ASTEROID RETRIEVAL BY ROTARY ROCKET*
Jerome Pearson**
U. S. Air Force Flight Dynamics Laboratory
Wright-Patterson Air Force Base, Ohio 45433
Abstract
A new technique is proposed for recovering
asteroid resources. The target asteroid would
be equipped with a "rotary rocket" propulsion
system consisting of a rapidly spinning, tapered
tube of high strength material driven electrically
by solar or nuclear power. Pellets of asteroid
material would be released from the tube ends
with velocities of a few km/s, achieving specific
impulses comparable with the best chemical
rockets. The ejected pellets could be launched on
trajectories to near-earth space for capture, or
they could be expended as reaction mass to bring
the remainder of the asteroid to high earth orbit
for the construction of solar power satellites or
space habitats.
I. Introduction
In the development of large-scale structures
for solar power generation, space communication,
or space habitats, large quantities of mass are
required for structure, shielding, atmosphere,
and propellant. Advanced launch vehicles
have been proposed to lift large payloads
from the earth, but environmental concerns may
limit the rate at which earth resources can be
orbited. The use of lunar resources has also
been suggested, either by the use of a linear
motor mass driver
1
or by anchored lunar satellite
launch.
2
The use of lunar resources would enjoy
the advantages of the shallower gravitational
well of the moon; however, the moon lacks
appreciable quantities of carbon and hydrogen,
both of which are vital to space habitation.
The newest potential source for space resources
is the class of asteroids known as carbonaceous
chondrites, which are relatively rich in
hydrocarbons. Asteroidal resources have been
considered in the past, and renewed interest
has been elicited by the recent discovery of new
members of the class of earth-orbit-approaching
bodies belonging to the Apollo and Amor
groups. Some of these asteroids are in orbits
that require less energy to reach than the
lunar surface. Considering the lack of lunar
hydrocarbons and the pollution inherent in
orbiting earth resources, the Apollo asteroids
represent an attractive potential resource.
Their value may be gauged from the fact that
the Apollo asteroid that hit the earth 1.8
* Copyright
c
1979 by Jerome Pearson, with
release to AIAA to publish in all forms.
This research is not part of the author’s
official duties.
** Aerospace Engineer; Associate Fellow AIAA.
billion years ago near present-day Sudbury,
Ontario, provides half the world’s nickel and
has produced billions of dollars of value in
copper, nickel, silver, platinum, and gold.
Various techniques have been suggested for
the recovery of asteroid resources. Niehoff
3
examined the round-trip requiren nts to return
samples of 1976 AA and 1973 EC using conventional
rockets and the space shuttle. To retrieve
entire asteroids, Cole and Cox proposed the use
of nuclear explosives
4
, and O’Leary et al, have
analyzed the use of the mass driver.
5
This paper
proposes a new technique, the use of the rotary
rocket, for asteroid recovery.
II. The Rotary Rocket Concept
The rotary rocket is a derivative of the
orbital tower, an extremely elongated synchronous
satellite which extends from the surface of a
planet to some distance beyond the synchronous
altitude balance point. This structure is
theoretically capable of extracting energy from
the rotation of the planet and transferring
it to the linear motion of payloads released
from the upper end, providing them with escape
velocity.
6
A rapidly rotating asteroid could be
fitted with a synchronous orbital tower, but
an enormous length would be required to achieve
a reasonable tip velocity. An asteroid with
a typical period of 4 hours would require a
tower 6900 km long to attain a tip speed of 3
km/s. At this slow rotational rate, payloads
could be launched only twice each 4 hours.
A better approach may be to use a rapidly
spinning tower with its axis attached to the
asteroid. The concept is shown in Figure 1
as pairs of relatively short, high-velocity,
dual- ended launch tubes attached to an asteroid
and contra-rotated to prevent a net angular
momentum. The launch tubes are maintained in
rapid rotation by an electric motor whose power
may be solar or nuclear. Small masses to be
expelled are fed into the hollow interior of
the launch tubes and are then released from the
ends. The exhaust velocity of this rotary rocket
is a function of the tip velocity attainable.
For a material with a density and a working
stress o, the maximum tip velocity attainable for
a uniform rod may be found by equating the load-
carrying capacity, σA, to the centrifugal force
due to the rotating mass:
σA =
Z
R
0
v(r)
2
r
dm (1)
1
Fig. 1 Rotary rocket concept.
