3 Celestial Mechanics
Assign: Read Chapter 2 of Carrol and Ostlie (2006)
3.1 Kepler’s Laws
OJTA: 3. The Copernican Revolution/Kepler
– (4) Kepler’s First Law
b r' r εa θ a
r C r0 D 2a b2 D a2.1 2/ (1) a.1 2/ r D Area D ab (2) 1 C cos
Example: For Mars a D 1:5237 AU D 0:0934 At perihelion D 0ı and a.1 2/ .1:5237 AU/.1 0:09342/ r D D D 1:3814 AU 1 C cos 1 C 0:0934 cos.0ı/ At aphelion D 180ı and .1:5237 AU/.1 0:09342/ r D D 1:6660 AU 1 C 0:0934 cos.180ı/
5 Difference = 19%
– (5) Kepler’s Second Law – (6) Kepler’s Third Law
P 2.yr/ D a3.AU/
Example: For Venus, a D 0:7233 AU. Then the period is
P D a3=2 D .0:7233/3=2 D 0:6151 yr D 224:7 d
3.2 Galileo
OJTA: 4. The Modern Synthesis/Galileo
– (3) New Telescopic Observations – (4) Inertia
3.3 Mathematical Interlude: Vectors
Vectors are objects that have a magnitude (length) and a direction.
Direction
V h Lengt
6 Therefore, they require more than one number to specify them. In contrast a scalar is specified by only one number. One way to specifiy a vector is to give its components:
y Components θ
in -1 θ = tan (Vy /V x )
|s V V 2 2 2 |V| = V x + V y + V z = | y θ V x Vx = |V|cos θ
The direction of a vector can be specified by orientation angles (one in two dimensions and two in three dimensions). Its length is given by the Pythagorean theorem in terms of its components. For a 3-dimensional vector, q 2 2 2 jV jD Vx C Vy C Vx :
The components are related to the angles by basic trigonometry. In the 2-D example above, Â Ã Vy D tan 1 Vx
Some important vector quantities include the position, velocity, momentum, and accelera- tion of objects. Let us illustrate in 2-D for simplicity.
Position: The position of a point can be specified by a vector r
7 y Position
θ -1
in θ = tan (r /r ) r y x |s r 2 2 |r| = r x + r y = |
y θ r x rx = | r |cos θ where q 2 2 rx D r cos ry D r sin rjrjD rx C ry :
Velocity: The velocity vector can be defined in terms of the time derivative of the position vector, dr r r r v ' D 1 2 dt t t1 t2
y Velocity θ -1 in θ = tan (v /v ) v y x |s v 2 2 |v| = v x + v y = |
y θ v x vx = |v|cos θ
The standard units of velocity in the SI system are m s 1.
Momentum: The momentum vector is defined to be the mass m times the velocity vector, dr p mv D m : dt Its standard units are kg m s 1.
8 Acceleration: The acceleration vector a is defined to be the time derivative of the velocity vector, dv d 2r a D : dt dt2
y Acceleration θ -1 in θ = tan (a /a ) a y x |s a 2 2 |a| = a x + a y = | y θ a x ax = |a|cos θ
The standard units of acceleration in the SI system are m s 2.
Example: Uniform circular motion
Consider the case of uniform circular motion
v
The magnitude of the velocity is constant, but the direction is changing, so there is an acceleration
Is this accelerated motion? Yes, because there is a continuous change in the direction of the velocity, even though its magnitude is constant.
9 Addition of vectors: Vectors can be added graphically by a head–to–tail rule:
A B C = A + B
C
Unit vectors: It is convenient to define unit vectors that point along the coordinate system axis and have unit length. For cartesian coordinates,
z
z x y y
x
In terms of components, we can write for a vector A
A D AxxO C AyyO C AzzO:
Scalar product of vectors: There are two kinds of vector products of interest to us. The scalar product of two vectors A and B is a number (a scalar), defined by
AB jAjjBj cos D AB cos ; (3) where is the angle between the two vectors. The order does not matter in the scalar product: AB D BA. That is, the scalar product commutes. The scalar product is often called the dot product.
