3 position by time.pdf

Orbital Position as a Function
of Time
3
CHAPTER OUTLINE
3.1 Introduction ..................................................................................................................................145
3.2 Time since periapsis .....................................................................................................................145
3.3 Circular orbits (e [ 0)...................................................................................................................147
3.4 Elliptical orbits (e < 1)...................................................................................................................147
3.5 Parabolic trajectories (e [ 1)........................................................................................................163
3.6 Hyperbolic trajectories (e > 1)........................................................................................................165
3.7 Universal variables .......................................................................................................................173
Problems.............................................................................................................................................183
Section 3.2 .........................................................................................................................................183
Section 3.4 .........................................................................................................................................184
Section 3.5 .........................................................................................................................................186
Section 3.6 .........................................................................................................................................186
Section 3.7 .........................................................................................................................................186
3.1 Introduction
In Chapter 2, we found the relationship between position and true anomaly for the two-body problem.
The only place time appeared explicitly was in the expression for the period of an ellipse. Obtaining
position as a function of time is a simple matter for circular orbits. For elliptical, parabolic, and
hyperbolic paths, we are led to the various forms of Kepler’s equation relating position to time. These
transcendental equations must be solved iteratively using a procedure like Newton’s method, which is
presented and illustrated in this chapter.
The different forms of Kepler’s equation are combined into a single universal Kepler’s equation by
introducing universal variables. Implementation of this appealing notion is accompanied by the
introduction of an unfamiliar class of functions known as Stumpff functions. The universal variable
formulation is required for the Lambert and Gauss orbit determination algorithms in Chapter 5.
The road map of Appendix B may aid in grasping how the material presented here depends on that
of Chapter 2.
3.2 Time since periapsis
The orbit formula, r ¼ ðh2=mÞ=ð1 þ ecos qÞ, gives the position of body m2 in its orbit around m1 as a
function of the true anomaly. For many practical reasons, we need to be able to determine the position
CHAPTER
Orbital Mechanics for Engineering Students. http://dx.doi.org/10.1016/B978-0-08-097747-8.00003-7
Copyright Ó 2014 Elsevier Ltd. All rights reserved.
145

of m2 as a function of time. For elliptical orbits, we have a formula for the period T (Eqn (2.82)), but we
cannot yet calculate the time required to fly between any two true anomalies. The purpose of this
section is to come up with the formulas that allow us to do that calculation.
The one equation we have that relates true anomaly directly to time is Eqn (2.47), h ¼ r2 _
q, which
can be written as
dq
dt
¼
h
r2
Substituting r ¼ ðh2=mÞ=ð1 þ ecos qÞ we find, after separating variables,
m2
h3
dt ¼
dq
ð1 þ ecos qÞ2
Integrating both sides of this equation yields
m2
h3

t tp

¼
Zq
0
dw
ð1 þ e cos wÞ2
(3.1)
where the constant of integration tp is the time at periapsis passage, where by definition q ¼ 0. tp is
the sixth constant of the motion that was missing in Chapter 2. The origin of time is arbitrary. It is
convenient to measure time from periapsis passage, so we will usually set tp ¼ 0. In that case
we have
m2
h3
t ¼
Zq
0
dw
ð1 þ e cos wÞ2
(3.2)
The integral on the right may be found in any standard mathematical handbook, such as Beyer (1991),
in which we find
Z
dx
ða þ bcos xÞ2
¼
1
ða2 b2Þ
3=2
2atan1
ffiffiffiffiffiffiffiffiffiffiffi
a b
a þ b
r
tan
x
2

b
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 b2
p
sin x
a þ bcos x
!
ðb aÞ (3.3)
Z
dx
ða þ bcos xÞ2
¼
1
a2

1
2
tan
x
2
þ
1
6
tan3 x
2

ðb ¼ aÞ (3.4)
Z
dx
ða þ bcos xÞ2
¼
1
ðb2 a2Þ3=2
2
6
4
b
b2 a2
p
sin x
a þ bcos x
aln
b þ a
p
þ
b a
p
tan x
2
b þ a
p

b a
p
tan x
2
!
3
7
5 ðb aÞ (3.5)
146 CHAPTER 3 Orbital Position as a Function of Time

3.3 Circular orbits (e [ 0)
If e ¼ 0, the integral in Eqn (3.2) is simply
Z q
0
dw, which means
t ¼
h3
m2
q
Recall that for a circle (Eqn (2.62)), r ¼ h2
/m. Therefore h3 ¼ r
3
2m
3
2, so that
t ¼
r
3
2
ffiffiffi
m
p q
Finally, substituting the formula (Eqn (2.64)) for the period T of a circular orbit, T ¼ 2pr
3
2=
ffiffiffi
m
p
, yields
t ¼
q
2p
T
or
q ¼
2p
T
t
The reason that t is directly proportional to q in a circular orbit is simply that the angular velocity 2p/T
is constant. Therefore, the time Dt to fly through a true anomaly of Dq is (Dq/2p)T.
Because the circle is symmetric about any diameter, the apse linedand therefore the periapsisd
can be chosen arbitrarily.
C, F P
D
t = 0 Apse
line
r
FIGURE 3.1
Time since periapsis is directly proportional to true anomaly in a circular orbit.
3.4 Elliptical orbits (e 1)
Set a ¼ 1 and b ¼ e in Eqn (3.3) to obtain
Zq
0
dw
ð1 þ ecos wÞ2
¼
1
ð1 e2Þ
3
2

2tan1
1 e
1 þ e
r
tan
q
2
!

e
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 e2
p
sin q
1 þ ecos q
#
3.4 Elliptical orbits (e 1) 147

Therefore, Eqn (3.2) in this case becomes
m2
h3
t ¼
1
ð1 e2Þ
3
2

2tan1
1 e
1 þ e
r
tan
q
2
!

e
1 e2
p
sin q
1 þ ecos q
#
or
Me ¼ 2tan1
1 e
1 þ e
r
tan
q
2
!

e
1 e2
p
sin q
1 þ ecos q
(3.6)
where
Me ¼
m2
h3

1 e2
3
2
t (3.7)
Me is called the mean anomaly. The subscript e reminds us that this is for an ellipse and not for pa-
rabolas and hyperbolas, which have their own “mean anomaly” formulas. Equation (3.6) is plotted in
Figure 3.2. Observe that for all values of the eccentricity e, Me is a monotonically increasing function
of the true anomaly q.
From Eqn (2.82), the formula for the period T of an elliptical orbit, we have
m2ð1 e2Þ
3
2=h3 ¼ 2p=T, so that the mean anomaly can be written much more simply as
Me ¼
2p
T
t (3.8)
0
e = 0
e = 0.2
e = 0.5
e = 0.8
e = 0.9
Mean anomaly, Me
FIGURE 3.2
Mean anomaly vs true anomaly for ellipses of various eccentricities.

The angular velocity of the position vector of an elliptical orbit is not constant, but since 2p radians are
swept out per period T, the ratio 2p/T is the average angular velocity, which is given the symbol n and
called the mean motion,
n ¼
2p
T
(3.9)
In terms of the mean motion, Eqn (3.5) can be written simpler still,
Me ¼ nt
The mean anomaly is the azimuth position (in radians) of a fictitious body moving around the ellipse at
the constant angular speed n. For a circular orbit, the mean anomaly Me and the true anomaly q are
identical.
It is convenient to simplify Eqn (3.6) by introducing an auxiliary angle E called the eccentric
anomaly, which is shown in Figure 3.3. This is done by circumscribing the ellipse with a concentric
auxiliary circle having a radius equal to the semimajor axis a of the ellipse. Let S be that point on the
ellipse whose true anomaly is q. Through point S we pass a perpendicular to the apse line, intersecting
the auxiliary circle at point Q and the apse line at point V. The angle between the apse line and the
radius drawn from the center of the circle to Q on its circumference is the eccentric anomaly E.
Observe that E lags q from periapsis P to apoapsis Að0 q 180Þ, whereas it leads q from A to P
ð180 q 360Þ.
To find E as a function of q, we first observe from Figure 3.3 that, in terms of the eccentric anomaly,
OV ¼ acos E, whereas in terms of the true anomaly, OV ¼ ae þ rcos q. Thus,
acos E ¼ ae þ rcos q
Q
F V P
E
S
r
a O
A
b
a
B
D
ae
FIGURE 3.3
Ellipse and the circumscribed auxiliary circle.

