2. Apsidal Angles & Precession A Brief Discussion! If a particle undergoes bounded, non-circular motion in a central force field  There will always be radial turning points r 1 & r 2 such that r 1 < r < r 2. So, from the V(r) vs r curve we can see that only 2 apsidal distances exist for bounded, noncircular motion. V min < E 2 gives oscillatory radial motion. μ moves back & forth between radial turning points r 2 & r 4. These turning points are the Apsidal Distances of the orbit (the r max & r min from before). r = r 2  E = E 2 (½)μr 2  V min r = r 4 

3. One possible motion: The particle makes one complete revolution in θ but doesn’t return to original position (r & θ). That is, the orbit is not closed! The angle between any 2 consecutive apsides is φ  The Apsidal Angle A closed orbit must be symmetric about any apsis  All apsidal angles must be equal for a closed orbit. For an ellipse the Apsidal angle = π. φ -  

4. If the orbit is not closed  The mass gets to the apsidal distances at different θ in each time around so that the Apsidal angle is not a rational fraction of 2π. If the orbit is almost closed  The Apsides Precess. That is they rotate slowly in the plane of motion. For the 1/r 2 force  All elliptic orbits must be EXACTLY closed!  The apsides must stay fixed in space for all time. If the apsides move with time, no matter how slowly  This is an indication that the force law is not exactly the inverse square law ! Newton: “The advance or regression of a planet’s perihelion would require deviation of the force from 1/r 2.” This is a sensitive test of Newton’s Law of Gravitation!

5. Precession Observational FACT: In planetary motion: The total force experienced by a planet deviates from 1/r 2 (r measured from the sun), because of perturbations due to gravitational attractions to other planets, etc. For most planets, this effect is very small. Celestial (classical) dynamics calculations account for this very accurately! by perturbation theory. The largest effect is for Mercury: It’s observed perihelion advances 574  of arc length PER CENTURY! Accurate classical dynamics calculations using perturbation theory, as mentioned, predict 531  of arc length per century! 1° = 60´ = 3600´´  574´´  0.159° 531´´  0.1475°

6. Discrepancy between observation & classical dynamics calculations: 574  - 531  = 43  (  0.01194°) arc length/century! Neither calculational nor observational uncertainties can account for this difference! Until the early 1900’s this was the major outstanding difficulty with Newtonian Theory! Einstein in the early 1900’s: Showed that General Relativity (GR) accounts (VERY WELL) for this difference! A major triumph of GR! Instead of doing GR, we can approximately account for this (in ad-hoc manner) by inserting GR (plus Special Relativity) effects into Newtonian equations.

7. Back a few lectures: For a general central force F(r), we found the differential orbit equation for u(θ)  1/r(θ). Replacing mass μ with the planetary mass m gives: (d 2 u/dθ 2 ) + u = - (m/ 2 )u -2 F(1/u) (1) Alternatively: (d 2 [1/r]/dθ 2 ) + (1/r) = - (m/ 2 )r 2 F(r) Put the gravitation law, F g (r) = -(GMm)/(r 2 ), into it & get: (d 2 u/dθ 2 ) + u = (Gm 2 M)/( 2 ) (2) Now, add an additional force to F g (r) varying as (1/r 4 ) = u 4 : F GR (r) = -(3GMm 2 )/(c 2 r 4 ) = -(3GMm 2 )u 4 /(c 2 ) Put F g (r) + F GR (r) into (1):  (2) is replaced by: (d 2 u/dθ 2 ) + u = (Gm 2 M)/( 2 ) + (3GM)u 2 /(c 2 ) (3)

8.  The differential orbit equation now is: (d 2 u/dθ 2 ) + u = (Gm 2 M)/( 2 ) + (3GM)u 2 /(c 2 ) (3) Let: (1/α)  (Gm 2 M)/( 2 ), δ  (3GM)/(c 2 )  (3) becomes: (d 2 u/dθ 2 ) + u = (1/α) + δu 2 (4) Now do some math to get an approximate soln to (4): (Like a nonlinear harmonic oscillator equation!) Note that the nonlinear term δu 2 is very small, since δ  (1/c 2 ) Use perturbation theory as outlined on pages 319-320 of the textbook.

9. After a lot of work (!), we find an apsidal angle (perihelion precession angle) of θ = (2π)/[1-(δ/α)]  (2π)[1+(δ/α)]  The effect of the GR term is to displace the perihelion in each revolution by Δ  2π(δ/α) = 6π[(GmM)/(c )] 2 (5) Further results for elliptic orbits (μ = m): ε = eccentricity, 2 = mka(1- ε 2 ); k = GMm, a = semimajor axis  (5) becomes: Δ  [6πGM]/[ac 2 (1- ε 2 )] (6)

10. Mercury: Calc.: Δ  43.03  /century. Obs.: Δ  43.11  /century! Prediction for elliptic orbits, ε = eccentricity, a = semimajor axis is Δ  [6πGM]/[ac 2 (1- ε 2 )] (6) (6)  An enhanced Δ is expected for small semimajor axis & large eccentricity.  From an earlier table: Mercury (ε = 0.2056, a = 0.3871 AU) should have the largest effect! Get: