Continue reading pp. 51 - 88 from The Great Physicists from Galileo to Einstein. The material is on Isaac Newton.

F = - G m₁ m₂ / r₁₂²,

where F is a force of attraction between two objects having masses m₁ and m₂, r₁₂ is the distance between the centers of the two masses (their centers-of-mass), and G is a proportionality constant 6.67 x 10^-11 N m² / kg².

We said that by Newton's Third Law of Motion (for every action, there is an equal and oppposite reaction) that this formula would, for example, not only provide the force of attraction on the moon from the earth (that is responsible for the moon's revolving around the earth), but also provide the force of attraction of the earth from the moon (that is responsible for the tides).

We discussed there being a "high tide" on the the side of the earth closest to the moon (because the water is closer and pulled more strongly by the moon) as well as on the opposite side of the earth (because that water is farther away and pulled less than everything else). On a given shore, there is a high tide every 12 hours and 24 minutes. The "extra" 24 minutes is associated with the moon's revolution around the earth about once a month.

The force of gravitation is also responsible for things falling, so we looked to recover the more familiar g=9.8 m/s² acceleration due to gravity. Using Newton's second law (F=ma) and the law of graivtation, we arrived at

g = G M_earth / R_earth².

If the object is on the earth then the distance separating the two masses' centers is the radius of the earth. It was suggested that the radius of the earth was 4000 miles. We further justified this by considering that Washington state is about 3000 mile away and requires changing one's watch by 3 hrs, suggesting that it is about 1/8 of the way around the world. This gave an estimated circumference about 24000; then using that the circumference is 2 p radius gave an estimate of approximately 4000 miles. (We ignored the fact that we weren't at the equator and used 3 instead of p in this "back of the envelope" caclculation.) We muliplied that by 1600 which is the number of yards in a mile, and ignored the difference between meters and yards to arrive at 6,400,000 meters for the radius of the earth. We then compred this value to one looked up in a physics text book(by Randall Knight) 6. 37 x 10⁶ m.

Substituting this and the mass of the earth (5.98 x 10²⁴ kg, just looked up) yielded

g = (6.67 x 10^-11)(5.98 x 10²⁴) / (6. 37 x 10⁶)²

or

g = (6.67)(5.98) /(6. 37)² 10 = 9.83