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 m m R ≡ 2 r + 1 r CM M 2 M 1 m1 r ≡ r − r r 2 1 1 R CM m2 r 2 2‐body problem is “solvable” CoM m m R ≡ 2 r + 1 r CM M 2 M 1 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; r F LTOT = ri × pi ≡ × τ = × ∑ 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) .