GRAVITATIONAL FIELD: CHAPTER 11. The groundwork for Newton s great contribution to understanding gravity was laid by three majors players:


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1 CHAPT 11 TH GAVITATIONAL FILD (GAVITY) GAVITATIONAL FILD: The goundwok fo Newton s geat contibution to undestanding gavity was laid by thee majos playes: Newton s Law of Gavitation o gavitational and inetial mass Gavitational potential enegy o satellites o Keple s d Law o escape speed Gavitational field inside sphees* o hollow sphee o solid sphee * You study using notes povided Copenicus ( ) Galileo ( ) Keple ( ) shape of planetay obits speed aound an obit elating time aound obit to distance fom the Sun 1
2 m 1 m F G mm 1 distance between cm s A value fo G, the univesal gavitational constant, can be obtained fom measuements by Heny Cavendish ( ) of the specific gavity of the ath, as we shall show shotly Cavendish s value fo the specific gavity of the ath was 548, ie, the density of the ath is 548 times the density of wate: ρ ath kg m 5480kg m 4 m volume density π ρ ( ) π kg So, Cavendish s expeiment allowed the fist detemination of the mass of the ath; it is within 1% of today s accepted value Conside the apple that allegedly fell in Newton s ochad m Newton s d Law tells us that: FA FA Using the nd Law we have: m a m a A A F A m A F A But m A 01 kg and m kg a aa Since nea the ath s suface, a A g, a m/s So, the ath also expeiences an acceleation towads the apple, but it is insignificant!
3 The foce on the apple (ie, its weight) F G m m G m m A A A ( + h), and the acceleation of the apple is F m a A G ma ( + h) Vey close to ath s suface, h << and a g, so G m g whee is the ath s adius We know g 981 m/s, m kg and m, Let s take a close look m G N m /kg m A h+ eaanging the expession fo the univesal gavitational constant, gives us a method fo detemining the acceleation at the suface g G m of a massive object, is valid univesally So, if we know the mass (m p ) and adius ( p ) of a constant! Hee ae some examples planet we can detemine the acceleation due to gavity at the suface because G is a univesal g G m p p p ath mass kg adius 678km g 98 m/s
4 Mas mass kg adius 9km gmas 7 m/s 08 g But look at the ath! eq p g G m Moon mass kg adius 178km gmoon 16 m/s 0165 g In deiving the expession fo g we have made an assumption that the ath and planets ae spheical! it s eally a flattened sphee with eq > p g < g eq p In fact, g p ~ 9 8m/s and g eq ~ 9 78m/s Since ones weight is mg, one weighs less up a mountain, in an aiplane, than at sealevel! At 0,000ft: w w ~ 0 % 4
5 Gavitational and inetial mass So fa, without stating it explicitly we have used the same mass (m) in Newton s Law of gavitation as was defined by Newton s nd Law in chapte 4 Note that in the definition of the inetial mass of an object, F inetial mass minet, a thee is no mention of gavity The mass used in Newton s Law of gavitation is called the gavitational mass, which can be detemined by measuing the attactive foce exeted on it by anothe mass (M) a distance away, viz: F gavitational mass mgav Mm GM Thee is no mention of acceleation in this expession Although these ae two diffeent concepts of mass, Newton asseted that, fo the same object, these masses ae identical, ie, mgav mine This is known as the pinciple of equivalence Gavitational potential enegy The acceleation of gavity at a distance h above the h+ ath s suface is: g G m m G ( h+ ) (Note: it is neve zeo so you ae neve completely weightless!!) g ( m/s ) m 60 m kg Height above ath s suface h ( 10 km) 5
6 Nea the ath s suface the gavitational potential enegy of an object of mass m is Ug mgh mg( ), whee U g 0 at the ath s suface Fa fom the ath we must take into account the fact that g In chapte 6 we defined the change in potential enegy as du F ds, whee F is a (consevative) foce acting on the object and ds is a geneal displacement m ˆ d φ F g ds m d ds cos φ ˆ ds Fo the gavitational foce, du F ds F ds F ds G m m g ( gˆ) gˆ d U () G m m d Gm m d G m m + Uo, whee U o is an integation constant We choose U o 0, ie, we define the gavitational potential enegy to be zeo at U () 0 U ( ) U () G m m G m m mg We obtain a negative value fo U( ) because we defined U( ) 0 In ealie chaptes we usually defined U ( ) 0 Howeve, it does not cause difficulties as we nomally wok with changes in potential enegy, ie, U, which is independent of ou choice of U o 6