Note some special cases of the scalar product:
10 1. If A and B point in the same direction, cos D 1 and AB D AB. 2. If A and B point in the opposite direction, cos D 1 and AB D AB. 3. If A and B are perpendicular, cos D 0 and AB D 0.
Example: Scalar Product of Two Vectors
Consider the following two vectors
A 5 | = |A θ = 30o B |B| = 8
Their scalar product is AB D BA D AB cos D .5/.8/ cos.30ı/ D 34:6:
Cross product of vectors: The second kind of vector product is called the cross product or the vector product. It differs from the scalar product in that it produces a new vector, not a scalar like the scalar product. The cross product is defined by AB D .AB sin /IO; (4) where is again the angle between the vectors and IO is a unit vector that is perpendicular to the plane containing the vectors A and B, with its direction (up or down) given by the right-hand rule:
A B
B Rotate A into B with the right hand. The thumb points in the direction (up or down) of A the new vector A x B.
Right hand
11 Unlike for the scalar product, the order in the cross product matters: A B D B A
(easily seen from the right-hand rule: try to rotate B into A in the preceding diagram and note the direction that your thumb points).
Note some special cases of the vector product:
1. If A and B point in the same or opposite directions, sin D 0 and jABjD0. 2. If A and B are perpendicular, sin D 1 and jABjDAB.
Example: Angular Momentum
Angular momentum L is a vector that measures the tendency of a body in angular motion to remain in that motion. It is conserved (the reason an ice skater spins faster if the arms are drawn in is conservation of angular momentum).
Angular momentum with respect to some coordinate system is the cross product of the position vector with the momentum vector:
L r p D r .mv/: (5) For example, consider the angular momentum associated with uniform circular motion
p L = r p 90o r p
r
Right hand
The magnitude of the angular momentum is L D rp sin 90ı D rp
12 and the direction is out of the paper, as illustrated by the right-hand rule in the figure. The SI units for angular momentum are kg m2s 1.
3.4 Conservation Laws
The case of angular momentum just considered is an example of a quantity that is conserved by all interactions in Newtonian physics. Conservation of angular momentum is an example of a conservation law. In Newtonian physics, we believe that
Energy Momentum Mass
Angular momentum are always conserved in isolated systems. Conservation laws are very important. Since they must be obeyed, no matter what, they often can be used to simplify the solution of problems. We will see specific examples shortly.
3.5 Newton’s Three Laws of Motion
Newton’s 1st Law: Objects in a state of uniform motion remain in that state of motion unless an external force acts on them (The law of inertia).
Newton’s 2nd Law: If an external force F acts on an object, the acceleration a experienced by the object is given by the force divided by the mass, so that:
F D ma
This permits the change in velocity (acceleration) to be computed.
13 The force F in this case is the vector sum of all forces acting on the object: Xn F D F 1 C F 2 C :::C F n D F i : iD1 Assuming constant mass, Newton’s 2nd law may be written in the equivalent forms dv d.mv/ dp F D ma D m D D : (6) dt dt dt These vector equations are equivalent to three simultaneous equations in the components (in 3-D). For example 8 < Fx D max F D ma ! F D ma : y y (7) Fz D maz:
The standard unit of force in the SI system is the Newton: 1 Newton 1 N D 1 kg m s 2:
Newton’s 3rd Law: For every reaction, there is an equal and opposite reaction.
Notice that in Newton’s 3rd law the action and reaction are forces that always act on differ- ent objects (never on the same object):
1 2 F12 F21
If object 1 exerts a force F 21 on object 2, then object 2 exerts a force F 12 D F21 on object 1. These forces are equal in magnitude but opposite in direction.
3.6 Newton’s Universal Law of Gravitation
Newton reasoned that gravity was a force, obeying his three laws of motion.
