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Problem 5.149
The optimal way (from an energy standpoint) to transfer from one
circular orbit about a primary body
B
to another circular orbit is via
the so-called Hohmann transfer, which involves transferring from one
circular orbit to another using an elliptical orbit that is tangent to both
at the periapsis and apoapsis of the ellipse. The ellipse is uniquely
defined because we know
rP
(the radius of the inner circular orbit) and
rA
(the radius of the outer circular orbit), and therefore we know the
semimajor axis
a
by Eq. (5.88) and the eccentricity
e
by Eq. (5.87) or
Eqs. (5.90). Performing a Hohmann transfer requires two maneuvers,
the first to leave the inner (outer) circular orbit and enter the transfer
ellipse and the second to leave the transfer ellipse and enter the outer
(inner) circular orbit.
A spacecraft
S2
must transfer from a circular Earth orbit whose
period is
12
h (i.e., it is overhead twice per day) to a low Earth circular
orbit with an altitude of
110 mi
. Determine the change in speed
vA
required at apogee
A
of the elliptical transfer orbit and the change
in speed
vP
required at perigee
P
. In addition, compute the time
required for the orbital transfer. Assume that the changes in speed are
impulsive; that is, they occur instantaneously.
Solution
Next, the speed
vAc
of the satellite while on the initial circular orbit, and the speed
vAe
that the satellite must
have to get onto the transfer orbit at Apogee are computed using Eq. (5.82) on p. 388 of the textbook and
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Dynamics 2e 1081
Eq. (5.105) on p. 391, respectively, which give
e
Therefore, the change in speed at apogee is vADvAe vAc, whose value is
Replacing
rA
in Eqs. (5) with
rP
we can compute the speeds at perigee corresponding to the final circular
destination orbit and the transfer orbit, respectively. This gives
e
Finally, observe that the spacecraft only completes half of the transfer orbit. Hence, the time
t
needed for
the orbital transfer is
⌧e=2
, where
⌧e
is the period of the elliptical transfer orbit. Letting
aDae
in Eq. (1),
solving for the corresponding ⌧e, and dividing by 2, we have
e
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Problem 5.150
The optimal way (from an energy standpoint) to transfer from one
circular orbit about a primary body
B
to another circular orbit is via
the so-called Hohmann transfer, which involves transferring from one
circular orbit to another using an elliptical orbit that is tangent to both
at the periapsis and apoapsis of the ellipse. The ellipse is uniquely
defined because we know
rP
(the radius of the inner circular orbit) and
rA
(the radius of the outer circular orbit), and therefore we know the
semimajor axis
a
by Eq. (5.88) and the eccentricity
e
by Eq. (5.87) or
Eqs. (5.90). Performing a Hohmann transfer requires two maneuvers,
the first to leave the inner (outer) circular orbit and enter the transfer
ellipse and the second to leave the transfer ellipse and enter the outer
(inner) circular orbit.
A spacecraft
S1
is transferring from circular low Earth parking
orbit with altitude
100 mi
to a circular orbit with radius
rA
. Plot, as
a function of
rA
for
rPrA100rP
, the change in speed
vP
required at perigee of the elliptical transfer orbit, as well as the change
in speed
vA
required at apogee. In addition, plot the time as a
function of
rA
, again for
rPrA100rP
, required for the orbital
transfer. Assume that the changes in speed are impulsive; that is, they
occur instantaneously.
Next, let
v1
,
v2
,
v3
and
v4
denote the speeds of the satellite corresponding to the initial circular orbit, the
speed needed at perigee to move onto the the transfer orbit, the speed on the transfer orbit at apogee, and
e
reD3959 mi
is the radius of the Earth and
gD32:2 ft=s2
is the acceleration due to gravity on the surface of
the Earth. Hence, applying Eq. (2), we can determine the speeds v1,v2,v3and v4as follows:
e
e
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1083
The above results allow us to derive expressions for
vPDv2v1
and
vADv4v3
. Specifically, using
the second of Eqs. (1), we have
e
e
Now that we have derived relations for
vP
and
vA
, we need to derive a relation for the time
t
needed for
the orbital transfer. Since the spacecraft only completes one half of the elliptical orbit during transfer, then
t
is half of the orbital period
⌧
. In turn, the orbital period is found using Eq. (5.97) on p. 390 of the textbook,
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permission of McGraw-Hill, is prohibited.