7 We have U () G m m G m m + h G ( ) + Fo small values of h, h << 1, then mm h ( 1 ) U G m m 1 h G m m h () 1+ 1 G m m + G m m h But G m m U ( ) is constant and G m g U() U( ) + mgh, ie, in moving an object a height h above the ath s suface inceases the gavitational potential enegy by U mgh, poviding h <<, a esult we used in chaptes 6 and 7 Note that the weight of an object is Fg mg, ie, g is a vecto and is called the gavitational field Since g is a vecto field, the esultant gavitational field of a system of masses is a vecto sum of the individual contibutions Question 1: Five objects, each of mass 00kg ae ae equally spaced on the ac of a semicicle of adius 100cm An object of mass 00kg is located at the cente of cuvatue of the ac (a) What is the gavitational foce on the 00kg mass? (b) What is the gavitational field at the cente of cuvatue of the ac? 00 kg 00 kg 00 kg 00 kg 00 kg 10 0cm 00 kg 7
8 y M ĵ M M î M m M x (a) The foce acting on m due to each of the masses M is F G mm G mm G mm G mm x + o o sin45 sin45 0, and F G mm G mm G mm y o + + o sin45 sin sin45 o G mm ( ) ( 0 100) (b) The gavitational foce acting on m is F mg + F Fi ˆ Fj ˆ 8 x y ˆj g m m ĵ m/s N (in the ĵ diection) 8
9 Question : A solid sphee of adius and density ρ o has its cente at the oigin Thee is a hollow, spheical cavity of adius within the sphee and centeed at x, as shown Find the gavitational field at points on the xaxis fo x > y x y x Then g( x) g ( x) + g ( x) solid cavity G M ( m) + G x ( x ) We will model the system as a sphee of mass M 4 π ρ o and a sphee of adius with negative mass ( m) 4 4 G G ρ o π ρ o π x ( x ) + ( ) 4πρ G o 1 1 x 8( x ) When x πρ g ( ) G o 9
10 Question : An object is fied upwad fom the suface of the ath with an initial speed of 40km/s Find the maximum height it achieves Compae that height with the height it would have achieved if the gavitational field wee constant We use consevation of enegy, ie, Ki + Ui Kf + Uf Using the expession fo K f 0:U( + h) gavitational potential enegy, we get h 1 mv G m m i Ki 1 mv i :U ( ) 0 G m m ( + h) Solving fo h we find h, Gm g 1 1 v v i whee g Gm is the gavitational field at the aths s suface h ( ) i m (95km) 10
11 If the gavitational field (g) wee constant, then K U mgh, v ( ) ie, h g 981 i 815 5km Satellites M v m In the fist pat we found the coect height was h g v 1 i But, fom above, v i h, whee h is the height fo a g constant gavitational field h 1 h h h ( ) Check h( km) km ( ) 95 Conside a satellite, mass m, in a cicula obit, adius, aound the ath (o a planet aound the Sun) The gavitational foce between the satellite and the ath povides the centipetal acceleation G m m mv So, in a stable obit Gm v The time fo one obit T π v 4π ie, v G m g, T whee g is the gavitational field acting on the satellite 11
12 4π T Gm constant ie, T ~ Keple s d Law of planetay motion ~ Note also v G m ie, the velocity of a satellite in a stable obit does not depend on its mass only the mass of the cental object! so what? well, it povides us a way of woking out the mass of a planet Take the athmoon system fo example Conside the Moon as the satellite aound the ath 840 km The speed of the Moon is distance aound obit π v time of obit T π m/s Fom above we have: m v G kg, which the the same as the esult we got befoe Note: we didn t need to know the mass of the Moon, only its distance fom the ath and its obital peiod v v G m 1
13 Using a simila appoach we can detemine the mass of the Sun by thinking of the ath as a satellite! Look Question 4: If K is the kinetic enegy of a satellite in a km π v T G m S stable ath obit and U g is the potential enegy of the athsatellite system, what is the elationship between K and U g? 11 π v m/s So m v S ( ) G kg 1
14 We saw that the centipetal foce of an athsatellite system is a esult of the gavitational attaction between the ath and the satellite, ie, F g G m m mv, so v G m 1 K mv 1 m m G But, the potential enegy is U G m m g, U K g Theefoe, the total mechanical enegy is Ug K+ Ug 1 G m m Note that < 0 so the satellite is bound Question 5: uopa obits Jupite with a peiod of 55d at an aveage distance of m fom Jupite s cente (a) Assuming the obit is cicla, what is the mass of Jupite? (b) Anothe moon, Callisto, obits Jupite with a peiod of 167d What is its aveage distance fom Jupite? negy K U g 14
15 (a) In algebaic fom, Keple s d Law states that T 4 e π GM e, J whee T e and e ae the obital peiod and the obital adius of uopa, espectively, and M J is the mass of Jupite 8 4π M 4π ( ) J e GT 11 e ( ) kg (b) Also, T 4 c π GM c, whee subscipt c efes to the J moon Callisto Tc c T, ie, c c e Te e Te m ( ) Wok done to put a satellite in obit m d M The wok done against the gavitational foce in moving an object a distance d is dw F d G M m d ( F ) (Note, if d ( h) is vey small compaed with, eg, taking as the ath s adius (6400 km) and h a few metes, so g is constant, then dw m G m h mgh, a esult we obtained befoe fo the wok done in lifting an object a shot distance) 15
