Gravity Chapter 8 Homework answers (Dec. 2009)

Transcription

1 Gravity Chaptr 8 Howork answrs (Dc 2009) 1 Givn th valu of littl-g at th quator is /s 2, what is th valu of gravity at th North Pol or South Pol? Gravity varis with latitud according to this Intrnational Gravity Forula (IGF) as g( θ) = γ (1 + γ sin θ + γ sin 2 θ), γ = 97801, γ = 5024, γ = Th latitud at th North Pol is Ѳ=+90 and th latitud at th South Pol is Ѳ=-90 Thrfor, th prdictd gravity at th North Pol is: g( + 90 ) = γ (1 + γ sin (90 ) + γ sin (2 * 90 ) = γ (1 + γ 1 + γ 0 ) = γ (1 + γ ) g( + 90 ) = ( ) = s 1Gal 5 g( + 90 ) = 9822 * = 9822 gals = 98, 220gals s 10 / s Q: Stat why th gravity ay or ay not b diffrnt at th north and South Pol according to th IGF? Q: Stat th nubr of significant figur for ach of th fiv nubrs in th IGF? γ,γ,γ Q: What KS units dos th trs 1 2 hav? Q: What units dos Ѳ hav and what is th diffrnc btwn radian and dgr units? Q: What two gravitational ffcts dos th IGF account for? 2 A horizontal sill that xtnds wll outsid th survy ara has a thicknss of 0 and dnsity of 05 g/ in xcss of th rocks it intruds Estiat th axiu dpth at which it would b dtctabl using a gravitr that can asur to 01 Gal g( ρ,) t = 2πG ρt Gals Th quation for a flat infinit sht is: whr G is gravitational constant, ρ is dnsity contrast, and t is th thicknss of th infinit layr First, w not that th infinit sht quation dos not vary with th dpth of th sht Hnc, th answr is that th dpth of th infinit sht is irrlvant to whthr w can dtct th gravitational anoaly with a gravitr accurat to 01 Gal Scond, lt us calculat th gravitational pull of th infinit sht ass anoaly 8 g 6 2 g( ρ, t) = 2πG ρt = 2 π(6672 )(05000 )(000 ) = 6288 / s = Gals 2 g s Thrfor, our gravitr with a 01 Gal accuracy could dtct this anoaly But, th iportant point is that th gravity fild for an infinit sht dos not chang anywhr, it just aks th absolut lvl of th gravity fild 06 Gal gratr EVERYWHERE! Thrfor, th gravity anoaly fro th infinit sht cannot b dtctd 1

2 An xtnsiv dolrit sill was intrudd at th intrfac btwn horizontal sandstons Sktch th gravity profils xpctd if th sill and bds shav bn displacd by: (a) A stply dipping noral fault (b) A shallow thrust fault (c) A strik-slip fault (d) Rpat (c) whn th bds dip at about 40' 4 Calculat how uch gravity changs, and whthr it is an incras or dcras, on going on k north fro th following starting latituds: (a) quator (b) 45 N (c) 45 S What lvation changs in air would giv th sa chang in g? Th siplifid latitudinal gravity quation for sall (<20 k) polward ovnts is approxiatd as: glat ( λ) = 0812 sin(2 λ) Gal / k polwards An iportant point to undrstand is that polwards in northrn hisphr is ovnt toward th North Pol and polwards in southrn hisphr is ovnt towards th South Pol If you ar in ithr hisphr and ov towards th quator that is anti-polward otion 2

