PHYS 4110 – Dynamics of Space Vehicles
Chapter 7: Orbital Maneuvers
Earth, Moon, Mars, and Beyond
Dr. Jinjun Shan, Professor of Space Engineering
Department of Earth and Space Science and Engineering
Room 255, Petrie Science and Engineering Building
Tel: 416-736 2100 ext. 33854
Email: [email protected]
Homepage: http://www.yorku.ca/jjshan
Impulsive Maneuvers
n 
n 
n 
Impulsive maneuvers are those in which brief firings of onboard
rocket motors change the magnitude and direction of the velocity
vector instantaneously.
The position of the
spacecraft is
considered to be
fixed during the
maneuvers.
This is true for highthrust rockets with
burn times short
compared with
the coasting of
the spacecraft.
Prof. Jinjun Shan
Orbital Maneuvers - 2
Hohmann Transfer - Definition
n 
n 
The Hohmann transfer is an elliptical orbit tangent to both
circles at its apse line. The periapse and apoapse of the
transfer ellipse are the radii of the inner and outer circles.
The Hohmann transfer is the most energy efficient twoimpulse maneuver for transferring between two coplanar
circular orbits sharing a common focus. [Conditional]
Prof. Jinjun Shan
Orbital Maneuvers - 3
Hohmann Transfer - 1
Example 2 in “Two-Body Problem”: Orbits 1 and 2 are two circular orbits
with altitudes of 300 km and 35,786 km, respectively. An elliptical orbit is
tangent to both circles at its apse line. Determine the velocities of spacecraft
at point A and B on both elliptical and circular orbits.
Prof. Jinjun Shan
Orbital Maneuvers - 4
Hohmann Transfer - 2
v1 =
µ
rp
vp =
µ 2ra
rp (ra + rp )
va =
µ 2rp
ra (ra + rp )
v2 =
µ
ra
Δv = Δv A + Δv B = (v p − v1 ) + (v2 − va )
µ
ε1 = −
2rp
µ
ε2 = −
2ra
µ
µ
ε3 = − = −
2a
ra + rp
ra > rp
ε1 < ε 3 < ε 2
Prof. Jinjun Shan
=
µ 2ra
µ
µ
µ 2rp
−
+
−
rp (ra + rp )
rp
ra
ra (ra + rp )
⎞
2rp ⎞
µ ⎛⎜
2ra
µ ⎛⎜
⎟
⎟
=
−1 +
1−
rp ⎜⎝ (ra + rp ) ⎟⎠
ra ⎜⎝
(ra + rp ) ⎟⎠
µ ⎡ 2ra / rp ⎛ rp ⎞ rp ⎤
⎜⎜1 − ⎟⎟ +
=
− 1⎥
⎢
rp ⎢⎣ (1 + ra / rp ) ⎝ ra ⎠
ra ⎥⎦
Orbital Maneuvers - 5
Hohmann Transfer - 3
.
rp
µ + 2ra /rp % rp (
Δv =
⋅−10
'1 − * +
rp -, (1+ ra /rp ) & ra )
ra
0/
2ra /rp % rp (
rp
Δv
Δv =˙
=
−1
'1 − * +
v1
(1+ ra /rp ) & ra )
ra
ra
r=
rp
2r % 1 (
1
Δv =
−1
'1 − * +
(1+ r ) & r )
r
3
dΔv
= 3 2r + 2 − (1+ r ) 2
dr
€
Prof. Jinjun Shan
Orbital Maneuvers - 6
Hohmann Transfer - Bi-Elliptical 1
n 
n 
The bi-elliptic transfer consists of
two half elliptic orbits. From the
initial orbit, a delta-v is applied
boosting the spacecraft into the first
transfer orbit with an apoapsis at
some point B away from the central
body. At this point, a second delta-v
is applied sending the spacecraft
into the second elliptical orbit with
periapsis at the radius of the final
desired orbit where a third delta-v is
performed injecting the spacecraft
into the desired orbit.
Bi-elliptic transfer may, in certain
situations, require less delta-v than
a standard Hohmann transfer.
Prof. Jinjun Shan
Orbital Maneuvers - 7
Hohmann Transfer - Bi-Elliptical 2
rC
α=
rA
rB
β=
rA
Δv Hohmann
µ & 1
2(1 − α ) )
=
⋅(
−
−1+
rA ' α
α (1+ α ) *
Δv bi-elliptical
*
µ ' 2(α + β) 1+ α
2
=
⋅)
−
−
⋅ (1 − β) ,
rA (
αβ
β(1+ β)
α
+
Δv Hohmann − Δv bi-elliptical = δv
Prof. Jinjun Shan
#> 0
%
⇒ $= 0
%
&< 0
Orbital Maneuvers - 8
Hohmann Transfer - Bi-Elliptical 3
Prof. Jinjun Shan
Orbital Maneuvers - 9
Hohmann Transfer - Example 1
n 
Example 6.3 in textbook:
Find the total delta-v
requirement for a bielliptical Hohmann transfer
from a geocentric circular
orbit of 7000 km radius to
one of 105 000 km radius.
