Math 1302, Week 3
Polar coordinates and orbital motion
1
Motion under a central force
We start by considering the motion of the earth E around the (fixed) sun (figure 1). The key
point here is that the force (here gravitation) is directed towards the fixed sun. Taking the
origin O at the sun and r as the position vector of the earth, the gravitation pull acts in the
direction EO, i.e. in the same direction as −r. One could try to write the equations of motion
Figure 1: Motion round sun under influence of gravity
in cartesian form:
mr̈ = F
becomes
m(ẍi + ÿj) = Fx i + Fy j.
Since the force is central (towards O),
F =F
1
(−r),
|r|
where F = |F | and r = OE. Now
Fx = F · i = −F cos θ = −F × p
x
x2
Fy = F · j = −F sin θ = −F × p
1
+ y2
y
x2 + y 2
Hence the equations of motion read
xF (r)
ẍ = − p
m x2 + y 2
yF (r)
ÿ = − p
m x2 + y 2
(1)
(2)
where now I have replaced F by F (r) to emphasise that F varies with x, y. It is not immediately
obvious how to tackle these equations! The cartesian description does not lend itself ideally to
this model.
To deal with motion under a central force it pays dividends to use a different coordinate system.
The force points along the position vector, so a natural way to write any vector is in terms of a
unit vector r̂ pointing along r and another unit vector θ̂ pointing perpendicular to r̂ (see figure
2). The force F then takes the very simple form1
F = |F |(−r̂).
To apply Newton’s law, the problem now becomes how to express the acceleration a in terms
Figure 2: Polar coordinates: choice of orthonormal basis vectors
of the basis vectors r̂, θ̂, and these basis vectors are different at different points in space! Thus
although we have made the force easier to deal with, the acceleration is trickier to calculate...but
the effort, it turns out, is worth it, since Newton’s law will immediately tell us that the component
of acceleration in the direction of θ̂ is zero, since the force acts purely along r̂. This will tell us
that the angular momentum is conserved.
2
Acceleration in polar coordinates
Consider figure 3 where a particle P is at r = xi + yj in cartesian coordinates. Let r = |r| be
the distance of the particle from the origin and r̂ = r/r the unit vector pointing along OP. Let
1
|F |(−r̂) for an attractive force, |F |r̂ for a repulsive force
2
Figure 3: Polar coordinates
θ̂ be a unit vector perpendicular to r̂ as shown. Then
r = rr̂,
so that the velocity of P is
˙
ṙ = ṙr̂ + rr̂.
˙ But r̂ = cos θi + sin θj, so that r̂˙ = − sin θθ̇i + cos θθ̇j = (− sin θi + cos θj)θ̇. Now
So we need r̂.
by the geometry, − sin θi + cos θj = θ̂. Thus r̂˙ = θ̂θ̇ and we get
ṙ = ṙr̂ + rθ̇θ̂
Thus the transverse (tangential) velocity component is rθ̇ and the radial component is ṙ.
Now for the acceleration. Differentiate again:
r̈ = r̈r̂ + ṙ
d
d
d
d
r̂ + (rθ̇)θ̂ + rθ̇ θ̂ = r̈r̂ + ṙθ̂θ̇ + (ṙθ̇ + rθ̈)θ̂ + rθ̇ θ̂.
dt
dt
dt
dt
To complete the formula, we thus need
d
d
θ̂ = (− sin θi + cos θj) = − cos θθ̇i − sin θθ̇j = −r̂θ̇
dt
dt
and hence we have r̈ = r̈r̂ + ṙθ̂θ̇ + (ṙθ̇ + rθ̈)θ̂ − rθ̇2 r̂ = (r̈ − rθ̇2 )r̂ + (2ṙθ̇ + rθ̈)θ̂. Sometimes it is
convenient to rewrite 2ṙθ̇ + rθ̈ = 1r dtd (r2 θ̇). To summarise:
Velocity: ṙ = ṙr̂ + rθ̇θ̂
Acceleration: r̈ =
(r̈ − rθ̇2 ) r̂ +
| {z }
radial component
3
1d 2
(r θ̇)
|r dt{z }
tangential component
θ̂.
Figure 4: Motion under an attractive central force in the plane
3
Motion under a central force in polar coordinates
Now instead of using −|F |r̂ for an attractive central force and +|F |r̂ for a repulsive central force
we will write the force as f (r)r̂, where f < 0 for attractive and f > 0 for repulsive forces.