Using Eqn (2.72), r ¼ að1 e2Þ=ð1 þ ecos qÞ, we can write this as
acos E ¼ ae þ
a

1 e2

cos q
1 þ ecos q
Simplifying the right-hand side, we get
cos E ¼
e þ cos q
1 þ ecos q
(3.10a)
Solving this for cos q we obtain the inverse relation,
cos q ¼
e cos E
ecos E 1
(3.10b)
Substituting Eqn (3.10a) into the trigonometric identity sin2
E þ cos2
E ¼ 1 and solving for sin E
yields
sin E ¼
1 e2
p
sin q
1 þ ecos q
(3.11)
Equation (3.10a) would be fine for obtaining E from q, except that, given a value of cos E between 1
and 1, there are two values of E between 0 and 360, as illustrated in Figure 3.4. The same comments
hold for Eqn (3.11). To resolve this quadrant ambiguity, we use the following trigonometric identity:
tan2 E
2
¼
sin2
E=2
cos2 E=2
¼
1 cos E
2
1 þ cos E
2
¼
1 cos E
1 þ cos E
(3.12)
From Eqn (3.10a)
1 cos E ¼
1 cos q
1 þ ecos q
ð1 eÞ and 1 þ cos E ¼
1 þ cos q
1 þ ecos q
ð1 þ eÞ
0
1
1
2 2
cos E
EI EIV
E
FIGURE 3.4
For 0 cos E 1, E can lie in the first or fourth quadrant. For 1 cos E 0, E can lie in the second or third
quadrant.

Therefore,
tan2 E
2
¼
1 e
1 þ e
1 cos q
1 þ cos q
¼
1 e
1 þ e
tan2 q
2
where the last step required applying the trig identity in Eqn (3.12) to the term (1 cos q)/(1 þ cos q).
Finally, therefore, we obtain
tan
E
2
¼
1 e
1 þ e
r
tan
q
2
(3.13a)
or
E ¼ 2tan1
1 e
1 þ e
r
tan
q
2
!
(3.13b)
Observe from Figure 3.5 that for any value of tan (E/2), there is only one value of E between 0 and
360. There is no quadrant ambiguity.
Substituting Eqns (3.11) and (3.13b) into Eqn (3.6) yields Kepler’s equation,
Me ¼ E esin E (3.14)
This monotonically increasing relationship between mean anomaly and eccentric anomaly is plotted
for several values of eccentricity in Figure 3.6.
Given the true anomaly q, we calculate the eccentric anomaly E using Eqn (3.13). Substituting E
into Kepler’s formula, Eqn (3.14), yields the mean anomaly directly. From the mean anomaly and the
period T we find the time (since periapsis) from Eqn (3.5),
t ¼
Me
2p
T (3.15)
tan
E
2
E
0
FIGURE 3.5
For any value of tan (E/2), there is a corresponding unique value of e in the range 0 to 2p.

On the other hand, if we are given the time, then Eqn (3.15) yields the mean anomaly Me. Substituting
Me into Kepler’s equation, we get the following expression for the eccentric anomaly:
E esin E ¼ Me
We cannot solve this transcendental equation directly for E. A rough value of E might be read from
Figure 3.6. However, an accurate solution requires an iterative, “trial and error” procedure.
Newton’s method, or one of its variants, is one of the more common and efficient ways of finding
the root of a well-behaved function. To find a root of the equation f(x) ¼ 0 in Figure 3.7, we estimate it
to be xi and evaluate the function f(x) and its first derivative f 0(x) at that point. We then extend the
tangent to the curve at f(xi) until it intersects the x-axis at xiþ1, which becomes our updated estimate of
the root. The intercept xiþ1 is found by setting the slope of the tangent line equal to the slope of the
curve at xi, that is,
f0
ðxiÞ ¼
0 fðxiÞ
xiþ1 xi
from which we obtain
xiþ1 ¼ xi
fðxiÞ
f0ðxiÞ
(3.16)
The process is repeated, using xiþ1 to estimate xiþ2, and so on, until the root has been found to the
desired level of precision.
To apply Newton’s method to the solution of Kepler’s equation, we form the function
fðEÞ ¼ E esin E Me
0
Eccentric anomaly, E
Mean anomaly, Me
e = 0
e = 0.2
e = 0.4
e = 0.6
e = 0.8
e = 1.0
FIGURE 3.6
Plot of Kepler’s equation for an elliptical orbit.

and seek the value of eccentric anomaly that makes f(E) ¼ 0. Since
f0
ðEÞ ¼ 1 ecos E
for this problem, Eqn (3.16) becomes
Eiþ1 ¼ Ei
Ei esin Ei Me
1 ecos Ei
(3.17)
n
ALGORITHM 3.1
Solve Kepler’s equation for the eccentric anomaly E given the eccentricity e and the mean anomaly
Me. See Appendix D.11 for the implementation of this algorithm in MATLAB.
1. Choose an initial estimate of the root E as follows (Prussing and Conway, 1993). If Me p, then
E ¼ Me þ e/2. If Me p, then E ¼ Me e/2. Remember that the angles E and Me are in radians.
(When using a hand held calculator, be sure that it is in the radian mode.)
2. At any given step, having obtained Ei from the previous step, calculate
fðEiÞ ¼ Ei esin Ei Me and f0
ðEiÞ ¼ 1 ecos Ei.
3. Calculate ratioi ¼ fðEiÞ=f0
ðEiÞ.
4. If jratioij exceeds the chosen tolerance (e.g., 108
), then calculate an updated value of E,
Eiþ1 ¼ Ei ratioi
Return to Step 2.
5. If jratioij is less than the tolerance, then accept Ei as the solution to within the chosen accuracy.
n
Root
x
f
xi+1
f(xi)
xi
Slope =
df
dx
f
x
=
d
dx
x
x=xi
FIGURE 3.7
Newton’s method for finding a root of f(x) ¼ 0.

EXAMPLE 3.1
A geocentric elliptical orbit has a perigee radius of 9600 km and an apogee radius of 21,000 km, as shown in
Figure 3.8. Calculate the time to fly from perigee P to a true anomaly of 120.
Solution
Before anything else, let us find the primary orbital parameters e and h. The eccentricity is readily obtained from
the perigee and apogee radii by means of Eqn (2.84),
e ¼
ra rp
ra þ rp
¼
21;000 9600
21;000 þ 9600
¼ 0:37255 (a)
We find the angular momentum using the orbit equation,
9600 ¼
h2
398;600
1
1 þ 0:37255 cosð0Þ
0 h ¼ 72;472 km2
=s
With h and e, the period of the orbit is obtained from Eqn (2.82),
T ¼
2p
m2

h
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
1 e2
p
3
¼
2p
398;6002

72; 472
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 0:372552
p
3
¼ 18;834 s (b)
Equation (3.13a) yields the eccentric anomaly from the true anomaly,
tan
E
2
¼
ffiffiffiffiffiffiffiffiffiffiffiffi
1 e
1 þ e
r
tan
q
2
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 0:37255
1 þ 0:37255
r
tan
120
2
¼ 1:1711 0 E ¼ 1:7281 rad
Then Kepler’s equation, Eqn (3.14), is used to find the mean anomaly,
Me ¼ 1:7281 0:37255 sin 1:7281 ¼ 1:3601 rad
120°
Earth
21,000 km
9600 km
P
A
B
FIGURE 3.8
Geocentric elliptical orbit.

Finally, the time follows from Eqn (3.12),
t ¼
Me
2p
T ¼
1:3601
2p
18;834 ¼ 4077s ð1:132 hÞ
EXAMPLE 3.2
In the previous example, find the true anomaly at three hours after perigee passage.
Solution
Since the time (10,800 s) is greater than one-half the period, the true anomaly must be greater than 180.
First, we use Eqn (3.12) to calculate the mean anomaly for t ¼ 10,800 s.
Me ¼ 2p
t
T
¼ 2p
10;800
18;830
¼ 3:6029 rad (a)
Kepler’s equation, E esin (E) ¼ Me (with all angles in radians), is then employed to find the eccentric anomaly.
This transcendental equation will be solved using Algorithm 3.1 with an error tolerance of 106
. Since Me p, a
good starting value for the iteration is E0 ¼ Me e/2 ¼ 3.4166. Executing the algorithm yields the following
steps:
Step 1:
E0 ¼ 3:4166
fðE0Þ ¼ 0:085124
f0
ðE0Þ ¼ 1:3585
ratio ¼
0:085124
1:3585
¼ 0:062658
jratioj 106
; so repeat:
Step 2:
E1 ¼ 3:4166 ð0:062658Þ ¼ 3:4793
fðE1Þ ¼ 0:0002134
f0
ðE1Þ ¼ 1:3515
ratio ¼
0:0002134
1:3515
¼ 1:5778 104
jratioj 106
; so repeat:

Step 3:
E2 ¼ 3:4793

1:5778 104

¼ 3:4794
fðE2Þ ¼ 1:5366 109
f0
ðE2Þ ¼ 1:3515
ratio ¼
1:5366 109
1:3515
1:137 109
jratioj 106
; so accept E ¼ 3:4794 as the solution:
Convergence to even more than the desired accuracy occurred after just two iterations. With this value of the
eccentric anomaly, the true anomaly is found from Eqn (3.13a) to be
tan
q
2
¼
1 þ e
1 e
r
tan
E
2
¼
1 þ 0:37255
1 0:37255
r
tan
3:4794
2
¼ 8:67210q ¼ 193:2
EXAMPLE 3.3
Let a satellite be in a 500 km by 5000 km orbit with its apse line parallel to the line from the earth to the sun, as
shown in Figure 3.9. Find the time that the satellite is in the earth’s shadow if: (a) the apogee is toward the sun and
(b) the perigee is toward the sun.
Solution
We start by using the given data to find the primary orbital parameters, e and h. The eccentricity is obtained from
Eqn (2.84),
e ¼
ra rp
ra þ rp
¼
ð6378 þ 5000Þ ð6378 þ 500Þ
ð6378 þ 5000Þ þ ð6378 þ 500Þ
¼ 0:24649 (a)
P
A
To the sun
a
b
c
d
r
Earth
RE
FIGURE 3.9
Satellite passing in and out of the earth’s shadow.

