18. Newton's 2nd Law
18a. The Third Law
18d. Work against
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)
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 circle--rather than wandering off--the 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 know--it 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 Greece--see 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/R2 (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 "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.
Calculating the Moon's Motion
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 RE be the average radius of the Earth (first estimated by Erathosthenes)
RE= 6 371 km
The distance R to the Moon is then about 60 RE. 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 602 = 3600 times weaker and would equal
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 RE its length is
2 π R = 120 π RE
Suppose the time required for one orbit is T seconds. The velocity v of the motion is then
v = distance/time = 120 π RE/T
The centripetal force holding the Moon in its orbit must therefore equal
mv2/R = mv2/(60 RE)
and if the Earth's gravity provides that force, then
mg/3600 = mv2/(60 RE)
dividing both sides by m and then multiplying by 60 simplifies things to
g/60 = v2/RE = (120 π RE)2/(T2 RE)
Canceling one factor of RE , multiplying both sides by 60 T2 and dividing them by g leaves
T2 = (864 000 π2 RE)/g = 864 000 RE (π2/g)
Providentially, in the units we use g ~ 9.81 is very close to π2 ~ 9.87, so that the term in parentheses is close to 1 and may be dropped. That leaves (the two parentheses are multiplied)
T2 = (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 4-figure accuracy)
864 000 = (929.5)2 6 371 000 = (2524)2
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