3 a) Equator is Ѳ=0 Thrfor, oving 1 k towards North pol is Gal glat ( λ = 0 ) = 0812 sin(2*0 ) * 1 k north = 0 Gal k polward b) Ѳ= 45 in North hisphr Thrfor, oving 1 k toward North pol is otion toward pol Gal glat ( λ = 45 ) = 0812 sin(2*45 ) * 1 k north = Gal k polward c) Ѳ= 45 in South hisphr Thrfor, oving 1 k towards North pol is otion toward quator Gal glat ( λ = 45 ) = 0812 sin(2*45 ) * 1 k north = Gal k polward Now for th tricky part, gravity is a iniu at th quator du to axial outward dirctd cntrifugal forc at th quator du to daily rotation of plant AND th fact that this cntrifugal forc aks th shap (calld figur) of th Earth a flattnd llipsoid Not that th quatorial radius is about 6,78 k and th polar radius is about 6,56 k So, to corrct for ths two factors that chang gravity, on ust kp track of which way on is oving (i, towards pol or towards quator) Th gravitational corrction is subtractd if otion is polward and th gravitational corrct is addd if otion is towards th quator Thrfor, (b) answr is Gal and (c) answr is Gal This logic is iportant Q: Why is th cntrifugal forc dirctd outwards and why dos this dcras gravity? Q: Why is th cntrifugal forc calld a non-inrtial rfrnc fra forc (also calld ficticous forc)? Q: Why is th arth not a sphr, but a flattnd llipsoid with an quatorial bulg? 5 Why is it or corrct to talk of dtrining th ass of th Earth rathr than wighing th Earth? Dos (a) a spring balanc, and (b) a pair of scals asur ass or wight? Stat what th balanc and scal will asur at th Earth s surfac and in intr-galactic spac whr w will assu gravity is zro A spring-balanc usd in th Earth s gravitational fild (g, on th surfac) asurs th forc of gravitational attraction btwn th Plant s ass and th objct s ass bing wighd Th wight of an objct is thus ONLY dfind in a gravitational fild (g, th arth and/or nar any larg assiv objct such as th Sun and Plants) Whr th gravity is nar zro (i, intr-galactic spac), an objct s wight is zro! Thrfor, an objcts wight changs dpnding on whr th objct is wighd A spring-balanc works by using Hook s law of lasticity which stats that th forc on a spring is siply qual to th displacnt of th spring fro its rst stat tis th stiffnss of th spring Th gravitational forc btwn two objcts, whr is th Earth s ass and o is th objct s ass is dfind by Nwton s gravitational law o kg F() r = G r Forc in Nwtons ( ) r s W can dfin a sphr s gravitational acclration fild outsid its surfac siply by laving out th ass of th objct to b wighd

4 g() r = G r acclration ( ) r s Evaluation this quation at th Earth s an radius (r=671 k) givs: g( r) = G r ( ) = (667 )*(597 kg)*( ) = r s kg s (671 ) s So, if you drop a ball, and ignor air friction, th vlocity of th ball incrass by 98 /s vry scond Of cours, if you know th gravitational acclration, you can calculat your ass, but ONLY to th accuracy that you know th gravitational fild of th plant (which varis by hundrds of Gal) Howvr, a ass-balanc can ovrco th wighing probl by NOT asuring th wight of an objct, but dirctly asuring its ass This is don by balancing an objct s ass against known asss placd upon th opposit balanc-pan A ass-balanc can prfor this iraculous fat of asuring ass, not wight, bcaus th gravitational acclration that pulls on th objct s ass in on pan and th known asss placd in th othr pan CANCEL OUT! Not: w ar assuing that th gravity fild dos NOT vary btwn th two pans of th ass-balanc So wight is just th gravitational forc that any ass fls and your wight dpnds on whr you ar with rspct to othr larg asss or quantitativly, th wight of an objct dpnds on th invrs squard distanc btwn two objcts cntr of asss So, now that I know what wight is, thn what is ass? ass is a asur of an objcts inrtia And, inrtia is a proprty of all ass (attr) which aks a ass ithr rain at rst (zro vlocity) or rain at a constant vlocity UNLESS actd upon by an unbalancd nt forc Th fact that ass has inrtia was only undrstood in 1594 by Galilo and bfor that all th physicists (calld natural philosophrs thn) had it wrong! Thy thought that a ball thrown in th air had to hav a forc that continud to push on it to kp it oving: g, Aristotl said th air around th ball was oving with th ball and that this forc applid to th ball, was what kp th ball oving WRONG! As a corollary to this wrong-hadd notion of otion, Aristotl also prdictd that th or assiv a ball, th quickr it would fall This is what Galilo finally tstd with his xprint that droppd balls fro th Towr of Pisa and provd Aristotl to b wrong If you rally want to know what ass is, th particl physics odl QCD-lit prdicts that 99% of ass is th thr quarks that fly-around ach othr to ak th protons and nutrons that ak th atoic nuclus Sinc rlativity rquirs that E=c 2, ass is just spatially localizd (condnsd) nrgy Hnc, ost of th ass of th quarks that ak th protons and nutrons is actually angular ontu nrgy that kps th quarks ALWAYS bound togthr and swirling around ach othr at a scal of Anothr dfinition of ass is siply th nubr of atos prsnt in a givn volu That is calld th olar ass W can calculat this approxiatly bcaus th atoic asss ar wll-know: just add up th ass of th protons and nutrons that ak th nuclus (th lctrons wigh nothing copard to th nuclons) But, ost iportantly, rbr that ass is a asur of inrtia; and, inrtia ans that you nd to apply an unbalancd nt forc to an objct to chang its stat of otion, whthr that otion b at rst or oving along a straight lin at constant vlocity So, what thn is a forc? 4