If a particle mass m moves in the plane under an attractive force F we have mr̈ = F (see figure
4). Suppose that F = f (r)r̂. We have for such a central force,
mr̈ = m(r̈ − rθ̇2 )r̂ +
md 2
(r θ̇)θ̂ = f (r)r̂.
r dt
Since r̂, θ̂ are an orthonormal pair of vectors, we may equate coefficients of r̂, θ̂ on both side to
obtain
m(r̈ − rθ̇2 ) = f (r)
(1a, b)
md 2
(r θ̇) = 0.
r dt
The second equation (1b) tells us the very important fact: r2 θ̇ is conserved during the motion.
It is traditional to write r2 θ̇ = h where h is constant (it is angular momentum per unit mass).
Thus (1b) leads to conservation of angular momentum, i.e. r2 θ̇ = h, and this results from
the fact that the force is central (if not central, there would be a non-zero component of θ̂ in
the force, so that the right hand side of (1b) would be non-zero and r2 θ̇ would no longer be
constant).
From (1a,b) have m(r̈ − rθ̇2 ) = f (r) and r2 θ̇ = h. Rearranging the second equation, θ̇ = h/r2 ,
so that
h2
m(r̈ − 3 ) = f (r).
r
Unlike when we used cartesian coordinates, where the x, y coordinates were tied up in two
equations (1) and (2), we now have just one equation in r.
4
1
Conservation of Energy
From Newton’s second law,
mr̈ = F (r).
Thus
Z
Z
mṙ · r̈ dt =
so that
Z
ṙ · F (r) dt =
1
m|ṙ|2 −
2
dr
F (r) ·
dt =
dt
Z
F (r) · dr.
Z
F (r) · dr = constant.
If the force F is conservative (so that the work done in going between two points is independent
of the path), then there is a potential function V such that F = −∇V and we have from the
previous equation
1
m|ṙ|2 + V (r) = E,
| {z }
|2 {z }
potential energy
kinetic energy
where E is the total conserved energy. Since r = ṙr̂ + rθ̇θ̂ = ṙr̂ + hr θ̂ we have
mṙ2 mh2
+
+ V (r) = E.
2
2r2
Here E is the conserved total energy. When the force is gravitational, with gravitational constant
GM m
G we get V (r) = −
.
r
A very useful trick is to change from using r to using u = 1/r. With this change we find:
du
dt du
1d 1
1 −1
ṙ
=
=
=
ṙ = − .
(4)
2
dθ
dθ dt
h
θ̇ dt r
θ̇ r
Differentiation w.r.t. θ yields:
d2 u
dt d
=
dθ2
dθ dt
2
2
−ṙ
h
=−
1
r̈.
hθ̇
2
Thus r̈ = −hθ̇ ddθu2 = −h(hu2 ) ddθu2 = −h2 u2 ddθu2 , since h = r2 θ̇ = θ̇/u2 gives θ̇ = hu2 . Hence from
(1a),
2
2 2d u
2 2
m −h u
− (1/u)(hu ) = f (1/u)
dθ2
which rearranges to give
d2 u
f (1/u)
+u=−
.
2
dθ
mh2 u2
5
(5)
4
Planetary motion
We wish to establish the following laws that Johannes Kepler (1605) extracted from the observations of Tycho Brahe:
Kepler’s Laws of motion
First Law Planets move in ellipses with the sun at a focus;;
Second Law The area swept out per unit time by the radius vector joining a planet and the sun is constant;
Third Law The ratio of the square of the period of orbit to the
cube of the semi-major axis is constant.
The planets orbit the sun constrained planes that pass through the sun (see Question sheet 3,
Qu 2). Thus we may describe the position of a planet using polar coordinates in the appropriate
plane, and the motion is given by (1a,b). Often it is more convenient to work with equation
(5). When the force is due to gravity, we have f (r) = −GmM/r2 = −µ/r2 where µ = GmM
and G > 0 is the gravitational constant, m is mass of the planet and M the mass of the Sun.
Equation (5) then simplifies to
(−µu2 )
µ
d2 u
+
u
=
−
=
.
dθ2
mh2 u2
mh2
The general solution of this is u = C cos θ + D sin θ + particular integral, and it is easy to see
µ
that the particular integral is
. Thus
mh2
µ
.
u = C cos θ + D sin θ +
mh2
Alternatively we can write the general solution more conveniently as
µ
1
u = = A cos(θ − δ) +
.
r
mh2
Now define
mh2
Amh2
`=
,e =
,
µ
µ
so that the previous equation becomes
`
= 1 + e cos(θ − δ).
r
(6)
Equation (6) is the equation2 for a conic section in polar coordinates, and you need to know
that
0 < e < 1 ellipse
e = 0 circle
(7)
e = 1 parabola
e > 1 hyperbola
2
Normally in geometry textbooks the formula is given when δ = 0. The effect of δ is to rotate the curve by δ
about the origin.