Since dm = ρAdr and v(r) = v
R
r/R, where v
R
is the
tip velocity, we have:
σA = ρAv
2
R
Z
R
0
r
R
2
dr =
ρAv
2
R
2
(2)
The maximum tip velocity can be called the
"critical velocity" of the material:
v
2
C
= v
2
R
= 2σ/ρ (3)
This is a surprisingly simple relationship
between the strength-to-density ratio of the
material and its maximum tip velocity for a
uniform rod, The critical velocity of a material
is a useful measure of its performance in a
rotating launcher. The parameters of a few
materials of interest are shown in Table 1.
In general, conventional metals show v
c
of
about 0.5 km/s. composites show v
c
of about
1.25 to 2 km/s, and whisker materials show the
potential of achieving v
c
= 3 to 4 km/s. Table
1 also shows the strength/density parameter
in terms of the characteristic height, σ/ρg
0
,
which was used as a design parameter for
orbital towers
6
and lunar anchored satellites.
2
Fig. 2 Launch tube characteristics.
Table 1 Construction material parameters
WORKING CHARACTERISTIC CRITICAL
DENSITY STRESS HEIGHT VELOCITY
MATERIAL kg/m
3
GP a km km/s
METALS:
ALUMINUM 2700 0.3 11.3 0.47
STEEL 7900 4.2 54.2 1.03
COMPOSITES:
GRAPHITE/EPOXY 1550 1.24 81.6 1.26
E-GLASS 2550 3.5 140 1.66
KEVLAR 49 1400 2.7 197 1.96
WHISKERS:
SIC 3200 20 638 3.54
GRAPHITE 2200 19 881 4.16
These values indicate that the electrically
driven rotary rocket has the potential
capability of providing a high level of
performance. A tip velocity of 4 km/s
corresponds roughly to the specific impulse of
a liquid-hydrogen/liquid-oxygen chemical rocket.
III. Rotating Launcher Design
A uniform rod is limited in tip velocity to its
critical velocity, as we have seen. Higher tip
velocities can be achieved for the same material,
however, by tapering the rod. The design of the
tapered launch tube is optimized by using a cross-
sectional area that varies along the length so
as to produce a constant stress. The details are
shown in Figure 2, where the tube of radius R has
a cross-sectional area A(r) at distance r from the
axis. The tube has a cylindrical hole in which
the propellant masses move outward. The tip area
of the tube is sized to be able to hold the mass m
against the tip velocity ωR. From Figure 2, the
radial force at any point r, with positive toward
the axis, is:
dF = σdA = −ρA(r) ω
2
rdr (4)
dA
A
= −
ρω
2
r
σ
dr (5)
Integrating and substituting A
0
= A(r = 0) gives:
A(r) = A
0
exp[−
ρω
2
2σ
r
2
] (6)
Using v
2
c
= 2σ/ρ, v
2
R
= ω
2
R
2
gives:
A(r) = A
0
exp[−
v
2
R
v
2
c
r
2
R
2
] (7)
The area taper ratio is defined as the ratio of
the cross-sectional area of the tube material on
the axis to that at the tip:
2
T R = A
0
/A
R
= exp(v
2
R
/v
2
c
) (8)
This tapered tube can support a force at the
end of σA
R
. If a small mass m is attached at the
end, it experiences a force of mω
2
R or mv
2
R
. The
cross-sectional area of the tube at any point r
can be expressed in terms of this mass m as:
A(r) =
mv
2
R
σR
exp[
v
2
v
2
c
(1 −
r
2
R
2
)] (9)
The total mass of the launch tube, from r = o to
r = R, is then:
M =
Z
R
0
A(r)dr =
mv
R
v
c
√
πexp(
v
2
R
v
2
c
)erf (
v
R
v
c
) (10)
where erf is the error function. This result was
obtained by symbolic integration using the MACSYMA
computer program.
7
The error function is related
to the area under the normal curve with mean µ and
standard deviation s by the equation:
1
s
√
2π
Z
X
−∞
exp[
−(t − µ)
2
2s
2
]dt (11)
=
1
2
[1 + erf (
x − µ
s
√
2
)] (12)
This is to be expected, because A(r) is an
e
−r
2
function, a normal probability curve in
the square of the ratio of tip velocity to
the critical velocity. The total mass is
therefore a function of the area under this
gaussian probability curve from x = 0 to x = R.
