Optional — Central Forces

Section 1.9 Optional — Central Forces

One of the great triumphs of Newtonian mechanics was the explanation of Kepler’s laws

The German astronomer Johannes Kepler (1571–1630) developed these laws during the course of an attempt to relate the five extraterrestrial planets then known to the five Platonic solids. He based the laws on a great number of careful measurements made by the Danish Astronomer Tycho Brahe (1546–1601). Then Isaac Newton (English, 1642–1727) provided the explanation in 1687. Kepler also wrote a paper entitled “On the Six-Cornered Snowflake”. Tycho Brahe lost his nose in a sword duel and wore a prosthetic nose from then on. The story is that Brahe died from a burst bladder that resulted from his refusing to leave the dinner table before his host.

, which said

The planets trace out ellipses about the sun as focus.
The radius vector $r$ sweeps out equal areas in equal times.
The square of the period of each planet is proportional to the cube of the major axis of the planet’s orbit.

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Newton showed that all of these behaviours follow from the assumption that the acceleration

a (t)

of each planet obeys the law of motion

m a = F

where

m

is the mass of the planet and

F = - \frac{G M m}{r^{3}} r

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is the “gravitational force” applied on the planet by the sun. Here

G

is a constant

Its value is about

6.67408 \times 10^{- 11} m^{3} {kg}^{- 1} {sec}^{- 2} .

, called the “gravitational constant” or the “universal gravitational constant”,

M

is the mass of the sun,

r

is the vector from the sun to the planet and

r = | r | .

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In this section, we’ll show that some of these properties follow from the weaker assumption that the acceleration

a (t)

of each planet obeys the law of motion

m a = F

with

F

being a central force. That is, the assumption that

F

is parallel to

r .

The verification that the other properties follow from the specific form of the gravitational force, proportional to

r^{- 2},

will be delayed until the optional §1.10.

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So, in this section, we assume that we have a parametrized curve

r (t)

and that this curve obeys

m \frac{d^{2} r}{d t^{2}} (t) = F (r (t))

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where, for all

r \in R^{3},

F (r)

is parallel to

r .

We shall show that

The path $r (t)$ lies in a plane through the origin and that
the radius vector $r$ sweeps out equal areas in equal times.

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We’ll start by trying to guess what the plane is. Pretend that we know that

r (t)

lies in a fixed plane through the origin. Then

v (t) = \frac{d r}{d t} (t)

lies in the same plane and

r (t) \times v (t)

is perpendicular to the plane. If our path really does lie in a fixed plane,

r (t) \times v (t)

cannot change direction — it must always be parallel to the normal vector to the plane. So let’s define

Ω Ω (t) = r (t) \times v (t)

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and check how it depends on time. By the product rule,

\begin{aligned} \frac{d Ω Ω}{d t} (t) & = \frac{d}{d t} (r (t) \times v (t)) = v (t) \times v (t) + r (t) \times a (t) \\ = \frac{1}{m} r (t) \times F (r (t)) \\ (A) & = 0 \end{aligned}

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because

r (t)

and

F (r (t))

are parallel. So

Ω Ω (t)

Physicists call

m Ω Ω (t)

the angular momentum at time

t

and refer to (A) as (an example of) conservation of angular momentum. Conservation of angular momentum is exploited in gyro-compasses and by ice skaters (to spin faster/slower).

in fact independent of

t .

It is a constant vector that we’ll just denote

Ω Ω .

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r (t) \times v (t) = Ω Ω,

we have that

r (t)

is always perpendicular to

Ω Ω

and

r (t) \cdot Ω Ω = 0

If $Ω Ω \neq 0,$ this is exactly the statement that $r (t)$ always lies in the plane through the origin with normal vector $Ω Ω .$
If $Ω Ω = 0,$ then $r (t)$ is always parallel to $v (t)$ and there is some function $α (t)$ such that
$\frac{d r}{d t} (t) = v (t) = α (t) r (t)$
This is a first order, linear, ordinary differential equation that we can solve by using an integrating factor. Set
$β (t) = \int_{0}^{t} α (t) d t$
Then
$\begin{aligned} \frac{d r}{d t} (t) = α (t) r (t) & ⟺ e^{- β (t)} \frac{d r}{d t} (t) - α (t) e^{- β (t)} r (t) = 0 \\ ⟺ \frac{d}{d t} [e^{- β (t)} r (t)] = 0 \\ ⟺ e^{- β (t)} r (t) = r (0) \\ ⟺ r (t) = e^{β (t)} r (0) \end{aligned}$
so that $r (t)$ lies on a line through the origin. This makes sense — the particle is always moving parallel to its radius vector.

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This completes the verification that

r (t)

lies in a plane through the origin.

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Now we show that the radius vector

r (t)

sweeps out equal areas in equal times. In other words, we now verify that the rate at which

r (t)

sweeps out area is independent of time. To do so we rewrite the statement that

| r (t) \times v (t) |

is constant in polar coordinates. Writing

r (t) = r (t) \hat{r} (θ (t))

and then applying Lemma 1.8.2.b gives that

\begin{aligned} constant = | r \times v | & = | r \hat{r} \times (\frac{d r}{d t} \hat{r} + r \frac{d θ}{d t} \hat{θ θ}) | = r^{2} \frac{d θ}{d t} \\ since | \hat{r} \times \hat{r} | = 0, | \hat{r} \times \hat{θ θ} | = 1 \end{aligned}

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is constant. It now suffices to observe that

r (t)^{2} \frac{d θ}{d t} (t)

is exactly twice the rate at which

r (t)

sweeps out area. To see this, just look at the figure below. The shaded area is essentially a wedge of a circular disk of radius

r .

(If

r (t)

were independent of

t,

it would be exactly a wedge of a circular disk.) Its area is the fraction

\frac{d θ}{2 π}

of the area of the full disk, which is

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\frac{d θ}{2 π} π r^{2} = \frac{1}{2} r^{2} d θ

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Exercises Exercises

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Exercise Group.

Exercises — Stage 3

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1. (✳).

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Let

r (t) = x (t) \hat{ı ı} + y (t) \hat{ȷ ȷ} + z (t) \hat{k}

be the position of a particle at time

t

. Suppose the motion of the particle satisfies the differential equation

\frac{d^{2} r}{d t^{2}} = f (r) r

where

r = | r |

Suppose $f (r)$ is an arbitrary function of $r$ . Prove or disprove each of the following statements.
1. The motion of the particle is planar.
2. The path of the particle sweeps out equal areas in equal times.
Find all forms of $f (r)$ for which the motion of the particle always lies on a straight line.
Give a specific form of $f (r)$ for which the motion of the particle could lie on an ellipse.

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2. (✳).

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An object moves along a curve in the

x y

-plane having polar equation

r = \frac{1}{θ + α}

(where

α

is a constant) under the influence of a central force so that the object has no transverse acceleration.

Verify that $r^{2} \dot{θ} = h$ remains constant as the object moves.
Express the magnitude of the acceleration of the object as a function of $r$ and $h .$

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