6
It is important to appreciate that the radius r in (6) is the distance of a point on the curve
from a focus. You are referred to the Appendix for a discussion of the conic sections for various
eccentricities e > 0.
For the planets f (r) = −µ/r2 where µ = GmM , G is the gravitational constant, m is the mass
of the planet, M the mass of the Sun. We find that
`=
h2
Amh2
Amh2
Ah2
mh2
=
, e=
=
=
.
µ
GM
µ
GM m
GM
One finds that for all the planets in the solar system that e < 1 (in fact most of them have
e 1 so that their motion is close to circular). Thus from (7) we have Kepler’s first law that
the planets move in ellipse with the Sun at a focus.
For the second law, suppose that in time T the radius vector sweeps out an area A as θ changes
from θ0 to θT . We use that r2 θ̇ = h so that (see figure 5)
Z T
Z θT
1
1
dθ(t)
1
2
r(θ) dθ =
r(t)2
dt = hT.
A=
dt
2
0 2
θ0 2
Thus the area swept out by the radius per unit time is A/T = h/2 a constant. This is Kepler’s
second law.
area A
θΤ
r
θ0
θ
ellipse
focus
Figure 5: Kepler’s second law
Finally for the third law. We have, from Kepler’s second law,
Area of ellipse
h
= .
Period of orbit
2
But area of an ellipse with semi-major axis b and semi-minor axis a is πab. Hence
Tperiod =
2πab
.
h
We also have ` = a(1 − e2 ), b2 = a2 (1 − e2 ) and ` = h2 /GM . Thus
2
Tperiod
=
4π 2 a2 b2
4πa2 b2
4πa2 b2
4πa2 b2
4πa3
=
=
=
=
.
h2
GM `
GM a(1 − e2 )
GM a(b2 /a2 )
GM
7
Thus
2
Tperiod
4π
=
a3
GM
is independent of the planet, and hence we have Kepler’s third law.
Before reading the example problems you might want to read the appendix on ellipses.
5
Some example problems for central forces
Example 1
Consider a particle moving under a central force f (r)r̂ per unit mass where f (r) = k/r3 with
k > 0 constant [thus force is repulsive]. Find u(θ) given that the particle is projected from the
point r = a, θ = 0 with radial and transverse velocities U, V respectively. Show that the particle
moves off the infinity when tan(qθ) → qV /U where q 2 = 1 + k/(a2 V 2 ). Solution: We have from
Figure 6: Example 1
equation (5) that
(ku3 )
ku
d2 u
k
d2 u
+ u = − 2 2 = − 2 ⇒ 2 + 1 + 2 u = 0.
dθ2
hu
h
dθ
h
This has general solution u = A cos ωθ + B sin ωθ where ω 2 = 1 + k/h2 . When faced with such a
solution we need two conditions to get the two constants A, B. The first condition comes from
knowing the position of the particle at some point, and the second from knowing something
about its speed at some point.
Thus θ = 0, u−1 = r = a gives A = 1/a. Also θ = 0, ṙ = U, aθ̇ = V . Now we have
du = −ωA sin ω0 + Bω cos ω0 = Bω.
dθ θ=0
8
To find h we use r2 θ̇ = h at the point r = a, since then h = a2 θ̇ = a(aθ̇) = aV . But ṙ = −h du
,
dθ
du
so that at t = 0, U = ṙ = −(aV ) dθ . This gives that B = −U/(ωaV ) and hence that
!
!
r
r
1
k
k
U
u(θ) = cos
1+ 2 2 θ− √
1 + 2 2 θ.
sin
a
aV
aV
k + a2 V 2
For the second
goes off toq
infinity when
qpart, the particle
r(θ) → ∞, i.e. u(θ) → 0. But then this
means a1 cos
required.
1+
k
a2 V 2
U
θ − √k+a
2 V 2 sin
1+
k
a2 V 2
√
θ → 0, i.e. tan qθ →
k+a2 V 2
aU
= qV /U as
Example 2
(i.e. a force
Suppose particle moves under an attractive force per unit mass rα2 + rβ3 where β = aα
2
directed towards the origin). If at t p
= 0 the particle is distance a from the origin and moving
purely tangentially with speed V = α/a find the equation for its path and the furthest and
closest it comes to the origin.
