3 Celestial Mechanics

Transcription

1 3 Celestial Mechanics 3.1 Keple s Laws Assign: Read Chapte 2 of Caol and Ostlie (2006) OJTA: 3. The Copenican Revolution/Keple (4) Keple s Fist Law b ' εa θ a C 0 D 2a b 2 D a / (1) D a.1 2 / Aea D ab (2) 1 C cos Example: Fo Mas a D 1:5237 AU D 0:0934 At peihelion D 0 ı and D a.1 2 / 1 C cos D.1:5237 AU/.1 0:09342 / D 1:3814 AU 1 C 0:0934 cos.0 ı / At aphelion D 180 ı and D.1:5237 AU/.1 0:09342 / 1 C 0:0934 cos.180 ı / D 1:6660 AU 5

2 Diffeence = 19% (5) Keple s Second Law (6) Keple s Thid Law P 2.y/ D a 3.AU/ Example: Fo Venus, a D 0:7233 AU. Then the peiod is P D a 3=2 D.0:7233/ 3=2 D 0:6151 y D 224:7 d 3.2 Galileo OJTA: 4. The Moden Synthesis/Galileo (3) New Telescopic Obsevations (4) Inetia 3.3 Mathematical Intelude: Vectos Vectos ae objects that have a magnitude (length) and a diection. Diection V Length 6

3 Theefoe, they equie moe than one numbe to specify them. In contast a scala is specified by only one numbe. One way to specifiy a vecto is to give its components: y Components Vy = V sin θ V θ Vx = V cos θ θ = tan -1 (Vy/Vx) V = V x 2 + V y 2 + V z 2 x The diection of a vecto can be specified by oientation angles (one in two dimensions and two in thee dimensions). Its length is given by the Pythagoean theoem in tems of its components. Fo a 3-dimensional vecto, q jv jd Vx 2 C V y 2 C V x 2: The components ae elated to the angles by basic tigonomety. In the 2-D example above, D tan 1 Vy V x Some impotant vecto quantities include the position, velocity, momentum, and acceleation of objects. Let us illustate in 2-D fo simplicity. Position: The position of a point can be specified by a vecto 7

4 y Position y = sin θ θ x = cos θ θ = tan -1 ( y/x) = x 2 y 2 + x whee x D cos y D sin jj D q 2 x C 2 y : Velocity: vecto, The velocity vecto can be defined in tems of the time deivative of the position v d dt ' t D 1 2 t 1 t 2 y Velocity vy = v sin θ v θ vx = v cos θ θ = tan -1 (v y/vx) v = v x 2 v y 2 + x The standad units of velocity in the SI system ae m s 1. Momentum: The momentum vecto is defined to be the mass m times the velocity vecto, Its standad units ae kg m s 1. p mv D m d dt : 8

5 Acceleation: vecto, The acceleation vecto a is defined to be the time deivative of the velocity a dv dt D d 2 dt 2 : y Acceleation ay = a sin θ a θ ax = a cos θ θ = tan -1 (a y/ax) a = a x 2 a y 2 + x The standad units of acceleation in the SI system ae m s 2. Example: Unifom cicula motion Conside the case of unifom cicula motion v The magnitude of the velocity is constant, but the diection is changing, so thee is an acceleation Is this acceleated motion? Yes, because thee is a continuous change in the diection of the velocity, even though its magnitude is constant. 9

6 Addition of vectos: Vectos can be added gaphically by a head to tail ule: A B C = A + B C Unit vectos: It is convenient to define unit vectos that point along the coodinate system axis and have unit length. Fo catesian coodinates, z z x y y x In tems of components, we can wite fo a vecto A A D A x Ox C A y Oy C A z Oz: Scala poduct of vectos: Thee ae two kinds of vecto poducts of inteest to us. The scala poduct of two vectos A and B is a numbe (a scala), defined by AB jajjbj cos D AB cos ; (3) whee is the angle between the two vectos. The ode does not matte in the scala poduct: AB D B A. That is, the scala poduct commutes. The scala poduct is often called the dot poduct. Note some special cases of the scala poduct: 10

