Chapter 17 Appendices
The appendices presented here involve more mathematical sophistication than is
commonly presented in freshman-level subjects. Analytic consideration of the equations
resulting from Newton’s Second Law and Newton’s Law of Universal Gravitation are
presented in these notes in a manner that should be understood and appreciated by
students who have a more than casual interest in the subject.
The topics of the appendices are:
17.A: Derivation of the Orbit Equation
Two presentations of the result of Equation (17.3.11)
17.B: Properties of Elliptical Orbits
The dynamical properties of objects in elliptical orbits
17.C: Analytic Geometric Properties of Ellipses
Demonstrating how the results of Appendices 17.A and 17.B are consistent with
more familiar representations of ellipses
17.D: A Proof that Bounded Keplerian Orbits are Ellipses
Including a parameterization of the orbit radius as a function of time, and a
demonstration of the consistency with Newton’s Law of Universal Gravitation
17.E: Orbit Radius as a Function of Time by Direct Integration
An Extension of Appendix 17.D, starting from Equation (17.3.8)
(Warning! Serious but very useful calculus techniques)
17.F: Even More on Kepler Orbits
Using geometry, vector algebra, but minimal calculus to find the orbit equation,
introducing the Laplace-Runge-Lenz vector.
Appendix 17.A: Derivation of the Orbit Equation
Start from Equation (17.3.10) in the form
(1/ r )
L
dr .
dθ =
1/ 2
2μ ⎛
L2
G m1 m2 ⎞
+
⎜E−
⎟
r
2μ r 2
⎝
⎠
2
(17.A.1)
What follows involves a good deal of hindsight, allowing selection of convenient
substitutions in the math in order to get a clean result. First, note the many factors of the
reciprocal of r . Multiplying numerator and denominator by a factor of r , while well10/15/2009 17App - 1
intended, leaves a factor of 1/ r in the numerator. So, we’ll try the substitution u = 1/ r ,
du = −(1/ r 2 ) dr , with the result
dθ = −
L
du
.
1/ 2
2
2μ ⎛
⎞
L 2
u + G m1 m2 u ⎟
⎜E −
2μ
⎝
⎠
(17.A.2)
Experience in evaluating integrals suggests that we make the absolute value of the factor
multiplying u 2 inside the square root equal to unity. That is, multiplying numerator and
denominator by 2 μ / L ,
dθ = −
( 2μ E / L − u
2
du
+ 2 ( μ G m1 m2 / L2 ) u
2
)
1/ 2
.
(17.A.3)
As both a check and a motivation for the next steps, note that the left side (dθ ) of
Equation (17.A.3) is dimensionless, and so the right side must be. This means that the
factor of μ G m1 m2 / L2 in the square root must have the same dimensions as u , or
length −1 ; so, define r0 ≡ L2 / μ G m1 m2 . This is of course the semilatus rectum as defined
in Equation (17.3.12), and it’s no coincidence; this is part of the “hindsight” mentioned
above. The differential equation then becomes
dθ = −
=−
du
( 2μ E / L − u
2
2
+ 2u / r0 )
1/ 2
( 2μ E / L + 1/ r
2
du
2
0
=−
1/ 2
du
( 2μ E / L + 1/ r
2
0
=−
− u 2 + 2u / r0 − 1/ r0 2 )
2
(17.A.4)
− (u − 1/ r0 )2 )
1/ 2
r0 du
( 2μ Er02 / L2 + 1 − (r0 u − 1)2 )
1/ 2
.
Next, we note that the combination of terms 2 μ Er0 2 / L2 + 1 is dimensionless, and is in
fact equal to the square of the eccentricity ε as defined in Equation (17.3.13); more
hindsight. The last expression in (17.A.4) is then
dθ = −
(ε
r0 du
2
− (r0 u − 1) 2 )
1/ 2
.
(17.A.5)
From here, we’ll combine a few calculus steps, going immediately to the substitution
10/15/2009 17App - 2
r0 u − 1 = ε cos α , r0 du = −ε sin α dα , with the final integral as
dθ = −
−ε sin α dα
(ε 2 − ε 2 cos2 α )
1/ 2
= dα ,
(17.A.6)
or, θ = α + constant . We have a choice in selecting the constant, and if we pick
θ = α − π , α = θ + π , cosα = − cosθ , the result is
r=
1
r0
=
u 1 − ε cosθ
(17.A.7)
which is Equation (17.3.11).
Note that if we chose the constant of integration to be zero, the result would be
r=
1
r0
=
u 1 + ε cosθ
(17.A.8)
which is the same trajectory reflected about the “vertical” axis in Figure 17.3, indeed the
same as rotating by π .
The derivation of Equation (17.A.7) in the form
u=
1
(1 − ε cosθ )
r0
(17.A.9)
suggests that the equation of motion for the reduced one-body problem might be
manipulated to obtain a simple differential equation. That is, start from
G
G
F = μa
2
⎛ d 2r
m1m2
dθ ⎞ ⎞
⎛
−G 2 rˆ = μ ⎜ 2 − r ⎜
⎟ ⎟ rˆ.
⎜ dt
r
⎝ dt ⎠ ⎟⎠
⎝
(17.A.10)
Setting the components equal, using the constant of motion L = μ r 2 ( dθ / dt ) and
rearranging,
d 2r
L2 Gm1m2
μ 2 = 3−
.
μr
dt
r2
(17.A.11)
What we will do is use the same substitution u = 1/ r and change the independent
variable from t to r , using the chain rule twice, since Equation (17.A.11) is a secondorder equation. That is, the first time derivative is
10/15/2009 17App - 3
dr dr du dr du dθ
=
=
.
dt du dt du dθ dt
(17.A.12)
From r = 1/ u we have dr / du = −1/ u 2 . Combining with the angular velocity dθ / dt in
terms of L and u , dθ / dt = L u 2 / μ , Equation (17.A.12) becomes
dr
1 du L u 2
du L
=− 2
=−
,
dt
u dθ μ
dθ μ
(17.A.13)
a very tidy result, with the variable u appearing linearly. Taking the second derivative
with respect to t ,
d 2 r d ⎛ dr ⎞ d ⎛ dr ⎞ dθ
= ⎜ ⎟=
⎜ ⎟
dt 2 dt ⎝ dt ⎠ dθ ⎝ dt ⎠ dt
d 2u L ⎛ L ⎞
= − 2 ⎜ u2 ⎟
dθ μ ⎝ μ ⎠
=−
(17.A.14)
d 2u 2 L2
.
u
dθ 2 μ 2
Substituting into Equation (17.A.11), and using r = 1/ u ,
−
d 2u 2 L2 L2 3
u
= u − Gm1m2 u 2 .
2
dθ
μ μ
(17.A.15)
Canceling the common factor of u 2 and rearranging,
−
d 2u
μ Gm1m2
=u−
.
2
dθ
L2
(17.A.16)
Equation (17.A.16) is mathematically equivalent to the Harmonic Oscillator Equation
with a constant term. The solution consists of two parts: the angle-independent solution
u0 =
μ Gm1m2
L2
(17.A.17)
and a sinusoidally varying term of the form
uH = A cos (θ − θ 0 ) ,
(17.A.18)
where A and θ 0 are constants determined by the form of the orbit. The expression in
Equation (17.A.17) is the inhomogeneous solution and represents a circular orbit. The
10/15/2009 17App - 4
expression in Equation (17.A.18) is the homogeneous solution (as hinted by the subscript)
and must have two independent constants. We can readily identify 1/ u0 as the semilatus
rectum r0 , with the result that
u = u0 + u H =
1
(1 + r0 A ( cosθ − θ 0 ) )
r0
r0
.
r=
1 + r0 A ( cosθ − θ 0 )
(17.A.19)
Choosing the product r0 A to be the eccentricity ε and θ 0 = π (much as was done leading
to Equation (17.A.7) above), Equation (17.A.7) is reproduced.
10/15/2009 17App - 5
Appendix 17.B: Properties of an Elliptical Orbit
We now consider the special case of an elliptical orbit.
Figure 17.B.1 Ellipse.