14 3.6.1 The gravitational force
From the observed properties of gravity, Newton deduced his Universal Law of Gravita- tion:
Universal Law of Gravitation Every mass in the Universe exerts a force on every other mass that is at- tractive and directed along the line of centers for the two masses, with the magnitude of the force given by
m1m2 F jF jDG ; (8) r2
F -F m1m2 m1 m2 F = G r2
r
where the universal gravitational constant G is measured to be
G D 6:673 10 11 Nm2kg 2 D 6:673 10 11 kg 1m3s 2. (9)
An important property of the gravitational force is that we can prove (see Carrol and Ostlie) that for a spherical mass distribution exerting a gravitational force on a point mass outside the mass distribution, the gravitational force is exactly as if all the mass of the mass distri- bution were concentrated at its center.
Extended masses Point masses
r r m1 m2 m2 m1
15 For an object of mass m at the surface of the Earth or a height h above it
m h
R
M + the magnitude of the gravitational force acting on m is M m F D G ˚ : .R C h/2 But by Newton’s 2nd law F D ma, so by comparing the local acceleration due to gravity is given by M g G ˚ ; .R C h/2 and we can write F D mg Typically, near the surface of the Earth (h 0) we measure that g ' 9:8 ms 2:
Let us check this explicitly:
24 6 M˚ D 5:97 10 kg R˚ D 6:38 10 m:
Therefore, the gravitational acceleration should be M g D G ˚ 2 R Â Ã 5:97 1024 kg D 6:673 10 11 kg 1m3s 2 .6:38 106 m/2 D 9:79 ms 2:
16 Example: Acceleration Due to Earth’s Gravity at the Moon’s Orbit
What is the the local gravitational acceleration due to Earth at a distance equal to the Moon’s orbit (ignoring the gravity of the Moon)? The Moon’s orbit is about 384,000 km from the center of the Earth, so M g D G ˚ .R C h/2 Â Ã 5:97 1024 kg D 6:673 10 11 kg 1m3s 2 .3:84 108 m/2 D 0:0027 ms 2: Thus, the ratio of the gravitational force exerted by the Earth on a mass at its surface to the force exerted on that same mass at the distance of the Moon is F mg g 9:79 1 D 1 D 1 D ' 3626: F2 mg2 g2 0:0027
3.6.2 Weight and mass
Weight and mass are not the same thing. Weight is the gravitational force exerted on a mass, Weight D Force D mg: Its SI units are Newtons (N). In the English system the unit of weight is the pound (lb), with the conversion 1 lb = 4.448 N.
The mass of an object is constant but its weight depends on its location (because it depends on the local gravitational accelera- tion).
3.6.3 Gravitational potential energy
Energy is conserved in physical processes. The total energy of an object is generally a sum of a kinetic energy (energy of motion) and a potential energy. If an object is in a gravitational field, its gravitational potential energy can change if it changes its location.
17 Consider a mass m that moves from a position ri to rf in a gravitational field that is gener- ated by a mass M at the origin of the coordinate system:
z dr
F m
rf
ri
M y
x
The gravitational force F exerted on m is directed toward the origin and the definition of the change in potential energy is Z rf U D Uf Ui D F dr: ri Let’s take a simple case of a mass m moving vertically along the z axis.
z
m ri dr F
rf
M y
x
Then the scalar product is easy since the position vectors and the force vector are pointed
18 in opposite directions ( D ) and Z Z Z rf rf rf U D F dr D F cos./dr D Fdr ri ri ri Z r ˇr f Mm GM m ˇ f D G dr D ˇ 2 ˇ r r r r i  à i 1 1 D GM m : : rf ri This is the change in the gravitational potential energy. We make three general remarks about it
1. A more general derivation would have shown that the result is independent of path, depending only on the endpoints ri and rf. 2. Generally, only changes in the potential are relevant and we can define an arbitrary zero for the gravitational potential energy scale. It is conventional to choose
U ! 0 as ri !1:
Then 1=ri ! 0 and (dropping subscripts) we may write for the gravitational poten- tial GM m U : (10) r Conventional: we could choose any zero for the scale if we wished.
