Conic Sections and the Inverse Square Force Law
We start with 1/r = B + A cos(-c) where we have defined B = -mK/L2 ,
and L = mr2(d/dt) = constant. Normally we use the initial conditions to determine L, B,
A and c. We have also defined: ε = A / │B│.
Since A multiplies the cosine function, we can treat A as being a positive number since
there will always be a value of c to make the term negative if needed.
1. A=0
Let’s start with A=0. From the definition of ε = A / │B│, we get ε = 0 .
From the basic equation
1/r = B + A cos(-c) ,
1/r = B = constant;
or
with A=0 we get
r = 1/B = constant.
This is the equation of a circle.
Note that since r can’t be negative, B must be greater than zero. Since B = -mK/L2 this
implies that K must be less than zero, and hence the force must be attractive.
The area of a circle should be readily known: S = πr2. This can be used to determine the
period of the orbit, T, using the equation we developed in the General Central Force
section in part 8:
S = (L/2m)T .
2. 0 < A < │B│
In this case, from the definition of
ε = A / │B│, we get 0 < ε < 1 .
From the basic equation
1/r = B + A cos(-c) , with 0 < A < │B│, we see that
B must be positive and so K must be less than zero as in the case of circular motion.
We also see that since A < B, we will never have 1/r = 0, so we will have a closed (finite)
orbit which we identify as an ellipse.
We can easily see that the quantity B+A cos(-c) will be a maximum (and hence r a
minimum) when  = c, and B+A cos(-c) will be a minimum (and hence r a
maximum) when  = c+180o .
1/rmin = B + A
and
1/rmax = B – A
Note that if we know both rmin and rmax, we can get both B and A, and knowing B we can
get L, and knowing both B and A we can get .
Below we plot a picture of an example of the orbit with c = 0.
rmax =180o
rmin =0o
We could draw the exact same orbit if we flipped everything by 180o . This is really just
choosing c = 180o .
rmin
rmax
Let’s now see if we can show that this is in fact an ellipse by the geometrical definition,
which is r1 + r2 = 2a for all points on the curve. First we define the distance “a”
(semimajor axis) as:
2a = rmin +rmax.
We can do this mechanically by using a string of length 2a, attaching the two ends to the
two foci, and then have a pencil push the string taught and move it around in an orbit.
r2
r1
Here we will start with the orbit as: 1/r = B + A cos(-c) and show that this leads
to the geometrical definition of an ellipse as r1 + r2 = 2a.
First, lets define 2x as the distance between the foci: 2x = rmax – rmin .
rmax
rmin
2x
x
x
a
Note that a+x = rmax and a-x = rmin . Note also that 1/rmax = B–A and 1/rmin = B+A,
or , with ε = A / B,
rmax = a + x = 1/(B-A) = 1/(B – εB) = 1/[B(1-ε)]
rmin = a – x = 1/(B+A) = 1/(B + εB) = 1/[B(1+ε)]
rmax + rmin = 2a = 1/[B(1-ε)] + 1/[B(1+ε)] =
(1+ε) /[B(1-ε)(1+ε)] + (1-ε)/[B(1+ε)(1-ε)] = 2/[B(1-ε2)]
rmax - rmin = 2x = 1/[B(1-ε)] - 1/[B(1+ε)] =
(1+ε) /[B(1-ε)(1+ε)] - (1-ε)/[B(1+ε)(1-ε)] = 2ε/[B(1-ε2)]
From these two expressions, we see that x = εa , and a = 1/[B(1-ε2)] . Solving this last
expression for B gives:
B(1-ε2)] = 1/a,
or B = 1/[a(1-ε2)] .
Below we choose 1c = 180o and 2c = 0o .
r1
r2
1 - 1o
2
rmin
2x = 2εa
rmin
For the left focus point, using the law of cosines we have:
r22 = r12 + (2εa)2 – 2r1(2εa)cos(180o – [1-θ1c]) .
For the right focus point, using the law of cosines we have:
r12 = r22 + (2εa)2 – 2r2(2εa)cos(180o-[2-θ2c]) .
Using the trig identity that cos(180o-) = -cos(), we get
r22 = r12 + (2εa)2 + 2r1(2εa)cos(1-θ1c) ,
r12 = r22 + (2εa)2 + 2r2(2εa)cos(2-θ2c) .
Now using our trajectory equations:
1/r = B + A cos(-θc) ,
or
cos(-θo) = [(1/r) – B] / A ,
we get:
r22 = r12 + (2εa)2 + 2r1(2εa) [(1/r1) – B] / A
,
r12 = r22 + (2εa)2 + 2r2(2εa) [(1/r2) – B] / A .
