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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.”
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