16 Hence, the wok done lifting a satellite to a height h is + h W G M m h + d GMm 1 d + h GM m 1 GM m h + + GM m ( h ) m GM h ( h) ( + h) h mg ( + h ) By the wokenegy theoem, the launch speed equied fo the satellite to each a height h is given by 1 mv ie, v h mg ( + h), h g ( + h) The escape speed (v e ) is defined as the minimum speed equied fo an object to escape the ath s gavitational field Then K f 0: U( ) 0 Ki + Ui Kf + Uf, ie, 1 mv e U( ) 0 1 mv G M m e, K i 1 mv i :U ( ) GM so, v e g Fo the ath, we find 4 v e m/s 11 km/s Fo an object launched fom ath, if v e < 11 km/s the object will be bound and will obit the ath in a cicle o ellipse, if v e > 11 km/s the object will be unbound and will follow a hypebolic path and leave the ath and neve etun 16
17 Question 6: One way of thinking about a black hole is to conside a spheical object whose density is so lage that the escape speed at the suface is geate than the speed of light, c The citical adius fo the fomation of a black hole is called the Schwazschild adius, S (a) GM Show that S, whee M is the mass of the black c hole (b) Calculate the Schwazschild adius fo a black hole whose mass is equal to 10 sola masses (a) The escape speed is given by GM ve Fo a black hole, v e c and S, GM GM ie, c S S c (b) If M 10 M kg, Sun S 8 ( 10 ) m (95km) 17
18 Gavitational field inside a hollow sphee The gavitational field inside a hollow sphee is zeo, ie, g 0 ( < ) We can show this esult by A 1 m m o 1 consideing a small (point) 1 A mass m o situated at some φ m point inside the sphee If the distance fom the mass to points on the shell diametically opposite ae 1 and, then the aeas A 1 and A and the volumes V1 A1 and V A ae popotional to 1 and, espectively If the density of the shell is ρ, then m1 V1ρ 1 and m Vρ, ie, m 1 m 1 and G mm o 1 G mm o 1 Thus, the net foce on m o is zeo That agument can be made fo all diameteically opposite aeas A 1 and A ove the sphee So, the gavitational field inside a hollow sphee is zeo In fact, the esult is tue fo any thickness of shell We can apply the pevious esult by dividing the shell into a continuum of concentic shells and summing the fields at the the point of inteest Thus, g 0, inside all hollow sphees Gavitational field inside a solid sphee To detemine the gavitational field inside a solid sphee, a distance fom the cente, we use the pevious esult The mass of the pat of the sphee outside makes no contibution to the gavitational field at o inside Only the mass M inside adius contibutes to the gavitational field at This mass poduces a field equal to that of a point mass M at the cente of the sphee If M is the mass of the sphee, then 18
19 4 π M M 4 π M Then, using an ealie expession fo g, we find g ( ) G M GM, inside the sphee g G M g () The magnitude of the gavitational field is zeo at the oigin and inceases linealy with But, as befoe, it is diected towads the cente of the sphee g G M Question 7: Agamemnon is a planet of adius 4850km and density 5500kg/m A staight shaft is dilled completely though the planet Agamemnon If a stone is dopped fom the top of the shaft, what would its speed be (a) at the cente, and (b) at a distance of 500km above the cente? (c) Would the stone each the opening at the othe end of the shaft? The stipulation fo this esult is that the sphee is of unifom mass density 19
20 (a) If we let a mass m fall down the shaft, to a adius, the wok done by the gavitational field W is F d F d mg d, g g whee g is the gavitational field at adius But g G M M G, whee M is the mass contained within the sphee of adius W GMm GMm d [ ] 4 4 But M π ρ π( ) kg, and fom the wokenegy theoem, W K With 4850 km and 0, W K GM 11 4 m m (J) (b) With ie, 1 7 mv m 7 v m/s 4850 km and 500 km, W K m (J) 7 v m/s (c) Yes, the stone would just each the opposite opening Since the scenaio is symmetical, its speed at the opening would be zeo, having done W m Joules of wok against the gavitational field v(m/s) 6014m/s m/s 1 0
21 Question 8: When fathest fom ath, ie, at apogee, the Moonath distance is 406,400km When closest to ath, ie, at peigee, the Moonath distance is 57,600km If the mass of the ath is kg, what ae the obital speeds of the Moon at apogee and peigee? In the pevious chapte we saw that poviding thee ae no extenal foces/toques, then the angula momentum of a system like the athmoon is conseved The angula momentum of the Moon is L mv But at peigee and apogee v p p a v, va so L mv mvaa mvpp, ie, v v a p p a Also, mechanical enegy is conseved, ie, 1 1 mv G M m mv G M m a p a v G M v G M a p a p Hence v p G M v G M p p, a a p p p ie, v 1 1 p 1 GM a p a 1
22 v a p GM p a p a a p so v a + p GM p, a p ie, v p GMa p( a + p) ( ) 1090m/s (1090km/s) v v a p p a m/s (0959km/s)
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