3. The magnitude of the gravitational force is obtained from the derivative of the gravi- tational potential, Â Ã dU d GM m GM m F D D D ; grav dr dr r r2
where the minus sign indicates that it is attractive.
3.6.4 Escape velocity
The escape velocity from a gravitational field is a useful concept:
19 Escape Velocity The initial vertical component of velocity from a given location that gives v ! 0 as r !1.
We may derive a formula for escape velocity simply by using conservation of energy. The total energy of some mass m moving in a gravitational field is 1 Mm E D mv2 G D E C E : 2 r kinetic potential As r !1, by definition v ! 0, so at infinity,
Ekinetic ! 0Epotential !1 E ! 0:
Thus, at infinity the total energy is zero. But since energy is conserved, the total energy must be zero at any r. Setting E D 0 in the preceding equation gives Mm 1 mv2 G D 0; 2 r which may be solved for v to give r 2GM v D : (11) esc r
Notice that
The mass of the object m has cancelled out. The escape velocity depends only on the properties of the gravitational field, not on the mass of the object that is escaping.
The escape velocity depends on where we start from (r in the preceding formula). The escape velocity from the surface of the Earth is greater than the escape velocity from an orbit 200 km above the surface of the Earth, for example.
20 Example: Escape velocity from Earth’s surface r r 2GM 2GM r p v D D D 2gr esc 2 q r r D 2.9:8 ms 2/.6:38 106 m/ D 11; 182 ms 1 D 11:2 km s 1:
Example: Escape velocity from Jupiter’s surface r 2GM vesc D s r 2.6:673 10 11 1 3 2 /.1:9 1027 / D kg m s kg 7:149 107 m D 59; 556 ms 1 D 59:6 km s 1:
Example: Escape velocity from Sun’s surface r 2GM vesc D s r 2.6:673 10 11 1 3 2 /.1:99 1030 / D kg m s kg 6:96 108 m D 617; 728 ms 1 D 618 km s 1:
Example: Escape velocity from surface of Phobos
The Martian moon Phobos is not spherical but its average radius is about 11 km. Using this and its mass of 1:08 1016 kg, we may estimate that r 2GM vesc D s r 2.6:673 10 11 1 3 2 /.1:08 1016 / D kg m s kg 11 103 m D 11:4 ms 1:
21 That’s not very much. Could a basketball player with Michael Jordan leaping ability attain escape velocity on Phobos just by jumping straight up?
Suppose a basketball player can leap vertically by 40 inches (about one meter) on Earth. By energy conservation again we have
1 mv2 C mgy D 1 mv2 C mgy ; 2 1 1 2 2 2 where quantities on the left side refer to the player on the floor, and the quantities on the right side to the player at the top of his leap. Dividing through by m and rearranging,
2 2 v1 v2 D 2g.y2 y1/:
But v2 is at the top of the leap so it is equal to zero, and y2 y1 D y is just the vertical leap of 1 meter. Therefore, solving for the initial velocity v1, p q 2 1 v1 D 2gy D 2.9:8 ms /.1 m/ D 4:4 ms :
So Michael Jordan could not launch into orbit by jumping from Phobos, but he wouldn’t miss it by very far!
For further reference, a world-class sprinter can attain a speed of a little over 10 m s 1, UT softball pitcher Monica Abbot’s 70 mph fastball corresponds to about 31 m s 1, and a hard kick in a world cup football (soccer) match, or the serve of a top tennis player, can reach initial speeds in the vicinity of 50–60 m s 1. (Convenient conversion: 1 mph is 0.447 m s 1.)
3.6.5 Center of mass reference frame
Consider a collection of masses mi at position coordinates ri ,
22 z
m2 m1
r1 r2
r3 y
m3 x
Define a position vector R that is the weighted average Xn mi ri R D iD1 : Xn mi iD1
Xn The position R is termed the center of mass. The total mass is M D mi so iD1 Xn M R D mi ri : iD1
Assume the masses to be constant and differentiate dR Xn dr M D m i dt i dt iD1 which is equivalent to Xn M V D mi vi iD1 and also to Xn P D pi ; iD1
23 where V in the CM velocity and P is the CM momentum. Thus, the system behaves as if all mass were concentrated at the CM R, moving with the CM velocity V and CM momentum P.
Differentiate with respect to t, dP Xn dp D i : dt dt iD1
But if no external forces act on the masses (all forces are internal between the masses) the total force must be zero, by Newton’s third law applied to any two interacting pairs (equal and opposite forces for each pair). Therefore dP d 2R F D D M D 0: dt dt2
The center of mass does not accelerate if there are no external forces acting on the system of masses.
This implies that we may simplify the manybody problem by choosing a coordinate system for which R D 0 V D 0: This is called the CM frame. It is an inertial frame (one in which Newton’s first law is valid).
3.6.6 Center of mass for a binary system
Consider the important special case of two masses (binary system):
24 z M r m2
m1
R r2 r1
y
x
r1 C r D r2 ! r D r2 r1:
Then choosing the center of mass as the origin, m r C m r R D 1 1 2 2 D 0 m1 C m2 Therefore, m1r1 C m2r2 D 0; and since r2 D r1 C r,
m1r1 C m2.r1 C r/ D 0
m1r1 C m2r1 D m2 r
m2 m2 r1 D r D r: m1 C m2 M
By a similar proof, m1 m1 r2 D r D r: m1 C m2 M
Introducing the reduced mass m m 1 2 ; m1 C m2 we can write m2 m2m1 r1 D r D r m1 C m2 m1.m1 C m2/ D r: m1
25 By a similar proof, r2 D r: m2
Total energy: The utility of the CM system can be seen in writing the total energy of the binary system, m m E D 1 m jv j2 C 1 m jv j2 G 1 2 : 2 1 1 2 2 2 jr2 r1j Substituting and rearranging, M E D 1 v2 G ; 2 r where r DjrjDjr 2 r1j and v Djvj, with dr d v D D .r r / D v v : dt dt 2 1 2 1 The total energy is now the sum of the kinetic energy of the reduced mass and the potential energy of the reduced mass moving about the total mass M D m1 C m2 at the origin.
Orbital angular momentum: The orbital angular momentum for the binary is
L D r1 p1 C r2 p2
D m1r1 v1 C m2r2 v2:
Substituting gives  à  à L D m1 r v1 C m2 r v2 m1 m2
D r v1 C r v2
D r .v2 v1/ D r v; so the total L is the angular momentum of the reduced mass only.
26 The binary problem of calculating the motion of two bodies has been replaced by the calculation of the motion of a single effec- tive mass (the reduced mass ) about a stationary point contain- ing the total mass M of the system, with the separation between M and given by the separation between m1 and m2.
Binary center of mass: Choosing the CM as the origin for the binary,
r1 r2
m1 CM m2
Take the line of centers as the x axis. From the preceding equations, r .=m /r m rx 1 D 1 D 2 D 1 : x r2 .=m2/r m1 r2 But from the diagram x x r1 D jr1jD r1 r2 Djr2jDr2 Substituting these gives the seesaw equation,
r1m1 D r2m2 where r1 C r2 D r D distance between masses:
Two special cases are of interest.
Suppose that m1 D m2. Then r m D r m r D r D 1 r; 1 1 2 2 2 1 2 and the CM lies halfway between the masses.
27 Suppose one mass much larger than the other, m1 >> m2. Then r m 1 D 2 ' 0: r2 m1 Therefore, the CM almost coincides with the center of the large mass.
