
Index
18. Newton's 2nd Law 18a. The Third Law 18b. Momentum 18c. Work 18d. Work against Electric Forces 19.Motion in a Circle 20. Newton's Gravity 21. Kepler's 3rd Law 21a.Applying 3rd Law 21b. Fly to Mars! (1) 21c. Fly to Mars! (2) 21d. Fly to Mars! (3) 22.Reference Frames 22a.Starlight Aberration 22b. Relativity 22c. Flight (1) 
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 circlerather than wandering offthe Earth must exert a pull on the Moon, and Newton named that pulling force gravity. Was that the same force which pulled all falling objects downward? 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:
If it was the same force, then a connection would exist between the way objects fell and the motion of the Moon around Earth, that is, its distance and orbital period. The orbital period we knowit is the lunar month, corrected for the motion of the Earth around the Sun, which also affects the length of time between one "new moon" and the next. The distance was first estimated in ancient Greecesee here and here. 
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 R was proportional to 1/R^{2} (varied like the "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 "inversesquare 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, Nobelprize 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 handheld calculator is helpful. Calculating the Moon's MotionWe 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_{E} be the average radius of the Earth (first estimated by Erathosthenes) R_{E}= 6 371 km The distance R to the Moon is then about 60 R_{E} (see here). If a mass m on Earth is pulled by a force mg, and if Newton's "inverse square law" holds, then the pull on the same mass at the Moon's distance would be 60^{2} = 3600 times weaker and would equal mg/3600 If m is the mass of the Moon, that is the force which keeps the Moon in its orbit. If the Moon's orbit is a circle, since R = 60 R_{E} its length is 2 π R = 120 π R_{E} Suppose the time required for one orbit is T seconds. The velocity v of the motion is then v = distance/time = 120 π R_{E}/T
The centripetal force holding the Moon in its orbit must therefore equal mv^{2}/R = mv^{2}/(60 R_{E}) and if the Earth's gravity provides that force, then mg/3600 = mv^{2}/(60 R_{E}) dividing both sides by m and then multiplying by 60 simplifies things to g/60 = v^{2}/R_{E} = (120 π R_{E})^{2}/(T^{2} R_{E}) Canceling one factor of R_{E} , multiplying both sides by 60 T^{2} and dividing them by g leaves T^{2} = (864 000 π^{2} R_{E})/g = 864 000 R_{E} (π^{2}/g) Providentially, in the units we use g ~ 9.81 is very close to π^{2} ~ 9.87, so that the term in parentheses is close to 1 and may be dropped. That leaves (the two parentheses are multiplied) T^{2} = (864 000) (6 371 000) With a hand held calculator, it is easy to find the square roots of the two terms. We get (to 4figure accuracy) 864 000 = (929.5)^{2} 6 371 000 = (2524)^{2} 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 T = 27.153 days pretty close to the accepted value T = 27.3217 days

Next Stop: #21 Kepler's Third Law
Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Reformatted 24 March 2006 ; Updated 13 October 2016
Curators: Robert Candey, Alex Young, Tamara Kovalick
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