L1 – Center of mass, reduced mass, angular momentum

Read course notes secons 2‐6 Taylor 3.3‐3.5 Newton’s Law for 3 bodies (or n)

 d 2r    m 1 = F + 0 + f + f 1 dt 2 1 12 13  d 2r    m 2 = F + f + 0 + f 2 dt 2 2 21 23 2  d r3    m3 2 = F3 + f31 + f32 + 0 dt

   What are m ? r ? F ? f ? i i i ij Center of mass & its moon

• Posion of center of mass is the (mass) weighted average of individual mass posions

 m1  m1  m1  mn  RCM ≡ r1 + r2 + r3 + ... = ∑ rn M M M n M  d 2r  m n = F ∑ n dt 2 ∑ n n  n  d 2 R  d 2 R M = F 2 ∑ n M 2 = 0 ⇒ ? dt n dt 2‐body problem is “solvable” CoM

m2  m1  RCM ≡ r2 + r1 M M m1     r r ≡ r2 − r1 1  R CM  m2 r 2 2‐body problem is “solvable” CoM

m2  m1  RCM ≡ r2 + r1 M M m1 CoM moves as if all the mass is located there.

In absence of ext forces, VCM = constant  We can choose const vel = 0 r R = const and we can choose = 0. 1 CM  R CM  m2 r 2 2‐body problem is “solvable”

relave coordinates can choose if Fext=0 m  m   R ≡ 2 r + 1 r = 0 REDUCED MASS CM M 2 M 1  m    m1m2 r 1 r ; m r µr µ ≡ 2 = 2 2 = m1 + m2 m1 + m2  m2    r1 = − r ; m1r1 = − µr m1 m1 + m2     r ≡ r − r r 2 1 Check limits, direcons for sense; 1  Students to derive mathemacally R at home. CM  m r 2 2 2‐body problem is “solvable” relave coordinates m1m2     µ ≡ m2r2 = µr ; m1r1 = − µr m + m 1 2  d 2r  m 1 = f 1 dt 2 12 2  d r2  m m2 2 = f21 1    dt  r r r 2  r ≡ 2 − 1 d r  1 ˆ  µ 2 = f21 = f (r)r for central force R dt CM  Relave moon obeys a single‐parcle equaon. m2 r2 Ficous parcle, mass µ moving in a central force field. Vector r is measures RELATIVE displacement; we have to “undo” to get actual moon of each of parcles 1 and 2. Oen m1>>m2. Central force

• Define central force   Gm m ˆ 1 2 f21 = − f12 = f (r)r; gravity: f (r) = − 2 r • f12 reads “the force on 1 caused by 2” • Depends on magnitude of separaon and not orientaon • Points towards origin in a 1‐parcle system • Derivable from potenal (conservave)  Gm1m2 f12 = −∇U (r); gravity: U(r) = − r • Work done is path independent Angular momentum • Define (must specify an origin!)          L = r × p L ≡ r × p; τ = r × F TOT ∑ i i   i     dr1  dr2 LTOT = ∑ri × pi = r1 × m1 + r2 × m2 i dt dt    ⎛ dr ⎞  dr = r × − µ + r × µ 1 ⎝⎜ dt ⎠⎟ 2 dt    dr = (r1 − r2 ) × µ dt Show the last statement is true for  central force.    dLTOT Angular momentum is conserved if LTOT = r × µv; = 0 no external TORQUE. dt Conservaon of AM & Kepler’s 2nd law

• Equal areas equal mes (cons. angular mom.) • Later (once we’ve explored polar coordinates). Summary

• Define RCM; total mass M

• Define r, relave coord; µ = m1m2/M • No ext forces => CoM constant momentum • No external torque => AM conserved • AM conserved ‐> what is value? (see later) • AM conserved ‐> Orbit in a plane (hwk)