Download Seminar 7: CENTRAL FORCE PROBLEM Problem 26 A particle of

Seminar 7: CENTRAL FORCE PROBLEM
Problem 26
A particle of mass m is attracted to a force center by a force which varies inversely
as the cube of its distance from the center. Solve the equations of motion for the orbits and discuss how the nature of the orbits depends on the parameters of the system.
Solution
From Lecture 2, we already know the equations of motion for a particle in the
presence of an arbitrary central force f (r). These equations are as follows:
(
m(r̈ − rθ̇2 ) = f (r)
d
mr2 θ̇ = 0.
dt
(1)
The second equation has the first integral,
mr2 θ̇ = l,
(2)
which is the angular momentum of the particle about the origin since it can be
represented as a product
l = r · mrθ̇ ≡ r · mvθ ,
(3)
where vθ is the angular velocity of the particle.
For the force
f =−
kr
,
r4
(4)
the first of Eqs. (1.1) is
k
,
r3
or, by substituting θ̇ expressed in terms of l in accordance with Eq. (1.3),
m(r̈ − rθ̇2 ) = −
k
l2
mk
l2
+
=
mr̈
−
mr̈ −
1− 2
3
3
3
mr
r
mr
l
(5)
!
= 0.
(6)
We know that r = r(θ), and that this equation can be essentially simplified if we
change this variable letting
r=
1
.
u
l
mr2
=
(7)
Indeed, then we have
θ̇ =
ṙ =
d 1
dt u
lu2
m
2
lu
= − u12 du
θ̇ = − u12 du
= − ml du
dθ
dθ m
dθ
2
2
2
2
2
d du
r̈ = − ml dθ
= − ml ddθu2 θ̇ = − ml ddθu2 lum = − ml 2 u2 ddθu2 .
dt
(8)
With these transformations, Eq. (6) becomes
mk
l2 d2 u l2 u3
1− 2
−m 2 u2 2 −
m
dθ
m
l
or
!
= 0,
mk
d2 u
+ 1 − 2 u = 0.
2
dθ
l
(9)
!
(10)
As we see, this is the linear homogeneous differential equation of the second order.
The form of its solution depends critically on the relation between the parameters
involved. Namely, if
l2 > mk,
this solution is
(11)
"s
1
u = cos
r0
#
mk
1 − 2 (θ − θ0 ) ,
l
(12)
where (r0 , θ0 ) is a point on the orbit. In the opposite case, when
l2 < mk,
(13)
the form of solution is changed to
1
u = cosh
r0
"s
#
mk
− 1 (θ − θ0 ) .
l2
(14)
The corresponding equations for the orbits are
!#−1
s
"
r = r0 cos
and
mk
1 − 2 (θ − θ0 )
l
s
"
r = r0 cosh
,
(15)
!#−1
mk
− 1 (θ − θ0 )
l2
,
(16)
respectively.
Problem 27
Consider a particle of mass m moving in a plane under a central force
f (r) = −
k
k0
+
.
r2 r3
(17)
Find the equation for the orbit in the polar coordinates (r, θ), assuming that k > 0
and l2 > −mk 0 , where l is the orbital angular momentum.
Solution
The force (17) differs from the one given by Eq. (4) in the preceeding problem by
the presence of the inverse-square term. Hence the equation of motion (6) is modified
to
l2
mk 0
mr̈ −
1
+
mr3
l2
!
+
k
= 0,
r2
(18)
or, with the use of the transformations (8),
d2 u
mk 0
mk
+
1
+
u= 2 .
2
2
dθ
l
l
!
(19)
Opposite to our previous case [see: Eq. (10)], Eq. (19) contains in its r.h.s. the
driving term
mk
.
l2
This term is constant, and hence an obvious particular solution of
the inhomogeneous Eq.(19) is also constant,
mk
mk 0
u= 2 1+ 2
l
l
!
mk
.
l2 + mk 0
=
(20)
This particular solution must be add to the general solution of the corresponding
homogeneous equation,
mk 0
d2 u
+
1
+
u = 0.
dθ2
l2
!
(21)
where l2 > −mk 0 ,by assumption. Therefore we may write the solution of this equation
in the form given by Eq. (12), that is
s
u = A cos
!
mk 0
1 + 2 (θ − θ0 ) ,
l
(22)
The general solution of the inhomogeneous equation (19) is thus
s
!
mk 0
mk
1 + 2 (θ − θ0 ) + 2
,
l
l + mk 0
u = A cos
(23)
where A and θ0 are constants. By a suitable choice of coordinates, θ0 can be put to
zero, and hence the equation for the orbit can be written as
1
r=
q
A cos
mk0
l2
1+
θ +
mk
,
(24)
l2 +mk0
or
r=
1
h
q
C 1 + ε cos
i ,
(25)
l2 + mk 0
.
mk
(26)
1+
mk0
l2
θ
where C and e are new constants,
C=
mk
l2 + mk 0
ε=A
To understand the character of this orbit, let first consider a particular case, when
k 0 = 0. Then Eq. (25) becomes
1
,
C(1 + ε cos θ)
r=
(27)
where
mk
l2
.
(28)
,
ε
=
A
l2
mk
We already met with this orbit in our Lectures [see: Eq. (4.125) in Lecture Notes],
C=
and we know that this orbit is an ellipse with one of the foci at the origin of the plane
polar coordinates and with the eccentricity ε. Indeed, you know that an ellipse is
defined as the curve traced by a particle moving so that the sum of its distances from
two fixed points F, F 0 is constant. These points are the foci of the ellipse. Using the
notation indicated in Figure (it will be demonstrated during Seminar), we have
r0 + r = 2a,
(29)
where a is a half the largest diameter (major axis) of the ellipse. In terms of the
polar coordinates with center at the focus F and with the negative x-axis through
the focus F 0 , the cosine law gives
r02 = r2 + 4a2 ε2 + 4raε cos θ,
(30)
where aε is the distance from the center of the ellipse to the focus. Combining Eqs.
(29) and (30), we find
r=
a(1 − ε2 )
.
(1 + ε cos θ)
(31)
Comparing this equation with our orbit given by Eq. (27), we see that these expressions are identical, if
C=
1
.
a(1 − ε2 )
(32)
But this is really the case because for the perihelion point θ = 0 we have [see Figure
3.38]
r = a − aε = a(1 − ε),
(33)
while Eq. (27) with θ = 0 reduces to
r=
1
,
C(1 + ε)
(34)
Therefore
a(1 − ε) =
1
,
C(1 + ε)
(35)
which yields
C=
1
,
a(1 − ε2 )
(36)
that is the expression for C given by the standard equation (32). Note that the constant A which determines the eccentricity ε according to Eq. (28) remains unknown:
it can be found from the formal solution of the equations of motion or from the energy
consideration but not from the differential equations for the orbit which we used in
our problem.
Now it is not difficult to guess that if the constant k 0 is not zero but small enough,
the orbit will be a modified ellipse whose perihelion will be shifted a little from its
conventional value given by Eq. (33). If necessary, this shift can be easily calculated
from the equation of orbit with k 0 6= 0, that is, Eq. (25). This problem is in no way
only of academic interest. Say, it can be shown that under the influence of the Earth
attraction Mercury moves in the force
F =−
GM m
α
− 3,
2
r
r
(37)
where α is a positive constant, that is, just in the force of the type, given by Eq. (17).