In Figure 17.B.1, let a denote the semimajor axis, b denote the semiminor axis and x0
denote the distance from the center of the ellipse to the origin of our coordinate system
( r ,θ ) . We shall now express the parameters a , b and x0 in terms of the constants of the
motion L , E , μ , m1 and m2 .
Appendix 17.B.1: The semimajor axis:
See Equation (17.A.7) above. The major axis A = 2a is given by
A = 2 a = rmax + rmin
(17.B.1.1)
where the distance of furthest approach occurs when θ = 0 , hence
rmax = r (θ = 0 ) =
r0
1− ε
(17.B.1.2)
and the distance of nearest approach occurs when θ = π , hence
rmin = r (θ = π ) =
r0
.
1+ ε
(17.B.1.3)
10/15/2009 17App - 6
Figure 17.B.2: nearest and furthest approach
Thus
1⎛ r
r ⎞
r
a= ⎜ 0 + 0 ⎟= 0 2 .
2 ⎝ 1− ε 1+ ε ⎠ 1− ε
(17.B.1.4)
The semilatus rectum r0 can be re-expressed in terms of the semimajor axis and the
eccentricity,
r0 = a (1 − ε 2 ) .
(17.B.1.5)
We can now express the distance of nearest approach, Equation (17.B.1.3), in terms of
the semimajor axis and the eccentricity,
rmin
a (1 − ε 2 )
r0
=
=
= a (1 − ε ) .
1+ ε
1+ ε
(17.B.1.6)
In a similar fashion the distance of furthest approach is
rmax =
a (1 − ε 2 )
r0
=
= a (1 + ε ) .
1− ε
1− ε
(17.B.1.7)
Figure 17.B.2 shows the distances of nearest and furthest approach. Using our results for
r0 and ε from Equations (17.3.12) and (17.3.13), we have for the semimajor axis
a=
L2
1
2
μ Gm1 m2 1 − 1 + 2 E L / μ ( G m1 m2 )2
=−
( (
)) .
(17.B.1.8)
G m1 m2
2E
and so the energy is determined by the semimajor axis,
10/15/2009 17App - 7
E=−
G m1 m2
.
2a
(17.B.1.9)
The angular momentum is related to the semilatus rectum r0 by Equation (17.3.12).
Using Equation (17.B.1.5) for r0 , we can express the angular momentum (17.B.1.3) in
terms of the semimajor axis and the eccentricity,
L = μ Gm1 m2 r0 = μ Gm1 m2 a (1 − ε 2 ) .
(17.B.1.10)
Note that
(1 − ε ) =
L
,
μ Gm1 m2 a
2
(17.B.1.11)
a result we will return to later.
Appendix 17.B.2: The location x0 of the center of the ellipse:
From Figure 17.B.1, the distance from a focal point to the center of the ellipse is
x0 = rmax − a .
(17.B.2.1)
Using Equation (17.B.1.7) for rmax , we have that
x0 = a (1 + ε ) − ε = ε a.
(17.B.2.2)
Thus, from Equations (17.3.12), (17.B.2.2) and (17.B.1.8),
x0 = ε a = −
G m1 m2
2E
(1 + 2E L / μ (G m m ) ) .
2
2
1
2
(17.B.2.3)
Appendix 17.B.3: The semi-minor axis:
From Figure 17.B.1,
b=
(r
b
2
− x0 2 )
(17.B.3.1)
where
rb =
r0
,
1 − ε cosθ b
(17.B.3.2)
10/15/2009 17App - 8
which can be rewritten as
rb − rb ε cosθ b = r0 .
(17.B.3.3)
x0 = rb cosθ b ,
(17.B.3.4)
rb = r0 + ε x0 .
(17.B.3.5)
Note that from Figure 17.B.1,
so that
Substituting Equation (17.B.2.2) for
Equation (17.B.3.5) yields
x0
and Equation (17.B.1.5) for r0
rb = a (1 − ε 2 ) + aε 2 = a .
into
(17.B.3.6)
The fact that rb = a is a well-known property of an ellipse reflected in the geometric
construction, that the sum of the distances from the two foci to any point on the ellipse is
a constant. Thus the semi-minor axis b becomes
b=
(r
b
2
− x0 2 ) = a 2 − ε 2 a 2 = a 1 − ε 2 .
(17.B.3.7)
Using Equation (17.B.1.11) for 1 − ε 2 , we have for the semi-minor axis
b = aL2 / μ Gm1 m2 .
(17.B.3.8)
We can now use Equation (17.B.1.8) for a in the above expression, yielding
b = aL2 / μ Gm1 m2 = L −
1
G m1 m2
(17.B.3.9)
/ μ Gm1 m2 = L −
2E
2μ E
Appendix 17.B.4: Speeds at nearest and furthest distances:
At nearest approach the velocity vector is tangent to the orbit, so the angular
momentum is
L = μ rmin v p
(17.B.4.1)
and the speed at nearest approach is
10/15/2009 17App - 9
v p = L / μ rmin .
(17.B.4.2)
Figure 17.B.3 Speeds at nearest and furthest approach
Using Equation (17.B.1.10) for the angular momentum and Equation (17.B.1.6) for rmin ,
Equation (17.B.4.2) becomes
L
vp =
=
μ rmin
μ Gm1 m2 (1 − ε 2 )
μ a (1 − ε )
Gm1 m2 (1 − ε 2 )
=
μ a (1 − ε )
2
Gm1 m2 (1 + ε )
.(17.B.4.3)
μ a (1 − ε )
=
A similar calculation show that the speed va at furthest approach,
L
va =
=
μ rmax
μ Gm1 m2 (1 − ε 2 )
μ a (1 + ε )
Gm1 m2 (1 − ε 2 )
=
μ a (1 + ε )
2
=
Gm1 m2 (1 − ε )
.(17.B.4.4)
μ a (1 + ε )
Appendix 17.C: Analytic Geometric Properties of Ellipses
Consider Equation 17.3.18, and for now take ε < 1 , so that the equation is that of an
ellipse. It takes some, but not a great deal, of algebra to put this into the more
familiar form
( x − x0 )
a2
2
+
y2
=1
b2
(17.C.1)
where the ellipse has axes parallel to the x and y coordinate axes, center at ( x0 , 0 ) ,
semimajor axis a and semiminor axis b .
We rewrite Equation 17.3.18 as
2ε r0
y2
r02
.
=
x −
x+
1− ε 2
1− ε 2 1− ε 2
2
(17.C.2)
We next complete the square,
10/15/2009 17App - 10
x2 −
2ε r0
ε 2r02
ε 2r02
y2
r02
+
+
=
+
x
2
2
2
2
1− ε 2
(1 − ε 2 ) 1 − ε 1 − ε (1 − ε 2 )
ε r0 ⎞
y2
r02
⎛
−
+
=
x
⎜
2
2 ⎟
2
⎝ 1− ε ⎠ 1− ε
(1 − ε 2 )
2
ε r0 ⎞
⎛
⎜x−
2 ⎟
y2
⎝ 1− ε ⎠ +
2
r0 / (1 − ε 2 )
r0 / 1 − ε 2
(17.C.3)
2
) (
(
)
2
= 1.
The last expression in (17.C.3) is the equation of an ellipse with semimajor axis
a=
r0
,
1− ε 2
(17.C.4)
semiminor axis
b=
(
r0
1− ε
2
= a 1− ε 2
(17.C.5)
)
and center at ε r0 / (1 − ε 2 ) , 0 = ( ε a, 0 ) , as found in Equation (17.B.2.2).
Appendix 17.D: A Proof (in Cartesian Coordinates) that
Bounded Keplerian Orbits are Ellipses1
Introduction: The Question:
Kepler’s First Law, which states that the orbits of planets are ellipses with the sun at one
of the foci, is usually presented in first-year subjects in Newtonian Mechanics, but is
rarely proved. There are many alternatives to presenting a proof, but they often require
mathematics that first-year students find unfamiliar – usually calculus in polar
coordinates, as is done in the Course Notes, Sections 17.2 and 17.3. Here we set out to
demonstrate that an elliptical orbit for which equal areas are swept out in equal times
(which is quantified in Kepler’s Second Law) does indeed satisfy Newton’s inversesquare law of Universal Gravitation.