5 A forc is ALWAYS and intraction btwn two (or or) objcts that changs thir otions Forc is not consrvd lik nrgy or ontu During th intraction (think of two billiard balls colliding), forc is ALWAYS an action-raction pair That is, th forc of objct 1 ON objct two is qual and opposit (in a vctor sns) to th forc of objct 2 ON objct 1 This UST b tru, othrwis th ontu (ass * vlocity) of th two balls is NOT consrvd (i, consrvd ans th ontu rains constant) This is Nwton s third law of otion publishd in 1687 Physicist s now know that thr ar only four fundantal forcs: gravity, lctro-agntic, and th two forcs that ar only significant in th nuclus of atos (th strong and wak forcs) Fro th gravity and lctro-agntic fundantal forcs w can driv th concpt of prssur and strss in a solid/liquid/gas which ar calld contact forcs For xapl, whn two balls collid, th fundantal forc at work is th lctro-agntic rpulsion btwn th lctron clouds that surround th atoic nucli at th two balls surfacs BUT, w can ignor this fundantal forc viw and just calld th forc btwn th lctrons a contact forc which is th prssur/strss that is iprssd during th two balls collision Forc at distanc also occurs for gravity and lctro-agntis In ssnc, a forc-fild dos xtnd through both attr (th arth) and pty spac allowing objcts to xchang forc without vr touching ach othr! This concpt of forcs without contact gav th physicist s ral fits for a long ti and was not accptd until th 1890 s Q: if an objct with ass 1=1 kg with a finit spd, collids with a vry havy ass 2=1 12 kg at rst, what happns? (Us Nwton s scond law) Q: So, if y body ass has th forc of th arth s gravity pulling down, why I a not acclrating? You ar not acclrating downwards bcaus th floor is pushing back against gravity s forc to ak th nt Forc zro Of cours, whn you jup in th air, and find that it is th gravitational fild of th arth that dclrats your otion on th way up and acclrats your otion on th way down Q: Whn I drop a ball at th Earth surfac, th ball clarly falls (acclrats) towards th arth s cntr of ass, but th Earth dos not s to ov towards (acclrat) th ball Nwton s gravitational quation says th Forc of th arth on th ball is qual and opposit to th forc of th ball on th arth Q: Prov athatically using Nwton s gravity law (Eq 82) and Nwton s 2 nd law ( F = *a) that two balls of diffrnt ass fall at that sa rat (ignor air friction) Q: Assu you ar in a plac whr gravity is zro, how could you asur an objcts ass? (Hint: us Nwton s scond law) Q: Is th inrtial ass usd in Nwton s scond law th sa as th gravitational ass usd in Nwton s gravitational quation? Q: What is th valu of gravity at th cntr of th arth and why? 6 A sphrical cavity of radius 8 has its cntr 15 blow th surfac If th cavity is full of watr and is in rocks of dnsity 24 g/, what is th axiu siz of th gravitational anoaly? Not, w ar NOT calculating th absolut valu of th forc intraction btwn th Earth s ass and th ass anoaly associatd with th watr in th burid sphrical cavity W ar just calculating th variation in th gravity fild associatd with rplacing th rock in th sphrical cavity with watr Thus, w can driv a sipl approxiation for th gravitational variation (Eqn 85) δ g = G d 2 whr is th chang in ass fro th rfrnc stat (all rock) and d is th distanc btwn th cntr of ass of th watr filld cavity and th plac whr th gravity asurnts is 5

6 ad at th arth s surfac To calculat, w not that th volu of a sphr is V= 4/ * π * r and hnc th ass anoaly = V * ρ whr ρ is th dnsity diffrnc btwn th watr and rock (140 g/ ) Plugging in th nubr, w gt V* ρ * δ g = G = G = 667 = 89 / s or 0089Gal 2 d d Which on of th following is NOT tru Th valu of littl-g varis ovr th surfac of th arth: (i) (ii) (iii) (iv) (v) Tru If dnsity, hnc ass distribution varis, gravity will vary Tru Th llipsoidal figur of th arth and topography aks variabl gravity Fals All othr things qual, thr is NO systatic variation of gravity with longitud Tru Th gravity fild systatically varis du to two ffcts: 1) th arth is NOT a sphr, but a flattnd llipsoid, i, th quatorial radius is 40 k largr than th polar radius; 2) th arth is a non-inrtial rfrnc fra du to its daily rotation with rspct to th stars Any objct (g, ass in gravitr) on this rotating body will xprinc th non-inrtial rfrnc fra cntrifugal forc dirctd away fro th rotational axis If an objct is in-otion with rspct to th plant, it xprincs th non-inrtial rfrnc fra Coriolis forc Sa answr as (iv) On s distanc fro th pols (north or south) is just th dfinition of latitud 8 How dos littl-g vary btwn th surfac (r1=6400 k) and on kilotr up (r2=6401 k)? Lt us subtract th two littl-g acclration valus at distanc fro th plants cntr of ass for radius r1 and r2 Lt us ignor th vctor proprty of gravity and do a 1-dinsional probl whos only indpndnt variabl is radius (r) g() r = G acclration ( ) r s g= gr ( 1) gr ( 2) = 6672 *597 ( ) 2 2 (6400 ) (6401 ) g = 98 * 7627 = 000 / s = 00 Gal This is an inaccurat calculation bcaus w nd to us or than four significant figurs givn th hug variations in th diffrnt tr xponnts (10 14 and ) A or accurat calculation is providd siply by using th fr air gravity corrction forula = = = g( z) 0086 Gal / * hight ( ) 0086* Gal This positiv valu is th nubr addd to th asurd gravity to CORRECT for th 1 k incras abov th arth s surfac So, th asurd gravity at an altitud of 1 k would b 08 gal lss than as asurd at th arth s surfac 9 If you took a gravitr 1 k down a in in rocks of dnsity 2 g/, th gravity would chang by how uch? 6