Example: Center of mass for the Earth–Sun system
We have
30 24 11 Mˇ D 1:99 10 kg M˚ D 5:97 10 kg r D 1:496 10 m and we must solve simultaneously
r2 M˚ D r1 D r r2: r1 Mˇ
Substituting the right equation into the left, r M 2 D ˚ ; r r2 Mˇ which may be solved for r2 to give
M˚=Mˇ r2 D r: 1 M˚=Mˇ
For the Earth–Sun system,
24 M˚ 5:97 10 kg D D 3 10 6: 30 Mˇ 1:99 10 kg Neglecting this term in the denominator,
M˚ 6 11 5 r2 ' r D .3 10 /.1:496 10 m/ ' 4:5 10 m: Mˇ
8 For reference, Rˇ ' 7 10 m, so the CM of the Earth–Sun system is well inside the Sun.
Elliptical Kepler motion in the CM system for a binary star:
28 Orbit of Star 1
Center of Mass Star 1 r1 r2
Star 2 Center of Orbit of Mass Star 2
(See the Java applet OJTA 4.25 for binary motion.)
Actual example of ellipical motion for the Sirius B system:
1990 5 arcsec
Sirius A
Center of Mass
Sirius B
3.7 Kepler’s Laws from Newton’s Law of Gravitation
We now outline how Kepler’s laws follow from Newton’s law of gravitation
29 3.7.1 Kepler’s first law
By considering the effect of gravity on the orbit of a planet in the CM sysem (see Ostlie and Carrol, pp. 43–45), we can prove L2=2 r D : (12) GM.1 C cos / But this is the equation of a conic section, which corresponds to equations for parabolas, ellipses, and hyperbolas:
L2=2 r D GM.1 C / 8 cos  à ˆ 2p L2=2 ˆ parabola;D 1; 2p D ˆ 1 C cos GM ˆ <ˆ  à a.1 2/ L2=2 D ellipse; < 1; a.1 2/ D (13) ˆ 1 C cos GM ˆ  à ˆ 2 2 2 ˆ a. 1/ 2 L = : hyperbola; > 1; a. 1/ D 1 C cos GM
The geometrical definition of a conic section is summarized in the following figure
Conic section g/h = constant P h ε = g/h = eccentricity g Cases: Parabola (ε = 1) Ellipse (ε < 1) Focus F Hyperbola (ε > 1)
Directrix d
The three general types of conic sections are parabolas, ellipses, and hyperbolas (the cir- cle is a special case of an ellipse). The geometrical properties of parabolas, ellipses, and hyperbolas are summarized in the following figure.
30 F θ p p Parabola
d
b r F’ F θ Ellipse aε a ε (1-ε) a(1-ε) b
a d’ d
asymptot e
e asymptot
θ F F’ Hyperbola
a(ε -1)
d d’ a/ε a
So “orbits of the planets are ellipses” generalizes to “orbits in gravitational fields are conic sections”, with the ellipse as a special case for a bound orbit.
Examples of conic-section gravitational orbits are shown in the following figure,
31 20
Ellipse Parabola e = 0.5 e = 1 a = 10 p = 5
10 Circle e = 0 Hyperbola a = 5 e = 1.3 a = 16.7
-20 20
Conic orbits with same focus and -10 same vertex which shows conic section orbits having the same focus and the same vertex (distance of closest approach to the focus).
3.7.2 Kepler’s second law
Consider the following ellipse:
v r θ
rdθ
dθ dr
32 The differential area of the shaded strip is rd, so the differential area swept out by the angle d is Z r ˇ dA D d rdr D d.1 r2/ˇr D 1 r2d 2 0 2 0 and the rate of change of the area swept out in a time dt is dA d D 1 r2 : dt 2 dt
From the diagram, the velocity vector v can be resolved into components vr along the radial O drawn from the focus and v perpendicular to the radial. In terms of unit vectors rO and in these directions, dr d v D v rO C v O D rO C r O r  dt dt
Therefore, v D rd=dt and substituting d=dt D v =r into the earlier equation gives dA v D 1 r2  D 1 rv : dt 2 r 2 Â
O ı Since rO and are orthogonal, jrvjDrv sin.90 / D rv and 1 rv Djr vjD jr .v/ j  „ƒ‚… p 1 L D j r p jD ; „ƒ‚… L where L is the magnitude of the orbital angular momentum. Therefore, the change in area is dA 1 L D rv D D constant dt 2 2 since angular momentum and the reduced mass are conserved. But constant dA=dt is just Kepler’s 2nd law:
The line joining the planet to the focus sweeps out equal areas in equal times.