7 1. If A and B point in the same diection, cos D 1 and AB D AB. 2. If A and B point in the opposite diection, cos D 1 and AB D AB. 3. If A and B ae pependicula, cos D 0 and AB D 0. Example: Scala Poduct of Two Vectos Conside the following two vectos A = 5 A θ = 30 o B = 8 B Thei scala poduct is AB D BA D AB cos D.5/.8/ cos.30 ı / D 34:6: Coss poduct of vectos: The second kind of vecto poduct is called the coss poduct o the vecto poduct. It diffes fom the scala poduct in that it poduces a new vecto, not a scala like the scala poduct. The coss poduct is defined by AB D.AB sin /OI; (4) whee is again the angle between the vectos and OI is a unit vecto that is pependicula to the plane containing the vectos A and B, with its diection (up o down) given by the ight-hand ule: A B B A Rotate A into B with the ight hand. The thumb points in the diection (up o down) of the new vecto A x B. Right hand 11

8 Unlike fo the scala poduct, the ode in the coss poduct mattes: A B D B A (easily seen fom the ight-hand ule: ty to otate B into A in the peceding diagam and note the diection that you thumb points). Note some special cases of the vecto poduct: 1. If A and B point in the same o opposite diections, sin D 0 and jabj D0. 2. If A and B ae pependicula, sin D 1 and jabj DAB. Example: Angula Momentum Angula momentum L is a vecto that measues the tendency of a body in angula motion to emain in that motion. It is conseved (the eason an ice skate spins faste if the ams ae dawn in is consevation of angula momentum). Angula momentum with espect to some coodinate system is the coss poduct of the position vecto with the momentum vecto: L p D.mv/: (5) Fo example, conside the angula momentum associated with unifom cicula motion p L = p 90 o p Right hand The magnitude of the angula momentum is L D p sin 90 ı D p 12

9 and the diection is out of the pape, as illustated by the ight-hand ule in the figue. The SI units fo angula momentum ae kg m 2 s Consevation Laws The case of angula momentum just consideed is an example of a quantity that is conseved by all inteactions in Newtonian physics. Consevation of angula momentum is an example of a consevation law. In Newtonian physics, we believe that Enegy Momentum Mass Angula momentum ae always conseved in isolated systems. Consevation laws ae vey impotant. Since they must be obeyed, no matte what, they often can be used to simplify the solution of poblems. We will see specific examples shotly. 3.5 Newton s Thee Laws of Motion Newton s 1st Law: Objects in a state of unifom motion emain in that state of motion unless an extenal foce acts on them (The law of inetia). Newton s 2nd Law: If an extenal foce F acts on an object, the acceleation a expeienced by the object is given by the foce divided by the mass, so that: F D ma This pemits the change in velocity (acceleation) to be computed. 13

10 The foce F in this case is the vecto sum of all foces acting on the object: nx F D F 1 C F 2 C :::CF n D F i : Assuming constant mass, Newton s 2nd law may be witten in the equivalent foms F D ma D m dv dt D d.mv/ dt id1 D dp dt : (6) These vecto equations ae equivalent to thee simultaneous equations in the components (in 3-D). Fo example 8 < F x D ma x F D ma! F y D ma y (7) : F z D ma z : The standad unit of foce in the SI system is the Newton: 1 Newton 1 N D 1 kg m s 2 : Newton s 3d Law: Fo evey eaction, thee is an equal and opposite eaction. Notice that in Newton s 3d law the action and eaction ae foces that always act on diffeent objects (neve on the same object): 1 F12 F21 2 If object 1 exets a foce F 21 on object 2, then object 2 exets a foce F 12 D F 21 on object 1. These foces ae equal in magnitude but opposite in diection. 3.6 Newton s Univesal Law of Gavitation Newton easoned that gavity was a foce, obeying his thee laws of motion. 14

11 3.6.1 The gavitational foce Fom the obseved popeties of gavity, Newton deduced his Univesal Law of Gavitation: Univesal Law of Gavitation Evey mass in the Univese exets a foce on evey othe mass that is attactive and diected along the line of centes fo the two masses, with the magnitude of the foce given by F jf jdg m 1m 2 2 ; (8) m1 F -F m2 F = G m1m2 2 whee the univesal gavitational constant G is measued to be G D 6: Nm 2 kg 2 D 6: kg 1 m 3 s 2. (9) An impotant popety of the gavitational foce is that we can pove (see Caol and Ostlie) that fo a spheical mass distibution exeting a gavitational foce on a point mass outside the mass distibution, the gavitational foce is exactly as if all the mass of the mass distibution wee concentated at its cente. Extended masses Point masses m 1 m 2 m 1 m 2 15