The orbit equation can then be used to find the angular momentum
rp ¼
h2
m
1
1 þ ecosð0Þ
0 6878 ¼
h2
398;600
1
1 þ 0:24649
0 h ¼ 58;458km2
=s (b)
The semimajor axis may be found from Eqn (2.71),
a ¼
h2
m
1
1 e2
¼
58;4582
398;600
1
1 0:246492
¼ 9128 km (c)
or from the fact that a ¼ ðrp þ raÞ=2. The period of the orbit follows from Eqn (2.83),
T ¼
2p
ffiffiffi
m
p a3=2
¼
2p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
398;600
p 91283=2
¼ 8679:1 s ð2:4109 hÞ (d)
(a) If the apogee is toward the sun, as in Figure 3.9, then the satellite is in earth’s shadow between points a and b
on its orbit. These are two of the four points of intersection of the orbit with the lines that are parallel to the
earthesun line and lie at a distance RE from the center of the earth. The true anomaly of b is therefore given by
sin q ¼ RE/r, where r is the radial position of the satellite. It follows that the radius of b is
r ¼
RE
sin q
(e)
From Eqn (2.72), we also have
r ¼
a

1 e2

1 þ ecos q
(f)
Substituting Eqn (e) into Eqn (f), collecting terms and simplifying yields an equation in q,
ecos q

1 e2
a
RE
sin q þ 1 ¼ 0 (g)
Substituting Eqns (a) and (c) together with RE ¼ 6378 km into Eqn (g) yields
0:24649cos q 1:3442sin q ¼ 1 (h)
This equation is of the form
A cos q þ B sin q ¼ C (i)
It has two roots, which are given by (see Exercise 3.9)
q ¼ tan1 B
A
cos1 C
A
cos

tan1 B
A

(j)
For the case at hand,
q ¼ tan1 1:3442
0:24649
cos1 1
0:24649
cos

tan1 1:3442
0:24649

¼ 79:607
137:03

That is
qb ¼ 57:423
qc ¼ 216:64 ðþ143:36Þ
(k)
For apogee toward the sun, the flight from perigee to point b will be in shadow. To find the time of flight from
perigee to point b, we first compute the eccentric anomaly of b using Eqn (3.13b):
Eb ¼ 2 tan1
1 e
1 þ e
r
tan
qb
2
!
¼ 2tan1
1 0:24649
1 þ 0:24649
r
tan
1:0022
2
!
¼ 0:80521 rad (l)
From this we find the mean anomaly using Kepler’s equation,
Me ¼ E esin E ¼ 0:80521 0:24649 sin 0:80521 ¼ 0:62749 rad (m)
Finally, Eqn (3.5) yields the time at b,
tb ¼
Me
2p
T ¼
0:62749
2p
8679:1 ¼ 866:77 s (n)
The total time in shadow, from a to b, during which the satellite passes through perigee, is
t ¼ 2tb ¼ 1734 s ð28:98 minÞ (o)
(b) If the perigee is toward the sun, then the satellite is in shadow near apogee, from point c (qc ¼ 143.36
)
to d on the orbit. Following the same procedure as above, we obtain (see Exercise 3.11),
Ec ¼ 2:3364 rad
Mc ¼ 2:1587 rad
tc ¼ 2981:8 s
(p)
The total time in shadow, from c to d, is
t ¼ T 2tc ¼ 8679:1 2$2891:8 ¼ 2716 s ð45:26 minÞ (q)
The time is longer than that given by Eqn (o) since the satellite travels slower near apogee.
We have observed that there is no closed-form solution for the eccentric anomaly E in Kepler’s
equation, E esin E ¼ Me. However, there exist infinite series solutions. One of these, due to Lagrange
(Battin, 1999), is a power series in the eccentricity e,
E ¼ Me þ
X
N
n¼1
anen
(3.18)
where the coefficient an is given by the somewhat intimidating expression
an ¼
1
2n1
X
floorðn=2Þ
k¼0
ð1Þk 1
ðn kÞ!k!
ðn 2kÞn1
sin½ðn 2kÞMe (3.19)

Here, floor(x) means x rounded to the next lowest integer [e.g., floor(0.5) ¼ 0, floor(p) ¼ 3]. If e is
sufficiently small, then the Lagrange series converges. This means that by including enough terms in the
summation, we can obtain E to any desired degree of precision. Unfortunately, if e exceeds 0.6627434193,
the series diverges, which means taking more and more terms yields worse and worse results for some
values of M.
The limiting value for the eccentricity was discovered by the French mathematician Pierre-Simone
Laplace (1749–1827) and is called the Laplace limit.
In practice, we must truncate the Lagrange series to a finite number of terms N, so that
E ¼ Me þ
X
N
n¼1
anen
(3.20)
For example, setting N ¼ 3 and calculating each an by means of Eqn (3.19) leads to
E ¼ Me þ esin Me þ
e2
2
sin2 Me þ
e3
8
ð3sin3 Me sin MeÞ (3.21)
For small values of the eccentricity e, this yields good agreement with the exact solution of Kepler’s
equation (plotted in Figure 3.6). However, as we approach the Laplace limit, the accuracy degrades
unless more terms of the series are included. Figure 3.10 shows that for an eccentricity of 0.65, just
below the Laplace limit, Eqn (3.21) (N ¼ 3) yields a solution that oscillates around the exact solution
but is fairly close to it everywhere. Setting N ¼ 10 in Eqn (3.20) produces a curve that, at the given
e = 0.65
Exact and N = 10
N = 3
Mean
anomaly,
M
e
π
FIGURE 3.10
Comparison of the exact solution of Kepler’s equation with the truncated Lagrange series solution (N ¼ 3 and
N ¼ 10) for an eccentricity of 0.65.

scale, is indistinguishable from the exact solution. On the other hand, for an eccentricity of 0.90, far
above the Laplace limit, Figure 3.11 reveals that Eqn (3.21) is a poor approximation to the exact
solution, and using N ¼ 10 makes matters even worse.
Another infinite series for E (Battin, 1999) is given by
E ¼ Me þ
X
N
n¼1
2
n
JnðneÞsin n Me (3.22)
where the coefficients Jn are Bessel functions of the first kind, defined by the infinite series
JnðxÞ ¼
X
N
k¼0
ð1Þk
k!ðn þ kÞ!
x
2
nþ2k
(3.23)
J1 through J5 are plotted in Figure 3.12. Clearly, they are oscillatory in appearance and tend toward
zero with increasing x.
It turns out that, unlike the Lagrange series, the Bessel function series solution converges for all
values of the eccentricity less than 1. Figure 3.13 shows how the truncated Bessel series solutions
E ¼ Me þ
X
N
n¼1
2
n
JnðneÞsin n Me (3.24)
Exact
N = 10
N = 3
0
e = 0.9
Mean
anomaly,
M
e
FIGURE 3.11
Comparison of the exact solution of Kepler’s equation with the truncated Lagrange series solution (N ¼ 3 and

FIGURE 3.12
Bessel functions of the first kind.
0
e = 0.99
Exact
N = 3
N = 10
Mean
anomaly,
M
e
FIGURE 3.13
Comparison of the exact solution of Kepler’s equation with the truncated Bessel series solution (N ¼ 3 and

for N ¼ 3 and N ¼ 10 compare to the exact solution of Kepler’s equation for the very large elliptical
eccentricity of e ¼ 0.99. It can be seen that the case N ¼ 3 yields a poor approximation for all but a few
values of Me. Increasing the number of terms in the series to N ¼ 10 obviously improves the
approximation, and adding even more terms will make the truncated series solution indistinguishable
from the exact solution at the given scale.
Observe that we can combine Eqn (3.10) and Eqn (2.72) as follows to obtain the orbit equation for
the ellipse in terms of the eccentric anomaly:
r ¼
a