First we introduce a perhaps unfamiliar parameterization of the elliptical orbit, showing
that this parameterization is equivalent to the more familiar Cartesian representation. We
1
This Appendix is adapted from material provided by Prof. Paul Schechter of MIT.
10/15/2009 17App - 11
then locate the foci of the ellipse, introducing the concept of eccentricity as a
consequence of the parameterization.
Next we use the equal-areas law (Kepler’s Second Law) to derive an expression for time
as a function of the introduced parameter (the eccentric anomaly).
With both position and time expressed as functions of this parameter, we then use basic
calculus to take second time derivatives of the position and show that the acceleration is
that given by Newton’s Second Law and Newton’s law of Universal Gravitation.
Appendix 17.D.1: Step 1: A Parametric Representation of the Ellipse
Most students’ first mathematical encounter with the ellipse is in the Cartesian form
x2 y 2
+
= 1,
a 2 b2
(17.D.1.1)
representing an ellipse with semimajor axis a and semiminor axis b , and center at the
origin of the Cartesian coordinate system.
If we define the parameter η such that
⎛ ay ⎞
⎟,
⎝ bx ⎠
η = arctan ⎜
(17.D.1.2)
we can infer the relations
x = a cosη , y = b sinη ,
(17.D.1.3)
which are seen to satisfy Equation (17.D.1.1), the Cartesian equation for an ellipse. The
representation in (17.D.1.3) has a straightforward geometric interpretation. The ellipse
(shown as a solid curve in Figure 17.D.1) may be thought of as the projection of a circle
(shown as the dashed curve), with Cartesian equation
x′2 y′2
+
=1.
a2 a2
(17.D.1.4)
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Figure 17.D.1 - Properties of an ellipse
Introducing new variables y = (b / a) y′ and x = x′ gives the projected ellipse, as given in
Equation (17.D.1.1). The magnitude of the y -coordinate of every point P on the ellipse
is smaller by the factor b / a < 1 than the y′ -coordinate of the corresponding point P′ on
the circumscribed circle. In the Figure 17.D.2 below we show the parameter η as an arc
A′P , but it is also the angle made by OP′ and OA .
Appendix 17.D.2: Step 2: Where is the Focus?
While we’re at it, where’s the other one?
Our approach will be to differentiate the position twice with respect to time to
demonstrate that the acceleration is given by Newton’s Universal Law of Gravitation.
We know both x and y as functions of η , as in Equations (17.D.1.3), but not as
functions of time. A direct consequence of the conservation of angular momentum is
Kepler’s Second Law, which states that equal areas are swept out in equal times by the
vector denoting the position of the planet with respect to the sun. But the sun sits not at
the center O of the ellipse but at S , one of the foci.
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Figure 17D.2 – Parameters of an Ellipse
The foci are the two points lying on the major axis such that the sum of the lengths of the
two vectors drawn from the foci to any point on the ellipse is the same. The two foci are
shown as S and S ′ in Figure 17D.2. The sum of the distances SA and S ′A is equal to
2a , twice the length of the semimajor axis. The foci are both a distance aε from the
center of the ellipse. The dimensionless number ε can range from zero to unity, and is
called the eccentricity, or “off-centeredness” of the ellipse.
Consider the point B where the semiminor axis intercepts the ellipse. The segment OB
has length b . The segment S ′B has length a 2ε 2 + b 2 . But twice the length of S ′B
must be equal to 2a , allowing us to solve for the eccentricity ε in terms of the lengths of
the semimajor and semiminor axes a and b ,
ε = 1−
a2
.
b2
(17.D.2.1)
10/15/2009 17App - 14
Appendix 17.D.3: Step 3: Parameterizing the Time
In what follows, we take the sun to be at focus S . If we took the focus to be at point S ′ ,
as was done in Appendix 17.B, we would need to replace ε by −ε , but the results would
be unchanged.
Recalling Kepler’s Second Law, the area swept out by the vector from the focus to the
points on the ellipse is proportional to time. The total area of the ellipse is π ab , a factor
b / a less than the area π a 2 of the circumscribed circle because all the y -coordinates of
the points on the ellipse are smaller than the corresponding y′ coordinates on the circle
by the same factor b / a .
As the planet moves from point A to point P , in Figure 17.D.2, the position vector
sweeps out the area bounded by the segments SA and SP , and the arc p
AP . The time t
the planet takes to go from A to P can be expressed as a fraction of the orbital
period T ,
t area ( SAP )
=
.
π ab
T
(17.D.3.1)
The area swept out in time t can also be expressed as the difference between two areas,
area ( SAP ) = area ( OAP ) − area ( OSP ) .
(17.D.3.2)
The triangle formed by the points O , S and P has base aε and height b sinη , the y coordinate of point P , and hence area ( aε )( b sin η ) / 2 . The oddly-shaped region
AP has an area smaller by a factor
bounded by the segments OA and OP , and the arc p
b / a than the area of the corresponding sector of the circumscribed circle bounded by the
AP′ . The area of that circular sector is just η a 2 / 2 .
segments OA and OP′ , and the arc q
Combining this with Equations (17.D.3.1) and (17.D.3.2), we have
t η ab / 2 − ε sinη ab / 2 η − ε sin η
=
=
.
T
π ab
2π
(17.D.3.3)
Appendix 17.D.4: Step 4: The Position Vector with Respect to
the Focus
G
We will obtain the acceleration a by differentiating the x - and y -coordinates of the
planet with respect to time t , and show that this is consistent with Newton’s Law of
Universal Gravitation,
GM G
G
a=− 3: r
r
(17.D.4.1)
10/15/2009 17App - 15
where M : is conventional notation for the solar mass. In Equation (17.D.4.1), note the
“extra” power of r in the denominator required give an overall dependence on 1/ r 2 ,
G
since we use the vector r instead of r̂ to determine direction. This will soon be seen to
be quite convenient for the purpose of taking derivatives.
G
The vector r from the sun to the planet is given by
G
r = ( a cosη − aε ) ˆi + b sinη ˆj .
(17.D.4.2)
The first term a cosη in the x -component represents the position of the planet with
respect to the center O of the ellipse, while the second term − aε accounts for the offset
of the focus with respect to the center of the ellipse.
G
The length r of the vector r is obtained by squaring and adding the components,
r 2 = a 2 ( cosη − ε ) + b 2 sin 2 η
2
= a 2 ( cos 2 η − 2ε cosη + ε 2 ) + a 2 (1 − ε 2 ) sin 2 η
= a 2 ⎡⎣cos 2 η + sin 2 η − 2ε cosη + ε 2 − ε 2 (1 − cos 2 η ) ⎤⎦
= a 2 ⎡⎣1 − 2ε cosη + ε 2 cos 2 η ⎤⎦
= ⎡⎣ a (1 − ε cosη ) ⎤⎦
(17.D.4.3)
2
r = a (1 − ε cosη ) .
Appendix 17.D.5: Step 5: Differentiating Twice with Respect to Time
G
The acceleration a is given by
G d 2x ˆ d 2 y
a = 2 i + 2 ĵ ,
dt
dt
(17.D.5.1)
but we know, from Equations (17.D.1.3), x and y as explicit functions of the eccentric
anomaly η rather than the time t . We will need to use either implicit differentiation or
the chain rule. Choosing the latter (they of course yield the same result),
dx dx dη dx / dη
=
=
.
dt dη dt dt / dη
(17.D.5.2)
The first of the equations in (17.D.1.3) leads quite readily to dx / dη = − a sin η and
Equation (17.D.3.3) gives dt / dη = (T / 2π )(1 − ε cosη ) , with the result
10/15/2009 17App - 16
dx
2π sinη
= −a
.
dt
T 1 − ε cosη
(17.D.5.3)
dy
2π cosη
=b
.
dt
T 1 − ε cosη
(17.D.5.4)
A similar calculation gives
In finding the second derivatives with respect to time, doing some preliminary
calculations will, we hope, help in the long run. Specifically, we have, starting from
Equation (17.D.5.2) (the corresponding expressions for the derivatives of y follow
immediately),
dx dx / dη
=
dt dt / dη
d ⎛ dx ⎞
d x d ⎛ dx ⎞ dη ⎜⎝ dt ⎟⎠
= ⎜ ⎟=
dt 2 dt ⎝ dt ⎠ ( dt / dη )
2
=
1
d ⎛ dx / dη ⎞
( dt / dη ) dη ⎜⎝ dt / dη ⎟⎠
(17.D.5.5)
2
2
2
1 ( d x / dη ) ( dt / dη ) − ( dx / dη ) ( d t / dη )
=
2
dt / dη
( dt / dη )
⎡ d 2 x dt dx d 2t ⎤
=
−
⎥.
3 ⎢
( dt / dη ) ⎣ dη 2 dη dη dη 2 ⎦
1
The advantage is in the fairly simple form for the terms
dt
T
T r
d 2t
T
=
1
−
ε
cos
η
=
,
=
ε sin η .
(
)
2
2π a dη
2π
dη 2π
(17.D.5.6)
d 2x
d2y
= − a cosη ,
= −b sinη .
dη 2
dη 2
(17.D.5.7)
We also have
Combining,
10/15/2009 17App - 17
d 2 x 4π 2
= 2
dt 2
T
4π 2
= 2
T
a
⎡ − cosη (1 − ε cosη ) − ( − sinη )( ε sinη ) ⎤⎦
r3 ⎣
4π 2 x
a
ε
η
−
=
−
cos
.
[
]
r3
T 2 r3
(17.D.5.8)
b
⎡ − sin η (1 − ε cosη ) − ( cosη )( ε sin η ) ⎤⎦
r3 ⎣
4π 2 y
b
η
−
=
−
sin
.
[
]
r3
T 2 r3
(17.D.5.9)
Similarly,
d 2 y 4π 2
= 2
dt 2
T
4π 2
= 2
T
Combining the results given in (17.D.5.8) and (17.D.5.9), we have
G
⎛ 2π ⎞ G
a = −⎜
⎟ r.
⎝ T ⎠
2
(17.D.5.10)
Kepler’s Third Law predicts T 2 = 4π 2a 3 / GM : (note that a is NOT the magnitude of the
acceleration), and so Equation (17.D.5.10) becomes
GM G
G
a = − 3: r,
r
(17.D.5.11)
which is the combination of Newton’s Second Law and Newton’s Law of Universal
Gravitation, and is Equation (17.D.4.1).
Appendix 17.D.6 – Fringe Benefits
Most of the proofs that the orbits of planets are ellipses give the distance r from the sun
as a function of the polar angle θ rather than as a function of time (see the derivations in
Chapter 17). The above proof gives r and t as functions of the eccentric anomaly η .
For many purposes one needs to know position as a function of time, not just distance as
a function of polar angle. We would like to complete our description of planetary motion
by deriving θ as a function of η .
The x -component of the vector from the sun to the planet is given by
r cos θ = a ( cosη − ε ) .
(17.D.6.1)
Substituting r = a (1 − ε cosη ) from (17.D.4.3), we find
10/15/2009 17App - 18
cos θ =
cosη − ε
.
1 − ε cosη
(17.D.6.2)
We then get the two expressions
1 − cos θ =
(1 + ε )(1 − cosη ) ,
1 − ε cosη
(17.D.6.3)
(1 − ε )(1 + cosη ) .
1 + cos θ =
1 − ε cosη
Then, using the trigonometric half-angle identities,
tan 2
θ
2
=
1 − cos θ (1 + ε )(1 − cosη )
=
,
1 + cos θ (1 − ε )(1 + cosη )
(17.D.6.4)
which gives us
tan
θ
2
=
1+ ε
η
tan ,
1− ε
2
(17.D.6.5)
the relation between the polar angle and the eccentric anomaly.
A further fringe benefit to using the eccentric anomaly is in calculating integrals in polar
coordinates, specifically the area of an ellipse. We have found the standard result,
area = π ab , by using the known form for the area in Cartesian coordinates and
multiplying by the ratio b / a .
However, in polar coordinates, the expression for the area,
a 2 (1 − ε 2 ) 2π
dθ
1 2π 2
area = ∫ r dθ =
,
2
∫
0
2 0
2
(1 + ε cosθ )
(17.D.6.6)
is an integral is not often done in freshman-level calculus (although the technique is
included in the 18.013 and 18.014 texts at MIT). Using the eccentric anomaly turns out
to simplify the integral greatly. Note that in this Appendix, the origin is taken to be on the
positive x -axis, and so we need to replace ε by −ε in Section 17.3.
To start, re-express the radius r , the distance from the sun to the planet, in the first
integral in terms of η , as given in Equation(17.D.4.3). Then,
10/15/2009 17App - 19
area =
a2
2
a
2
2 dθ
∫0 (1 − ε cosη ) dθ = 2 ∫0 (1 − ε cosη ) dη dη .
2
2π
2π
(17.D.6.7)
From the half-angle formulas, we have
1 − ε cosη = 1 − ε
cos 2 (η / 2 ) − sin 2 (η / 2 )
cos 2 (η / 2 ) + sin 2 (η / 2 )
1 − tan 2 (η / 2 )
= 1− ε
1 + tan 2 (η / 2 )
=
(17.D.6.8)
(1 − ε ) + (1 − ε ) tan 2 (η / 2 ) .
1 + tan 2 (η / 2 )
The term from the change of variables, dθ / dη , is found by differentiating
Equation (17.D.6.5),
dθ
θ dη 1 + ε
η
sec 2 =
sec2
2
2
2 1− ε
2
θ⎞
1+ ε ⎛
⎛
2η ⎞
dθ ⎜ 1 + tan 2 ⎟ = dη
⎜ 1 + tan ⎟
2⎠
2⎠
1− ε ⎝
⎝
η⎞
1+ ε
⎛ 1+ ε
dθ ⎜ 1 +
tan 2 ⎟ = dη
2⎠
1− ε
⎝ 1− ε
(17.D.6.9)
⎛
2η ⎞
⎜ 1 + tan ⎟ .
2⎠
⎝
Thus we see that, after some minor algebra,
η⎞
⎛ 1+ ε
tan 2 ⎟
⎜1 +
1+ ε
1− ε
2⎠
dθ ⎝
= dη
⎛
1− ε
2η ⎞
⎜ 1 + tan ⎟
2⎠
⎝
dθ
(1 − ε ) + (1 + ε ) tan
⎛
2η ⎞
⎜ 1 + tan ⎟
2⎠
⎝
2
(17.D.6.10)
η
2 = dη 1 − ε 2 ,
so that the second integral in Equation (17.D.6.7) becomes
a 2 2π
area = ∫ (1 − ε cosη ) 1 − ε 2 dη
2 0
=πa
2
(17.D.6.11)
1 − ε = π ab,
2
as determined as described above.
10/15/2009 17App - 20
Appendix 17.E: Orbit Radius as a Function of Time by
Direct Integration
Introduction
In previous presentations, in Chapter 17 and the Chapter 17 Appendices, we have only
calculated the distance of a planet from the sun as a function of angle, reproducing the
classical forms for conic sections, but have only hinted at the possibility of finding the
distance as a function of time. This latter dependence can be of great utility, but at the
cost of a great deal of math. This will not stop us.
In what follows, we’ll try to reproduce the notation of Chapter 17 as closely as possible,
including that of Appendix 17.D, and introduce new notation that should be consistent
with other sources (of course, there’s no way we can guarantee consistency with all
sources). Although the problem may be generalized, we’ll make reference to the case of
a single object moving subject to the gravitation of the sun. This object is likely to be a
planet or a comet, but for the purposes of this discussion, the motion of other objects,
specifically objects in unbound orbits, and objects subject to the earth’s gravitation, could
be found by extension. In fact, in an example that will not be presented here, the
expansion of the universe could be modeled by equations of the same mathematical form.
As a disclaimer at the beginning, we will see that we can find explicit forms r = r ( t ) for
only two very special cases: circular motion (necessarily uniform) and the case of an
object with escape velocity on a linear trajectory. For all other cases, our solutions will
give the distance r as an implicit function of the time t .
For a given solar mass and object mass, the trajectory may be determined by the total
mechanical energy and the angular momentum. Using these parameters exclusively leads
to rather cumbersome expressions, and we will find it convenient to use other parameters,
specifically the semilatus rectum and eccentricity for non-zero angular momentum and
the semimajor axis for bound orbits. We will also use common notation in evaluating
needed integrals, which will allow us to compare solutions in limiting cases.
