978-0078112829 Chapter 1 Solution Manual Part 1

SOLUTIONS MANUAL

to accompany

ORBITAL MECHANICS FOR ENGINEERING STUDENTS

Third Edition

Howard D. Curtis

Embry-Riddle Aeronautical University

Daytona Beach, Florida

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 1

Problem 1.1 Given the three vectors

AAx

ˆ

iAy

ˆ

jAz

ˆ

k

,

BBx

ˆ

iBy

ˆ

jBz

ˆ

k

and

CCx

ˆ

iCy

ˆ

jCz

ˆ

k

,

show analytically that

(a)

AAA2

(b)

ABC

 AB

 C

(c)

ABC

 B A C

 C A B

 

฀

ABC

 Ax

ˆ

iAy

ˆ

jAz

ˆ

k

 

ˆ

iˆ

jˆ

k

BxByBz

CxCyCz



ˆ

iˆ

jˆ

k

AxAyAz

ByCzBzCyBzCxBxCyBxCyByCx

AyBxCyByCx

 AzBzCxBxCz

 





ˆ

iAzByCzBzCy

 AxBxCyByCx

 





ˆ

j

AxBzCxBxCz

 AyByCzBzCy

 





ˆ

k

AyBxCyAzBxCzAyByCxAzBzCx

 ˆ

iAxByCxAzByCzAxBxCyAzBzCy

 ˆ

j

AxBzCxAyBzCyAxBxCzAyByCz

 ˆ

k

BxAyCyAzCz

 CxAyByAzBz

 





ˆ

iByAxCxAzCz

 CyAxBxAzBz

 





ˆ

j

BzAxCxAyCy

 CzAxBxAyBy

 





ˆ

k

Add and subtract the underlined terms to get

ABC

 BxAyCyAzCzAxCx

 CxAyByAzBzAxBx

 





ˆ

i

ByAxCxAzCzAyCy

 CyAxBxAzBzAyBy

 







ˆ

j

BzAxCxAyCyAzCz

 CzAxBxAyByAzBz

 





ˆ

k

Bx

ˆ

iBy

ˆ

jBz

ˆ

k

 AxCxAyCyAzCz

 Cx

ˆ

iCy

ˆ

jCz

ˆ

k

 AxBxAyByAzBz

 

Bx

ˆ

iBy

ˆ

jBz

ˆ

k

 AC

 Cx

ˆ

iCy

ˆ

jCz

ˆ

k

 AB

 

Or,

ABC

 B A C

 C A B

 

Problem 1.2 Use just the vector identities in Problem 1.1 to show that

AB

 CD

 AC

  BD

 AD

  BC

 

AB

CAB 11.575

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 1

Problem 1.4 Since

ˆ

ut

and

ˆ

un

are perpendicular and

ˆ

utˆ

unˆ

ub

, use the bac-cab rule to show that

ˆ

ubˆ

utˆ

un

and

ˆ

unˆ

ubˆ

ut

, thereby verifying Equation 1.29.

Problem 1.5 The x, y and z coordinates (in meters) of a particle as a function of time (in seconds) are

xsin 3t

,

ycost

and

zsin 2t

. At

t3s

determine:

(a) The velocity v, in Cartesian coordinates.

(b) The speed v.

(c) The unit tangent

ˆ

ut

.

(d) The angles



x

,



y

and



z

that v makes with the x, y and z axes.

(e) The acceleration a in Cartesian coordinates.

(f) The unit binormal vector

ˆ

ub

.

(g) The unit normal vector

ˆ

un

.

(h) The angles



x

,



y

and



z

that a makes with the x, y and z axes.

(i) The tangential component

at

of the acceleration.

(j) The normal component

an

of the acceleration.

(k) The radius of curvature of the path of P.

(l) The Cartesian coordinates of the center of curvature of the path.

Solution

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 1



y92.601



zcos1az

a









cos12.1761

6.2505













z69.626

(i)

ataˆ

ut 5.8526ˆ

i0.28366ˆ

j2.1761 ˆ

k

  0.76267ˆ

i0.32089ˆ

j0.56157 ˆ

k

 

at3.1505 m s2

(j)

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 1

anaˆ

un 5.8526ˆ

i0.28366ˆ

j2.1761 ˆ

k

  0.63904ˆ

i0.23982ˆ

j0.73083 ˆ

k

 

r

C 0.40678ˆ

Problem 1.6 An 80 kg man and 50 kg woman stand 0.5 meter from each other. What is the force of

gravitational attraction between the couple?

