Some Fundamental Questions
About a Sounding Rocket
How high does it go?
What maximum speed does it achieve going up?
What maximum speed does it achieve coming down?
Deriving the Equations for
Sounding Rocket Motion: Basic Approach
Break the rocket trajectory into three parts:
(1) high-thrust segment
Key simplifying assumptions:
Trajectory is purely vertical (perfect control)
Rocket thrust is so large and operates over such a short period of time that gravity
and drag can be ignored during the high-thrust segment.
Definition of Terms
=
=
=
m
V
h
Rocket altitude
Rocket vertical velocity
Rocket mass
High-thrust segment
Conservation of mass and momentum implies:
c
dt
dm
dt
dV
m−=
c
dt
dm
mdt
dV 1
−=
We have three variables here: time, mass and velocity;
but in this case we can eliminate time from the equation using
or
High-thrust segment (2)
m
dm
cdV −=
Equation 1
Equation 1 is a (very simple) example of a separable first order ODE.
Its solution, recognizing the initial conditions are the launch conditions
(V = 0 and mass ), is
0
m
High-Thrust Segment
To summarize:
Starting from
c
dt
dm
dt
dV
m−=
Phase 2: No-thrust segment going up
VSCgm
dt
dV
mDbb
−−=
2
2
1
Objective of phase 2: Determine how high the sounding rocket will go.
Newton’s second law: Force = mass times acceleration
Here force = weight plus drag in downward direction:
No-thrust segment going up (2)
dt
kVg
dV −=
+2
This is an example of a “separable” first order DE.
Recall the general form of a separable first order DE is:
No-thrust segment going up (3)
dt
kVg
dV −=
+2
We want to make a substitution to get the left-hand side into the form
Equation 1
No-thrust segment going up (4)
After some algebra Equation 1 becomes
dtgk
v
dv −=
+2
1
Integrating Equation 2 from t = 0 to t, and, correspondingly,
Equation 2
v
from
No-thrust segment going up (5)
From Equation 3 we see that apogee (point of highest altitude, i.e.,
when vertical velocity has fallen to zero) occurs at time
gk
ta
1
=
arctan(
b
Vgk
)
No-thrust segment going up (6)
a
t
and obtainTo simplify Equation 4 we use the equation for
Now let
tgk
=
)()tan(
1
)()( 00 −== t
a
ttgkdtgktgk
k
dttVth
No-thrust segment going up (7)
The maximum altitude is reached when
a
a
a
tgktgk
tt
=
=
=
No-thrust segment going up (8)
1
a
b
Vgk
Recall
)arctan( baa Vgktgk ==
From the diagram:
How to Check
A check on this result:
For small x
xx + )1ln(
which is the classical drag-free result.
No-thrust segment going up (9)
dt
kVg
dV
−=
+
2
To summarize:
Starting with
Homework Assignment 3
Prepare for short quiz in next class covering all material presented so
far, not including this class
In the text: