4. Central Forces
In this section we will study the three-dimensional motion of a particle in a central
force potential. Such a system obeys the equation of motion
mẍ = rV (r) (4.1)
where the potential depends only on r = |x|. Since both gravitational and electrostatic
forces are of this form, solutions to this equation contain some of the most important
results in classical physics.
Our first line of attack in solving (4.1) is to use angular momentum. Recall that this
is defined as
L = mx ⇥ ẋ
We already saw in Section 2.2.2 that angular momentum is conserved in a central
potential. The proof is straightforward:
dL
= mx ⇥ ẍ = x ⇥ rV = 0
dt
where the final equality follows because rV is parallel to x.
The conservation of angular momentum has an important consequence: all motion
takes place in a plane. This follows because L is a fixed, unchanging vector which, by
construction, obeys
L·x=0
So the position of the particle always lies in a plane perpendicular to L. By the same
argument, L · ẋ = 0 so the velocity of the particle also lies in the same plane. In this
way the three-dimensional dynamics is reduced to dynamics on a plane.
4.1 Polar Coordinates in the Plane
We’ve learned that the motion lies in a plane. It will turn out to be much easier if we
work with polar coordinates on the plane rather than Cartesian coordinates. For this
reason, we take a brief detour to explain some relevant aspects of polar coordinates.
To start, we rotate our coordinate system so that the angular momentum points in
the z-direction and all motion takes place in the (x, y) plane. We then define the usual
polar coordinates
x = r cos ✓ , y = r sin ✓
– 48 –
,Our goal is to express both the velocity and acceleration y ^ ^
θ r
in polar coordinates. We introduce two unit vectors, r̂
ˆ in the direction of increasing r and ✓ respectively
and ✓
as shown in the diagram. Written in Cartesian form, θ x
these vectors are
! !
cos ✓ ˆ= sin ✓
r̂ = , ✓
sin ✓ cos ✓
Figure 11:
These vectors form an orthornormal basis at every point
on the plane. But the basis itself depends on which angle ✓ we sit at. Moving in the
radial direction doesn’t change the basis, but moving in the angular direction we have
! !
dr̂ sin ✓ d ˆ
✓ cos ✓
= =✓ˆ , = = r̂
d✓ cos ✓ d✓ sin ✓
This means that if the particle moves in a way such that ✓ changes with time, then the
basis vectors themselves will also change with time. Let’s see what this means for the
velocity expressed in these polar coordinates. The position of a particle is written as
the simple, if somewhat ugly, equation
x = rr̂
From this we can compute the velocity, remembering that both r and the basis vector
r̂ can change with time. We get
dr̂ ˙
ẋ = ṙr̂ + r ✓
d✓
= ṙr̂ + r✓˙✓
ˆ (4.2)
The second term in the above expression arises because the basis vectors change with
˙ (Strictly speaking, this is the angular
time and is proportional to the angular velocity, ✓.
speed. In the next section, we will introduce a vector quantity which is the angular
velocity).
Di↵erentiating once more gives us the expression for acceleration in polar coordinates,
dr̂ ˙ ˆ
ẍ = r̈r̂ + ṙ ✓ + ṙ✓˙✓ ˆ + r✓˙ d✓ ✓˙
ˆ + r✓¨✓
d✓ d✓
= (r̈ ˙ 2 ¨ ˙
r✓ )r̂ + (r✓ + 2ṙ✓)✓ ˆ (4.3)
The two expressions (4.2) and (4.3) will be important in what follows.
– 49 –
, An Example: Circular Motion
Let’s look at an example that we’re already all familiar with. A particle moving in a
circle has ṙ = 0. If the particle travels with constant angular velocity ✓˙ = ! then the
velocity in the plane is
ˆ
ẋ = r! ✓
so the speed in the plane in v = |ẋ| = r!. Similarly, the acceleration in the plane is
ẍ = r! 2 r̂
The magnitude of the acceleration is a = |ẍ| = r! 2 = v 2 /r. From Newton’s second
law, if we want a particle to travel in a circle, we need to supply a force F = mv 2 /r
towards the origin. This is known as a centripetal force.
4.2 Back to Central Forces
We’ve already seen that the three-dimensional motion in a central force potential ac-
tually takes place in a plane. Let’s write the equation of motion (4.1) using the plane
polar coordinates that we’ve just introduced. Since V = V (r), the force itself can be
written using
dV
rV = r̂
dr
and, from (4.3) the equation of motion becomes
dV
m(r̈ r✓˙2 )r̂ + m(r✓¨ + 2ṙ✓)
˙ ✓ˆ= r̂ (4.4)
dr
ˆ component of this is particularly simple. It is
The ✓
1 d ⇣ 2 ˙⌘
r✓¨ + 2ṙ✓˙ = 0 ) r ✓ =0
r dt
It looks as if we’ve found a new conserved quantity since we’ve learnt that
l = r2 ✓˙ (4.5)
does not change with time. However, we shouldn’t get too excited. This is something
that we already know. To see this, let’s look again at the angular momentum L. We
already used the fact that the direction of L is conserved when restricting motion to
the plane. But what about the magnitude of L? Using (4.2), we write
⇣ ⌘ ⇣ ⌘
L = mx ⇥ ẋ = mrr̂ ⇥ ṙr̂ + r✓˙✓ˆ = mr2 ✓˙ r̂ ⇥ ✓
ˆ
– 50 –
In this section we will study the three-dimensional motion of a particle in a central
force potential. Such a system obeys the equation of motion
mẍ = rV (r) (4.1)
where the potential depends only on r = |x|. Since both gravitational and electrostatic
forces are of this form, solutions to this equation contain some of the most important
results in classical physics.
