Newton’s evidence for the inverse square law
Isaac Newton knew that the distance from the Earth to the Moon was about 240,000 miles, about
60 times the Earth’s radius which is about 4,000 miles.
Assuming that the Moon was held in its orbit by gravity which hypothetically obeyed an inverse
square law, then the force of gravity acting on the Moon would be (1/60)2 = 1/3600 as strong as
the force of gravity on the surface of the Earth. On the surface of Earth, the acceleration due to
gravity is about 32 feet per second per second. So on Earth, an object falls 16 feet in one second
according to Galileo’s Law. The moon should “fall” about 1/3600*16 feet each second or
about 0.05 inch each second.
But we can calculate this quantity using the known distance to the moon and known speed of
moon in its orbit!
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In the diagram, r represents the radius of the Moon’s orbit (240,000 miles). d represents the
path the Moon would take if it weren’t being pulled into its orbit. x represents the distance that
the Moon has “fallen” in the time it would have taken the Moon to travel a distance of d.
Then (r  x)2  r 2  d 2 . Take the square root of both sides and subtract r to get
x  r2  d 2  r
Now d is approximately how far the Moon travels in 1 second. We know the Moon travels 2πr
miles in about 27.3 days (sidereal month) or 2358720 seconds. So
d
2π(240, 000)
miles  0.639 miles  4.04 104 inches .
2358720
r  240,000 miles  1.52 1010 inches
x  (1.52 1010 ) 2  (4.04 104 ) 2  1.52 1010  0.05 inch
This result is an extraordinary agreement with the theory of an inverse square law, and it was the
experimental confirmation that initially convinced Newton of the validity of the inverse square
law.
Brief note: If we expand (r  x)2  r 2  d 2 , we get r 2  2rx  x 2  r 2  d 2 , so 2rx  x 2  d 2 . It
turns out that x2 is so small we could safely ignore it. Here is a more formal justification.
We will use a power series expansion. Note d is extremely small compared to r (roughly 1
millionth).
r2  d 2  r  r 1
 1 d2 
d2
d2
d2

r

r
1


r

r


r


2 
r2
2r
2r
 2r 
which is exactly what you would get if you ignored the x2 term in the equation 2rx  x 2  d 2 in
the first place. Some presentations of this material will make this simplification, and it is in fact
correct.
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