3.7. DIFFERENTIAL EQUATION FOR TRAJECTORY OF APARTICLEUNDER ACENTRAL FORCE 41
Now, consider the rate of change of angular momentum of a particle. Taking the derivative of
Equation (3.16) w.r.t. t, one obtains
d
*
H
o
dt
D
d
*
r
dt
m
*
v C
*
r m
d
*
v
dt
D
*
v m
*
v C
*
r m
*
a : (3.19)
e first term on the rhs is zero because
*
v and m
*
v are collinear. On the other hand, m
*
a D
P
*
F
by Newton’s second law of motion. erefore, Equation (3.19) reduces to
d
*
H
o
dt
D
*
r m
*
a D
*
r
X
*
F D
X
*
M
o
or
X
*
M
o
D
P
*
H
o
: (3.20)
is equation expresses the sum of the moments of the forces acting on the particle about the origin O
as the rate of change of angular momentum of the particle about O.
3.7 DIFFERENTIAL EQUATION FOR TRAJECTORY OF A
PARTICLE UNDER A CENTRAL FORCE
e objective of this section is to derive a differential equation that predicts or provides the
trajectory of the particle under a central force. Areas of application are in space mechanics in
which the orbit of a satellite is of great importance.
To begin with, one considers a particle P under a central force
*
F , as shown in Figure 3.6.
y
F
⃑
xO
P
r⃑
Figure 3.6: Particle under a central force.
Recall that the position vector
*
r measures positive from the origin O of the NFR to the
particle P . Note that the central force as indicated in the figure is in the opposite direction
and therefore the central force is considered negative. Recall also that the radial and transverse
components of the force of the particle in polar coordinates are given by Equation (3.8) as
X
F
r
D m
Rr r
P
2
;
X
F
D m
r
R
C 2 Pr
P
:
42 3. DYNAMICS OF PARTICLES
With reference to Figure 3.6,
X
F
r
D F D m
Rr r
P
2
;
X
F
D 0: (3.21)
For the particle under a central Equation (3.20) gives
*
H D 0 since
X
*
M
o
D 0:
Upon integration w.r.t. t,
*
H
o
D h
where h is a constant and by making use of Equation (3.15)
*
H
o
D h D mr
2
P
:
Rewriting this equation,
P
D
d
dt
D
h
mr
2
D
c
r
2
(3.22)
in which c is a constant since h and m are constant. us, c is a constant representing the angular
momentum per unit mass.
By definition,
Pr D
dr
dt
D
dr
d
d
dt
D
dr
d
c
r
2
: (3.23)
Also,
d
1
r
d
D
d
1
r
dr
dr
d
D
1
r
2
dr
d
: (3.24)
Substituting this equation into (3.23), one obtains
Pr D
dr
d
c
r
2
D r
2
d
1
r
d
c
r
2
D c
d
1
r
d
: (3.25)
Taking the time derivative of Pr and applying Equation (3.22),
Rr D
d Pr
dt
D
d Pr
d
d
dt
D
d Pr
d
c
r
2
:
Substituting for Equation (3.25), one has
Rr D
d Pr
d
c
r
2
D
d
h
c
d
.
1=r
/
d
i
d
c
r
2
D
c
r
2
d
2
1
r
d
2
: (3.26)
Now, writing u D 1=r and substituting Equations (3.22) and (3.26) into the first equation of
(3.21), one obtains
F D m
c
2
u
2
d
2
u
d
2
r
cu
2
2
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