Appendix 17.E.1 – Basics
Take the mass of the sun as M sun = M : = m2 and let the mass of the planet be m1 , with
reduced mass μ = m1m2 / ( m1 + m2 ) . In many cases, it would be possible to go to the limit
m2 m1 , μ → m1 ; this presentation will use the more general form for the reduced mass.
The semilatus rectum of the orbit trajectories in terms of the angular momentum
magnitude L is
10/15/2009 17App - 21
r0 =
L2
.
μ G m1 m2
(17.E.1.1)
The eccentricity ε ≥ 0 of a conic section is given by
ε = 1+
2
2 EL2
μ ( Gm1m2 )
2
.
(17.E.1.2)
For a bound orbit, the semimajor axis a > 0 is related to the semilatus rectum r0 by
a=
r0
1− ε 2
(17.E.1.3)
and the total mechanical energy for a bound orbit is
E=−
Gm1m2
< 0.
2a
(17.E.1.4)
We can then proceed to Equation (17.3.8), reproduced here as
1
dr
=
dt
2⎛
1 L2 G m1 m2 ⎞ 2
−
+
E
⎜
⎟ .
r
μ⎝
2 μ r2
⎠
(17.E.1.5)
Equation (17.E.1.5) is a separable equation in r and t , and may in general be
expressed as
dr
2
1 L
G m1 m2
E−
+
2
r
2 μr
=
2
μ
dt .
(17.E.1.6)
Equation (17.E.1.6) could be formidable, depending on the relative values of the
parameters, specifically the total mechanical energy E and the orbital angular
momentum L .
We’ll start by looking at the special cases that make the integral simple, primarily to
demonstrate the calculus techniques involved and to introduce some useful notation.
Appendix 17.E.2 – No Angular Momentum
10/15/2009 17App - 22
Setting L = 0 in Equation (17.E.1.6) manifestly simplifies the argument of the
square root,
dr
2
dt .
=
μ
G m1 m2
E+
r
(17.E.2.1)
Before jumping into the integral without a life vest, however, we would be wise to
consider the three cases E = 0 , E < 0 and E > 0 separately.
The trajectories of such objects would be “degenerate” conic sections, parabolas, ellipses
or hyperbolas extended so far that they become confined to straight segments or rays.
E.2.1 – Zero Angular Momentum, Zero Mechanical Energy.
We should recognize this situation as that of an object (not likely to be a planet) launched
or thrown with exactly the needed escape velocity. Setting E = 0 in Equation (17.E.2.1)
yields, almost immediately,
1
2
dr r1/ 2 =
dt
μ
Gm1m2
(17.E.2.2)
1
2 3/ 2
2
r =
t
μ
Gm1m2 3
where any constants of integration have been adjusted so that r ( 0 ) = 0 .
Unlike most of the expressions arising from the more general situations,
Equation (17.E.2.2) can be readily solved for r ( t ) ,
1/ 3
⎛ 9Gm1m2 ⎞ 2 / 3
r (t ) = ⎜
⎟ t .
⎝ 2μ ⎠
(17.E.2.3)
dr
→ ∞ as t → 0 . This indicates that for
dt
launching an object from a planet, or anything expelled from the surface of the sun, the
dr
limit
→ ∞ is unphysical for classical physics. However, cosmologists do consider
dt
dr
this limit. We also see, as a simple check, that as t → ∞ and r → ∞ ,
→ 0 , consistent
dt
with zero net mechanical energy.
It should be noted that in Equation (17.E.2.3),
10/15/2009 17App - 23
An alternative method of doing the integral in Equation (17.E.2.3), one that allows
comparison to the other degenerate orbits, will be given below (Subsection E.2.4).
E.2.2 – Zero Angular Momentum, Negative Mechanical Energy
This would then be the case where an object is launched vertically upward, but with
insufficient energy to escape. The object (planets are not likely to do this – comets come
close) would then move some distance away, and then return. Or, an object dropped
from rest in an inverse-square gravitational field (but with no initial tangential velocity)
would plummet.
To simplify the calculations, use Equation (17.E.1.4), reproduced here;
E=−
Gm1m2
,
2a
(17.E.2.4)
where a is the semimajor axis of the bound orbit, a degenerate ellipse, as described
above. We can recognize that for a projectile launched from a (nonrotating) surface, 2a
is the furthest distance the projectile rises from the center of the attracting body.
Using Equation (17.E.2.4) in Equation (17.E.2.1) leads to
dr
1/ r − 1/ ( 2a )
dr
=
2Gm1m2
μ
dt
(17.E.2.5)
r
Gm1m2
dt.
=
aμ
2a − r
The integral arising from the second equation in (17.E.2.5) may not be standard, and is
not in all tables (and sometimes gives pause to computer integration programs), but is not
difficult. Clearly, it would help to make a substitution so that both numerator and
denominator in the square root on the left are squares. One choice is to make the
substitution
r = 2a sin 2
η
2
.
(17.E.2.6)
The factor of 2 in the angle η / 2 will turn out to be a slight convenience, and will be
seen to match conventional usage, specifically that in Appendix 17.D. For starters,
we have
η
η
dr = 2a sin cos dη ,
2
2
(17.E.2.7)
from which
10/15/2009 17App - 24
2a sin 2 (η / 2 )
r
η
η
dr
= 2a sin cos dη
2a − r
2
2
2a − 2a sin 2 (η / 2 )
(17.E.2.8)
= 2a sin 2 (η / 2 ) dη .
This is easily integrated, using standard methods and the half-angle formula
sin 2
η
2
=
1
(1 − cosη ) ,
2
(17.E.2.9)
Gm1m2
t.
a3μ
(17.E.2.10)
which leads to
η − sinη =
Equations (17.E.2.6) and (17.E.2.10) then give both r and t as functions of η , and
include the parameter a . We cannot solve for r ( t ) , but we can solve for t ( r ) in terms
of the energy (recall that E < 0 ),
t=
1
Gm1m2
⎛
rE
−E3 ⎛
−1
⎜ 2sin ⎜⎜ −
μ ⎜⎝
2Gm1m2
⎝
⎞
rE
rE
1+
⎟⎟ − 2 −
2Gm1m2
2Gm1m2
⎠
⎞
⎟ .(17.E.2.11)
⎟
⎠
Equation (17.E.2.11) is a terrible mess, and not really of great use. From now on in this
presentation, such substitutions will not be made.
Of far more use is to consider Equations (17.E.2.6) and (17.E.2.10) together,
reproduced here;
r = 2a sin 2
η
2
(17.E.2.12)
Gm1m2
t.
η − sinη =
a3μ
With the choice of integration constants such that r (η = 0 ) = 0 , t (η = 0 ) = 0 , we see that
the projectile would return to r = 0 at η = 2π , and
t (η = 2π ) =
4π 2 μ a 3
=T ,
Gm1m2
(17.E.2.13)
the period predicted by Kepler’s Third Law as given in Equation (17.6.13).
10/15/2009 17App - 25
Before moving on, anticipating later sections, it is indeed possible to perform the needed
integral in (17.E.2.5) by other methods, specifically by rewriting the second expression as
r dr
2ar − r 2
=
r dr
a 2 − ( a 2 − 2ar + r 2 )
=
r dr
a2 − ( a − r )
=
2
Gm1m2
dt. (17.E.2.14)
aμ
In the last expression in (17.E.2.14), make the standard substitution
a − r = a cosη , r = a (1 − cosη ) , dr = a sin η dη
(17.E.2.15)
(which is, of course just Equation (17.E.2.6) rewritten) to obtain
r dr
a − (a − r )
2
2
= a (1 − cosη ) dη ,
(17.E.2.16)
equivalent to Equation (17.E.2.10). At this point, note that the substitution in (17.E.2.15)
is that given by Equation (17.D.4.3) with ε = 1 . This is not a coincidence, and we are not
making this up.
E.2.3 – Zero Angular Momentum, Positive Mechanical Energy
We anticipate that for this case, the trajectory will be unbound, a degenerate hyperbola.
As presented here, we’ll have to use a slightly different calculus technique, and we won’t
be able to consider the period of the non-periodic orbit or the semimajor axis. However,
for mathematical convenience we will introduce a parameter with dimensions of length
and call it a .