1 e2

1 þ ecos q
¼
a

1 e2

1 þ e

e cos E
ecos E 1

From this it is easy to see that
r ¼ að1 ecos EÞ (3.25)
In Eqn (2.86), we defined the true-anomaly-averaged radius rq of an elliptical orbit. Alternatively, the
time-averaged radius rt of an elliptical orbit is defined as
rt ¼
1
T
ZT
0
rdt (3.26)
According to Eqns (3.12) and (3.14),
t ¼
T
2p
ðE esin EÞ
Therefore,
dt ¼
T
2p
ð1 ecos EÞdE
On using this relationship to change the variable of integration from t to E and substituting Eqn (3.25),
Eqn (3.26) becomes
rt ¼
1
T
Z
2p
0
½að1 ecos EÞ
T
2p
ð1 ecos EÞ dE
¼
a
2p
Z
2p
0
ð1 ecos EÞ2
dE
¼
a
2p
Z
2p
0

1 2ecos E þ e2
cos2
E

dE
¼
a
2p

2p 0 þ e2
p


so that
rt ¼ a

1 þ
e2
2

Time-averaged radius of an elliptical orbit (3.27)
Comparing this result with Eqn (2.87) reveals, as we should have expected, that rt rq. In fact,
combining Eqn (2.87) and Eqn (3.27) yields
rq ¼ a
3 2
rt
a
r
(3.28)
3.5 Parabolic trajectories (e [ 1)
For the parabola, Eqn (3.2) becomes
m2
h3
t ¼
Zq
0
dw
ð1 þ cos wÞ2
(3.29)
Setting a ¼ b ¼ 1 in Eqn (3.4) yields
Zq
0
dw
ð1 þ cos wÞ2
¼
1
2
tan
q
2
þ
1
6
tan3 q
2
Therefore, Eqn (3.29) may be written as
Mp ¼
1
2
tan
q
2
þ
1
6
tan3 q
2
(3.30)
where
Mp ¼
m2t
h3
(3.31)
Mp is dimensionless, and it may be thought of as the “mean anomaly” for the parabola. Equation (3.30)
is plotted in Figure 3.14. Equation (3.30) is also known as Barker’s equation.
Given the true anomaly q, we find the time directly from Eqns (3.30) and (3.31). If time is the given
variable, then we must solve the cubic equation
1
6

tan
q
2
3
þ
1
2
tan
q
2
Mp ¼ 0
which has but one real root, namely,
tan
q
2
¼

3Mp þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi

3Mp
2
þ 1
q 1
3

3Mp þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi

3Mp
2
þ 1
q
1
3
(3.32)
3.5 Parabolic trajectories (e ¼ 1) 163

EXAMPLE 3.4
A geocentric parabola has a perigee velocity of 10 km/s. How far is the satellite from the center of the earth
six hours after perigee passage?
Solution
The first step is to find the orbital parameters e and h. Of course we know that e ¼ 1. To get the angular momentum,
we can use the given perigee speed and Eqn (2.80) (the energy equation) to find the perigee radius,
rp ¼
2m
v2
p
¼
2$398;600
102
¼ 7972 km
It follows from Eqn (2.31) that the angular momentum is
h ¼ rpvp ¼ 7972$10 ¼ 79;720 km2
=s
We can now calculate the parabolic mean anomaly by means of Eqn (3.31),
Mp ¼
m2t
h3
¼
398;6002$ð6$3600Þ
79;7203
¼ 6:7737 rad
Therefore, 3Mp ¼ 20.321 rad, which, when substituted into Eqn (3.32), yields the true anomaly,
tan
q
2
¼

20:321 þ
20:3212 þ 1
p 1
3

20:321 þ
20:3212 þ 1
p 1
3
¼ 3:1481 0 q ¼ 144:75
Finally, we substitute the true anomaly into the orbit equation to find the radius,
r ¼
79;7202
398;600
1
1 þ cos ð144:75

Þ
¼ 86;899 km
2
Mean
anomaly,
M
p
2
FIGURE 3.14
Graph of Eqn (3.30).

3.6 Hyperbolic trajectories (e 1)
Setting a ¼ 1 and b ¼ e in Eqn (3.5) yields
Zq
0
dw
ð1 þ ecos wÞ2
¼
1
e2 1

esin q
1 þ ecos q

1
e2 1
p ln
e þ 1
p
þ
e 1
p
tanðq=2Þ
e þ 1
p

e 1
p
tanðq=2Þ
!#
Therefore, for the hyperbola, Eqn (3.1) becomes
m2
h3
t ¼
1
e2 1
esin q
1 þ ecos q

1
ðe2 1Þ
3
2
ln
e þ 1
p
þ
e 1
p
tanðq=2Þ
e þ 1
p

e 1
p
tanðq=2Þ
!
Multiplying both sides by ðe2 1Þ
3
2, we get
Mh ¼
e
e2 1
p
sin q
1 þ ecos q
ln
e þ 1
p
þ
e 1
p
tanðq=2Þ
e þ 1
p

e 1
p
tanðq=2Þ
!
(3.33)
where
Mh ¼
m2
h3

e2
1
3
2
t (3.34)
Mh is the hyperbolic mean anomaly. Equation (3.33) is plotted in Figure 3.15. Recall that q cannot
exceed qN (Eqn (2.97)).
We can simplify Eqn (3.33) by introducing an auxiliary angle analogous to the eccentric anomaly E
for the ellipse. Consider a point on a hyperbola whose polar coordinates are r and q. Referring to
Figure 3.16, let x be the horizontal distance of the point from the center C of the hyperbola, and let y be
its distance above the apse line. The ratio y/b defines the hyperbolic sine of the dimensionless variable
0.01
1
100
10,000
2
e = 1.1
e = 1.5
e = 2.0
e = 3.0
e = 5.0
Mean
anomaly,
M
h
FIGURE 3.15
Plots of Eqn (3.33) for several different eccentricities.
3.6 Hyperbolic trajectories (e 1) 165

F that we will use as the hyperbolic eccentric anomaly. That is, we define F to be such that
sinh F ¼
y
b
(3.35)
In view of the equation of a hyperbola,
x2
a2

y2
b2
¼ 1
it is consistent with the definition of sinh F to define the hyperbolic cosine as
cosh F ¼
x
a
(3.36)
(It should be recalled that sinhx ¼ ðex exÞ=2 and cosh x ¼ ðex þ exÞ=2 and, therefore, that
cosh2
x sinh2
x ¼ 1.)
From Figure 3.16 we see that y ¼ r sin q. Substituting this into Eqn (3.35), along with
r ¼ aðe2 1Þ=ð1 þ ecos qÞ (Eqn (2.104)) and b ¼ a
e2 1
p
(Eqn (2.106)), we get
sinh F ¼
1
b
rsin q ¼
1
a
e2 1
p
a

e2 1

1 þ ecos q
sin q
Focus
P
rp a
Apse
line
C
b
M
x
y
A
s
y
m
p
t
o
t
e
FIGURE 3.16
Hyperbolic parameters.

so that
sinh F ¼
e2 1
p
sin q
1 þ ecos q
(3.37)
This can be used to solve for F in terms of the true anomaly,
F ¼ sinh1
e2 1
p
sin q
1 þ ecos q
!
(3.38)
Using the formula sinh1
x ¼ lnðx þ
x2 þ 1
p
Þ, we can, after simplifying the algebra, write
Eqn (3.38) as
F ¼ ln
sin q
e2 1
p
þ cos q þ e
1 þ ecos q
!
Substituting the trigonometric identities,
sin q ¼
2tan ðq=2Þ
1 þ tan2 ðq=2Þ
cos q ¼
1 tan2ðq=2Þ
1 þ tan2ðq=2Þ
and doing some more algebra yields
F ¼ ln

1 þ e þ ðe 1Þ tan2ðq=2Þ þ 2 tanðq=2Þ
e2 1
p
1 þ e þ ð1 eÞ tan2ðq=2Þ
#
Fortunately, but not too obviously, the numerator and the denominator in the brackets have a common
factor, so that this expression for the hyperbolic eccentric anomaly reduces to
F ¼ ln
e þ 1
p
þ
e 1
p
tanðq=2Þ
e þ 1
p

e 1
p
tanðq=2Þ
#
(3.39)
Substituting Eqns (3.37) and (3.39) into Eqn (3.33) yields Kepler’s equation for the hyperbola,
Mh ¼ esinh F F (3.40)
This equation is plotted for several different eccentricities in Figure 3.17.
If we substitute the expression for sinh F, Eqn (3.37), into the hyperbolic trig identity
cosh2
F sinh2
F ¼ 1
we get
cosh2
F ¼ 1 þ
e2 1
p
sin q
1 þ ecos q
!2

A few steps of algebra lead to
cosh2
F ¼

cos q þ e
1 þ ecos q
2
so that
cosh F ¼
cos q þ e
1 þ ecos q
(3.41a)
Solving this for cos q, we obtain the inverse relation,
cos q ¼
cosh F e
1 ecosh F
(3.41b)
The hyperbolic tangent is found in terms of the hyperbolic sine and cosine by the formula
tanh F ¼
sinh F
cosh F
In mathematical handbooks, we can find the hyperbolic trig identity,
tanh
F
2
¼
sinh F
1 þ cosh F
(3.42)
Substituting Eqns (3.37) and (3.41a) into this formula and simplifying yields
tanh
F
2
¼
e 1
e þ 1
r
sin q
1 þ cos q
(3.43)
Interestingly enough, Eqn (3.42) holds for ordinary trig functions, too; that is,
tan
q
2
¼
sin q
1 þ cos q
0.01
1
100
10,000
2 4
1 3 6
5
Eccentric anomaly, F
e = 1.1
e = 1.5
e = 2.0
e = 3.0
e = 5.0
Mean
anomaly,
M
h
FIGURE 3.17
Plot of Kepler’s equation for the hyperbola.