F1.0679 106 N

Problem 1.7 If a person’s weight is W on the surface of the earth, calculate the earth’s gravitational pull

on that person at a distance equal to the moon’s orbit.

F0.0002753W

Problem 1.8 If a person’s weight is W on the surface of the earth, calculate what it would be, in terms

of W, at the surface of

(a) the moon

(b) Mars

(c) Jupiter

Solution

WMars 0.3790W

Problem 1.9 A satellite of mass m is in a circular orbit around the earth, whose mass is M. The orbital

radius from the center of the earth is r. Use Newton’s Second Law of motion, together with Equations

1.25 and 1.40, to calculate the speed v of the satellite in terns of M, r and the gravitational constant G.

Solution

r

Problem 1.10 If the earth takes 365.25 days to complete its circular orbit of radius

149.6 106 km

around the sun, use the result of Problem 1.9 to calculate the mass of the sun.

2



r

Problem 1.11 F is a force vector of fixed magnitude embedded on a rigid body in plane motion (in the

xy plane). At a given instant

2ˆ

k rad s

,

&

 5ˆ

k rad s2

,

&&

3ˆ

k rad s3

and

F15ˆ

i10ˆ

j N

. At that

instant calculate

&&&

F

.

Problem 1.12 The absolute position, velocity and acceleration of O are

r

O 16ˆ

I84ˆ

J59 ˆ

K (m)

vO7ˆ

I9ˆ

J4ˆ

K (m s)

aO3ˆ

I7ˆ

J4ˆ

K (m s2)

The angular velocity and acceleration of the moving frame are

 0.8ˆ

I0.7ˆ

J0.4 ˆ

K (rad s) &

 0.4ˆ

I0.9ˆ

J1.0 ˆ

K (rad s2)

The unit vectors of the moving frame are

ˆ

i 0.15617ˆ

I0.31235ˆ

J0.93704 ˆ

K

ˆ

j 0.12940I0.94698ˆ

J0.29409 ˆ

K

ˆ

k 0.97922I0.075324ˆ

J0.18831 ˆ

K

The absolute position of P is

r51ˆ

I45ˆ

J36 ˆ

K (m)

The velocity and acceleration of P relative to the moving frame are

vrel 31ˆ

i68ˆ

j77 ˆ

k (m s) arel 2ˆ

i6ˆ

j5ˆ

k (m s)

Calculate the absolute velocity

vP

and acceleration

aP

of P.

vPvOr

rel vrel

aO3ˆ

&

ˆ

Iˆ

Jˆ

K

79.358 68.277 23.55

arel 2ˆ

i6ˆ

j5ˆ

k

20.15617ˆ

I0.31235ˆ

J0.93704 ˆ

K

 

60.12940ˆ

I0.94698ˆ

J0.29409 ˆ

K

 

50.97922ˆ

I0.075324ˆ

J0.18831 ˆ

K

 

arel  4.432ˆ

I6.6832ˆ

J0.83202 ˆ

K

(11)

Substituting (7), (8), (9), (10) and (11) into (6) yields

a3ˆ

I7ˆ

J4ˆ

K

 

a0

6 744 84 4

 149.7ˆ

I76.2ˆ

J8.7 ˆ

K

 

&

r

rel

6 74 4 4 4 84 4 4 4

36.05ˆ

I59.24ˆ

J31.57 ˆ

K

 

r

rel

 

6 74 4 4 44 84 4 4 4 4

87.592ˆ

I101.17ˆ

J1.857 ˆ

K

 

2vrel

6 74 4 4 4 4 84 4 4 4 4

 4.432ˆ

I6.6832ˆ

J0.83202 ˆ

K

 

arel

6 74 4 4 4 44 84 4 4 4 4 4

a85.129 0.32292ˆ

I0.82842ˆ

J0.45765 ˆ

K

 

m s2

 