7 Thr ar two gravitational ffcts (changs) to add togthr: th fr air ffct du to oving closr to th arth s cntr of ass and th Bougur ffct of th ass abov th gravitr whos ass would pull upwards Th tricky part of this probl is gtting th sign corrct and rbring that w ar asking for what th gravitational ffct is not what th gravitational corrction is Th agnitud of th fr air ffct is providd in Qustion 8 (09 Gal round off) Th sign of this ffct will b positiv (incras gravity) Th Bougur gravity ffct is givn by th Bougur forula: g t = G t Gal = G = Gal b ( ρ, ) 2π ρ 2 π * * (2 00) *1 96 Bcaus th ass is abov th gravitr which is 1 k down a wll, th ass pulls th gravitr-ass upwards, opposit to th arth s gravity pull which is down Thrfor, th Bougur anoaly should b ngativ Adding th two ffcts, th total chang in gravity asurd by th down-wll gravitr is (-96) Gal which quals a +21 Gal INCREASE in th asurd gravity 10 A prson having carrid out a icrogravity survy to locat a lost shaft (filld with air), crats a gravity profil that shows a sall gravitational dip in it (i, th asurd gravity gos down or gts lss and th anoaly has a ngativ sign) Th survyor nots that th dip in th gravity coincids with a dip (dprssion) in th othrwis lvl ground surfac and thus sh says sh has locatd th in shaft But, thn you find out that sh has not corrctd hr data for topography (both fr air and Bougur ffcts) Discuss whthr applying th topography corrctions ight rsult in th gravity dip disapparing Th cobind fr air and Bougur gravity quation 811 shows that fr-air gravitational ffcts (ovnt abov or blow on s datu) will doinat th quation and dtrin its sign Thus, using th flat surfac as a datu, th Bougur gravity corrction will hav a ngativ valu bcaus th topographic dip is blow th flat surfac datu This could ak th asurd dip in th gravity fild that is coincidnt with th topographic dip gt biggr Said anothr way, whn on corrcts for th fact that th gravity asurd in th topographic dip will b highr bcaus th asurnt point is closr to th arth s cntr of ass, thn th fr-air corrction (086 Gal/) will b subtractd fro th asurd data This will ak th Bougur anoaly vn or ngativ 11 Th an radius of th Earth is 671 k On taking a gravitr 1 k abov th arth s surfac in a balloon, you would xpct th valu of littl-g to dcras by how uch? For a ratio of th gravity at r (671 k) and r +1 and thn turn th ratio into a prcnt by ultiplying by 100 gr ( ) r 671 g( r ) = G, g( r ) = G, ratio : = G / G = = = r1 r2 gr ( 2 ) r1 r2 r1 672 To convrt th ratio to prcnt chang, ultipl by 100 Thus, th gravity at 1 k hight is 9997% of its valus asurd at th surfac So, th gravity has dcras by 00% ( )% 12 Th Intrnational Gravity forula dscribs gravity: 7

8 (i) (ii) (iii) (iv) (v) Only at th sa surfac Wrong Th IGF dscribs gravity only with rspct to latitud and dos not car whthr on is ovr sa or land On a surfac siplifid to b a sphr approxiating th arth Wrong Th IGF uss th bst fitting llipsoid for th arth s figur To ONLY allow for th quatorial bulg Wrong It allows for th bulg and cntrifugal forcs To allow for th cntrifugal forcs associatd with th arth s rotation Tru To allow for both th arth s rotation and th quatorial bulg Exactly what th IGF accounts for 1 An ancint burial chabr is to b found using a icrogravity survy Th chabr is about 4 across and is covrd by about of atrial of dnsity 2 g/ Estiat: (a) th axiu agnitud of th anoaly; (b) a suitabl grid spacing assuing th total rror in asurnt is 01 Gal Assu th burial chabr is a sphr of radius 2 with th sphr s cntr at its burial dpth ( ) plus th sphr s radius (2 ) to giv a sphrical cntr dpth of 5 Us th gravity forula (Eqn 85) that givs th axiu gravity valu whn asurd dirctly ovr th cntr of a sphr V* ρ 64 δ π ρ π d d 75 δ = = = µ 2 g = G = G = G*(4/)* * r * / d = * * G 7 2 g 18 / s 002 Gal 20 Gal Th rlation btwn th dpth of th cntr of a sphrical ass anoaly and th gravitational half-width is: d = 1 * half-width () Thrfor, th half-width is qual to d/1 which for d=5 is about 4 trs Thus, if on dsirs to asur th gravity anoaly associatd with th burial chabr, on should sapl at last vry tr so that th gravity anoaly is not issd 14 Dscrib how you would carry out a icrogravity survy to dtrin th latral position of dns in workings (dnsity of 26 g/ ) that is burid bnath 20 of rocks of dnsity 21 g/ Assu th gravitr is accurat to 5 μgal Estiat th accuracy in hight and position rquird for ths rrors to b lss than th gravitrs accuracy Littl inforation is givn, which is intntional and a ral-world situation All w know is that thr is so unknown volu of high dnsity in workings blow 20 But, w do know that th in working dnsity is 26 g/ and th ovrlying rock s dnsity is 21 g/ Thus, w know that w ar sking to find a positiv gravity anoaly Inspction of th dpth ruls (Fig 819 pag 121) shows th gravity anoaly half-width rlations for diffrnt ass anoaly gotris: sphr, cylindr, dipping sht, irrgular body ost usful for this probl is figur (d) for an irrgular body whr th rlation btwn th dpth to th top of th ass anoaly and th pak and slop of th anoaly ar: d 086* δ g ax dg ( dx ) ax whr d is axiu dpth to top surfac (20 ), δg ax is th pak hight (in Gal) of th gravity anoaly, and (dg/dx) ax is th axiu slop of th gravity anoaly (gals/) Substituting d=20 into th quation and rarranging givs: 8

