Newton and the

Spring 2007

His 3464

Newton at 83

Portrait by

Enoch Seeman

Sir James Thronhill’s

portrait of Newton

at 67

Newton in 1701 (at 59)

Newton at 46

?

Why does the moon orbit the earth?

The Problem:

??

??

Complicating the Problem

Galileo’s explanation:

inertial motioncircular

cosm

ic str

ing

"After dinner, the weather being warm, we went into the garden and drank tea, under the shade of some apple trees, only he and myself. Amidst other discourse, he told me he was in the same situation as when formerly the notion of gravitation came into his mind. It was occasioned by the fall of an apple as he sat in a contemplative mood."

William Stuckley, 1726

The legend of the apple

The structure of Newton’s argument is

p = The same force that affects apples also affects the moon

q = The moon falls 16 feet in one minute

Given the following statements:

A. If p then q

B. q

C. Therefore p

So he must proveboth A and B

Gravity likely weakens, but by how much?

Newton figures this out by combining his

own insights with those of his predecessors

What he already knew

a = 32 ft/sec2 (by measuring it)

d = 1/2 a t2 (from Galileo)

Inertial motion (his own “corrected” version)f a

f of tension in a string

The relation between a planet’s period and its distance from the sun

r1

r2 r

If r is the average distance of a planet from the sun

And T is the time it takes the planet to circle the sun

Then Kepler’s 3rd Law says: T2 r3

cosm

ic str

ing

To find the tension in the cosmic string Newtonexamined the case ofa rock on a string

f m v2/r

f mv2/r

v = d/t =

f m(2r/T)2 /r =

m42r2

T2x 1

rm42r

T2=

2r/T

m(42r2/T2) x 1/r

Now Newton applies this result to the moon case:

m42rT2

But, from Kepler’s Third LawT2 r3

m42rT2

= m42rr3

= m42

r2

So f

kThus f

2

r2

f is 1r2

At 1 earth radius: At 2 earth radii:

F F

1

(2r)2 r2

1= some amount F1 X F1

=

1

22F 1

r

2r

Newton’s next move: To show

that the moon (like apples)

falling body

can be considered a

From Newton’s Principia Mathematica

Two cases of projectile fall

Thrown slowlyThrown hard

d = 89.4 mi

1 mi

(400

1)

Throw at 4.92 mi/sec for 18.66 sec

(4001)2 + (89.4)2

= 4002

C

B

CB =

D

= CD

= 5280 ft

So CD = 4002 mi - 1 mi

= 4001 mi

(89.4)

CB - BD

BD = ½ x 32 ft/sec2 x (18.166)2

Since f a, a f

f is also 1r2

Therefore a is 1r2

At 1 earth radius: At 2 earth radii:

a= 32 ft/sec2 a= 1/22 x 32 = 8 ft/sec2

The moon is 60 earth radii away

Therefore at the distance of the moon

1602

x 32

or 32602

ft/sec2

a =

In one minute (60 sec) the moon will “fall”

d = 1/2 a t2 = 1/2 x 32602

x 602 =

16 feet

He has thus proven A (if p then q)

To prove B (q: the moon “falls” 16 in 1 min) Newton drew the following construction

A

F

C.

E

D

B

AEDADF

A

F

C.

E

D

B

His task is to find BD, the distance the moon “falls” in a given amount of time.

A

F

C.

E

D

B

To find BD Newton thinks like an engineer

A

F

C.

E

D

B

AED ADF

AEAD =

ADAF

SO AD2 = AE X AF

THUS AF = AD2

AE

A

D

represents 1 minute.

Therefore arc length AD is 1 min minutes in a month

x moon’s orbit

We now know AD

Arc AD

Earth

The whole circle is 1 month

A

F

C.

E

D

B

AF = AD2

AESince and since AE is known

Newton calculated AF to be ?

Recall the structure of Newton’s argument :

If p then q,

q:

therefore p

If the “apple force” is affecting the moon,then the moon “falls” 16 feet in 1 minute.

The moon does fall 16 feet in one minute.

Therefore the apple force affects the moon.

But, look at the following argument:

p: If you have enough money, you go to the movies

q: You go to the movies.

Therefore you have enough money.

Nature was more complicated than Newton ever dreamed.

"The most beautiful experience we can have is the mysterious. It is the fundamental emotion that stands at the cradle of true art and true science. Whoever does not know it and can no longer wonder, no longer marvel, is as good as dead.”