If we repeat the calculation for negative mechanical energy, but with E = Gm1m2 / 2a ,
a > 0 to represent a positive mechanical energy, we obtain
dr
1/ r + 1/ ( 2a )
=
2Gm1m2
μ
dt
(17.E.2.17)
r
Gm1m2
dr
dt.
=
aμ
2a + r
To employ similar means as used above, we would need the sum 2a + r to be the square
of some convenient function, and no circular trig function would suffice. However, if we
use hyperbolic functions, the ensuing calculus simplifies quite a bit.
A complete discussion of hyperbolic functions cannot be presented here. The features
that we will want to use are:
10/15/2009 17App - 26
d cosh u
d sinh u
= sinh u,
= cosh u, cosh 0 = 1, sinh 0 = 0
du
du
cosh 2 u − sinh 2 u = 1
(17.E.2.18)
sinh 2 u =
1
( cosh 2u − 1) .
2
Accordingly, if we make the substitution
η
η
η
r = 2a sinh 2 , dr = 2a cosh sinh dη
2
2
2
(17.E.2.19)
we obtain
dr
r
η
= 2a sinh 2 dη = a ( cosh η − 1) dη .
2a + r
2
(17.E.2.20)
Using Equation (17.E.2.20) in (17.E.2.17) and integrating yields
Gm1m2
t,
a3μ
sinhη − η =
(17.E.2.21)
where once again the constants of integration have been selected so that
r ( t = 0 ) = r (η = 0 ) = 0 . An equation similar to Equation (17.E.2.11) is not likely to help
anyone, and won’t be given here.
As a simple check, note that
dr
Gm1m2
=
dt
a3μ
2a + r
Gm1m2 cosh (η / 2 )
=
r
a 3μ sinh (η / 2 )
(17.E.2.22)
Gm1m2
=
coth (η / 2 )
a3μ
(yes, there really is such a thing as a “hyperbolic cotangent”). As η → ∞ , r → ∞ ,
t → ∞ , coth (η / 2 ) → 1 , dr / dt → Gm1m2 / aμ = 2 E / μ , indicating that in this limit
the energy is all kinetic.
We should also recognize that if we had used, in Equations (17.E.2.17)
r dr
r + 2ar
2
=
r dr
(r + a)
2
− a2
(17.E.2.23)
10/15/2009 17App - 27
and made the substitution r + a = a cosh η , we would obtain Equation (17.E.2.20)
directly.
E.2.4 – Limiting cases
We should expect that in the limit E → 0 , the results of Subsections E.2.2 and E.2.3
should reduce to that of Subsection E.2.1, and this is indeed the case. One way to see this
is to consider that in both cases of nonzero energy, in the limit as a is very large, η → 0
for finite r , and in this limit
1 − cosη →
η2
2
, cosh η − 1 →
⎛
⎝
1
⎞
⎠
η2
2
1
η − sin η → η − ⎜η − η 3 ⎟ = η 3
6
6
(17.E.2.24)
1 ⎞
1
⎛
sinh η − η → ⎜η + η 3 ⎟ − η = η 3
6 ⎠
6
⎝
r → aη 2 / 2,
and combining these results with either of the expressions in (17.E.2.12) or (17.E.2.21)
yields, after some algebra, the result of (17.E.2.3).
Equation (17.E.2.24) suggests that if we use the substitution r = aη 2 / 2 in the first
equation in (17.E.2.2), we may get a result similar to those in the expressions in
(17.E.2.12) and Equation (17.E.2.21). Indeed, in doing so, with dr = aη dη and doing the
fairly simple resulting integral leads to
η3
6
=
Gm1m2
t.
μ a3
(17.E.2.25)
The parameterizations that led to r = aη 2 / 2 and Equation (17.E.2.25) involve a
parameter a > 0 , but there is no length scale in Equation (17.E.2.2). It turns out that the
given parameterizations result in solutions to Equation (17.E.2.2), independent of a .
This is of course completely consistent; we can use any length scale we choose.
F’rheaven’s sake, lets’ draws a picture. Okay.
In Figure 17.E.1, the time axis is horizontal and the distance axis is vertical. The scales
are arbitrary. The lower curve, red if viewed in color, for E < 0 , has been extended for
two “bounces” or “collapses.” Whether or not this is useful depends on the situation.
Note that in this idealization, the object would pass through the origin without changing
its direction of motion, but in our notation r ≥ 0 , so the sign of r does not change, but
10/15/2009 17App - 28
the sign of the (infinite) velocity does. The upper (green) curve is for E > 0 and the
intermediate (blue) for E = 0 .
Figure 17.E.2 is the same, on an expanded scale near the origin. Part of the point in
including this figure is to show that if one wants to send a projectile in a prescribed
trajectory, one has to be careful; at the origin the curves are indistinguishable.
Figure 17.E.1
Figure 17.E.2
Appendix 17.E.3 – Non-Zero Angular Momentum
10/15/2009 17App - 29
As in Section E.2, we’ll start by considering the simplest mathematical case, that with
E = 0 . Again, in doing so we’ll gain some insight into the nature of the mathematical
techniques involved, and find some advantageous notation for the other cases.
E.3.1 – Non-Zero Angular Momentum, Zero Mechanical Energy
For E = 0 , we know that the trajectory will be a parabola. Equation (17.E.1.6) becomes
dr
2
G m1 m2 1 L
−
r
2 μ r2
=
2
μ
dt ,
(17.E.3.1)
which can be simplified as
dr
L2
1 1
1
−
r 2 μGm1m2 r 2
=
2Gm1m2
μ
dt
(17.E.3.2)
r dr
2Gm1m2
dt
=
μ
1
r − r0
2
where r0 =
L2
is the semilatus rectum (Equation (17.E.1.1) in this Appendix and
μ G m1 m2
Equation (17.3.12) from the Course Notes).
The needed integral is found by setting r = ( r − r0 / 2 ) + r0 / 2 , and
( r − r0 / 2 ) + r0 / 2 dr =
r − r0 / 2
r − r0 / 2 dr +
r0
2
1
dr ,
r − r0 / 2
(17.E.3.3)
which is readily integrated to obtain
2
2
2Gm1m2
3/ 2
t .(17.E.3.4)
( r − r0 / 2 ) + r0 r − r0 / 2 = r − r0 / 2 ( r + r0 ) =
μ
3
3
Equation (17.E.3.4) is perhaps hard to recognize as a parameterization in time of a
parabola in space, but we can identify some features that we would expect. In the limit
L → 0 , r0 → 0 , we reproduce the result found in (17.E.2.2). The constants of integration
give r = r0 / 2 at t = 0 , and this is the minimum value of r . For a parabola with
eccentricity ε = 1 , the minimum distance is rmin = r0 / (1 + ε ) = r0 / 2 , consistent with the
above result.
10/15/2009 17App - 30
Once more anticipating a result from Section E.4, we can perform the integration in the
second expression in (17.E.3.2) by setting
r − rmin = rmin
η2
⎛ η2 ⎞
2
, dr = rmin η dη , r dr = rmin
η ⎜1 + ⎟ dη ,
2
2⎠
⎝
(17.E.3.5)
leading to
r dr
η2 ⎞
2Gm1m2
3/ 2 ⎛
dt ,
= 2 rmin ⎜1 + ⎟ dη =
2⎠
μ
1
⎝
r − r0
2
(17.E.3.6)
from which
η3 ⎞
Gm1m2
3/ 2 ⎛
+
η
t.
rmin
⎜
⎟=
μ
6⎠
⎝
(17.E.3.7)
Note that the constants or integration are still set so that at t = 0 , η = 0 and r ( 0 ) = rmin .
Non-Zero Angular Momentum, Non-Zero Mechanical Energy
The two cases, E < 0 and E > 0 , could be considered in one calculation, at the expense
of introducing more sophisticated math (complex variables) than we need for the current
presentation. Accordingly, the cases will be considered separately.