33 3.7.3 Kepler’s third law
From the results for Kepler’s 2nd law, the total area of the ellipse is I I Â Ã dA A D dA D dt dt Zorbit orbit Z ˇ P P ˇP L L Lt ˇ D dt D dt D ˇ 0 2 2 0 2 0 L D P; 2 where P is the period for one orbit. But we also have from geometry that the area of an ellipse is A D ab [see Eq. (2)], so L ab D P; 2 which we can square and solve for P 2 to give. Â Ã 2 2 422a2b2 422a2Œa2.1 2/ P 2 D ab D D ; (14) L L2 L2 where in the last step we have used Eq. (1) for ellipses:
b2 D a2.1 2/: (15)
But we can also equate Eqs. (2) and (12) a.1 2/ L2=2 r D r D 1 C cos GM.1 cos / for ellipses and solve for the angular momentum L to give p L D GM a.1 2/
Inserting this into Eq. (14) for P 2 then gives 422a2Œa2.1 2/ P 2 D 2GM a.1 2/ Â Ã 42 D a3: GM
34 Thus, inserting explicitly that M D m1Cm2, we arrive at the most general form of Kepler’s 3rd law, 42 P 2 D a3: (16) G.m1 C m2/ This expression is valid for any appropriate units and can be written in the form 42 P 2 D ka3 k : (17) G.m1 C m2/
Example: Mass of the Earth determined from the Moon’s orbit
For the Moon, the period and semimajor axis of the orbit around Earth are
P D 27:322 d D 2:3606 106 s a D 3:844 108 m:
Then from Kepler’s 3rd law in the general form (17), 42 a3 M D m C m D 1 2 G P 2 Evaluating the constants 42 42 D D 5:916 1011 kg m 3s2 G 6:673 10 11 kg 1m3s 2 so in these units Á Â Ã2 a 3 seconds M D 5:9161 1011 meters P
Neglecting the mass of the Moon relative to that of the Earth, M D M˚ C MMoon ' M˚, Â Ã Á3 2 11 a seconds M˚ D 5:9161 10 meters P .3:844 108/3 D 5:9161 1011 .2:3606 106/2 D 6:03 1024 kg:
If we subtract from this the mass 7:349 1022 kg of the Moon, we obtain 5:97 1024 kg, which is almost exactly the mass of the Earth.
35 Let’s now demonstrate that we can choose a particular set of units such that numerically k D 1, so that we recover Kepler’s 3rd law in the form originally proposed by Kepler, if we neglect the mass of the planet relative to the mass of the Sun.
We adopt the following units for time, distance and mass:
Œtime D Earth years (yr) Œdistance D AU Œmass D Mˇ where we use the general notation [X] to denote the units of X. Converting the units of the gravitational constant we can write
2 2 4 4 2 D D 5:916 1011 kg m 3s G 6:673 10 11 1 3 2 kg m s à 2 1Mˇ D 5:916 1011 kg m 3s 1:989 1030 kg  à  à 1:496 1011 3 1 2 m yr 1 AU 3:156 107 s 2 3 D 1Mˇ yr AU 42 Therefore, employing these units for the factor G , Kepler’s 3rd law can be expressed as  à Á3 1Mˇ a P 2 D yr2; m1 C m2 AU and if the masses are measured in solar masses and the semimajor axis a in AU, the units of P will be years.
Finally, if we take M D mˇ C mplanet ' mˇ D 1Mˇ; the mass factor is just unity and we obtain
P 2 D a3 (Kepler’s 3rd Law) where a is in AU and P is in Earth years.
If we want to be explicit about the units that are required for Kepler’s 3rd law in this form, we could write  à a3 P 2 D yr2: AU3 for Kepler’s 3rd law.
36