The Moon orbits around the Earth. Since its size does not appear to change, its distance stays about the same, and hence its orbit must be close to a circle. To keep the Moon moving in that circle--rather than wandering off--the Earth must exert a Was that the Supposedly, the above question occured to Newton when he saw an apple falling from a tree. John Conduitt, Newton's assistant at the royal mint and husband of Newton's niece, had this to say about the event when he wrote about Newton's life: -
In the year 1666 he retired again from Cambridge ... to his mother in Lincolnshire & while he was musing in a garden it came into his thought that the power of gravity (which brought an apple from a tree to the ground) was not limited to a certain distance from earth, but that this power must extend much further than was usually thought. Why not as high as the Moon thought he to himself & that if so, that must influence her motion & perhaps retain her in her orbit, whereupon he fell a-calculating what would be the effect of that superposition...
If it To calculate the force of gravity on the Moon, one must also know how much weaker it was at the Moon's distance. Newton showed that if gravity at a distance inverse square of the distance"), then indeed the acceleration g measured at the Earth's surface would correctly predict the orbital period T of the Moon. Newton went further and proposed that gravity was a "universal" force, and that the Sun's gravity was what held planets in their orbits. He was then able to show that Kepler's laws were a natural consequence of the "inverse squares law" and today all calculations of the orbits of planets and satellites follow in his footsteps. Nowadays students who derive Kepler's laws from the "inverse-square law" use differential calculus, a mathematical tool in whose creation Newton had a large share. Interestingly, however, the proof which Newton published did not use calculus, but relied on intricate properties of ellipses and other conic sections. Richard Feynman, Nobel-prize winning maverick physicist, rederived such a proof (as have some distinguished predecessors); see reference at the end of the section. Here we will retrace the calculation, which linked the gravity observed on Earth with the Moon's motion across the sky, two seemingly unrelated observations. If you want to check the calculation, a hand-held calculator is helpful. We assume that the Moon's orbit is a circle, and that the Earth's pull is always directed toward's the Earth's center. Let R R The distance R to the Moon is then about 60 R mg If 2 p R = 120 p R Suppose the time required for one orbit is T seconds. The velocity v of the motion is then v = distance
The centripetal force holding the Moon in its orbit must therefore equal mv and if the Earth's gravity provides that force, then mg dividing both sides by g Canceling one factor of R T Providentially, in the units we use g ~ 9.81 is very close to p T With a hand held calculator, it is easy to find the square roots of the two terms. We get (to 4-figure accuracy) 864 000 = (929.5) Then T @ (929.5) (2524) = 2 346 058 seconds To get T in days we divide by 86400, the number of seconds in a day, to get
pretty close to the accepted value T = 27.3217 days Newton rightly saw this as a confirmation of the "inverse square law". More than a century later, in 1796, his countryman |

A detailed article: Keesing, R.G., **The History of Newton's apple tree**,
*Contemporary Physics, 39,* 377-91, 1998

Richard Feynman's calculations can be found in the book "**Feynman's
Lost Lecture: The Motion of Planets Around the Sun**" by D. L Goodstein
and J. R. Goodstein (Norton, 1996; reviewed by Paul Murdin in *Nature,*
vol. 380, p. 680, 25 April 1996).

**Next Stop: **** #21 Kepler's Third Law**

Author and curator: David P. Stern

Last updated 6 March 1999