2
3)
2
2
Solution: We have f (r) = − rα2 − rβ3 and so ddθu2 + u = (αuh2+βu
so that ddθu2 + 1 − hβ2 u = hα2 .
u2
Figure 7: Example 2
2 Now find h. Take θ = 0 where t = 0. Then h = r θ̇
= a2 θ̇(0) = aaθ̇(0) = aV . Putting this
t=0
value of h in the equation for u gives
1
αa
d2 u
α
2
+ 1− 2 2 u= 2 2
2
dθ
aV
aV
which tidies to
d2 u
dθ2
+ 12 u =
1
a
which has the general solution
u=
2
θ
θ
+ A cos( √ ) + B sin( √ ).
a
2
2
9
du
= −ṙh = 0 when
dθ
t = 0 (since the motion is then purely tangential). Thus √12 (B cos( √θ2 ) − A sin( √θ2 ))
= 0
θ=0
which gives B = 0 and hence
1
2 1
θ
u = = − cos √ .
r
a a
2
2
2
1
θ
1
−1
√
Thus r(θ) = ( a − a cos 2 ) and r is max when a − a cos √θ2 is min, i.e. when θ = 0 which
√
corresponds to r = a and r is min when a2 − a1 cos √θ2 is max, i.e. when θ = 2π which is
when r = a/3. The orbit is shown in figure 8 for a = 1. They are not ellipses!
But now use θ = 0, r = a to get
1
a
=
2
a
+ A cos 0 so that A = −1/a. Also
3
2
1
-3
-1
-2
1
2
3
-1
-2
-3
Figure 8: Orbits for example 2
6
Stability of circular orbits
First we need to find the conditions that a particle moves in a circular orbit. We take m = 1
2
for the mass. Recall that we have r̈ − hr3 = f (r). For circular motion radius b we have r = b,
θ̇ = ω0 and h = h0 say. Thus we have r̈ − (h2 /r3 ) = 0 − h20 /b3 = f (b) and also ub = h0 where u
is the constant speed u = bω0 . Hence we have for circular motion
−
h20
= f (b)
b3
h0
u =
b
Now consider small perturbations of the circular orbit. Thus suppose r = b + x and h = h0 + H
with x, H 1. Note that the new h of the perturbed motion is also constant by conservation
of angular momentum. Thus H is also constant. From r̈ − h2 /r3 = f (r) we obtain
d2
(h0 + H)2
(b
+
x)
−
= f (b + x).
dt2
(b + x)3
Now
1
1
1
1
3x
=
=
(1
−
+ · · · ),
x
(b + x)3
b3 (1 + b )3
b3
b
10
and by Taylor’s theorem
f (b + x) = f (b) + f 0 (b)x + · · · .
Hence we have
ẍ −
Hence
ẍ −
1 2
3x
+ · · · ) = f (b) + f 0 (b)x + · · · .
(h0 + 2h0 H + H 2 )(1 −
3
b
b
1 2
3xh20
) = f (b) + xf 0 (b) + higher orders in x, H
(h
+
2h
H
−
0
b3 0
b
h2
But for circular motion we had − b30 = f (b) and hence ẍ −
Letting q 2 = f 0 (b) −
3h20
b4
and x0 =
2h0 H
b3 q 2
2h0 H
b3
+
3xh20
b4
= xf 0 (b) to first order.
we have ẍ − q 2 (x − x0 ) = 0 which has general solution
x = x0 + Aeqt + Be−qt .
If q 2 > 0, so that q is real, either eqt or e−qt grows rapidly and hence the orbit is unstable. If
q 2 < 0 then q is complex and we get a stable “wobble” around the circular orbit:
q 2 = f 0 (b) −
q 2 = f 0 (b) −
7
3h20
b4
3h20
b4
> 0 ⇒ unstable
< 0 ⇒ stable
Appendix: The polar equation `/r = 1 + e cos θ
Here we consider the geometry of
`
= 1 + e cos θ,
r
for various e ≥ 0.
Since x = r cos θ we obtain
` = er cos θ + r
p
= ex + x2 + y 2 .
Thus
(` − ex)2 = x2 + y 2
`2 − 2`ex + e2 x2 = x2 + y 2
Rearranging we get
(1 − e2 )x2 + 2`ex + y 2 = `2 ,
and
el
x+
1 − e2
2
+
y2
`2
=
,
1 − e2
(1 − e2 )2
Now consider various values of e.
1. e = 0 Here r = `. Thus the conic is a circle;
11
(8)
x 2 y 2
`
`
2. 0 < e < 1 Let a =
> 0. Then
> 0 and b = √
+
= 1 and we get an
1 − e2
a
b
1 − e2
ellipse.
3. e = 1 Then y 2 = `2 − 2`x and we have a parabola.
4. e > 1 Then with a =
x 2 y 2
`
`
√
>
0
and
b
=
−
= 1 which is a
>
0
we
have
e2 − 1
a
b
e2 − 1
hyperbola.
Figure 9: Geometry of an ellipse e ∈ (0, 1).
For an introduction of conic sections, see pages 198-212 in “What is Mathematics?” by Richard
Courant and Herbert Robbins, OUP (1980).
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