Therefore, Eqn (3.43) can be written
tanh
F
2
¼
e 1
e þ 1
r
tan
q
2
(3.44a)
This is a somewhat simpler alternative to Eqn (3.39) for computing eccentric anomaly from true
anomaly, and it is a whole lot simpler to invert:
tan
q
2
¼
e þ 1
e 1
r
tanh
F
2
(3.44b)
If time is the given quantity, then Eqn (3.40)da transcendental equationdmust be solved for F by an
iterative procedure, as was the case for the ellipse. To apply Newton’s procedure to the solution of
Kepler’s equation for the hyperbola, we form the function
fðFÞ ¼ esinhF F Mh
and seek the value of F that makes f(F) ¼ 0. Since
f0
ðFÞ ¼ ecosh F 1
Equation (3.16) becomes
Fiþ1 ¼ Fi
esinh Fi Fi Mh
ecosh Fi 1
(3.45)
All quantities in this formula are dimensionless (radians, not degrees).
n
ALGORITHM 3.2
Solve Kepler’s equation for the hyperbola for the hyperbolic eccentric anomaly F given the ec-
centricity e and the hyperbolic mean anomaly Mh. See Appendix D.12 for the implementation of
this algorithm in MATLAB.
1. Choose an initial estimate of the root F.
a. For hand computations, read a rough value of F0 (no more than two significant figures) from
Figure 3.17 in order to keep the number of iterations to a minimum.
b. In computer software, let F0 ¼ Mh, an inelegant choice that may result in many iterations but
will nevertheless rapidly converge on today’s high-speed desktops and laptops.
2. At any given step, having obtained Fi from the previous step, calculate fðFiÞ ¼ esinh Fi Fi
Mh and f0
ðFiÞ ¼ ecosh Fi 1.
3. Calculate ratioi ¼ fðFiÞ=f0
ðFiÞ.
), then calculate an updated value of F,
Fiþ1 ¼ Fi ratioi
Return to Step 2.
5. If jratioij is less than the tolerance, then accept Fi as the solution to within the desired accuracy.
n

EXAMPLE 3.5
A geocentric trajectory has a perigee velocity of 15 km/s and a perigee altitude of 300 km. (a) Find the radius and
the time when the true anomaly is 100. (b) Find the position and speed 3 h later.
Solution
We first calculate the primary orbital parameters e and h. The angular momentum is calculated from Eqn (2.31)
and the given perigee data:
h ¼ rpvp ¼ ð6378 þ 300Þ$15 ¼ 100;170 km2
=s
The eccentricity is found by evaluating the orbit equation, r ¼ ðh2=mÞ½1=ð1 þ ecos qÞ, at perigee:
6378 þ 300 ¼
100;1702
398;600
1
1 þ e
0e ¼ 2:7696
(a) Since e 1, the trajectory is a hyperbola. Note that the true anomaly of the asymptote of the hyperbola is,
according to Eqn (2.97),
qN ¼ cos1

1
2:7696

¼ 111:17
Solving the orbit equation at q ¼ 100 yields
r ¼
100;1702
398;600
1
1 þ 2:7696 cos 100
¼ 48;497 km
To find the time since perigee passage at q ¼ 100, we first use Eqn (3.44a) to calculate the hyperbolic
eccentric anomaly,
tanh
F
2
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2:7696 1
2:7696 þ 1
r
tan
100
2
¼ 0:81653 0 F ¼ 2:2927 rad
Kepler’s equation for the hyperbola then yields the mean anomaly,
Mh ¼ esinh F F ¼ 2:7696 sinh 2:2927 2:2927 ¼ 11:279 rad
The time since perigee passage is found from Eqn (3.34),
t ¼
h3
m2
1

e2 1
3
2
Mh ¼
100;1703
398;6002
1

2:76962 1
3
2
11:279 ¼ 4141 s
(b) After 3 h, the time since perigee passage is
t ¼ 4141:4 þ 3$3600 ¼ 14;941 s ð4:15 hÞ
The corresponding mean anomaly, from Eqn (3.34), is
Mh ¼
398;6002
100;1703

2:76962
1
3
2
14;941 ¼ 40:690 rad
We will use Algorithm 3.2 with an error tolerance of 106
to find the hyperbolic eccentric anomaly F. Referring to
Figure 3.17, we see that for Mh ¼ 40.69 and e ¼ 2.7696, F lies between 3 and 4. Let us arbitrarily choose F0 ¼ 3
as our initial estimate of F. Executing the algorithm yields the following steps:
F0 ¼ 3

Step 1:
fðF0Þ ¼ 15:944494
f0
ðF0Þ ¼ 26:883397
ratio ¼ 0:59309818
F1 ¼ 3 ð0:59309818Þ ¼ 3:5930982
jratioj 106
; so repeat:
Step 2:
fðF1Þ ¼ 6:0114484
f0
ðF1Þ ¼ 49:370747
ratio ¼ 0:12176134
F2 ¼ 3:5930982 ð0:12176134Þ ¼ 3:4713368
jratioj 106
; so repeat:
Step 3:
fðF2Þ ¼ 0:35812370
f0
ðF2Þ ¼ 43:605527
ratio ¼ 8:2128052 103
F3 ¼ 3:4713368

8:2128052 103

¼ 3:4631240
jratioj 106
; so repeat:
Step 4:
fðF3Þ ¼ 1:4973128 103
f0
ðF3Þ ¼ 43:241398
ratio ¼ 3:4626836 105
F4 ¼ 3:4631240

3:4626836 105

¼ 3:4630894
jratioj 106
; so repeat:
Step 5:
fðF4Þ ¼ 2:6470781 103
f0
ðF4Þ ¼ 43:239869
ratio ¼ 6:1218459 1010
F5 ¼ 3:4630894

6:1218459 1010

¼ 3:4630894
jratioj 106
; so accept F ¼ 3:4631 as the solution:
We substitute this value of F into Eqn (3.44b) to find the true anomaly,
tan
q
2
¼
e þ 1
e 1
r
tanh
F
2
¼
2:7696 þ 1
2:7696 1
r
tanh
3:4631
2
¼ 1:3708 0 q ¼ 107:78
With the true anomaly, the orbital equation yields the radial coordinate at the final time
r ¼
h2
m
1
1 þ e cos q
¼
100;1702
398;600
1
1 þ 2:7696 cos 107:78
¼ 163;180 km

The velocity components are obtained from Eqn (2.31),
vt ¼
h
r
¼
100;170
163;180
¼ 0:61386 km=s
and Eqn (2.49),
vr ¼
m
h
esin q ¼
398;600
100;170
2:7696 sin 107:78
¼ 10:494 km=s
Therefore, the speed of the spacecraft is
v ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v2
r þ v2
t
q
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
10:4942 þ 0:613862
p
¼ 10:51 km=s
Note that the hyperbolic excess speed for this orbit is
vN ¼
m
h
e sin qN ¼
398;600
100;170
$2:7696$sin 111:7
¼ 10:277 km=s
The results of this analysis are shown in Figure 3.18.
100°
107.78°
Initial
position
Position three hours later
163,180 km
48,497 km
Perigee
Apse
line
FIGURE 3.18
Given and computed data for Example 3.5.

