978-0078112829 Chapter 3 Solution Manual Part 2

Problem 3.10 An earth-orbiting satellite has a period of 14 hours and a perigee radius of 10,000 km. At

time t = 10 hours after perigee passage, determine

(a) The radial position.

(b) The speed.

(c) The radial component of the velocity.

vr



hesin



398,600

81,366 0.66091sin 203.11  1.2708 km s

Problem 3.11 A satellite in earth orbit has perigee and apogee radii of

rp7500km

and

ra16,000km

, respectively. Find its true anomaly 40 minutes after passing true anomaly of 80°.

21e

1etan E

210.3617

10.3617 tan 3.0074

2 

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 3

Problem 3.12 Show that the solution to

acos



bsin



c

, where a, b and c are given, is:







cos1c

acos













where

tan



b a

.

acos





Problem 3.13 Verify the results of part (b) of Example 3.3.

2



2



Problem 3.14 Calculate the time required for a spacecraft launched into a parabolic trajectory at a

perigee altitude of 200 km to leave the earth’s sphere of influence (see Table A.2).

Problem 3.15 A spacecraft on a parabolic trajectory around the earth has a perigee radius of 6600 km.

(a) How long does it take to coast from



 90 degrees

to



 90 degrees

?

(b) How far is the spacecraft from the center of the earth 36 hours after passing through perigee?



1cos



398,600

1cos163.07304,700 km

Problem 3.16 A spacecraft on a hyperbolic trajectory around the earth has a perigee radius of 6600 km

and a perigee speed of

1.2vesc

.

(a) How long does it take to coast from



 90 degrees

to



 90 degrees

?

(b) How far is the spacecraft from the center of the earth 24 hours after passing through perigee?

Problem 3.17 A trajectory has a perigee velocity

1.1vesc

and a perigee altitude of 200 km. If at 10 AM

the satellite is traveling towards the earth with a speed of 8 km/s, how far will it be from the earth’s

surface at 5 PM the same day?

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 3



z142,630 6378 136,250 km

Problem 3.18 An incoming object is sighted at an altitude of 100,000 km with a speed of 6 km/s and a

flight path angle of –80°. (a) Will it impact the earth or fly by? (b) What is the time either to impact or

closest approach?

0.5064 hours until perigee passage

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 3

Problem 3.19 At a given instant the radial position of an earth-orbiting satellite is 7200 km, its radial

speed is 1 km/s. If the semimajor axis is 10 000 km, use Algorithm 3.3 to find the universal anomaly 60

minutes later. Check your result using Equation 3.58.

2

r 

2a

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 3

v2



1

r1

2a2398 600 1

7200 1

210000 8.41797 km s

vv2vr

28.417972128.35836 km s

hrv7200 8.35836 60180.2 km2s

vr



hesin



esin



hvr



60180.2 1

398 600 0.150979

(1)

฀

rh2



1

1ecos



ecos



h2



r160180.22

398 600 7200 10.261937

(2)

tan



esin



ecos



0.150979

0.261937 0.576395



29.9589

This is the true anomaly at the initial time. It lies in the first quadrant since, from (1) and (2), both

sin



and

cos



are positive.

From (1)

e0.150979

sin 29.95890.30233

Compute the time at the initial true anomaly as follows:

฀

tan E1

21e

1etan



1

210.30233

10.30233 tan 29.959

2 0.19584

฀

E10.38678 rad

฀

M1E

1esin E10.38678 0.30233 sin 0.38678 0.27273 rad

t1M1

2



T0.27273

2



9952.0 431.99 s

Obtain E one hour later.

t2t13600 4032 s

t2

4032

Problem 3.20 At a given instant a space object has the following position and velocity vectors relative

to an earth-centered inertial frame of reference:

r

020 000ˆ

i105 000ˆ

j19 000 ˆ

k km

 

v0=0.9000ˆ

i3.4000ˆ

j1.5000 ˆ

kkm s

 

Find

r

and

v

two hours later.