9 22* dg δ g ax dx Now, w know that th axiu gravity valu δg ax will b gratr than about ax 2 tis th axiu gravity slop (also known as th first drivativ of th function g with rspct to distanc x) W also know that th axiu gravity slop will b nar th dgs of th dns ins working ass Thrfor, if on can find th placs whr th gravity slop is a axiu, w can dtct th approxiat dgs of th in workings I would suggst that data b collctd along a sris of gravity asurnt travrss that ar guarantd to cross th gussd boundaris of th in works Fro ths travrss, th approxiat dgs of th in workings can b found Th vrtical accuracy can b assssd by using th cobind fr-air and Bougur anoaly quation 811: Substituting and siplifying givs: δg = h ρ Gal whr ρis g *( * ) 21 / δ g = h*02206 Gal / tr So for a 5 μgal gravitr accuracy, th gravitr hight will nd to b accurat to 5 µ Gal µ Gal 2206 / ( ) = 002 or 2c about on inch Th horizontal accuracy can b assssd by using quation 88 and noting that England has a latitud of about 50 N δglat ( λ) = 0812 sin(2 λ)( Gal / k polwards) = 0800 Gal / k polward So for a 5 μgal gravitr accuracy, th gravitr N-S position will nd to b accurat to 5 µ Gal Gal k 800 µ / = k or Calculat th avrag dnsity of th Earth, givn littl-g is 981 /s 2 and th arth s an radius is 671 k Dnsity is dfind as ass divid by volu Assuing a an radius of th arth s sphr as 671 k, th volu is: V = 4 / * π * r = k or Th total ass of th arth is found using quation 84 g() r = G acclration ( ) r s Rarranging th quation to solv for givs: 9

10 g( r = 671 k) * r 981( / s ) *(671 ( )) G 6672 ( / g s ) 2 21 ( kg) = = = 5968 g 8 2 Thrfor, th an arths dnsity is ρ = = 5512 g / which is about twic th dnsity valu of granit This ans th dp arth ust b uch dnsr than th surfac rocks with granit-lik dnsitis 16 Explain why portions (portions ans ass) wighd on th high Tibtan platau (>4 k an lvation) would b largr (gratr ass) than portions wightd at sa lvl, if th portions ar wightd using a spring balanc? But, th portions would NOT hav to b gratr on th Tibtan platau if a ass balanc is usd Th gravity on th Tibtan platau will b lss with rspct to sa-lvl bcaus th fr-air gravity dcras with altitud is biggr than th Bougur gravity fild ffct du to th xtra ass associatd with th topography (s quation 811) Thus, to gt th sa wight of portions asurd on th Platau and at sa lvl, th portions (ass) wighd on th Platau would hav to b largr than th portions (ass) wighd at sa lvl Q: What is th diffrnc btwn an objct s ass and wight (again)? 17 Th radius of th plant ars is 94 k and th ass is 0108 (108%) of th ass of th arth Calculat th valu of littl-g at th surfac of ars in KS units Th ass of ars is thus 0108 * ass-arth which quals 0108* g which quals g Thrfor, th gravitational acclration at ars surfac is: gr ( ) = G = 6672 * = 74 r (94 ) s Q: So if you xrt an upward forc with your lg uscls to jup into th ars air, why would th hight of your jup b diffrnt with rspct to th arth? 18 Copard to th Earth, th ass of th oon is about 1/80 and its radius is a quartr How dos th surfac gravity ratio copar btwn th Earth and th oon? 1 R 1 = and = 80 R 4 So, th calculatd gravity ratio is G g r r = = * = * = 02 ( no units) g r G 80 1 r 2 10

11 Convrt to prcntags by ultiplying by 100 givs th oon s gravity to b 20% of th valu of th arth s A laborious way to answr this qustion would b to calculat th oon and Earth gravity sparatly and thn ratio th valus 11