12 Fo an object of mass m at the suface of the Eath o a height h above it m h R M + the magnitude of the gavitational foce acting on m is F D G M m.r C h/ : 2 But by Newton s 2nd law F D ma, so by compaing the local acceleation due to gavity is given by M g G.R C h/ ; 2 and we can wite F D mg Typically, nea the suface of the Eath (h 0) we measue that g ' 9:8 ms 2 : Let us check this explicitly: M D 5: kg R D 6: m: Theefoe, the gavitational acceleation should be g D G M R 2 5:97 10 D 6: kg kg 1 m 3 s 2.6: m/ 2 D 9:79 ms 2 : 16

13 Example: Acceleation Due to Eath s Gavity at the Moon s Obit What is the the local gavitational acceleation due to Eath at a distance equal to the Moon s obit (ignoing the gavity of the Moon)? The Moon s obit is about 384,000 km fom the cente of the Eath, so M g D G.R C h/ 2 5:97 10 D 6: kg kg 1 m 3 s 2.3: m/ 2 D 0:0027 ms 2 : Thus, the atio of the gavitational foce exeted by the Eath on a mass at its suface to the foce exeted on that same mass at the distance of the Moon is F 1 D mg 1 D g 1 D 9:79 F 2 mg 2 g 2 0:0027 ' 3626: Weight and mass Weight and mass ae not the same thing. Weight is the gavitational foce exeted on a mass, Weight D Foce D mg: Its SI units ae Newtons (N). In the English system the unit of weight is the pound (lb), with the convesion 1 lb = N. The mass of an object is constant but its weight depends on its location (because it depends on the local gavitational acceleation) Gavitational potential enegy Enegy is conseved in physical pocesses. The total enegy of an object is geneally a sum of a kinetic enegy (enegy of motion) and a potential enegy. If an object is in a gavitational field, its gavitational potential enegy can change if it changes its location. 17

14 Conside a mass m that moves fom a position i to f in a gavitational field that is geneated by a mass M at the oigin of the coodinate system: z d m F f i M y x The gavitational foce F exeted on m is diected towad the oigin and the definition of the change in potential enegy is U D U f U i D Z f F d: i Let s take a simple case of a mass m moving vetically along the z axis. z m F i d f M y x Then the scala poduct is easy since the position vectos and the foce vecto ae pointed 18

15 in opposite diections ( D ) and Z f Z f U D F d D F cos./d D i i Z f D G Mm GM m f d D i 2 ˇ i 1 1 D GM m : : f i Z f i Fd This is the change in the gavitational potential enegy. We make thee geneal emaks about it 1. A moe geneal deivation would have shown that the esult is independent of path, depending only on the endpoints i and f. 2. Geneally, only changes in the potential ae elevant and we can define an abitay zeo fo the gavitational potential enegy scale. It is conventional to choose U! 0 as i!1: Then 1= i! 0 and (dopping subscipts) we may wite fo the gavitational potential U GM m : (10) Conventional: we could choose any zeo fo the scale if we wished. 3. The magnitude of the gavitational foce is obtained fom the deivative of the gavitational potential, F gav D du d D d GM m D GM m ; d 2 whee the minus sign indicates that it is attactive Escape velocity The escape velocity fom a gavitational field is a useful concept: 19

16 Escape Velocity The initial vetical component of velocity fom a given location that gives v! 0 as!1. We may deive a fomula fo escape velocity simply by using consevation of enegy. The total enegy of some mass m moving in a gavitational field is E D 1 2 mv2 G Mm D E kinetic C E potential : As!1, by definition v! 0, so at infinity, E kinetic! 0 E potential!1 E! 0: Thus, at infinity the total enegy is zeo. But since enegy is conseved, the total enegy must be zeo at any. Setting E D 0 in the peceding equation gives 1 2 mv2 G Mm D 0; which may be solved fo v to give 2GM v esc D : (11) Notice that The mass of the object m has cancelled out. The escape velocity depends only on the popeties of the gavitational field, not on the mass of the object that is escaping. The escape velocity depends on whee we stat fom ( in the peceding fomula). The escape velocity fom the suface of the Eath is geate than the escape velocity fom an obit 200 km above the suface of the Eath, fo example. 20