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
Dynamics 2e 1085
Problem 5.151
Referring to the description given for Probs. 5.148–5.150, for a
Hohmann transfer from an inner circular orbit to an outer circular
orbit, what would you expect to be the signs on the change in speed
at periapsis and at apoapsis?
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Problem 5.152
Referring to the description given for Probs. 5.148–5.150, for a
Hohmann transfer from an outer circular orbit to an inner circular
orbit, what would you expect to be the signs on the change in speed
at periapsis and at apoapsis?
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Dynamics 2e 1087
Problem 5.153
During the Apollo missions, while the astronauts were on the Moon
with the lunar module (LM), the command module (CM) would fly
in a circular orbit around the Moon at an altitude of
60 mi
. After
the astronauts were done exploring the Moon, the LM would launch
from the Moon’s surface (at
L
) and undergo powered flight until
burnout at
P
. This occurred when the LM was approximately
15 mi
above the surface of the Moon with its velocity
vbo
parallel to the
surface of the Moon (i.e., at periapsis). It would then fly under the
influence of the Moon’s gravity until reaching apoapsis
A
, at which
point it would rendezvous with the CM. The radius of the Moon
is
1079 mi
, and its mass is
0:0123
times that of the Earth. Assume
that the changes in speed occur instantaneously.
(a) Determine the required speed vbo at burnout P.
(b)
What is the change in speed
vLM
required of the LM at the
rendezvous point A?
(c) Determine the time it takes the LM to travel from Pto A.
(d)
In terms of the angle
✓
, where should the CM be when the LM
reaches Pso that they can rendezvous at A?
Solution
Part (a).
Using Eq. (5.105) on p. 391 of the textbook, the speed at burnout
vbo
is the speed at periapsis in
where we have used the following numerical data:
rmD1079 mi D1079.5280/ ft
,
hPD15 mi D
15.5280/ ft, and hAD60 mi D60.5280/ ft. Therefore, since
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Part (b).
Now the LM must go from its speed at apoapsis to the speed of the CM, which is in a circular
orbit about the Moon with an altitude of
60 mi
. Replacing
rP
with
rA
in Eq. (1) (and using Eq. (2) as well),
the speed of the LM at apoapsis is
Part (c).
Let the time it takes for the LM to go from
P
to
A
be denoted by
t
. This time is equal to half of
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Dynamics 2e 1089
Problem 5.154
Use the work-energy principle applied between periapsis
P
and
rD1
, along with the potential energy of
the force of gravity given in Eq. (4.25).
(a) Show that a satellite on a hyperbolic trajectory arrives at rD1with speed
v1DsrPv2
P2GmB
rP
:
(b)
In addition, using Eqs. (5.77) and (5.80), show that for a hyperbolic trajectory,
rPv2
P> 2GmB
, which
means that the square root in the above equation must always yield a real value.
Solution
Part (a). Applying the work-energy principle between periapsis and rD1in a hyperbolic orbit, we get
2mv2
P;T
2mv2
1;V
rP
Substituting the terms in Eq. (2) into Eq. (1), and the solving for for v1, we have
2v2
PGmB
rPD1
2v2
1)v1DsrPv2
rP
Part (b).
We start by repeating here for convenience Eq. (5.77) (on p. 387 of the textbook) and Eq. (5.80)
Now recall that, for a hyperbolic orbit e>1. Hence, from the first of Eqs. (3) we have
C2>Gm
B:
Using this result and the fact that
DrPvP
(see, for example, Eq. (5.79) on p. 388 of the textbook), the
second of Eqs. (3) can be written as
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