single point reduces as the distance between the 2 constituents of the 2 point system increases... I understand the vector math, but lost on the proper formula syntax. I could draw it and deduce it, but a singular mathematical statement eludes me.

It would do to consider the 2 point system to be static except for them receding from one another tangent to the single point.

BTW Thank you Swan.

2 hours ago, mathematic said:

Use inverse squarer relationship.  Your description doesn't describe the motion of the first two relative to the third.

I am using that relationship, however showing that the 2 point system influence diminishes with the distance between the 2 is what is eluding me... I suppose I need to reference the point equidistant from the 2 and apply the 1/x^2 there. Sorry people, probably making this more complex than need be.

And thank you mathematic.

Perhaps I can restate...

The resultant of 2 vectors with a common origin decreases as theta increases. I need a simple proof.

That is not exactly correct.

Edited by Butch
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If distance between 3rd object and group of 1-2 objects is significant, you can calculate center-of-mass of group 1-2, then use it like it would be one object.

 

If you would be also interested in direction:

If object A is at p0, object B is at p1, direction vector is normalized vector of result of subtraction. Pseudo code:

vector p0, p1;

vector delta = p1 - p0;

direction = normalize( delta );

 

normalize( vector ) is equal to:

double length = length( vector );

normal.x = vector.x / length;

normal.y = vector.y / length;

normal.z = vector.z / length;

(notice you can't divide by zero!)

 

length( vector ) is equal to:

double length = sqrt( x^2 + y^2 + z^2 );

 

Dot product of two normal vectors is cosine of angle between them so you can learn what is angle in radians/degrees using arc cosine.

 

So, if you have object A at position p0, object B at position p1 and object C at position p2. And AB makes a group of two close objects, you can calculate:

vector delta20 = p2-p0;

vector delta21 = p2-p1;

double cosine = dot( normalize(delta20), normalize(delta21));

The closer cosine variable is to 1.0 the smaller angle was between AC and BC (because arccos(1)=0)

The sharper angle the longer distance.

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One problem is that it’s not universally true.

You can see this in the limiting case of m1>>m2

When the orbital separation increases, m2 will get closer to m3 for part of its orbit, increasing the attraction.

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  • 2 weeks later...

If you view the above link, the resultant influence of "AB" and that of "CD" (rab and rac) follow an elliptical path... since the constituents of the systems are 180° out of phase and the influences are at same theta, I would expect total influence of both systems to remain on x axis... but I cannot find my error, perhaps I have been staring at it to long?

Okay, error was my expectations(thetas are not equal)... I think all is correct, I need to resolve the elliptical paths of rab and rcd according to Newton...

I would greatly appreciate another mind reviewing what I have to this point before proceeding, thank you.

Edited by Butch
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On 6/9/2021 at 12:10 AM, Sensei said:

If distance between 3rd object and group of 1-2 objects is significant, you can calculate center-of-mass of group 1-2, then use it like it would be one object.

 

If you would be also interested in direction:

If object A is at p0, object B is at p1, direction vector is normalized vector of result of subtraction. Pseudo code:

vector p0, p1;

vector delta = p1 - p0;

direction = normalize( delta );

 

normalize( vector ) is equal to:

double length = length( vector );

normal.x = vector.x / length;

normal.y = vector.y / length;

normal.z = vector.z / length;

(notice you can't divide by zero!)

 

length( vector ) is equal to:

double length = sqrt( x^2 + y^2 + z^2 );

 

Dot product of two normal vectors is cosine of angle between them so you can learn what is angle in radians/degrees using arc cosine.

 

So, if you have object A at position p0, object B at position p1 and object C at position p2. And AB makes a group of two close objects, you can calculate:

vector delta20 = p2-p0;

vector delta21 = p2-p1;

double cosine = dot( normalize(delta20), normalize(delta21));

The closer cosine variable is to 1.0 the smaller angle was between AC and BC (because arccos(1)=0)

The sharper angle the longer distance.

Thanks, very good, pretty much what I have so far... but just 2 dimensional.

On 6/9/2021 at 5:18 AM, swansont said:

One problem is that it’s not universally true.

You can see this in the limiting case of m1>>m2

When the orbital separation increases, m2 will get closer to m3 for part of its orbit, increasing the attraction.

I have to resolve to an average influence, I believe that means finding the point in the elliptical of "equal areas in equal time"... opinions? This is going to be tough one for me... any advice appreciated.

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