Orbital Mechanics IA: physical Interpretationof Kepler"s LawsKepler"s very first law statesthat the planets relocate in elliptical orbits about the Sun, through the Sunat one focus. Elliptical orbits are indeed a home of inversesquare law main forces, together we will present shortly.Let us study Kepler"s secondand 3rd laws in watch of Newton"s regulation of universal Gravitation.1. Legislation of Areas and also AngularMomentumKepler"s secondlaw states that the radius vector between the Sun and also an orbiting planetsweeps the end equal locations in equal times. Consider a planet moving alongits elliptical orbit in ~ a street r,with velocity v, together in the figurebelow.
so this is the suitable expressionfor the angular momentum of a planet around the Sun. The crucial is toexamine how the angular inert changes about the orbit, i.e.,
but v x p= mv x v = 0dL/dt = r x F(this is the talk on the planet)For any main force, in particularfor Newton"s law of global Gravitation, wherein F = -(GMm/r2) r,we are going to have r x F = 0also! Thus,dL/dt = 0; soL = constant.In fact, indigenous the above expression,L= mrvq.Finally, we watch that the declare of Kepler"s 2nd law is that sameas the explain of conservation of angular momentum:dA/dt = L/2m = constant2. Kepler"s third LawFor the generalcase of 2 masses connecting according come Newton"s regulation of global Gravitation,the 2 masses actually orbit around the facility of fixed of the system, notnecessarily the center of the more massive body. Recall the equation for centerof massrcm= Smiri/SmiFor a 2 mass system,we will refer to the separation of the two masses as a = r1+ r2,where r1is the street of massive m1from the center, and also r2is the street of massive m2from the center. Consider the instance when the two masses room in circularorbits. Throughout their motion, the two planets have to be plot on bya centripetal force given byF1= m1v12/r= 4p2m1r1/P2andF2= m2v22/r= 4p2m2r2/P2where we have used v= 2pr / P.Now, by Newton"s third law, these two pressures must be same in magnitude(and opposite in direction), which way m1r1=m2r2.This actually proves that the center of the circular activity is the centerof mass. Indigenous this and a = r1+ r2,we haver1= a.Also, through Newton"s legislation of UniversalGravitation we have actually the expression because that the force:F1= F2 = F= Gm1m2/a2,so using the expression forF1,we haveGm1m2/a2= 4p2m1r1/P2= (4p2m1/P2)a,orP2= <4p2/G(m1+m2)>a3which, together promised, is the expressioncorresponding come Kepler"s third law.Note the the center of massis likewise called the barycenter.The 2 masses orbit the barycenter through the same period--you use the separationbetween the masses, a, no thedistances the the masses r1and r2from the facility of mass, to recognize the period.3. Orbit VelocityWe will currently usethese results to derive a specifically simple equation because that the orbitalvelocity for any allude on one elliptical orbit.

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Due to the fact that mostorbits are elliptical, this will certainly be a really useful equation.We decompose the velocityinto its two components:
vr= dr/dt = r" andvq= r (dq/dt) = rq"
Going back to our equationfor one ellipse:r = a(1 -e2) / (1+ e cos q)we can explicitly take the derivativeand gain the radial component of the velocity asvr= dr/dt = a(1 -e2) d/dt <(1+ e cos q)>-1 = ae(1 -e2) sin qdq/dt/(1 + e cosq)2But keep in mind that earlier we hadrvq= r2dq/dt= 2pab/P = 2pa2(1-e2)1/2/P,sodq/dt =2pa2(1-e2)1/2/Pr2Substitution ofthis right into the equation because that vr,givesvr= ae(1 - e2)sin q <2pa2(1-e2)1/2/Pr2>/(1 + e cos q)2 = <2pa/ P(1 - e2)1/2>(e sin q).The matching perpendicularcomponent of the velocity isvq= r dq/dt = 2pa2(1-e2)1/2/Pr = <2pa/ P(1 - e2)1/2>(1 + e cos q).We simply sum the squares ofthese components to obtain the complete magnitude that the velocityv2= vr2+ vq2= (2pa / P)2(1 + 2e cos q + e2)/ (1 - e2).It is advantageous to instead of fromthe equation of one ellipse because that the quantity e cos q:e cos q= a(1 - e2)/r-1which gives:v2= (2pa / P)2<(2a/r)(1-e2)+ e2 -1>/ (1 - e2)= (2pa / P)2(2a/r-1).Finally, from Kepler"s thirdlaw, P2= <4p2/G(m1+m2)>a3,we havev2= <(4p2a2)G(m1+m2)/ 4p2a3>(2a/r-1)= G(m1+m2)(2/r - 1/a)This last equation for thevelocity of one elliptical orbit
v2=G(m1+m2)(2/r - 1/a)
is dubbed the visviva equation.What have we learned?We uncovered that Kepler"ssecond legislation (Law of equal Areas), is equivalent to conservation of angularmomentum l = mrvq,so that dL/dt = 0 for any kind of orbit.This is a repercussion of the central force nature that the gravitationalforce--only a perpendicular pressure could readjust a bodies" angular momentum, and also since over there is none, the angular momentum cannot change.We obtained basic expressions because that the rate of a world or other orbitingbody in ~ perihelion and also aphelion:vperi= (2pa/P)<(1 +e)/(1 -e)>1/2vap= (2pa/P)<(1-e)/(1+ e)>1/2.We also noted, utilizing Newton"sthird law (Law of equal activity and reaction), that 2 bodies orbit theircombined facility of fixed (the barycenter) quite than the centerof either body. From this and also Newton"s law of universal Gravitation(F = Gm1m2/a2),we proved Kepler"s third law in that is quantitative form:P2= <4p2/G(m1+m2)>a3.Applying Kepler"s third law,we were able to acquire a more general equation because that orbital speed, validat any allude in the orbit, the vis viva equation:v2=G(m1+m2)(2/r - 1/a).