17 Example: Escape velocity fom Eath s suface 2GM 2GM v esc D D D p 2g q 2 D 2.9:8 ms 2 /.6: m/ D 11; 182 ms 1 D 11:2 km s 1 : Example: Escape velocity fom Jupite s suface 2GM v esc D s 2.6: D 11 kg 1 m 3 s 2 /.1: kg/ 7: m D 59; 556 ms 1 D 59:6 km s 1 : Example: Escape velocity fom Sun s suface 2GM v esc D s 2.6: kg 1 m 3 s 2 /.1: kg/ D 6: m D 617; 728 ms 1 D 618 km s 1 : Example: Escape velocity fom suface of Phobos The Matian moon Phobos is not spheical but its aveage adius is about 11 km. Using this and its mass of 1: kg, we may estimate that 2GM v esc D s 2.6: kg 1 m 3 s 2 /.1: kg/ D m D 11:4 ms 1 : 21

18 That s not vey much. Could a basketball playe with Michael Jodan leaping ability attain escape velocity on Phobos just by jumping staight up? Suppose a basketball playe can leap vetically by 40 inches (about one mete) on Eath. By enegy consevation again we have 1 2 mv2 1 C mgy 1 D 1 2 mv2 2 C mgy 2; whee quantities on the left side efe to the playe on the floo, and the quantities on the ight side to the playe at the top of his leap. Dividing though by m and eaanging, v 2 1 v 2 2 D 2g.y 2 y 1 /: But v 2 is at the top of the leap so it is equal to zeo, and y 2 y 1 D y is just the vetical leap of 1 mete. Theefoe, solving fo the initial velocity v 1, v 1 D p q 2gy D 2.9:8 ms 2 /.1 m/ D 4:4 ms 1 : So Michael Jodan could not launch into obit by jumping fom Phobos, but he wouldn t miss it by vey fa! Fo futhe efeence, a wold-class spinte can attain a speed of a little ove 10 m s 1, UT softball pitche Monica Abbot s 70 mph fastball coesponds to about 31 m s 1, and a had kick in a wold cup football (socce) match, o the seve of a top tennis playe, can each initial speeds in the vicinity of m s 1. (Convenient convesion: 1 mph is m s 1.) Cente of mass efeence fame Conside a collection of masses m i at position coodinates i, 22

19 z m 2 m y m 3 x Define a position vecto R that is the weighted aveage R D nx m i i id1 : nx m i id1 The position R is temed the cente of mass. The total mass is M D nx m i so id1 M R D nx m i i : id1 Assume the masses to be constant and diffeentiate M dr dt D nx id1 m i d i dt which is equivalent to and also to M V D P D nx m i v i id1 nx p i ; id1 23

20 whee V in the CM velocity and P is the CM momentum. Thus, the system behaves as if all mass wee concentated at the CM R, moving with the CM velocity V and CM momentum P. Diffeentiate with espect to t, dp dt D nx id1 dp i dt : But if no extenal foces act on the masses (all foces ae intenal between the masses) the total foce must be zeo, by Newton s thid law applied to any two inteacting pais (equal and opposite foces fo each pai). Theefoe F D dp dt D M d 2 R dt 2 D 0: The cente of mass does not acceleate if thee ae no extenal foces acting on the system of masses. This implies that we may simplify the manybody poblem by choosing a coodinate system fo which R D 0 V D 0: This is called the CM fame. It is an inetial fame (one in which Newton s fist law is valid) Cente of mass fo a binay system Conside the impotant special case of two masses (binay system): 24

21 m 1 z M m 2 1 R 2 y x 1 C D 2! D 2 1 : Then choosing the cente of mass as the oigin, Theefoe, and since 2 D 1 C, R D m 1 1 C m 2 2 m 1 C m 2 D 0 m 1 1 C m 2 2 D 0; m 1 1 C m 2. 1 C / D 0 m 1 1 C m 2 1 D m 2 1 D m 2 m 1 C m 2 D m 2 M : By a simila poof, 2 D m 1 m 1 C m 2 D m 1 M : Intoducing the educed mass we can wite 1 D m 1m 2 m 1 C m 2 ; m 2 m 1 C m 2 D D m 1 : m 2 m 1 m 1.m 1 C m 2 / 25

22 By a simila poof, 2 D m 2 : Total enegy: binay system, Substituting and eaanging, whee Djj Dj 2 The utility of the CM system can be seen in witing the total enegy of the E D 1 2 m 1jv 1 j 2 C 1 2 m 2jv 2 j 2 G m 1m 2 j 2 1 j : E D 1 2 v2 1 j and v Djvj, with G M ; v D d D d dt dt. 2 1 / D v 2 v 1 : The total enegy is now the sum of the kinetic enegy of the educed mass and the potential enegy of the educed mass moving about the total mass M D m 1 C m 2 at the oigin. Obital angula momentum: The obital angula momentum fo the binay is L D 1 p 1 C 2 p 2 D m 1 1 v 1 C m 2 2 v 2 : Substituting gives L D m 1 m 1 v 1 C m 2 m 2 v 2 D v 1 C v 2 D.v 2 v 1 / D v; so the total L is the angula momentum of the educed mass only. 26