If we add both of these expressions, we get:
r22 + r12 = r12 + r22 + 2(2εa)2 + 4εar1 [(1/r1) – B] / A + 4εar2 [(1/r2) – B] / A
or, simplifying we get:
0 = 8ε2a2 + 4εa[(1/A) - Br1/A + (1/A) – Br2/A]
Now we divide out 4εa and note that B/A = 1/ε so that we get:
0 = 2εa + (2/A) – (1/ε)(r1+r2) .
If we multiply the 2/A by B/B (and again use B/A = 1/ε) and then multiply all terms by ε
we get:
r1+ r2 = 2ε2a + 2/B
Now recalling our expression for B from the top of the previous page:
B = 1/[a(1-ε2)] , or 1/B = a(1-ε2)
we get:
r1 + r2 = 2ε2a + 2[a(1-ε2)] = 2a .
This is just the geometric definition for an ellipse that we wanted to show.
To get the period, we can again use the equation we developed in the General Central
Force section in part 8:
S = (L/2m)T . The area of an ellipse is easily expressed
in terms of the semimajor axis, a, and the semiminor axis, b, as: S=πab. Thus we need
to determine the distance, b, since we have already determined the distance, a.
r2=a
r1=a
b
2x=2εa
εa
When r1 = r2, we get the situation pictured above. Also, since r1+r2 = 2a, when r1 = r2, we
have r = a, which is the hypotenuse of the right triangle. The base is εa, and the vertical
side is b. Thus, using the Pythagorean Theorem, we get
b = [a2-ε2a2]½ = a[1-ε2]½ , and we recall a = ½(rmax+rmin) .
Another nice relation is: x = εa, so ε = x/a, but x = ½(rmax-rmin), and a = ½(rmax+rmin), so
we get: ε = (rmax-rmin)/(rmax+rmin).
3) A = │B│
In this case, from the definition of ε = A / │B│, we get
ε=1.
From the basic equation 1/r = B + A cos(-c) , with A = │B│, we see that B must
be positive and so K must be less than zero (attractive force) as in the case of circular and
elliptical motion.
We also see that since A = B, we will have 1/r = 0 (and hence r approaches infinity) at
only one point in the motion (when -c = 180o) , so we will not have a closed (finite)
orbit and we identify this particular motion as that of a parabola.
Geometrically, the parabola is defined as (y-yo) = k(x-xo)2 . To see if our basic equation
gives us this, we start with B=A, and to make the parabola go up (in the y direction) we
will choose c = 270o. Note that cos(-270o) = cos(+90o) = -sin()
1/r = B + B cos(-270o) = B[1-sin()] .
We now use r = [x2+y2]½ and sin() = y/[x2+y2]½ to get:
1/[x2+y2]½ = B (1 - y/[x2+y2]½ )
Let’s now multiply through by [x2+y2]½ to get:
1 = B[x2+y2]½ - By , or
1 + By = B[x2+y2]½ .
If we square both sides, we get:
1 + 2By + B2y2 = B2x2 + B2y2 , or
1 + 2By = B2x2 ,
or
(y + 1/2B) = (B/2)(x2)
which is of the form that we wanted. (The choice of c being 270o resulted in xo being
zero in the above equation.) 1/rmin = B + A cos(θo-θo) = B + B = 2B, or rmin = 1/2B with
θ=270o. Note that this rmin agrees with the y = -1/2B when x=0 from the geometric form.
Since the area of a parabola is infinite, the period of the “orbit” is infinite.
r
rmin
4) A > │B│
In this case, from the definition of ε = A / │B│, we get
ε>1.
From the basic equation 1/r = B + A cos(-c) , with A > │B│, we see that B can be
either positive or negative, and so K can be either positive or negative, which means we
can have this kind of motion for either attractive or repulsive inverse square central
forces.
We also see that since A > │B│, we will have 1/r = 0 (and hence r approaches infinity)
at two points in the motion, so we will not have a closed (finite) orbit and we identify this
particular motion as that of a hyperbola.
Since the area of a hyperbola is infinite, the period of the “orbit” is infinite.
The closest approach (rmin) occurs when  = c.
A rough plot of this type of motion is below for the case where B is positive (so K is
negative, so an attractive force) with 1c = 180o.
r1
We can get this exact same plot (although starting at a different foci) if we have a case
where B is positive (repulsive force) and θc is different by 180o. To see this, consider the
critical angles involved. For the red case, r1 (attractive force, B<0, with 1c = 180o), the
critical angle, 1critical is when r1 goes to infinity and 1/r1 goes to zero:
0 = B + A cos(1critical -1c) = │B│+ A cos(1critical -180o),
or
cos(1critical) = - cos(1critical -180o) = │B│/ A = 1/ε .
