Butch 8 Posted June 8 Share Posted June 8 (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 June 8 by Butch more info Link to post Share on other sites

mathematic 104 Posted June 8 Share Posted June 8 Use inverse squarer relationship. Your description doesn't describe the motion of the first two relative to the third. Link to post Share on other sites

swansont 7597 Posted June 8 Share Posted June 8 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 Link to post Share on other sites

Butch 8 Posted June 8 Author Share Posted June 8 (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 June 9 by Butch Link to post Share on other sites

Butch 8 Posted June 9 Author Share Posted June 9 Sorry, folks... tired brain, will work more on question than solution. Link to post Share on other sites

Sensei 1068 Posted June 9 Share Posted June 9 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. Link to post Share on other sites

swansont 7597 Posted Wednesday at 09:18 AM Share Posted Wednesday at 09:18 AM 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. Link to post Share on other sites

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