Let the apogee of the first
ellipse be 210,000 km.
Compare the delta-v
schedule and the total
flight time with that for an
ordinary single Hohmann
transfer ellipse.
Prof. Jinjun Shan
Orbital Maneuvers - 10
Hohmann Transfer - Coaxial Elliptical Orbits 1
n 
Transfer between coaxial elliptical orbits
Δv)3 = Δv A + Δv B
⎛ µ 2rB
µ
2rAʹ′ ⎞
⎟ +
= ⎜⎜
−
rA rA + rAʹ′ ⎟⎠
⎝ rA rA + rB
⎛ µ 2rBʹ′
µ 2rA ⎞⎟
⎜
⎜ r r + r − r r + r ⎟
B A
B ⎠
⎝ B B Bʹ′
Δv)3ʹ′ = Δv Aʹ′ + Δv Bʹ′
⎛ µ 2rBʹ′
= ⎜⎜
−
⎝ rAʹ′ rAʹ′ + rBʹ′
⎛ µ 2rB
⎜
⎜ r r + r −
⎝ Bʹ′ B Bʹ′
Prof. Jinjun Shan
µ 2rA ⎞⎟
+
⎟
rAʹ′ rA + rAʹ′ ⎠
µ 2rAʹ′ ⎞⎟
rB rAʹ′ + rBʹ′ ⎟⎠
Orbital Maneuvers - 11
Hohmann Transfer - Coaxial Elliptical Orbits 2
Prof. Jinjun Shan
Orbital Maneuvers - 12
Hohmann Transfer - Example 2
n 
Example 6.1 in textbook: A spacecraft is in a 480 km by 800 km
earth orbit (orbit 1). Determine the most efficient transfer from
orbit 1 to a circular orbit of altitude 16 000 km (orbit 3), and the
required delta-v.
Prof. Jinjun Shan
Orbital Maneuvers - 13
Phasing Maneuvers
n 
n 
A phasing maneuver is a two-impulse Hohmann
transfer from and back to the same orbit.
Phasing maneuvers are used to change the position of
a S/C in its orbit.
Prof. Jinjun Shan
Orbital Maneuvers - 14
Phasing Maneuvers - Example 3
n 
Example 6.4 in textbook: S/C at A and B are in the same
orbit 1. At the instant shown, the chaser vehicle at A
executes a phasing maneuver so as to catch the target S/C
back at B after just one revolution of the chaser’s phasing
orbit 2. What is the required total delta-v.
Prof. Jinjun Shan
Orbital Maneuvers - 15
Phasing Maneuvers - Example 4
n 
It is desired to shift the longitude of a GEO satellite
from 99.1°W to 111.1°W in three revolutions of its
phasing orbit. Calculate the delta-v requirement.
Prof. Jinjun Shan
Orbital Maneuvers - 16
Non-Hohmann Transfers with A Common Apse Line - 1
n 
n 
Transfer between two coaxial elliptical orbits in which
the transfer trajectory shares the apse line but is not
necessarily tangent to either the initial or target orbit.
The problem is to determine if there exists such a
trajectory joining points A and B.
h32
1
rA =
µ 1+ e3 cosθ A
h32
1
rB =
µ 1+ e3 cosθ B
rB − rA
e3 =
rA cosθ A − rB cosθ B
h3 = µrA rB
Prof. Jinjun Shan
cos θ A − cos θ B
rA cos θ A − rB cos θ B
Orbital Maneuvers - 17
Non-Hohmann Transfers with A Common Apse Line - 2
rA =
rB =
e3 =
h32
1
µ 1 + e3 cosθ A
h32
1
µ 1 + e3 cosθ B
rB − rA
rA cosθ A − rB cosθ B
h3 = µrArB
cosθ A − cosθ B
rA cosθ A − rB cosθ B
Δv = v12 + v22 − 2v1 v2 cos Δγ
tan φ =
Δvr
Δv ⊥
Δε ≈ v1Δv cos Δγ
Prof. Jinjun Shan
Orbital Maneuvers - 18
Apse Line Rotation
n 
n 
n 
Transfers between two intersecting orbits which have
a common focus, but their apse line are not collinear.
A Hohmann transfer is impossible.
The opportunity for transfer from one orbit to the other
by a single impulsive maneuver occurs where they
intersect.