The results of equations 8 and 10 are shown
in Figure 3, where the area taper ratio A
0
/A
R
and the mass ratio M/m are shown in terms of
the ratio of the tip velocity to the critical
velocity. Both the mass ratio and the area
ratio are strong exponential functions of the
velocity ratio. This means that for a reasonable
mass of the launcher the exhaust velocity of
the rotary rocket is limited to a few times the
critical velocity of the launch tube material.
The exhaust velocities, c, and specific
impulses, I
sp
, achievable by the rotary rocket
are shown in Figure 4 as functions of the taper
ratio. In general, metal launch tubes can achieve
I
sp
in the range of 100-200 sec, composites can
provide 200-500 sec, and whisker materials, when
available, promise I
sp
of 400-1000 seconds. These
surprising results show that the rotary rocket
has the theoretical capability of exceeding
the performance of present liquid-fuel rockets
by using any inert material as a fuel. The
rotary rocket has no inherent limit on its
thrust duration, can be stopped and restarted
an arbitrary number of times, and can be throttled
down to zero exhaust velocity.
Fig. 3 Mass ratios and taper ratios for
single-ended launch tubes.
Fig. 4 Tip velocities and specific impulses
available using various materials.
One difficulty in the rotary rocket design is
to release the stream of pellets into a collimated
jet. This requires that a pellet be released once
each revolution, just as the launch tube reaches
the desired pointing direction. This may be done
by allowing one pellet at a time past a release
cam controlled by the tube rotational angle, as
3
shown in Figure 5. The cam is at the tip of the
tube, allowing the bore to be filled with pellets,
assumed to be roughly spherical. The cam can be
rotated electrically at the same angular velocity
as the launch tube, allowing one pellet to escape
at the same angular position each revolution. A
variation of up to ± 8
o
, however, will maintain
99% thrust.
Fig. 5 Pellet feed and release mechanism.
IV. Asteroid Retrieval Missions
Now we are ready to examine the application
of the rotary rocket to an asteroid recovery.
The easiest mission is to retrieve a small
member of the Apollo or Armor families of
earth-orbit-crossing asteroids. These objects
include many carbonaceous types rich in
hydrocarbons, ranging in size from a few
kilometers across to, presumably, a few meters
across.
8
Their orbits are not energetically
very distant from that of the earth. A minimum
A of about 3 km/s for return to earth is a
reasonable value from a low inclination orbit
and a return via Venus and moon gravity assists.
9
Using this velocity change as a design input,
we need to create a rotary rocket and power
system to bring back a small asteroid in a
reasonable amount of time. If we assume a
manned expedition, an upper limit of five years
for the round trip may be a reasonable choice.
The exhaust velocity of the rotary rocket
can be chosen to maximize the returned mass.
The fraction of the asteroid mass that can be
retrieved is given by the rocket equation:
m
i
m
f
= e
∆/c
(13)
where m
i
and m
f
are the initial and final
masses. The energy required is given by:
E = 1/2(m
i
− m
f
)c
2
(14)
Fig. 6 Specific energy vs exhaust velocity.
The energy invested per unit retrieved mass is
then:
E
m
f
=
m
i
− m
f
2m
f
c
2
(15)
After some substitution, this reduces to:
E
m
f
=
∆
2
(e
∆/c
− 1)
2(∆/c)
2
(16)
To minimize the energy invested per unit mass,
we define a non-dimensional energy efficiency, :
=
E
1/2m
f
∆
2
=
e
∆/c−1
(∆/c)
2
(17)
The minimum value of can be found easily as a
function of ∆/c. The result is seen in Figure
6, which shows the specific energy, in kWh/kg,
needed to retrieve a mass through various velocity
changes. Each curve is seen to have a minimum at
∆/c = 1.594. The minima are not sharp, however,
and other factors such as the available power may
require a higher value of ∆/c.
Choosing the minimum specific energy for a ∆
of 3 km/s gives an optimum exhaust velocity,
c, of 1.88 km/s. With this exhaust velocity, a
4