23 The binay poblem of calculating the motion of two bodies has been eplaced by the calculation of the motion of a single effective mass (the educed mass ) about a stationay point containing the total mass M of the system, with the sepaation between M and given by the sepaation between m 1 and m 2. Binay cente of mass: Choosing the CM as the oigin fo the binay, 1 2 m 1 CM m 2 Take the line of centes as the x axis. Fom the peceding equations, 1 D.=m 1/ 2.=m 2 / D m 2 D x 1 : m 1 x 2 But fom the diagam Substituting these gives the seesaw equation, x 1 D j 1jD 1 x 2 Dj 2jD 2 1 m 1 D 2 m 2 whee 1 C 2 D D distance between masses: Two special cases ae of inteest. Suppose that m 1 D m 2. Then 1 m 1 D 2 m 2 2 D 1 D 1 ; 2 and the CM lies halfway between the masses. 27

24 Suppose one mass much lage than the othe, m 1 >> m 2. Then 1 2 D m 2 m 1 ' 0: Theefoe, the CM almost coincides with the cente of the lage mass. Example: Cente of mass fo the Eath Sun system We have Mˇ D 1: kg M D 5: kg D 1: m and we must solve simultaneously 2 1 D M Mˇ 1 D 2 : Substituting the ight equation into the left, 2 D M ; 2 Mˇ which may be solved fo 2 to give 2 D M =Mˇ : 1 M =Mˇ Fo the Eath Sun system, M Mˇ D 5: kg 1: kg D : Neglecting this tem in the denominato, 2 ' M D /.1: m/ ' 4: m: Mˇ Fo efeence, Rˇ ' m, so the CM of the Eath Sun system is well inside the Sun. Elliptical Keple motion in the CM system fo a binay sta: 28

25 Obit of Sta 1 Sta Cente of Mass Cente of Mass Sta 2 Obit of Sta 2 (See the Java applet OJTA 4.25 fo binay motion.) Actual example of ellipical motion fo the Siius B system: acsec Siius A Cente of Mass Siius B 3.7 Keple s Laws fom Newton s Law of Gavitation We now outline how Keple s laws follow fom Newton s law of gavitation 29

26 3.7.1 Keple s fist law By consideing the effect of gavity on the obit of a planet in the CM sysem (see Ostlie and Caol, pp ), we can pove L 2 = 2 D GM.1 C cos / : (12) But this is the equation of a conic section, which coesponds to equations fo paabolas, ellipses, and hypebolas: D D L 2 = 2 GM.1 C cos / 8 2p paabola; D 1; 2p D L2 = 2 1 C cos GM ˆ< a.1 2 / ellipse; < 1; a.1 1 C cos 2 / D L2 = 2 GM ˆ: a. 2 1/ 1 C cos hypebola; > 1; a. 2 1/ D L2 = 2 GM (13) The geometical definition of a conic section is summaized in the following figue Conic section P h g/h = constant ε = g/h = eccenticity g Focus F Cases: Paabola (ε = 1) Ellipse (ε < 1) Hypebola (ε > 1) Diectix d The thee geneal types of conic sections ae paabolas, ellipses, and hypebolas (the cicle is a special case of an ellipse). The geometical popeties of paabolas, ellipses, and hypebolas ae summaized in the following figue. 30

27 F θ p p Paabola d F b b aε F a(1-ε) θ a ε (1-ε) Ellipse d a d asymptote asymptote F θ F Hypebola a(ε -1) d a a/ε d So obits of the planets ae ellipses genealizes to obits in gavitational fields ae conic sections, with the ellipse as a special case fo a bound obit. Examples of conic-section gavitational obits ae shown in the following figue, 31

28 20 Ellipse e = 0.5 a = 10 Paabola e = 1 p = 5 10 Hypebola e = 1.3 a = 16.7 Cicle e = 0 a = Conic obits with same focus and same vetex which shows conic section obits having the same focus and the same vetex (distance of closest appoach to the focus) Keple s second law Conside the following ellipse: v θ dθ dθ d 32