Our first line of attack in solving (4.1) is to use angular momentum. Recall that this
is defined as
L = mx ⇥ ẋ
We already saw in Section 2.2.2 that angular momentum is conserved in a central
potential. The proof is straightforward:
dL
= mx ⇥ ẍ = x ⇥ rV = 0
dt
where the final equality follows because rV is parallel to x.
The conservation of angular momentum has an important consequence: all motion
takes place in a plane. This follows because L is a fixed, unchanging vector which, by
construction, obeys
L·x=0
So the position of the particle always lies in a plane perpendicular to L. By the same
argument, L · ẋ = 0 so the velocity of the particle also lies in the same plane. In this
way the three-dimensional dynamics is reduced to dynamics on a plane.
4.1 Polar Coordinates in the Plane
We’ve learned that the motion lies in a plane. It will turn out to be much easier if we
work with polar coordinates on the plane rather than Cartesian coordinates. For this
reason, we take a brief detour to explain some relevant aspects of polar coordinates.
To start, we rotate our coordinate system so that the angular momentum points in
the z-direction and all motion takes place in the (x, y) plane. We then define the usual
polar coordinates
x = r cos ✓ , y = r sin ✓
– 48 –
,Our goal is to express both the velocity and acceleration y ^ ^
θ r
in polar coordinates. We introduce two unit vectors, r̂
ˆ in the direction of increasing r and ✓ respectively
and ✓
as shown in the diagram. Written in Cartesian form, θ x
these vectors are
! !
cos ✓ ˆ= sin ✓
r̂ = , ✓
sin ✓ cos ✓
Figure 11:
These vectors form an orthornormal basis at every point
on the plane. But the basis itself depends on which angle ✓ we sit at. Moving in the
radial direction doesn’t change the basis, but moving in the angular direction we have
! !
dr̂ sin ✓ d ˆ
✓ cos ✓
= =✓ˆ , = = r̂
d✓ cos ✓ d✓ sin ✓
This means that if the particle moves in a way such that ✓ changes with time, then the
basis vectors themselves will also change with time. Let’s see what this means for the
velocity expressed in these polar coordinates. The position of a particle is written as
the simple, if somewhat ugly, equation
x = rr̂
From this we can compute the velocity, remembering that both r and the basis vector
r̂ can change with time. We get
dr̂ ˙
ẋ = ṙr̂ + r ✓
d✓
= ṙr̂ + r✓˙✓
ˆ (4.2)
The second term in the above expression arises because the basis vectors change with
˙ (Strictly speaking, this is the angular
time and is proportional to the angular velocity, ✓.
speed. In the next section, we will introduce a vector quantity which is the angular
velocity).
Di↵erentiating once more gives us the expression for acceleration in polar coordinates,
dr̂ ˙ ˆ
ẍ = r̈r̂ + ṙ ✓ + ṙ✓˙✓ ˆ + r✓˙ d✓ ✓˙
ˆ + r✓¨✓
d✓ d✓
= (r̈ ˙ 2 ¨ ˙
r✓ )r̂ + (r✓ + 2ṙ✓)✓ ˆ (4.3)
The two expressions (4.2) and (4.3) will be important in what follows.
– 49 –
, An Example: Circular Motion
Let’s look at an example that we’re already all familiar with. A particle moving in a
circle has ṙ = 0. If the particle travels with constant angular velocity ✓˙ = ! then the
velocity in the plane is
ˆ
ẋ = r! ✓
so the speed in the plane in v = |ẋ| = r!. Similarly, the acceleration in the plane is
ẍ = r! 2 r̂
The magnitude of the acceleration is a = |ẍ| = r! 2 = v 2 /r. From Newton’s second
law, if we want a particle to travel in a circle, we need to supply a force F = mv 2 /r
towards the origin. This is known as a centripetal force.
4.2 Back to Central Forces
We’ve already seen that the three-dimensional motion in a central force potential ac-
tually takes place in a plane. Let’s write the equation of motion (4.1) using the plane
polar coordinates that we’ve just introduced. Since V = V (r), the force itself can be
written using
dV
rV = r̂
dr
and, from (4.3) the equation of motion becomes
dV
m(r̈ r✓˙2 )r̂ + m(r✓¨ + 2ṙ✓)
˙ ✓ˆ= r̂ (4.4)
dr
ˆ component of this is particularly simple. It is
The ✓
1 d ⇣ 2 ˙⌘
r✓¨ + 2ṙ✓˙ = 0 ) r ✓ =0
r dt
It looks as if we’ve found a new conserved quantity since we’ve learnt that
l = r2 ✓˙ (4.5)
does not change with time. However, we shouldn’t get too excited. This is something
that we already know. To see this, let’s look again at the angular momentum L. We
already used the fact that the direction of L is conserved when restricting motion to
the plane. But what about the magnitude of L? Using (4.2), we write
⇣ ⌘ ⇣ ⌘
L = mx ⇥ ẋ = mrr̂ ⇥ ṙr̂ + r✓˙✓ˆ = mr2 ✓˙ r̂ ⇥ ✓
ˆ
– 50 –