For the blue case (see next page), r2 (repulsive force, B<0, with 2c = 0o), the critical
angle, 2critical is:
0 = B + A cos(2critical -2c) = -│B│+ A cos(2critical - 0o),
cos(2critical) = │B│/ A = 1/ε .
or
Note that both give the same critical angle, 1critical = 2critical. Also note that the second
critical angle (part of the plot below the center line in the diagram below) will merely be
the negative of the first one, since cos() = cos(-) . These two angles then define two
lines that serve as asymptotes for the curve. We plot this in the figure below with r2
(repulsive force, 2c = 0o) in blue and r1 (attractive force, 1c = 180o) in red.
r2
r1
2critical
initial
Θ
final
If we consider the direction of the object, it changes from pointing in along the asymptote
initially to pointing out along the second asymptote finally. This angle, which we will
call Θ, is then the angle that the object deflects due to the central force located at
either focus. Note that Θ + 2θcritical = 180o , and so θcritical = 90o – Θ/2 .
The geometrical definition of a pair of hyperbolas is: r2 – r1 = 2a. Let’s see if our general
formula, 1/r = B + a cos(θ-θc) is consistent with this geometrical definition.
We start by looking at r1min and r2min. and we use the definition: ε = A / │B│,
1/r1min = B + A cos(θ1c-θ1c) where B > 0 (K < 0, attractive force)
or
1/r1min = B + εB, or r1min = 1/{B(ε+1)}; and
1/r2min = B + A cos(θ2c-θ2c) where B < 0 (K > 0, repulsive force)
or
1/r2min = -│B│ + ε│B│ , or r2min = 1/{│B│(ε-1)} .
Thus, we get:
r2min + r1min = 1/{│B│(ε-1)} + 1/{│B│(ε+1)} =
(ε+1)/ {│B│(ε2-1)} + (ε-1)/ {│B│(ε2-1)} = 2ε / {│B│(ε2-1)} ; and
:
r2mn – r1min = 1/{│B│(ε-1)} - 1/{│B│(ε+1)} =
(ε+1)/ {│B│(ε2-1) } - (ε-1)/ {│B│(ε2-1)} = 2 / {│B│(ε2-1)} = 2a
Note that : r2min + r1min = 2εa . Also note that we can define a in terms of │B│from
the above equation: a = 1 / {│B│(ε2-1) } . We’ll use this later in the geometric
definition part.
Let’s look at the diagram below to see what a and εa are.
2εa
2a
r2min
r1min
a
εa
To see that r2 – r1 = 2a for all points on the hyperbola, not just for the minimum radii,
we start with the law of cosines for each radius (attractive, r1, and repulsive, r2). See the
diagram below.
r2
r1
2εa
r22 = r12 + (2εa)2 - 2r1(2εa) cos(θ1-θ1c) where θ1c = 180o .
We will ASSUME that the geometric definition applies (that is, r2 = 2a + r1), and see if
we can then DERIVE from this our general formula for r: 1/r = B + A cos(θ-θc) . We
start with the law of cosines applied to the triangle shown in the above diagram.
(2a+r1)2 = r12 + (2εa)2 - 2r1(2εa) cos(θ1-θ1c)
which becomes on expanding the left side:
4a2 + 4ar1 + r12 = r12 + 4ε2a2 - 4r1εa cos(θ1-θ1c)
which reduces to (after canceling out the r12 terms and dividing the remaining terms by
4a:
a + r1 = ε2a - r1ε cos(θ-θc) ,
or
r1{1+ε cos(θ1-θ1c)} = a(ε2-1) , or
1/r1 = 1/[a(ε2-1)] + ε cos(θ1-θ1c) / [a(ε2-1)] ;
and if we let B = 1/[a(ε2-1)] (see the relation we developed earlier) and A = εB (from the
definition of ε), we get
1/r1 = B + A cos(θ1-θ1c)
which is the form we were trying to reach.
r1 → 
θcritical
Θ
initial
final
When r1 goes to infinity, (θ1– θ1c) goes to (180o - θcritical) and the above expression:
1/r1 = 1/[a(ε2-1)] + ε cos(θ1-θ1c) / [a(ε2-1)]
gives us the relation:
0 = 1/[a(ε2-1)] + ε cos(180o - θcritical) / [a(ε2-1)] , or
0 = 1 + ε cos(180o - θcritical),
or
cos(θcritical) = 1/ε .
Further, we have sin(θcritical) = [1 – cos2(θcritical)]½ = [1 – (1/ε)2]½ = [ε2 – 1]½ / ε .
From the definition of Θ and the figure above, we see that
Θ + 2θcritical = 180o,
or
θcritical = 90o – Θ/2.
cos(θcritical) = 1/ε = cos(90o – Θ/2) = sin(Θ/2)
sin(θcritical) = [ε2 – 1]½ / ε = sin(90o – Θ/2) = cos(Θ/2)
Therefore,
tan(Θ/2) = (1/ε) / {[ε2 – 1]½ / ε} = 1 / [ε2 – 1]½ .