E.3.2– Non-Zero Angular Momentum, Negative Mechanical Energy
We know that these trajectories should be ellipses with one focus at the sun. From the
previous result, we expect that using the semilatus rectum will be useful. However,
invoking a great deal of hindsight, using the semimajor axis a will simplify the
calculations, and reproduce more conventional notation. So, using Equation (17.E.2.4),
E = −Gm1m2 / 2a , and Equation (17.E.1.3), r0 = a (1 − ε 2 ) , Equation (17.E.1.6) becomes
dr
−1/ a − (1 − ε
2
)a / r
2
+ 2/r
=
Gm1m2
μ
dt
(17.E.3.8)
r dr
Gm1m2
=
dt.
2
2
2
μ
a
− a (1 − ε ) + 2ar − r
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The needed integral on the left above is done by completing the square in the argument of
the square root in the denominator,
− a 2 (1 − ε 2 ) + 2ar − r 2 = − a 2 (1 − ε 2 ) − ( a 2 − 2ar + r 2 ) + a 2
= a 2ε 2 − ( a − r ) .
2
(17.E.3.9)
This suggests, almost necessarily, the substitution
r = a (1 − ε cosη ) , dr = aε sin η dη ,
(17.E.3.10)
leading to
r dr
− a (1 − ε
2
2
) + 2ar − r
2
=
1 a (1 − ε cosη ) aε sin η dη
aε
sin η
(17.E.3.11)
= a (1 − ε cosη ) dη
and hence, from Equation (17.E.3.8),
η − ε sinη =
2π
Gm1m2
Gm1m2
t
t = 2π
t=
3
2
3
μa
T
4π μ a
(17.E.3.12)
where T is the period of the orbit.
It should be noted that the substitution used in Equation (17.E.3.10) and the result of
Equation (17.E.3.12) are those in Appendix 17.D; this is not a coincidence.
Note that the eccentricity does not appear on the right in the above result. If ε = 0 , the
case for a circle, the integral in Equation (17.E.3.11) would be improper, and this case
might be considered separately. However, since we have a result involving only the
semimajor axis a , with the period a function of a , and a simple function of the
eccentricity ε , there is no need to consider the special case separately.
The other limiting case, ε → 1 , L = 0 , is seen to reproduce the result of Section E.2
above.
E.3.3– Non-Zero Angular Momentum, Positive Mechanical Energy
We know that these trajectories will be hyperbolas with one focus at the sun. As in
Subsection E.2.3 above, introduce the parameter a > 0 so that the total mechanical
energy is E = Gm1m2 / 2a > 0 . The parameter a can no longer be interpreted as the
semimajor axis of a hyperbola. Equation (17.E.1.3) is no longer valid, but Equations
10/15/2009 17App - 32
(17.E.1.1) and (17.E.1.2) are, and Equation (17.E.1.3) becomes r0 = a ( ε 2 − 1) (recall that
ε > 1 ). Equations (17.E.3.8) are then
dr
1/ a − ( ε − 1) a / r + 2 / r
2
2
=
Gm1m2
μ
dt
(17.E.3.13)
r dr
Gm1m2
=
dt.
2
2
2
aμ
r + 2ar − a ( ε − 1)
Completing the square is slightly simpler for positive energy; adding and subtracting a 2
in the argument of the square root in the denominator on the left yields
r dr
(r + a)
2
− a 2ε 2
=
Gm1m2
dt .
aμ
(17.E.3.14)
As in Section E.2.3, a convenient substitution is
r = aε cosh η − a, dr = aε sinhη dη ,
(17.E.3.15)
from which
r dr
(r + a)
2
− a 2ε 2
=
a ( ε cosh η − 1) aε sinh η dη
aε sinh η
(17.E.3.16)
= a ( ε cosh η − 1) dη
and, in terms of the eccentricity, the parameter a and the parameter η ,
ε sinh η − η =
Gm1m2
t.
a3μ
(17.E.3.17)
Although the square root multiplying the time t does have dimensions of [ time ] , its
−1
physical interpretation might be hard to qualify, so we’ll leave it as presented.
Once again, plots are in order. The plot scales have been adjusted so that each radius is at
the same minimum value at t = 0 . One curve (red, if viewed in color) is clearly
oscillatory, corresponding to an elliptic orbit, E > 0 . The “lower” curve below (green)
corresponds to a hyperbolic orbit, E < 0 and the “middle” curve, (blue) corresponds to a
parabolic orbit, with E = 0 .
10/15/2009 17App - 33
It is not clear from this figure that as t → ∞ , the slope of the curve for the parabolic orbit
dr
approaches
→ 0 , but this is the case.
dt
Figure 17.E.3
d r
Note that as the energy increases, 2 at t = 0 decreases. This makes physical sense, in
dt
that at t = 0 the gravitational potential energy is the same for each, and hence trajectories
with higher energy must be moving faster (larger L ), and will not “fall in” as fast.
2
The plot on the left below shows, in an expanded scale, the behavior for smaller values
of t . The plot on the right shows that the hyperbolic trajectory will eventually intersect
the parabolic trajectory, and should give some indication that for the parabolic trajectory,
the slope does approach 0 .
Figures 17.E.4A and 17.E.4B
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E.3.3 – Limiting Cases
The limits we want here are E → 0 while maintaining L2 ≠ 0 . We have to be a bit tricky
for these cases. Consider Equation (17.E.3.10); if we simply set ε = 1 , we merely
reproduce Equation (17.E.2.15). What we do is separate terms that remain finite and
those that we’ve seen before. Such a vague statement needs explanation via an example.
What follows represents a great deal of both hindsight and foresight.
Rewrite the substitution used in Equation (17.E.3.10) as
r = a (1 − ε cosη ) = a (1 − ε ) + aε (1 − cosη ) .
(17.E.3.18)
The first term on the right will remain finite as a → ∞ and ε → 1 , and is equal to the
minimum distance (the perihelion), which in these notes we denote rmin . Expanding
cosη and keeping the lowest nonvanishing term yields
r → rmin + aε
η2
2
, η2 →
2
2
( r − rmin ) → ( r − rmin )
aε
a
where in the last step ε → 1 has been used.
Equation (17.E.3.12),
Gm1m2
μ
(17.E.3.19)
Using this same technique in
⎛
η3 ⎞
t → a 3/ 2 ⎜η − εη + ε ⎟
6⎠
⎝
= a 3/ 2 (1 − ε )η + ε a 3/ 2
= rmin a η + ε a
1/ 2
3/ 2
→ rmin a1/ 2η + a 3/ 2
η
η3
6
(17.E.3.20)
3
6
η3
6
.
Substituting the result of Equation (17.E.3.19) for η in terms of r and ε → 1 gives
Gm1m2
μ
t → 2 rmin r − rmin +
2 2
2
3/ 2
r − rmin ( r + 2rmin ) ,(17.E.3.21)
( r − rmin ) =
6
3
which is the same as Equation (17.E.3.4).
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The same limiting procedure applied to Equations (17.E.3.15) and (17.E.3.17) (we need
cosh η → 1 + η 2 / 2 , sinh η → η + η 3 / 6 ) will give the same result. A needed intermediate
calculation, to the effect that even though the parameter a is not physical, the product
a (1 − ε ) = rmin for a hyperbola is, as may be found from solving Equation (17.E.1.5) for
dr / dt = 0 with E = Gm1m2 / 2a .
The fact that we obtain the same expressions in the limiting cases is reflected in the graph
on the right above, following Subsection E.3.3.
Appendix 17.E.4 – Synopsis and a Few Other Points of Interest
The parameterizations for all six cases are summarized below:
E.2.1 & E.2.4 - Zero Angular Momentum, Zero Mechanical Energy
Trajectories are degenerate parabolas.
r=a
η2
2
,
η3
6
=
Gm1m2
t , a > 0 arbitrary .
μ a3
(17.E.4.1)
E.2.2 – Zero Angular Momentum, Negative Mechanical Energy
Trajectories are degenerate ellipses.
r = a (1 − cosη ) ,
Gm1m2
t = η − sin η , a = −Gm1m2 / 2 E > 0 .(17.E.4.2)
a3μ
E.2.3 – Zero Angular Momentum, Positive Mechanical Energy
Trajectories are degenerate hyperbolas.
r + a = a cosh η ,
Gm1m2
t = sinh η − η , a = Gm1m2 / 2 E > 0 .(17.E.4.3)
a3μ
E.3.1 – Non-Zero Angular Momentum, Zero Mechanical Energy
Trajectories are parabolas.
