From Newton's gravitational law to orbital motion, with geometric algebra

Channel: Peeter Joot's math and physics play Published: 2023-09-14 868 words Source: auto_caption
Advanced Mathematics & Geometric Physics

Transcript

in this video we'll use geometric algebra to show how we go from Newton's law of gravity f equals minus GMM over r squared to an orbital motion such as the elliptical orbital motion of the Earth traveling around the Sun lucium that the system we're trying to solve is like the Earth and Sun system with a solar mass being much much bigger than the earth Mass so that we can effectively ignore subtleties like whether or not we have to work with the center of a mass coordinate systems and reduce Mass in our last video without mentioning the application we did most of the work required by Computing R hat Prime equals one over R times the rejection of our hat from the velocity we can rewrite this writing P equals MV for the momentum x equals r r hat as before and an angular momentum bifector L equals x wedge P this gives us R hat Prime equals 1 over M times R hat over r squared times L now we'll write Newton's gravitational law as M DV DT equals minus GMM times R hat over R square we can solve this for R hat over r squared and substitute in finding R hat Prime equals minus one over GMM times DV DT times L we can now write DV DT times L as D DT of VL minus dldt angular momentum is conserved for the system which means that dldt is zero justifying that properly is beyond the scope of this video we're left with Dr hat DT equals -1 over GMM times DDT of VL this is a perfect derivative equation allowing us to integrate immediately finding R hat equals -1 over GMM times VL minus a vector constant e we use a negative Vector constant here for convenience knowing what's to come before we try to find the orbits associated with our apparent solution we need to understand a subtlety we have Vector terms R hat and E and we have a multi-factor term V times L where L is bi-vector in general product of a vector and a bi-vector has both vector and tri-factor components we'll start by rewriting VL as mvl so that it becomes momentum times the angular momentum this is a product that we can then expand product of the vector and bi-vector is the dot product of the vector in that bi-vector and the wedge product of that vector and the bi-vector however P wedged p is zero the drive Vector term is killed we're left with just the dot product of p with X wedge P that's a vector we can expand this explicitly if desired yielding P dot x times P minus P dot P times x we're now ready to put our solution into a standard conic equation form we're going to rewrite our velocity as p over M and then take dot products with our position Vector x equals R times R hat on the left we have X dotted with r hat which is r and on the right we have X started with E and we have X dotted with the pl Vector we'll write that X dotted with PL as the grade zero selection say a dot b is a grade zero selection of the product of a b this allows us to associate the XP product writing out XP equals x dot P plus X wedge p x wedge p is angular momentum X dotted with P all times L is a scalar times a bi Vector which is a bi Vector bi-factor has no grade zero selection so that product is killed we're left with the grade zero term of X which P which is the angular momentum times angular momentum angular momentum squared is a scalar a negative scalar in this case grouping our R terms we almost have the standard conic form R all times one plus e cos theta equals the scalar that scalar will write as e times D or D is the directrix and E is the eccentricity we used negative Vector e earlier so that we have e greater than zero in this final result e between 0 and 1 is an ellipse equals one is a parabola and E greater than one is a hyperbola to summarize we found that R hat Prime could be written as one over M times R hat over r squared all times L the angular we wrote Newton's gravitational law as DV DT equals minus GM R hat over r squared substituted that into our R hat Prime equation we then made use of the conserved angular momentum to find a perfect differential equation of first order we integrated that and put that into a standard conic form this video was made with Phantom Mathematica and DaVinci Resolve please like subscribe and share for more content of this nature for more geometric algebra content check out my blog peterno.com where you'll also find my book geometric algebra for electrical engineers and plenty of other math and physics related content including latex typeset notes for a number of undergrad and graduate physics and engineering classes