978-0073380308 Chapter 5 Solution Manual Part 24

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Problem 5.149

The optimal way (from an energy standpoint) to transfer from one

circular orbit about a primary body

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

(the radius of the inner circular orbit) and

(the radius of the outer circular orbit), and therefore we know the

semimajor axis

by Eq. (5.88) and the eccentricity

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

must transfer from a circular Earth orbit whose

period is

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

of the elliptical transfer orbit and the change

in speed

vP

required at perigee

. 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

Therefore, the change in speed at apogee is vADvAe vAc, whose value is

Replacing

in Eqs. (5) with

we can compute the speeds at perigee corresponding to the final circular

destination orbit and the transfer orbit, respectively. This gives

Finally, observe that the spacecraft only completes half of the transfer orbit. Hence, the time

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

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Problem 5.150

The optimal way (from an energy standpoint) to transfer from one

circular orbit about a primary body

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

(the radius of the inner circular orbit) and

(the radius of the outer circular orbit), and therefore we know the

semimajor axis

by Eq. (5.88) and the eccentricity

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

is transferring from circular low Earth parking

orbit with altitude

100 mi

to a circular orbit with radius

. Plot, as

a function of

for

rPrA100rP

, 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

, again for

rPrA100rP

, required for the orbital

transfer. Assume that the changes in speed are impulsive; that is, they

occur instantaneously.

Next, let

and

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

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:

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Dynamics 2e 1083

The above results allow us to derive expressions for

vPDv2v1

and

vADv4v3

. Specifically, using

the second of Eqs. (1), we have

Now that we have derived relations for

vP

and

vA

, we need to derive a relation for the time

needed for

the orbital transfer. Since the spacecraft only completes one half of the elliptical orbit during transfer, then

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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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

) and undergo powered flight until

burnout at

. 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

, at which

point it would rendezvous with the CM. The radius of the Moon

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?

(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

with

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

be denoted by

. 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

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

P2GmB

(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

Substituting the terms in Eq. (2) into Eq. (1), and the solving for for v1, we have

2v2

PGmB

rPD1

2v2

1)v1DsrPv2

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

C2>Gm

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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