η2
η3 ⎞
Gm1m2
3/ 2 ⎛
+
η
r − rmin = rmin ,
t = rmin
⎜
⎟.
μ
2
6⎠
⎝
(17.E.4.4)
E.3.2– Non-Zero Angular Momentum, Negative Mechanical Energy
Trajectories are ellipses.
r = a (1 − ε cosη ) ,
Gm1m2
2π
t=
t = η − ε sin η , a is semimajor axis .(17.E.4.5)
3
μa
T
10/15/2009 17App - 36
E.3.3– Non-Zero Angular Momentum, Positive Mechanical Energy
Trajectories are hyperbolas.
r = aε cosh η − a,
Gm1m2
t = ε sinh η − η , a = Gm1m2 / 2 E > 0 .(17.E.4.6)
a3μ
You should note the similarities and differences in the above parameterizations. The
parameter “η ” is used frequently in such calculations (but some sources use “ θ ,” which
we use for something completely different). This parameter is called many things; in
other notes we’ve used “eccentric anomaly;” “development angle” and “evolution angle
are also common.
aμ
dr / dη (or rmin instead of a
Gm1m2
for the ellipses). The radial component of velocity as a function of η is then
With these parameterizations, we see that dt / dη =
dr dr / dη
Gm1m2 1 dr
Gm1m2 d
=
=
=
ln r ,
dt dt / dη
aμ r dη
aμ dη
(17.E.4.7)
sometimes a convenient simplification.
In fact, to some extent the above
parameterizations were developed with this in mind.
In the above parameterizations, all have r (η ) an even function of η and t (η ) an odd
function. This makes easy allowance for negative times, and any of the graphs above
may be flipped about the vertical axis to account for earlier times.
Appendix 17.F: Even More on Kepler Orbits
We’ve seen so far that for a Kepler Orbit, we have two constants of the motion: the
angular momentum and the total energy. Since the angular momentum is a vector with
three components, these constitute a total of four scalar constants of the motion. The
Kepler Problem has six “degrees of freedom” (three position, three velocity), and so we
expect to be able to find two more scalar constants of the motion.
We might expect to be able to further identify any orbit by a vector in the plane of the
orbit, perpendicular to the angular momentum, and this is indeed the case. Symmetry
suggests that this vector would be along the major axis, and we’ll see that this is the case
as well. What follows uses a good deal of vector algebra, but minimal calculus, and leads
to the orbit equation in a surprisingly simple form.
We’ll need two results from vector algebra that we haven’t had to use yet. Specifically,
G G
G
for vectors a , b and c , we have
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G G G G G G G G G
a × b × c = b (a ⋅ c ) − c a ⋅ b
G G G G G G G G G
a ⋅ b × c = b ⋅ (c × a) = c ⋅ a × b .
( )
( )
( )
( )
(17.F.1)
These relations are not hard to derive in Cartesian coordinates; the derivations will not be
reproduced here. As a check, however, note that the vector on the right side of the first
G G
G
G G
G
relation is perpendicular to both a and b × c . If a , b and c are non-coplanar, the
common magnitude of the “triple products” in the second relation is the expression for
the volume of a parallelepiped with the three vectors forming the sides.
Let’s start with the known constant angular momentum,
G G G
G G
L = r ×p = μ r × v ,
(17.F.2)
and re-express this quantity in a way that will allow us to use Newton’s Laws.
Specifically, consider the velocity in terms of polar coordinates,
drˆ
G dr
v = rˆ + r
dt
dt
(17.F.3)
so that the angular momentum can be expressed as
from which
G
⎛ drˆ ⎞
⎛ drˆ ⎞
L = μ ( r rˆ ) × ⎜ r ⎟ = μ r 2 ⎜ rˆ × ⎟
dt ⎠
⎝ dt ⎠
⎝
(17.F.4)
G
drˆ 1 L
rˆ ×
=
.
dt r 2 μ
(17.F.5)
The advantage to this operation is that we now have an explicit scalar factor of 1/ r 2 ,
which can and should be related to the same factor that appears in Newton’s Law of
Gravitation. However, in order to use that law, we need a vector relation involving r̂ ,
and so we’ll cross r̂ into both sides of Equation (17.F.5), yielding
G
drˆ ⎞ rˆ L
⎛
rˆ × ⎜ rˆ × ⎟ = 2 ×
dt ⎠ r μ
⎝
drˆ
1 ⎛ d2 G⎞ G
− =−
r ⎟ × L.
⎜
dt
Gm1m2 ⎝ dt 2 ⎠
(17.F.6)
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In the above, the first relation in Equation (17.F.1) was used to simplify the left side, and
G
Newton’s Law of Gravitation, in the form μ ( d 2r / dt 2 ) − = ( Gm1m2 / r 2 ) rˆ was used on the
right side. Note the cancellation of the factor of the reduced mass μ .
Equation (17.F.6) may now be integrated to obtain
G G
v×L G
rˆ =
+A
Gm1m2
G G
G
v×L
,
A = rˆ −
Gm1m2
(17.F.7)
G
G
where A is a constant vector. Since r̂ is in the plane of the orbit, and L is
G
perpendicular to the plane of the orbit, A must lie in the plane of the orbit, as indicated
G
above. Further, by considering extreme points of the orbit, where v ⊥ rˆ , and hence
G
G G
v × L & rˆ , we see that at these points A is in the direction parallel to the major axis.
G
G
Since A is a constant vector, A must always be in this direction. By considering the
G
vector A at perihelion (or at any point on the orbit of a circular orbit), we can see that the
G
direction of A is that from the perihelion point to the focus; we’ll need this result below,
when we find the orbit equation.
G
The magnitude of A is readily found by calculating
G G
A2 = A ⋅ A = rˆ ⋅ rˆ −
2
1
G G
G G G G
rˆ ⋅ v × L +
v × L ⋅ v × L . (17.F.8)
2
Gm1m2
( Gm1m2 )
(
)
(
)(
)
The first dot product is manifestly 1 . The middle term, the “cross term,” is found using
the second relation in Equation (17.F.1),
G
1 G G
1 2
G G
G
G
rˆ ⋅ v × L = L ⋅ ( rˆ × v ) =
L ⋅ (r × μ v ) =
L.
μr
μr
(
)
(17.F.9)
G G
G G
The third term is most easily evaluated by recalling that v ⊥ L , so that v × L = vL and
G G G G
v × L ⋅ v × L = v 2 L2 . Combining, we see that
(
)(
)
10/15/2009 17App - 39
A2 = 1 −
= 1+
= 1+
2
1 2
1
L +
v 2 L2
2
Gm1m2 μ r
( Gm1m2 )
2 L2
( Gm1m2 )
2
⎛ 1 2 Gm1m2 ⎞
⎜ μv −
⎟
r ⎠
μ⎝2
2 L2 E
( Gm1m2 ) μ
2
(17.F.10)
= ε 2.
G
Thus, the constant vector A is directed along the major axis and has magnitude equal to
the eccentricity.
G
G
The orbit equation is now found algebraically by taking the dot product of A and r ;
G G
G
A ⋅ r = Ar cosθ = rˆ ⋅ r −
1 G G G
r⋅ v×L .
Gm1m2
(
)
(17.F.11)
The first term is merely the magnitude r of the position vector. The second term,
G
repeating the calculation of Equation (17.F.9) with r instead of r̂ , is
with the result
1
L2 = r0 ,
Gm1m2 μ
(17.F.12)
Ar cosθ = ε r cosθ = r − r0 .
(17.F.13)
Solving for r gives the orbit equation in the form
r=
r0
.
1 − ε cosθ
(17.F.14)
G
It should be noted that what we call the vector A is a negative scalar multiple of the
“Laplace-Runge-Lenz” vector (yes, it took three people to come up with this).
Specifically, the L-R-L vector is in many sources given as
G
G
G
G G
r
A LRL = p × L − μ Gm1m2 = − μ Gm1m2 A .
r
(17.F.15)
G
G
G
Our choice of the form for A allows A = ε and the direction of A to lead to
Equation (17.F.11) without introduction of extra minus signs.
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