Lecture 33: Rutherford s Formula, and Rocket Motion

Transcription

1 Lecture 33: Rutherrd s Frula, and Rcket Mtn Fr gravty and the electrc rce, S we have: U( r) Nte that we can deterne r n by ndng the dstance at whch the ttal energy equals the eectve ptental Ths ntegral s exactly the sae as the ne we dd n deternng the rbts planets under gravty. Rather than d t agan, I ll just gve the answer: k r / / b r dr Θ r ( ) n 1 / / b rdr r kr/ T b k rt r b r n ( κ / b) k cs Θ ; κ 1 + ( κ / b) T

2 Slvng r b gves: 1 + ( κ / b) cs Θ ( κ / b) ( ) [ ] κ / b 1 cs Θ cs Θ cs Θ ( κ / b) ct Θ sn Θ b κ tan Θ Therere, db db d Θ 1 κ sec Θ dθ d Θ dθ κ κ cs Θ cs ( π / θ / ) κ sn ( θ /) Nte that ths s negatve, as we expect

3 New we re set t nd the derental crss sectn: σ ( θ) b κ κ( κ tan Θ) snθ sn ( θ / ) snθsn ( θ /) κ ct ( θ / ) sn ( θ /)( sn ( θ /) cs ( θ /)) κ 4 4sn ( θ / ) Ths rula was derved by Rutherrd n 1911 (classcal echancs n the 0 th century!) By cncdence, the result als hlds true r a quantu echancal calculatn (nt true r ther types rces)

4 Ttal Crss Sectn Setes we want t knw the ttal prbablty r a cllsn t ccur when we re beas partcles at a target (r at each ther). The ttal crss sectn s deterned by ntegratng ver the derental crss sectn: σ ( ) ( ) T σ θ dω π σ θ snθdθ Fr k/r rces we have: π π πκ snθ ( ) ( ) sn θ / cs θ / σt dθ πκ d 4 4 sn ( /) θ θ sn ( θ /) 0 0 π 1 ( ) cs θ / du πκ dθ πκ ; u sn ( θ /) 3 3 sn ( θ / ) u πκ u 0! But st ths nnte crss sectn results n scatterng at very sall angles

5 Rcket Mtn The next syste we ll study s ne n whch ass s ejected r ne part the syste t change the tn the reander the syste A rcket s an exaple such a syste We ll rst cnsder the tn a rcket n ree space that s, wth n external rces actng n t Snce there are n external rces, the entu the syste (rcket plus uel) ust be cnstant Let s cpare the entu at a gven te wth the entu a sall te dt later: Velcty Intal Fnal exhausted v ( d )( v+ dv) + ved uel Mass rcket + uel Mass uel exhausted durng dt

6 Rcket engnes are typcally desgned t eject uel at a cnstant speed u (wth respect t the rcket). Ths eans that: ve v u S cnservatn entu nw tells us that: v ( d )( v + dv) + ( v u) d v vd + dv d dv + vd ud dv dv dv dt ud ud d u dt Neglgble In ters the change n ass the rcket, we have: dv d u dt dt 0 Ths s called the thrust the engne

7 Integratng the Equatn Mtn Snce the equatn tn s separable, we can slve t t nd: dv ud d dv u v v d dv u T axze nal v uln speed: 1. Eject uel at hghest speed pssble. Mnze rat ntal t nal asses

8 Engneerng Issues There are practcal lts t bth the speed uel exhaust and the rat ntal t nal asses Hgher exhaust speeds ean hgher teperatures lt reached when engne elts! Must bult the uel tank sturdy enugh t supprt the ntal uel weght A ultstage rcket ers a way t acheve hgher nal velcty: 1 a : Intal ass rcket+uel : Mass rcket ater rst-stage uel gne : Intal ass secnd stage (ncl. uel) : Mass ater secnd-stage uel gne Key pnt: a < 1 snce rst-stage uel tank s dscarded

9 Ater the rst stage uses up ts uel, the speed the rcket s: v1 v ln + u The rst-stage uel tank s then released (nte that ths desn t change the velcty the rcket), and secnd stage res. At the end: v v1 + uln v + uln + uln v1 + u ln + ln a v1 + uln a a a 1 Larger than snglestage versn: v1 + uln

10 Blast O In addtn t just lyng n ree space, we want t use rckets t lt r the grund In ths case, the external rce gravty s actng n the rcket We ll assue that the rcket hasn t lwn very hgh, s ths can be apprxated as a cnstant rce Newtn s Secnd Law tells us that: dp Fg dt gdt dp dp s the sae as we calculated bere, s: gdt dv + ud g v + u

11 Assue that the rate uel burn s cnstant: α Then the equatn tn s uα g v uα uα d g u dv g dt g d α α v v g dv α u g v + uln α gt + u ln d