# How do I demonstrate gravitational influence mathematically?

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Posted (edited)

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

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Write down the Newtonian gravity equations and do the algebra. You probably want to parameterize in terms of the distance between 1 and 2, if you can

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Posted (edited)

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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Sorry, folks... tired brain, will work more on question than solution.

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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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