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# Advanced Mechanics I PHYS 380

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This 3 page Class Notes was uploaded by Madyson Sporer on Monday October 5, 2015. The Class Notes belongs to PHYS 380 at Christian Brothers University taught by Johnny Holmes in Fall. Since its upload, it has received 16 views. For similar materials see /class/219454/phys-380-christian-brothers-university in Physics 2 at Christian Brothers University.

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Date Created: 10/05/15

The Restricted ThreeBody Problem M1 M2 are both big m is small in comparison all masses are in the same plane Example Earth Moon and space station If we neglect the small In M1 and M2 will orbit each other in circular or elliptical orbits or pass each other in parabolic or hyperbolic orbits Let s consider the easiest case circular orbits If we consider the 2 body system from the Rr coordinate system and if we neglect any external forces such as from the sun then Newton39s Second Law gives 2F ma which becomes GM1M2a2 un2a Eq 1 where the scalar a is the separation ofthe two big masses u is the reduced mass u M1M2M1M2 and 032a is the circular acceleration See section 47 in Symon andor the notes from PHYS 380 Part 3 section on 2 body collisions Rr coordinates From this we get the angular speed of the two masses as they orbit one another 03 VM1M2Ga3 Eq 2 This suggests that if we transform to a rotating coordinate system that has an angular frequency of 0 M1 and M2 will be stationa We can then more easily concentrate on the motion of the small mass m in this rotating system Let s picture the situation from this rotating frame We locate the center of the frame the center of mass at point C the origin of our frame From our de nition of the center of mass M1X1 M2X2 MLXC 0 and our de nition of a X1 X2 a we have solving the above two equations for two unknowns X1 and X2 X1 M2aM1M2 and X2 M1aM1M2 3 aX1X2 X1X2 Using Eq 10b from the Rotating Coordinate Systems section we can write Newton39s Second Law for the small mass m as maquot Flm F2m mwxcogtltr 2mcogtltv Eq 4 where a is the acceleration of the small mass m in the rotating system and v is the velocity of the small mass in the rotating system and 1 is the position of the mass in the system 1 r Flm is the gravitational force on m due to M1 and F2m is the gravitational force on m due to M2 Noting the following 1 xx yy so that with wgtltagtltxx m2Xzgtltzgtltx m2Xzgtlty m2Xx and wxwxyy w2yzxzxy w2yzXx w2yy and 0XV oz gtlt vxxVyy oavxy oavxx and recall from above that 032 M1MzGa3 we can write Newton39s Second Law in the rotating frame as m ax GM1mF1x rrl 392 GMzmF2x I rr2 392 mmzx 2mmvy or ax GM1XX1XX12y232 GM2XXzXXz2y23z 5a M1M2Gxa3 2va Eq ay GMlyXX12y23z GszXX22y232 M1M2Gya3 2va Eq 5b These are two coupled differential equations both y and vy in the ax equation and both X and VK in the ay equation The methods to solve these are not easily identified However there is more than one way to attack a problem We can also try using Conservation of Energy As viewed from the rotating frame there are four forces on the mass m the force of gravity due to M1 the force of gravity due to M2 the centrifugal force a fictitious force due to the rotating frame and the coriolis force another fictitious force due to the rotating frame If we consider the coriolis force we note that the force is perpendicular to the velocity This means that if we try to nd a potential energy for this force we will get zero since F J V and V is to dr then Fdr 0 and so the potential energy due to this force must be 0 If we consider the centrifugal force we note that the direction of the force is toward the center or in the r direction see previous page Since this force is a radial force it is a central force and so is a conservative force and so we can calculate a potential energy for it AVCXy ch dr Ichdx Fcydy Imm2x dX moazy dy 12m032X2y2 Eq 6 and we have the normal potential energies dues to the two gravities Vg1GM1mXX12y2 and vgz GM2nWxxz2y2 Eq 7 So the Conservation of Energy gives E KE PE or 12mVX2Vy2 GMlmWXX12y2 GMzmWXX22y2 12mGM1Mza3x2y2 E Eq 8 In general this equation does not look easy to solve either To nd the equilibrium points we could solve the two component equations of Newton39s Second Law for ax 0 and ay 0 with Vx 0 and Vy 0 or we could plot the potential energy Vgl ng Vc over an area of space and look at the equipotential lines From this we could then discover the equilibrium points In your book Symon 3rd edition on page 289 is a sample plot of this Note that there are ve points three along the X aXis and two off the aXis that correspond to equilibrium points WRITTENHOMEWORK PROBLEM 4 721 in Symon Consider two identical planets each of mass M and radius R that are separated by a distance a a Neglecting the orbital motion of the planets about each other what speed would you need to escape from one of the planets to reach the center point between the planets so you could fall down to the other planet b Now include the orbiting motion of the two planets assuming they are orbiting each other in a circle and find the speed you would need

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