1/31/13 ASEN 3200 Orbital Mechanics,
Attitude Dynamics and Control
Lecture 4: Example Problems
29 January 2013
Michael Croteau,
with input from Dr. Nerem
Announcements:
•  Homework 1 due today.
Place on side table at front before class begins.
•  Homework 2 posted.
1 1/31/13 Example 1: Finding E from Me.
Given e = 0.37255 and Me = 3.6029 rad, find E.
Kepler’s Equation:
E − esin E = M e
We cannot solve for E directly. Must iterate using:
M e − E n + esin E n
€ E n +1 = E n +
1 − ecos E n
We need an initial guess for E0.
Your book suggests E 0 = M e + e 2 for M e < π
and E€0 = M e − e 2 for M e > π . Note: E and Me are in rad
€
Now solve for E1, then use this as your new En and
€and again until €
iterate again
within a certain tolerance.
€
Example 1: Finding E from Me.
Given e = 0.37255 and Me = 3.6029 rad, find E.
Iteration 1:
E 0 = M e − e 2 = 3.6029 − 0.37255 2 = 3.4166
Me > π
M e − E 0 + esin E 0
1 − ecos E 0
€
3.6029 − 3.4166 + 0.37255sin 3.4166
E1 = 3.4166 +
1 − 0.37255cos 3.4166
E1 = 3.4793
E1 = E 0 +
€
check… is |E1 – E0| < tolerance? Say < 1×10-8?
E1 − E 0 = 3.4793 − 3.4166 = 0.0627
€
0.0627 < 1 × 10 −8 ?
No. Iterate again.
€
2 1/31/13 Example 1: Finding E from Me.
Given e = 0.37255 and Me = 3.6029 rad, find E.
Iteration 2:
M e − E1 + esin E1
1 − ecos E1
3.6029 − 3.4793 + 0.37255sin 3.4793
E 2 = 3.4793 +
1 − 0.37255cos 3.4793
E 2 = 3.4794
E 2 = E1 +
check… is |E2 – E1| < tolerance? Say < 1×10-8?
€
E 2 − E1 = 3.4794 − 3.4793 = 0.0001
0.0001 < 1 × 10 −8 ?
No. Iterate again.
€
Example 1: Finding E from Me.
Given e = 0.37255 and Me = 3.6029 rad, find E.
Iteration 3:
M e − E 2 + esin E 2
1 − ecos E 2
3.6029 − 3.4794 + 0.37255sin 3.4794
E 3 = 3.4794 +
1 − 0.37255cos 3.4794
E 3 = 3.4794
E3 = E2 +
check… is |E2 – E1| < tolerance? Say < 1×10-8?
€
Yes, so accept that
E = 3.4794 rad
3 1/31/13 Example 2:
An Earth satellite is observed to have a perigee
height of 100 km and an apogee height of 600 km.
Find the period (and e).
Example 2:
An Earth satellite is observed to have a perigee
height of 100 km and an apogee height of 600 km.
Find the period (and e).
4 1/31/13 Example 3:
How many days each year is the Earth farther from
the Sun that 1 AU? (1 AU = 149,597,870 km)
Example 3:
How many days each year is the Earth farther from
the Sun that 1 AU? (1 AU = 149,597,870 km)
5 1/31/13 Example 4:
Using Mean/Eccentric Anomalies
For a satellite in an Earth orbit with hA = 3000 km
and hP = 300 km, how long does it take to go from
an altitude of 1000 km to one of 2000 km?
Now, find time of flight between E1 and E2.
n(t 2 − t1 ) = M 2 − M1 = 42.94°
€
⇒ t 2 − t1 = 14.2min
€
Example 4:
Using Mean/Eccentric Anomalies
For a satellite in an Earth orbit with hA = 3000 km
and hP = 300 km, how long does it take to go from
an altitude of 1000 km to one of 2000 km?
Now, find time of flight between E1 and E2.
n(t 2 − t1 ) = M 2 − M1 = 42.94°
€
⇒ t 2 − t1 = 14.2min
€
6 1/31/13 Example 5:
Using Mean/Eccentric Anomalies
What is the altitude of this same satellite 10 minutes
past apogee?
à At apogee, M1 = 180°
à 10 minutes past apoapse:
# 10 &
M 2 = M1 +181° /hr⋅ % hr( = 210.16°
$ 60 '
M 2 = 210.16° = E 2 − esin E 2
e = 0.1682
Solve for E2 by iterating, then:
€
€
a = 8028km
r = a(1 − ecos E 2 )
h = r − 6378km
€
€
Example 5:
Using Mean/Eccentric Anomalies
What is the altitude of this same satellite 10 minutes
past apogee?
à At apogee, M1 = 180°
à 10 minutes past apoapse:
# 10 &
M 2 = M1 +181° /hr⋅ % hr( = 210.16°
$ 60 '
M 2 = 210.16° = E 2 − esin E 2
e = 0.1682
Solve for E2 by iterating, then:
€
€
a = 8028km
r = a(1 − ecos E 2 )
h = r − 6378km
€
€
7 1/31/13 aPluto = 39.48AU
ePluto = .25
Example 6:
aΨ = 30.07AU
eΨ = .00859
Determine the amount of time Pluto is within the
orbit of Neptune (Ψ) assuming Ψ’s orbit is circular.
€
€
1.  Determine the true anomaly where the orbits
cross, i.e. rPluto = aΨ.
$ a (1 − e 2 ) '
Solving for θ:
Pluto
) = aΨ
rPluto = &
&%1+ ecos θ Pluto )(
θ Pluto = 22.55°
2.  Find the time it takes Pluto to go from perihelion
to θP: M = n(t2 − t1 ) = E − esin E
€
where
€
1/ 2
1/ 2
E #1 − e &
θ #1 − 0.25 &
22.55
=%
( tan = %
( tan
2 $ 1+ e '
2 $ 1+ 0.25 '
2
E = 17.56°
tan
€
Example 6 continued:
2.  continued...
(t − t P ) = Δt =
E − esin E
n
n=
µSun
(aPluto )
3
n = 7.99 × 10 −10
rad
≈ 0.004 ° day
sec
Δt = 2.89 × 10 8 sec = 3,346days = 9.16years
3.  The time within the €
orbit of Ψ will be twice this time.
€
Time = 2Δt = 18.3years
Notes:
-  Period of Pluto’s orbit = 249.2 years
€
-  Actual perihelion
of Pluto 9/5/89, and Pluto crossed
Ψ orbit 2/11/99 à actual Δt = 9.4 years
-  Atmos. of Pluto will be frozen in 2020, making any
mission less interesting
8 1/31/13 Example 7:
A geocentric elliptical orbit has a perigee radius of 9600
km and an apogee radius of 21,000 km. Calculate the
time to fly from perigee P to a true anomaly of 120°.
t=
Me
T … so we need Me and T.
2π
T = Period =
€
2π $ h '
&
)
µ2 % 1 − e2 (
3
e=
ra − rp 21000 − 9600
=
= 0.37255
ra + rp 21000 + 9600
rp =
'3
2π $
72,472
T=
&
) = 18,834 sec
398,600 2 % 1 − 0.37255 2 (
'
h2 $
1
km 2
&
) ⇒ h = 72,472
µ %1+ ecos θ (
sec
Get Me from E:
€
€
€
E
1−e
θ
tan =
tan = 1.1711 ⇒ E = 1.7281rad
2
1+ e
2
M
1.3601
M e = E − esin E = 1.3601rad
t= eT=
18834 = 4,077sec = 1.132hr
2π
2π
€
9