29 The diffeential aea of the shaded stip is d, so the diffeential aea swept out by the angle d is Z da D d d D d /ˇˇ D d and the ate of change of the aea swept out in a time dt is 0 da dt D 1 d 2 2 dt : Fom the diagam, the velocity vecto v can be esolved into components v along the adial dawn fom the focus and v pependicula to the adial. In tems of unit vectos O and O in these diections, v D v O C v O d d D O C dt dt O Theefoe, v D d=dt and substituting d=dt D v = into the ealie equation gives da dt D v D 1 2 v : Since O and O ae othogonal, jvj Dv sin.90 ı / D v and v Djvj D 1 j.v/ ƒ p D 1 j ƒ p jd L ; L whee L is the magnitude of the obital angula momentum. Theefoe, the change in aea is da D 1 2 dt v D L 2 D constant since angula momentum and the educed mass ae conseved. But constant da=dt is just Keple s 2nd law: j The line joining the planet to the focus sweeps out equal aeas in equal times. 33

30 3.7.3 Keple s thid law Fom the esults fo Keple s 2nd law, the total aea of the ellipse is I I da A D da D dt obit obit dt Z P L D 2 dt D L Z P dt D Lt P 2 2ˇ 0 D L 2 P; whee P is the peiod fo one obit. But we also have fom geomety that the aea of an ellipse is A D ab [see Eq. (2)], so 0 ab D L 2 P; 0 which we can squae and solve fo P 2 to give. P 2 D 2 2 L ab D 42 2 a 2 b 2 D 42 2 a 2 Œa / ; (14) L 2 L 2 whee in the last step we have used Eq. (1) fo ellipses: But we can also equate Eqs. (2) and (12) b 2 D a /: (15) D a.1 2 / 1 C cos D L 2 = 2 GM.1 cos / fo ellipses and solve fo the angula momentum L to give Inseting this into Eq. (14) fo P 2 then gives L D p GMa.1 2 / P 2 D 42 2 a 2 Œa / 2 GMa.1 2 / 4 2 D a 3 : GM 34

31 Thus, inseting explicitly that M D m 1 Cm 2, we aive at the most geneal fom of Keple s 3d law, P D G.m 1 C m 2 / a3 : (16) This expession is valid fo any appopiate units and can be witten in the fom P 2 D ka 3 k 4 2 G.m 1 C m 2 / : (17) Example: Mass of the Eath detemined fom the Moon s obit Fo the Moon, the peiod and semimajo axis of the obit aound Eath ae P D 27:322 d D 2: s a D 3: m: Then fom Keple s 3d law in the geneal fom (17), Evaluating the constants M D m 1 C m 2 D 42 G a 3 P 2 so in these units 4 2 G D 4 2 6: kg 1 m 3 s D 5: kg m 3 s 2 M D 5: a 3 seconds metes P Neglecting the mass of the Moon elative to that of the Eath, M D M C M Moon ' M, M D 5: a 3 seconds 2 metes P D 5: : / 3.2: / 2 D 6: kg: If we subtact fom this the mass 7: kg of the Moon, we obtain 5: kg, which is almost exactly the mass of the Eath. 2 35

32 Let s now demonstate that we can choose a paticula set of units such that numeically k D 1, so that we ecove Keple s 3d law in the fom oiginally poposed by Keple, if we neglect the mass of the planet elative to the mass of the Sun. We adopt the following units fo time, distance and mass: Œtime D Eath yeas (y) Œdistance D AU Œmass D Mˇ whee we use the geneal notation [X] to denote the units of X. Conveting the units of the gavitational constant we can wite 4 2 G D 4 2 6: kg 1 m 3 s D 5: kg m 3 s 2 D 5: kg m 3 s 2 1Mˇ 1: kg 1: m 1 y 1 AU 3: s D 1Mˇ y 2 AU 3 Theefoe, employing these units fo the facto 42, Keple s 3d law can be expessed as G P 2 1Mˇ a 3 D y 2 ; m 1 C m 2 AU and if the masses ae measued in sola masses and the semimajo axis a in AU, the units of P will be yeas. Finally, if we take the mass facto is just unity and we obtain M D mˇ C m planet ' mˇ D 1Mˇ; P 2 D a 3 (Keple s 3d Law) whee a is in AU and P is in Eath yeas. If we want to be explicit about the units that ae equied fo Keple s 3d law in this fom, we could wite a P 2 3 D y 2 : AU 3 fo Keple s 3d law. 36