AE-777A (Optimal Space
Flight Control)
Quiz No. 6 (Solution)
Quiz No. 6 (Solution)
1. A spacecraft is in an orbit of eccentricity,
e = 0.7, around the Earth (µ = 398600.4 km3/s2)
When the radius is 6900 km and the speed
is 7.4 km/s, a velocity impulse of magnitude 500 m/s is applied at an angle 60◦
from the initial velocity direction, in order
to increase both the speed and flight-path
angle simultaneously, but without changing the orbital plane. Determine the new
orbit (a, e) of the spacecraft.
Ans.
We first determine the semi-major axis of
the original orbit as follows:
v2 µ
− = −30.38817 km2/s
=
2
r
µ
a = − = 6558.479 km
2
Next, the flight-path angle at the point of
application of the impulse is determined as
follows:
p = a(1 − e2) = 3344.8243 km
√
h =
µp = 36513.67274 km2/s
h
φ = cos−1
= 0.77401 rad. (44.3475◦)
rv
Now, from the velocity triangle of impulse
application yields the following sine law:
∆v
v
v0
=
=
sin β
sin ∆φ
sin(β − ∆φ)
where β = 60◦ is the angle at which the
impulse of magnitude ∆v = 0.5 km/s is
applied from the original velocity direction,
v 0 is the magnitude of the increased velocity, and ∆φ the increase in the flight-path
angle at the given radius, r. This sine law
yields ∆φ as follows:
∆v sin β
tan ∆φ =
= 0.056603
v + ∆v cos β
or ∆φ = 3.23965◦. Substituting this result into the sine law produces the following magnitude of increased velocity:
sin β
= 7.662245 km/s
sin ∆φ
The semi-major axis of the new orbit is
computed from the new orbital speed as
follows:
v 0 = ∆v
0 )2
µ
(v
0
− = −28.4132 km2/s
=
2
r
µ
= 7014.359 km
0
2
Now, the increased flight-path angle is used
to calculate the orbital eccentricity of the
new orbit as follows:
a0 = −
φ0 = φ + ∆φ = 47.5872◦
h0 = rv 0 cos φ0 = 35658.75288 km2/s
p0 = (h0)2/µ = 3190.0286 km
s
e0 =
p0
1 − 0 = 0.738386
a
2. Estimate the smallest total velocity impulse
magnitude required for sending a spacecraft from the Earth (orbital radius 1 AU =
1.495978 × 108 km) to Mars (orbital radius
1.524 AU). Assume that both the planets
are in coplanar circular orbits around the
Sun (µ = 1.327 × 1011 km3/s2), and that
the gravitational influence of the planets on
the spacecraft during the transfer is negligible.
Ans.
The optimal transfer between two circular orbits of radii r1 and r2 > r1, when
r2/r1 < 15.58 is Hohmann transfer. For a
Hohmann transfer between the given radii,
r1 = 1 AU and r2 = 1.524 AU, we have
the following semi-major axis of the transfer ellipse :
r1 + r2
a=
= 1.262 AU (188792423.6 km)
2
The two impulse magnitudes are then calculated by
s
∆v1 =
2µ µ
− −
r1
a
s
∆v2 =
µ
−
r2
s
s
µ
= 2.94605 km/s .
r1
µ
µ
− = 2.64998 km/s .
r2 a
Both the impulses are applied in the direction of motion. The total velocity impulse
magnitude required for the mission is the
following:
∆v = ∆v1 + ∆v2 = 5.59603 km/s .
3. Spacecraft A is orbiting the Moon (µ =
4902.8 km3/s2) in a circular orbit of radius 2000 km, and a lander spacecraft has
landed on the Moon’s surface (radius 1737.1
km) at a point B in the orbital plane of
A. Now the lander has to return to the
orbiting spacecraft A. What minimum velocity impulse (magnitude and direction)
should be provided to the lander such that
it reaches the orbit of A, when A is exactly
180◦ ahead of B?
Ans.
For a minimum-energy orbit between r1
and r2 with ∆θ = 180◦ we have c = r1 +r2,
or
1
1
a = (r1 + r2 + c) = (r1 + r2)
4
2
The orbit equation applied to the two positions gives
p
r1 =
1 + e cos θ1
p
p
r2 =
=
1 + e cos θ2
1 + e cos(θ1 + π)
p
=
1 − e cos θ1
which is solved to give
r − r1
e cos θ1 = 2
r1 + r2
Substituting this result into the orbit equation yields
2r1r2
p=
= a(1 − e2)
r1 + r2
Then substituting a = (r1 +r2)/2 and solving for e we get
s
e=
1−
r2 − r1
4r1r2
=
(r1 + r2)2
r1 + r2
This result implies that θ1 = 0 and θ2 = π,
hence r1 and r2 are the periapse and the
apoapse radii, respectively, of the transfer
ellipse.
