978-0078112829 Chapter 2 Solution Manual Part 3

Problem 2.33 –

Problem 2.34 If the perigee velocity is c times the apogee velocity, calculate the eccentricity of the orbit

in terms of c.

rarp

ecrprp

crprp

ec1

 rp

c1

 rp

ec1

 

c1

 

Problem 2.35 At what true anomaly does the speed on a parabolic trajectory equal



times the speed

at periapsis, where



1

?

r



2



h2



r

rh2

2



2

rh2



1

1cos



h2

2



2h2



1

1cos







cos12



21

 

Problem 2.36 What velocity, relative to the earth, is required to escape the solar system on a parabolic

path from earth’s orbit?

Problem 2.37 A hyperbolic earth departure trajectory has a perigee altitude of 250 km and a perigee

speed of 11 km/s.

(a) Calculate the hyperbolic excess speed (km/s).

(b) Find the radius (km) when the true anomaly is 100 degrees. {Ans.: 16,179 km}

(c) Find

vr

and

v

(km/s) when the true anomaly is 100 degrees.

฀

vh

r72, 908

16,179 4.5064 km s

Problem 2.38 A meteoroid is first observed approaching the earth when it is 402 000 km from the cen-

ter of the earth with a true anomaly of 150°. If the speed of the meteoroid at that time is 2.23 km/s, calcu-

late (a) the eccentricity of the trajectory; (b) the altitude at closest approach; (c) the speed at closest ap-

proach.

฀

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 2

Problem 2.39 If



is a number between 1 and

1e

  1e

 

, calculate the true anomaly at which the

speed on a hyperbolic trajectory is



times the hyperbolic excess speed.



e21



 

rh2



1

1ecos



ecos



h2



r1h2



2h2



e21

 



21

 













1

ecos



1

2



21

  e21

 1





cos1



21

  e21

 2

2e











Problem 2.40 For a hyperbolic orbit, find the eccentricity in terms of the radius at periapsis

rp

and the

hyperbolic excess speed

v

.

rp



e21

v











2



1

1e



v2

e21

e1



v2e1

 

e1rpv2



Problem 2.41 A space vehicle has a velocity of 10 km/s in the direction shown when it is 10 000 km

from the center of the earth. Calculate its true anomaly.

฀

ecos



Problem 2.42 A spacecraft at a radius

r

has a speed

v

and a flight path angle



. Find an expression

for the eccentricity of its orbit.

Problem 2.43 For an orbiting spacecraft,

rr

1

when



=



1

, and

rr2

when



=



2

. What is the eccen-

tricity?

r2cos



1r

1cos



2

Solutions Manual Orbital Mechanics for Engineering Students Third Edition Chapter 2

Problem 2.44 At a given instant, a spacecraft has the position and velocity vectors

r

07000 ˆ

i km

 

and

v07ˆ

i7ˆ

j km s

 

relative to an earth-centered non-rotating frame. (a) What is the position vector after

the true anomaly increases by 90°? (b) What is the true anomaly of the initial point?

hsin

49000 sin 90  6169.1 s