When determining orbital position as a function of time with the aid of Kepler’s equation, it is
convenient to have position r as a function of eccentric anomaly F. This is obtained by substituting Eqn
(3.41b) into Eqn (2.104),
r ¼
a

e2 1

1 þ ecos q
¼
a

e2 1

1 þ e

cos Fe
1ecos F

This reduces to
r ¼ aðecosh F 1Þ (3.46)
3.7 Universal variables
The equations for elliptical and hyperbolic trajectories are very similar, as can be seen from Table 3.1.
Observe, for example, that the hyperbolic mean anomaly is obtained from that of the ellipse as follows:
Mh ¼ m2
h3

e2 1
3
2
t
¼ m2
h3 ð1Þ

1 e2
3
2
t
¼ m2
h3 ð1Þ
3
2

1 e2
3
2
t
¼ m2
h3 ðiÞ

1 e2
3
2
t
¼ i m2
h3

1 e2
3
2
t
¼ iMe
In fact, the formulas for the hyperbola can all be obtained from those of the ellipse by replacing the
variables in the ellipse equations according to the following scheme, wherein “)” means “replace by”:
a) a
b)ib
Me) iMh
E)iF

i ¼
ffiffiffiffiffiffiffi
1
p
Note in this regard that sin (iF) ¼ i sinh F and cos (iF) ¼ cosh F. Relations among the circular and
hyperbolic trig functions are found in mathematics handbooks, such as Beyer (1991).
In the universal variable approach, the semimajor axis of the hyperbola is considered to have a
negative value, so that the energy equation (row 5 of Table 3.1) has the same form for any type of orbit,
including the parabola, for which a ¼ N. In this formulation, the semimajor axis of any orbit is found
using (row 3),
a ¼
h2
m
1
1 e2
(3.47)
3.7 Universal variables 173

If the position r and velocity v are known at a given point on the path, then the energy equation (row 5)
is convenient for finding the semimajor axis of any orbit,
a ¼
1
2
r v2
m
(3.48)
Kepler’s equation may also be written in terms of a universal variable, or universal “anomaly” c, that is
valid for all orbits. See, for example, Battin (1999), Bond and Allman (1993), and Prussing and
Conway (1993). If t0 is the time when the universal variable is 0, then the value of c at time t0 þ Dt is
found by iterative solution of the universal Kepler’s equation
ffiffiffi
m
p
Dt ¼
r0vr0
ffiffiffi
m
p c2
C

ac2

þ ð1 ar0Þc3
S

ac2

þ r0c (3.49)
in which r0 and vr0
are the radius and radial velocity, respectively, at t ¼ t0, and a is the reciprocal of the
semimajor axis
a ¼
1
a
(3.50)
a 0, a ¼ 0, and a 0 for hyperbolas, parabolas, and ellipses, respectively. The units of c are km1/2
(so ac2
is dimensionless). The functions C(z) and S(z) belong to the class known as Stumpff functions,
and they are defined by the infinite series,
SðzÞ ¼
X
N
k¼0
ð1Þk zk
ð2k þ 3Þ!
¼
1
6

z
120
þ
z2
5040

z3
362;880
þ / (3.51a)
Table 3.1 Comparison of Some of the Orbital Formulas for the Ellipse and Hyperbola
Equation Ellipse (e 1) Hyperbola (e 1)
1. Orbit Eqn (2.45) r ¼
h2
m
1
1 þ ecos q
Same
2. Conic equation in Cartesian coordinates (2.79)
and (2.109)
x2
a2
þ
y2
b2
¼ 1
x2
a2

y2
b2
¼ 1
3. Semimajor axis Eqns (2.71) and (2.103) a ¼
h2
m
1
1 e2
a ¼
h2
m
1
e2 1
4. Semiminor axis Eqns (2.76) and (2.06) b ¼ a
1 e2
p
b ¼ a
e2 1
p
5. Energy Eqns (2.81) and (2.111) v2
2

m
r
¼
m
2a
v2
2

m
r
¼
m
2a
6. Mean anomaly Eqns (3.7) and (3.34) Me ¼
m2
h3
ð1 e2
Þ
3
2t Mh ¼
m2
h3
ðe2
1Þ
3
2t
7. Kepler’s Eqns (3.14) and (3.40) Me ¼ E esin E Mh ¼ esinh F F
8. Orbit equation in terms of eccentric anomaly
(3.25) and (3.46)
r ¼ a(1 ecos E) r ¼ a(ecosh F 1)

CðzÞ ¼
X
N
k¼0
ð1Þk zk
ð2k þ 2Þ!
¼
1
2

z
24
þ
z2
720

z3
40;320
þ / (3.51b)
C(z) and S(z) are related to the circular and hyperbolic trig functions as follows:
SðzÞ ¼
8

:
ffiffi
z
p
sin
ffiffi
z
p
ð
ffiffi
z
p
Þ3
ðz 0Þ
sinh
ffiffiffiffiffiffi
z
p

ffiffiffiffiffiffi
z
p
ffiffiffiffiffiffi
z
p 3
ðz 0Þ
1
6
ðz ¼ 0Þ

z ¼ ac2

(3.52)
CðzÞ ¼
8

:
1 cos
ffiffi
z
p
z
ðz 0Þ
cosh
ffiffiffiffiffiffi
z
p
1
z
ðz 0Þ
1
2
ðz ¼ 0Þ

z ¼ ac2

(3.53)
Clearly, z 0, z ¼ 0, and z 0 for hyperbolas, parabolas, and ellipses, respectively. It should be
pointed out that if C(z) and S(z) are computed by the series expansions, Eqns (3.51a,b), then the forms
of C(z) and S(z), depending on the sign of z, are selected, so to speak, automatically. C(z) and S(z)
behave as shown in Figure 3.19. Both C(z) and S(z) are nonnegative functions of z. They increase
without bound as z approaches N and tend toward zero for large positive values of z. As can be seen
from Eqn (3.53), for z 0, C(z) ¼ 0 when cos
ffiffi
z
p
¼ 1, that is, when z ¼ (2p)2
, (4p)2
, (6p)2
, ..
–30 –20 –10
2
4
6
8
10
10 20
0.1
0.2
0.3
0.4
100 200 300 400
0.01
0.02
0.03
0
12 0.5
–50
0
–40
0
0 30 0 500
0.04
0
z
z
z
S(z)
C(z)
S(z)
C(z)
C(z)
S(z)
FIGURE 3.19
A plot of the Stumpff functions C(z) and S(z).

The price we pay for using the universal variable formulation is having to deal with the relatively
unknown Stumpff functions. However, Eqns (3.52) and (3.53) are easy to implement in both computer
programs and programmable calculators. See Appendix D.13 for the implementation of these ex-
pressions in MATLAB.
To gain some insight into how Eqn (3.49) represents the Kepler equations for all the conic sections,
let t0 be the time at periapse passage and let us set t0 ¼ 0, as we have assumed previously. Then Dt ¼ t,
vr0
¼ 0, and r0 equals rp, the periapsis radius. In that case Eqn (3.49) reduces to
ffiffiffi
m
p
t ¼

1 arp

c3
S

ac2

þ rpc ðt ¼ 0 at periapse passageÞ (3.54)
Consider first the parabola. In this case a ¼ 0, and S ¼ S(0) ¼ 1/6, so that Eqn (3.54) becomes a cubic
polynomial in c,
ffiffiffi
m
p
t ¼
1
6
c3
þ rpc
Multiply this equation through by ð
ffiffiffi
m
p
=hÞ3
to obtain
m2
h3
t ¼
1
6

c
ffiffiffi
m
p
h
3
þ rpc
ffiffiffi
m
p
h
3
Since rp ¼ h2=2m for a parabola, we can write this as
m2
h3
t ¼
1
6
ffiffiffi
m
p
h
c
3
þ
1
2
ffiffiffi
m
p
h
c

(3.55)
Upon setting c ¼ h tan ðq=2Þ=
ffiffiffi
m
p
, Eqn (3.55) becomes identical to Eqn (3.30), the time vs true
anomaly relation for the parabola.
Kepler’s equation for the ellipse can be obtained by multiplying Eqn (3.54) throughout by
ð
mð1 e2Þ
p
=hÞ3
:
m2
h3

1 e2
3
2
t ¼

c
ffiffiffi
m
p
h
1 e2
p 3
1 arp

SðzÞ þ rpc
ffiffiffi
m
p
h
1 e2
p 3
z ¼ ac2

(3.56)
Recall that for the ellipse, rp ¼ h2=½mð1 þ eÞ and a ¼ 1=a ¼ mð1 e2Þ=h2. Using these two
expressions in Eqn (3.56), along with SðzÞ ¼ ½
ffiffiffi
a
p
c sinð
ffiffiffi
a
p
cÞ=a
3
2c3 (from Eqn (3.52)), and working
through the algebra ultimately leads to
Me ¼
c
ffiffiffi
a
p esin

c
ffiffiffi
a
p

Comparing this with Kepler’s equation for an ellipse (Eqn (3.14)) reveals that the relationship between
the universal variable c and the eccentric anomaly E is c ¼
ffiffiffi
a
p
E. Similarly, it can be shown for
hyperbolic orbits that c ¼
a
p
F. In summary, the universal anomaly c is related to the previously
encountered anomalies as follows:
c ¼
8

:
h
ffiffiffi
m
p tan
q
2
parabola
ffiffiffi
a
p
E ellipse
a
p
F hyperbola
ðt0 ¼ 0; at periapseÞ (3.57)

When t0 is the time at a point other than periapsis, so that Eqn (3.49) applies, then Eqn (3.57) becomes
c ¼
8