Substituting the given numerical values, we
get a = 1868.55 km, and the following
velocity impulse magnitude applied to the
lander to send it along the transfer ellipse:
s
∆v1 =
2µ µ
− = 1.73809 km/s
r1
a
Since the periapse is at r1, the impulse is
applied tangentially to the transfer ellipse
at r1, i.e., in the horizontal direction (normal to the radius vector at r1).
4. A spacecraft flying in a spherical gravity
field of gravitational constant, µ, is powered by a continuous-thrust engine, which
has its thrust acceleration magnitude, a(t),
limited as follows:
0 ≤ a(t) ≤ am
where am µ/r2 is a constant. Formulate
the boundary-value problem to be solved
(differential equations, boundary conditions,
and control laws for both magnitude and
direction) for optimally transferring the spacecraft from an initial circular orbit of radius,
r1, to a final circular orbit of radius, r2, in
a fixed time, tf .
Ans.
The equations of motion of a spacecraft in
a spherical gravity field with thrust acceleration input, a(t) = a(t)n(t), where n is a
unit vector in the thrust direction, are the
following:
µ
v̇ = − 3 r + an
r
and
ṙ = v = ṙir + rθ̇iθ
where r = (r, θ)T and v = (ṙ, rθ̇)T are the
radius and velocity vectors for this coplanar
transfer.
The boundary conditions for transferring
the spacecraft from an initial circular orbit
of radius, r1, to a final coplanar, circular
orbit of radius, r2, in a fixed time, tf , are
given by:
r(0) = r1,
r(tf ) = r2
ṙ(0) = 0,
ṙ(tf ) = 0
s
v(0) = r1θ̇(0) =
µ
,
r1
s
v(tf ) = r2θ̇(tf ) =
µ
r2
The true anomaly at the beginning of the
transfer, θ(0), is fixed by the initial condition. The true anomaly for the final time,
θ(tf ), is given by:
s
θ(tf ) = θ(0) + tf
µ
r23
The final characteristic speed, c(tf ), is a
measure of the total propellant expenditure, m(0) − m(tf ), and therefore must be
minimized for an optimal transfer. Hence,
the objective function to be minimized for
the smallest possible propellant requirement
is the following:
J = c(tf ) =
Z t
f
0
a(t)dt
which results in the following Hamiltonian
for the optimal control problem:
T
H = λT
r v + λv [g(r) + an] + λc a
where g(r) = −µr/r3 is the acceleration
due to gravity, λr and λv are the twodimensional costate vectors corresponding
to r and v, respectively, and λc corresponds
to the characteristic speed.
The optimal thrust direction is given in
terms of the primer vector, p(t) = −λv (t),
as follows:
p
n̂ =
p
where p is governed by the following differential equation:
∂g T
p
p̈ =
∂r
µ
3µ
= − 3 p + 5 rrT p
r
r
The boundary conditions for the primer vector and its time derivative are the following:
ṗ(tf ) = 0,
p(tf ) = 0
The costate equation corresponding to the
characteristic speed is the following:
∂H T
∂H T
λ̇c = −
= −a
=0
∂c
∂t
where the Hamiltonian is time-invariant due
to the constant upper bound, am. Furthermore, the terminal boundary condition on
λc yields the following:
∂φ
λc(tf ) =
=1
∂c t=tf
which results in λc = λc(tf ) = 1, and yields
the following Hamiltonian:
H = a(1 − p) + ṗT v − pT g
Due to the linearity of H relative to a, the
optimal control problem is singular in acceleration magnitude. However, the bounds
on a allow the application of the Pontryagin’s minimum principle as follows:
â(1 − p) + ṗT v − pT g ≤ a(1 − p) + ṗT v − pT g
or
â(1 − p) ≤ a(1 − p)
where â refers to the optimal control, and
0 ≤ a ≤ am
The last two inequalities result in the following switching condition for the optimal
control magnitude:
â(t) =















am ,
(p − 1) > 0
0,
(p − 1) < 0
0 < â < am ,
p=1