978-0078112829 Chapter 11 Solution Manual

Problem 11.1 A two stage, solid-propellant sounding rocket has the following properties.

First stage:

m0249.5 kg mf170.1 kg &

me10.61 kg s Isp 235 s

Second stage:

m0113.4 kg mf58.97 kg &

me4.053 kg s Isp 235 s

Delay time between burnout of first stage and ignition of second stage: 3 seconds.

As a preliminary estimate, neglect drag and the variation of earth’s gravity with altitude to calculate the

maximum height reached by the second stage after burnout.

Solution

First stage:

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 11

hmax h0v0tmax 1

2gtmax228690 2400244.7 1

29.81244.72322300 m

Problem 11.2 A two-stage launch vehicle has the following properties:

First stage: two solid propellant rockets. Each one has a total mass of 525 000 kg, 450 000 kg of

which is propellant.

Isp 290 s

.

Second stage: two liquid rockets with

Isp 450 s

. Dry mass = 30 000 kg, propellant mass = 600 000

kg.

Calculate the payload mass to a 300 km orbit if launched due east from KSC. Let the total gravity and

drag loss be 2 km/s.



Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 11

mPL 110800 kg

Problem 11.3 Suppose a spacecraft in permanent orbit around the earth is to be used for delivering

payloads from low earth orbit (LEO) to geostationary equatorial orbit (GEO). Before each flight from

LEO, the spacecraft is refueled with propellant, which it uses up in its round trip to GEO. The outbound

leg requires four times as much propellant as the inbound return leg. The delta-v for transfer from LEO to

GEO is 4.22 km/s. The specific impulse of the propulsion system is 430 s. If the payload mass is 3500 kg,

calculate the empty mass of the vehicle.



0.5702 6.876me

 1.719me6018

me2733 kg

Problem 11.4 Consider a rocket comprising three similar stages (i.e., each stage has the same specific

impulse, structural ratio and payload ratio). The common specific impulse is 310 s. The total mass of the

vehicle is 150 000 kg, the total structural mass (empty mass) is 20 000 kg and the payload mass is 10 000

kg. Calculate

(a) The mass ratio n and the total



v

for the three-stage rocket.

(b)

mp1, mp2, and mp3

.

(c)

mE1, mE2 and mE3

.

(d)

mo1, mo2 and mo3

.



vIspgoln n33100.00981ln 2.03936.5 km s



PL1/ 3 mPL 10.066671/ 3

0.066671/ 3 10000 12570 kg



PL2/ 3 mPL 10.066671/ 3

0.066672/ 3 10000 5166 kg

mE31



PL1/ 3

 





PL1/ 3 mPL 10.066671/ 3

 0.1429

0.066671/ 3 10000 2095 kg

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 11

(d)

m03 mE3mp3mPL 2095 12570 10000 24660 kg

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 11

Problem 11.5 Find the extrema of the function

zxy

 2

subject to x and y lying on the circle

x1

 2y21

.

Solution

zx2y22xy

(1)

gx22xy2

(2)

hz



gx2y22xy 



x22xy2

 

(3)



0

: Local extrema are

x10

y10

z10

x21

y2 1

z20

If



0

, then (7) yields

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 11



x

y















1 1

1



1





 





1



0











1

 

2

 



11

1



1





 







0















1



2

1



2





















Substituting these values of x and y into (4), we get





1

 2



2

 22



1



21



2

 20

Multiply through by



2

 2

to get





1

 22



1

 



2

  10

or





24



20

The two roots are

0.5858

and

3.414

.





 0.5858:

The extremum is

x3



1



20.5858 1

0.5858 20.2929

y3  1



2  1

0.585 2 0.7071

z3x32y322x3y30.29292 0.7071

 220.2929 0.7071

 0.1716





 3.414:

The extremum is

x4



1



23.414 1

3.414 21.707

y4  1



2  1

3.414 20.7071

z4x42y422x4y41.70720.7071221.7070.7071 5.828

To determine whether the above four extrema are local minima or local maxima, we first note that

d2h



2z



x2



2g



x2









dx22



2z



x



y



2g



x



y









dxdy 



2z



y2



2g



y2













dy2

d2h2



2

 dx22 2 



0

 dxdy 2



2

 dy2

d2h2



1

  dx2dy2

 4dxdy

(9)

From (4) we get

2xdx 2dx 2ydy 0

0.7071dx 0.7071dy 0

dy  dx

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 11

Problem 11.6 A small two-stage vehicle is to propel a 10 kg payload to a speed of 6.2 km/s. The prop-

erties of the stage are:

First stage:

Isp 300 s

and



0.2

.

Second stage:

Isp 235 s

and



0.3

.

Estimate the optimum mass of the vehicle.



Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 11