:
h
ffiffiffi
m
p

tan
q
2
tan
q0
2

parabola
ffiffiffi
a
p
ðE E0Þ ellipse
a
p
ðF F0Þ hyperbola
(3.58)
As before, we can use Newton’s method to solve Eqn (3.49) for the universal anomaly c, given the time
interval Dt. To do so, we form the function
fðcÞ ¼
r0vr0
ffiffiffi
m
p c2
CðzÞ þ ð1 ar0Þc3
SðzÞ þ r0c
ffiffiffi
m
p
Dt (3.59)
and its derivative
dfðcÞ
dc
¼ 2
r0vr0
ffiffiffi
m
p cCðzÞ þ
r0vr0
ffiffiffi
m
p c2dCðzÞ
dz
dz
dc
þ 3ð1 ar0Þc2
SðzÞ þ ð1 r0aÞc3dSðzÞ
dz
dz
dc
þ r0 (3.60)
where it is to be recalled that
z ¼ ac2
(3.61)
which means of course that
dz
dc
¼ 2ac (3.62)
It turns out that
dSðzÞ
dz
¼
1
2z
½CðzÞ 3SðzÞ
dCðzÞ
dz
¼
1
2z
½1 zSðzÞ 2CðzÞ
(3.63)
Substituting Eqns (3.61)–(3.63) into Eqn (3.60) and simplifying the result yields
dfðcÞ
dc
¼
r0vr0
ffiffiffi
m
p c 1 ac2
SðzÞ þ ð1 ar0Þc2
CðzÞ þ r0 (3.64)
With Eqns (3.59) and (3.64), Newton’s algorithm (Eqn (3.16)) for the universal Kepler’s equation
becomes
ciþ1 ¼ ci
r0vr0
ffiffiffi
m
p c2
i CðziÞ þ ð1 ar0Þc3
i SðziÞ þ r0ci
ffiffiffi
m
p
Dt
r0vr0
ffiffiffi
m
p ci 1 ac2
i SðziÞ þ ð1 ar0Þc2
i CðziÞ þ r0

zi ¼ ac2
i

(3.65)
According to Chobotov (2002), a reasonable estimate for the starting value c0 is
c0 ¼
ffiffiffi
m
p
jajDt (3.66)

n
ALGORITHM 3.3
Solve the universal Kepler’s equation for the universal anomaly c given Dt, r0, vr0
, and a. See
Appendix D.14 for an implementation of this procedure in MATLAB.
1. Use Eqn (3.66) for an initial estimate of c0.
2. At any given step, having obtained ci from the previous step, calculate
fðciÞ ¼
r0vr0
ffiffiffi
m
p c2
i CðziÞ þ ð1 ar0Þc3
i SðziÞ þ r0ci
ffiffiffi
m
p
Dt
and
f0
ðciÞ ¼
r0vr0
ffiffiffi
m
p ci 1 ac2
i SðziÞ þ ð1 ar0Þc2
i CðziÞ þ r0
where zi ¼ ac2
i .
3. Calculate ratioi ¼ fðciÞ=f 0
ðciÞ.
), then calculate an updated value of c,
ciþ1 ¼ ci ratioi
Return to Step 2.
5. If jratioij is less than the tolerance, then accept ci as the solution to within the desired accuracy.
n
EXAMPLE 3.6
An earth satellite has an initial true anomaly of q0 ¼ 30, a radius of r0 ¼ 10,000 km, and a speed of v0 ¼ 10 km/s.
Use the universal Kepler’s equation to find the change in universal anomaly c after one hour and use that infor-
mation to determine the true anomaly q at that time.
Solution
Using the initial conditions, let us first determine the angular momentum and the eccentricity of the trajectory.
From the orbit formula, Eqn (2.45), we have
h ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
mr0ð1 þ e cos q0Þ
p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
398;600$10;000$ð1 þ e cos 30Þ
p
¼ 63;135
1 þ 0:86602e
p
(a)
This, together with the angular momentum formula, Eqn (2.31), yields
vt0 ¼
h
r0
¼
63;135
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ 0:86602e
p
10; 000
¼ 6:3135
1 þ 0:86602e
p
Using the radial velocity relation, Eqn (2.49), we find
vr0 ¼
m
h
esin q0 ¼
398;600
63;135
1 þ 0:86602e
p esin 30
¼ 3:1567
e
1 þ 0:86602e
p (b)
Since v2
r0 þ v2
t0 ¼ v2
0, it follows that

3:1567
e
1 þ 0:86602e
p
2
þ

6:3135
1 þ 0:86602e
p 2
¼ 102

which simplifies to become 39.86e2
17.563e 60.14 ¼ 0. The only positive root of this quadratic equation is
e ¼ 1:4682
Since e is greater than 1, the orbit is a hyperbola. Substituting this value of the eccentricity back into Eqns (a)
and (b) yields the angular momentum
h ¼ 95;154 km2
=s
as well as the initial radial speed
vr0 ¼ 3:0752 km=s
The hyperbolic eccentric anomaly F0 for the initial conditions may now be found from Eqn (3.44a),
tanh
F0
2
¼
e 1
e þ 1
r
tan
q0
2
¼
1:4682 1
1:4682 þ 1
r
tan
30
2
¼ 0:16670
Solving for F0 yields
F0 ¼ 0:23448 rad (c)
In the universal variable formulation, we calculate the semimajor axis of the orbit by means of Eqn (3.47),
a ¼
h2
m
1
1 e2
¼
95;1542
398;600
1
1 1:46822
¼ 19;655 km (d)
The negative value is consistent with the fact that the orbit is a hyperbola. From Eqn (3.50) we get
a ¼
1
a
¼
1
19;655
¼ 5:0878 105
km1
which appears throughout the universal Kepler’s equation.
We will use Algorithm 3.3 with an error tolerance of 106
to find the universal anomaly. From Eqn (3.66), our
initial estimate is
c0 ¼
398;600
p
$ 5:0878 106
$3600 ¼ 115:6
Executing the algorithm yields the following steps:
c0 ¼ 115:6
Step 1:
fðc0Þ ¼ 370;650:01
f0
ðc0Þ ¼ 26;956:300
ratio ¼ 13:750033
c1 ¼ 115:6 ð13:750033Þ ¼ 129:35003
jratioj 106
; so repeat:
Step 2:
fðc1Þ ¼ 25;729:002
f0
ðc1Þ ¼ 30;776:401
ratio ¼ 0:83599669
c2 ¼ 129:35003 0:83599669 ¼ 128:51404
jratioj 106
; so repeat:

Step 3:
fðc2Þ ¼ 102:83891
f0
ðc2Þ ¼ 30;530:672
ratio ¼ 3:3683800 103
c3 ¼ 128:51404 3:3683800 103
¼ 128:51067
jratioj 106
; so repeat:
Step 4:
fðc3Þ ¼ 1:6614116 103
f0
ðc3Þ ¼ 30;529:686
ratio ¼ 5:4419545 108
c4 ¼ 128:51067 5:4419545 108
¼ 128:51067
jratioj 106
So we accept
c ¼ 128:51km
1
2
as the solution after four iterations. Substituting this value of c together with the semimajor axis (Eqn (d)) into
Eqn (3.58) yields
F F0 ¼
c
a
p ¼
128:51
ð19;655Þ
p ¼ 0:91664
It follows from Eqn (b) that the hyperbolic eccentric anomaly after 1 h is
F ¼ 0:23448 þ 0:91664 ¼ 1:1511
Finally, we calculate the corresponding true anomaly using Eqn (3.44b),
tan
q
2
¼
e þ 1
e 1
r
tanh
F
2
¼
1:4682 þ 1
1:4682 1
r
tanh
1:1511
2
¼ 1:1926
which means that after one hour
q ¼ 100:04
Recall from Section 2.11 that the position r and velocity v on a trajectory at any time t can be found in
terms of the position r0 and velocity v0 at time t0 by means of the Lagrange f and g coefficients and
their first derivatives,
r ¼ fr0 þ gv0 (3.67)
v ¼ _
fr0 þ _
gv0 (3.68)
Eqns (2.158) give f, g, _
f, and _
g explicitly in terms of the change in true anomaly Dq over the time
interval Dt ¼ t t0. The Lagrange coefficients can also be derived in terms of changes in the eccentric
anomaly DE for elliptical orbits, DF for hyperbolas, or Dtan(q/2) for parabolas. However, if we take

advantage of the universal variable formulation, we can cover all these cases with the same set of
Lagrange coefficients. In terms of the universal anomaly c and the Stumpff functions C(z) and S(z), the
Lagrange coefficients are (Bond and Allman, 1996)
f ¼ 1
c2
r0
C

ac2

(3.69a)
g ¼ Dt
1
ffiffiffi
m
p c3
S

ac2

(3.69b)
_
f ¼
ffiffiffi
m
p
rr0
ac3
S

ac2

c (3.69c)
_
g ¼ 1
c2
r
C

ac2

(3.69d)
The implementation of these four functions in MATLAB is found in Appendix D.15.
n
ALGORITHM 3.4
Given r0 and v0, find r and v at a time Dt later. See Appendix D.16 for an implementation of this
procedure in MATLAB.
1. Use the initial conditions to find:
a. The magnitude of r0 and v0,
r0 ¼
r0$r0
p
v0 ¼
ffi
v0$v0
p
b. The radial component velocity of vr0
by projecting v0 onto the direction of r0,
vr0
¼
r0$v0
r0
c. The reciprocal a of the semimajor axis, using Eqn (3.48), is given by
a ¼
2
r0

v2
0
m
The sign of a determines whether the trajectory is an ellipse (a 0), parabola (a ¼ 0), or hyperbola
(a 0).
2. With r0, vr0
, a, and Dt, use Algorithm 3.3 to find the universal anomaly c.
3. Substitute a, r0, Dt, and c into Eqns (3.69a,b) to obtain f and g.
4. Use Eqn (3.67) to compute r and, from that, its magnitude r.
5. Substitute a, r0, r, and c into Eqns (3.69c,d) to obtain _
f and _
g.
6. Use Eqn (3.68) to compute v.
n
EXAMPLE 3.7
An earth satellite moves in the xy plane of an inertial frame with origin at the earth’s center. Relative to that frame,
the position and velocity of the satellite at time t0 are
r0 ¼ 7000:0^
i 12; 124^
jðkmÞ
v0 ¼ 2:6679^
i þ 4:6210^
jðkm=sÞ
(a)