Prof. Jinjun Shan
Orbital Maneuvers - 19
Apse Line Rotation - Case 1
n 
The apse line rotation is given as well as the orbital
parameters e and h of both orbits.
h12
1
rI 1 =
µ 1 + e1 cosθ1
h22
1
rI 2 =
µ 1 + e2 cosθ 2
e1h22 cosθ1 − e2h12 cosθ 2 = h12 − h22
η = θ1 − θ 2
a = e1h22 − e2 h12 cosη
b = −e2 h12 sin η
c = h12 − h22
ϕ = tan −1
b
a
⎛ c
⎞
θ1 = ϕ ± cos ⎜ cosϕ ⎟
⎝ a
⎠
−1
Prof. Jinjun Shan
Orbital Maneuvers - 20
Apse Line Rotation - Example 5
Prof. Jinjun Shan
Orbital Maneuvers - 21
Apse Line Rotation - Case 2
n 
Given the true anomaly of the point
of maneuver on orbit 1 and delta-V.
h12
1
rI1 =
µ 1 + e1 cos θ1
rI 2
h22
1
=
µ 1 + e2 cos θ 2
1 ( h1 + rΔv⊥)( µe1 sin θ1 + h1Δv r )
sin θ 2 =
e2
µh1
2
1 ( h1 + rΔv⊥) e1 cosθ1 + (2h1 + rΔv⊥) rΔv⊥
cosθ 2 =
e2
h12
h1 + rΔv⊥)( µe1 sin θ1 + h1Δv r )
(
h1
tan θ 2 =
µ ( h1 + rΔv⊥) 2 e1 cos θ1 + (2h1 + rΔv⊥) rΔv⊥
Prof. Jinjun Shan
Orbital Maneuvers - 22
Apse Line Rotation - Example 6
Prof. Jinjun Shan
Orbital Maneuvers - 23
Inclination and Launch Site
n 
Relationship between inclination and latitude of launch
site and launch direction
cosi = cosφ sin A
where
i - inclination;
φ - latitude of launch site;
A - launch azimuth, or the flight direction at
insertion, measured clockwise from north on
the local meridian.
Prof. Jinjun Shan
Orbital Maneuvers - 24
Inclination and Launch Site - Example 7
n 
Determine the required launch azimuth A for sun-synchronous
satellite of 98.43 ̊ inclination. If it is launched from Kennedy Space
Flight Center, which has a latitude of 28.6 ̊ N.
Prof. Jinjun Shan
Orbital Maneuvers - 25
Plane Change Maneuvers - 1
!
!
v1 = vr1uˆ r1 + v⊥1uˆ ⊥1
v2 = vr 2uˆ r 2 + v⊥2uˆ ⊥2
! ! !
Δv = v2 − v1 = (vr 2 − vr1 )uˆ r + v⊥2uˆ ⊥2 − v⊥1uˆ ⊥1
Prof. Jinjun Shan
Orbital Maneuvers - 26
Plane Change Maneuvers - 2
! ! !
Δv = v2 − v1 = (vr 2 − vr1 )uˆ r + v⊥2uˆ ⊥2 − v⊥1uˆ ⊥1
Δv = (vr 2 − vr1 )2 + v⊥21 + v⊥2 2 − 2v⊥1v⊥2 cos δ
Δv = v12 + v 22 − 2v1v 2 [cos Δγ − cosγ 2 cos γ1 (1 − cosδ )]
Δv = v12 + v22 − 2v1v2 cos Δγ
€
Δv = v12 + v22 − 2v1v2 cos δ
2
ΔvI = (v1 − v2 ) + 4v1v2 sin
Δvδ = 2v sin
Prof. Jinjun Shan
2
δ
2
δ
2
Orbital Maneuvers - 27
Plane Change Maneuvers - 4
2
ΔvI = (v1 − v2 ) + 4v1v2 sin
ΔvII = 2v1 sin
δ
2
+ v2 − v1
ΔvIII = v2 − v1 + 2v2 sin
Prof. Jinjun Shan
2
δ
2
> ΔvI
δ
2
> ΔvI
Orbital Maneuvers - 28
Plane Change Maneuvers - Example 8
n 
Example 6.11 in textbook: Find the delta-v required to transfer a
circular, 300 km altitude LEO of inclination of 28 deg to a GEO.
Compare that delta-v requirement with the one in which the
plane change is done in the LEO.
Prof. Jinjun Shan
Orbital Maneuvers - 29
Plane Change Maneuvers - Example 9
n 
Example 6.13 in textbook: A satellite is in a 500 km by 10 000
km altitude geocentric orbit which intersects the equatorial plane
at a true anomaly of 120 degrees. If the inclination to the
equatorial plane is 15 degrees, what is the minimum velocity
increment required to make this an equatorial orbit?
Prof. Jinjun Shan
Orbital Maneuvers - 30