Compute the position and velocity vectors of the satellite 60 min later using Algorithm 3.4.
Solution
Step 1:
r0 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
7000:02 þ ð12;124Þ2
q
¼ 14;000 km
v0 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2:66792 þ 4:62102
p
¼ 5:3359 km=s
vr0
¼
7000:0$2:6679 þ ð12;124Þ$4:6210
14;000
¼ 2:6679 km=s
a ¼
2
14;000

5:33592
398;600
¼ 7:1429 105
km1
The trajectory is an ellipse, because a is positive.
Step 2:
Using the results of Step 1, Algorithm 3.3 yields
c ¼ 253:53 km
1
2
which means
z ¼ ac2
¼ 7:1429 105
$253:532
¼ 4:5911
Step 3:
Substituting the above values of c and z into Eqns (3.69a,b), we find
f ¼ 1
c2
r0
C

ac2

¼ 1
253:532
14;000
Cð4:5911Þ
zfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflffl{
0:3357
¼ 0:54123
g ¼ Dt
1
ffiffiffi
m
p c3
S

ac2

¼ 3600
253:533
398;600
p Sð4:5911Þ
0:13233
¼ 184:35 s1
Step 4:
r ¼ fr0 þ gv0
¼ ð0:54123Þ

7000:0^
i 12:124^
j

þ 184:35

2:6679^
i þ 4:6210^
j

¼ 3296:8^
i þ 7413:9^
jðkmÞ
Therefore, the magnitude of r is
r ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð3296:8Þ2
þ 7413:92
q
¼ 8113:9 km
Step 5:
_
f ¼
ffiffiffi
m
p
rr0
ac3S

ac2

c
¼
398;600
p
8113:9$14;000

7:1429 105

$253:533$Sð4:5911Þ
0:13233
253:53
#
¼ 0:00055298 se1

_
g ¼ 1
c2
r
C

ac2

¼ 1
253:532
8113:9
Cð4:5911Þ
0:3357
¼ 1:6593
Step 6:
v ¼ _
fr0 þ _
gv0
¼ ð0:00055298Þ

7000:0^
i 12:124^
j

þ ð1:6593Þv0

2:6679^
i þ 4:6210^
j

¼ 8:2977^
i 0:96309^
j ðkm=sÞ
The initial and final position and velocity vectors, as well as the trajectory, are accurately illustrated in Figure 3.20.
Perigee
î
x
ĵ
y
r
t = t0 + 3600 s
t = t0
r0
O
v
v0
FIGURE 3.20
Initial and final points on the geocentric trajectory of Example 3.7.
PROBLEMS
Section 3.2
3.1 If f ¼
1
2
tan
x
2
þ
1
6
tan3x
2
, then show that df=dx ¼ 1=ð1 þ cos xÞ2
, thereby verifying the integral in
Eqn (3.4).
Section 3.2 183

Section 3.4
3.2 Find the three positive roots of the equation 10esin x
¼ x2
5x þ 4 to eight significant figures. Use
(a) Newton’s method.
(b) Bisection method.
3.3 Find the first four nonnegative roots of the equation tan (x) ¼ tanh (x) to eight significant
figures. Use
(a) Newton’s method.
(b) Bisection method.
3.4 In terms of the eccentricity e, the period T, and the angles a and b (in radians), find the time t
required to fly from point 1 to point 2 on the ellipse. C is the center of the ellipse.

Ans:: t ¼
T
2p
h
b a 2ecos
b þ a
2
sin
b a
2
i
P
A
F
C
a
α
β
2
1
3.5 Calculate the time required to fly from P to B, in terms of the eccentricity e and the period T. B lies
on the minor axis.

Ans::
1
4

e
2p

T

P
A
B
D
F
C
3.6 If the eccentricity of the elliptical orbit is 0.3, calculate, in terms of the period T, the time required
to fly from P to B.
fAns:: 0:157tg

3.7 If the eccentricity of the elliptical orbit is 0.5, calculate, in terms of the period T, the time required
to fly from P to B.
{Ans.: 0.170T}
P
A
B
rp
2rp
F
3.8 A satellite is in earth orbit for which the perigee altitude is 200 km and the apogee altitude is
600 km. Find the time interval during which the satellite remains above an altitude of 400 km.
{Ans.: 47.15 min}
3.9 An earth-orbiting satellite has a perigee radius of 7000 km and an apogee radius of 10,000 km. (a)
What true anomaly Dq is swept out between t ¼ 0.5 h and t ¼ 1.5 h after perigee passage? (b)
What area is swept out by the position vector during that time interval?
{Ans.: (a) 128.7; (b) 1.03 108
km2
}
3.10 An earth-orbiting satellite has a period of 14 h and a perigee radius of 10,000 km. At time
t ¼ 10 h after perigee passage, determine
(a) The radial position.
(b) The speed.
(c) The radial component of the velocity.
{Ans.: (a) 42,356 km; (b) 2.303 km/s; (c) 1.271 km/s}
3.11 A satellite in earth orbit has perigee and apogee radii of rp ¼ 7500 km and ra ¼ 16,000 km,
respectively. Find its true anomaly 40 min after passing the true anomaly of 80
.
{Ans.: 174.7}
3.12 Show that the solution to a cos q þ b sin q ¼ c, where a, b, and c are given, is
q ¼ f cos1
c
a
cos f

where tanf ¼ b=a.
3.13 Verify the results of part (b) of Example 3.3.
P
A
B
F
90°
Section 3.4 185

Section 3.5
3.14 Calculate the time required for a spacecraft launched into a parabolic trajectory at a perigee
altitude of 200 km to leave the earth’s sphere of influence (see Table A.2).
{Ans.: 7.77 days}
3.15 A spacecraft on a parabolic trajectory around the earth has a perigee radius of 6600 km.
(a) How long does it take to coast from q ¼ 90 degrees to q ¼ þ90 degrees?
(b) How far is the spacecraft from the center of the earth 36 h after passing through perigee?
{Ans.: (a) 0.8897 h; (b) 304,700 km}
Section 3.6
3.16 A spacecraft on a hyperbolic trajectory around the earth has a perigee radius of 6600 km and
a perigee speed of 1.2 vesc.
(a) How long does it take to coast from q ¼ 90
to q ¼ þ90
?
(b) How far is the spacecraft from the center of the earth 24 h after passing through perigee?
{Ans.: (a) 0.9992 h; (b) 656,610 km}
3.17 A trajectory has a perigee velocity 1.1 vesc and a perigee altitude of 200 km. If at 10 AM the
satellite is traveling toward the earth with a speed of 8 km/s, how far will it be from the earth’s
surface at 5 PM the same day?
{Ans.: 136,250 km}
3.18 An incoming object is sighted at an altitude of 100,000 km with a speed of 6 km/s and a flight
path angle of 80
. (a) Will it impact the earth or fly by? (b) What is the time either to impact or
closest approach?
{Partial Ans.: (b) 30.4 min}
Section 3.7
3.19 At a given instant, the radial position of an earth-orbiting satellite is 7200 km and its radial speed
is 1 km/s. If the semimajor axis is 10,000 km, use Algorithm 3.3 to find the universal anomaly
60 min later. Check your result using Eqn (3.58).
3.20 At a given instant, a space object has the following position and velocity vectors relative to an
earth-centered inertial frame of reference:
r0 ¼ 20; 000^
i 105; 000^
j 19; 000^
k ðkmÞ
v0 ¼ 0:9000^
i 3:4000^
j 1:5000^
k ðkm=sÞ
Use Algorithm 3.4 to find r and v 2 h later.
{Ans.: r ¼ 26; 338^
i 128; 750^
j 29; 656^
kðkmÞ; v ¼ 0:862; 800^
i 3:2116^
j 1:4613^
kðkm=sÞ}

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