| Week of | Aug 23
| Sep 6
| Oct 2
| Nov 6
| Phy 205 page |
| Aug 28 | Sep 11 | Oct 9 | Nov 13 | ||
| Sep 18 | Oct 16 | Nov 20 | Notes for Fall, 1999 | ||
| Sep 25 | Oct 23 | Nov 27 | |||
| Oct 30 |
27 Nov Emphasize that I can write down the same equation using many different-looking variables.
are all the same equation. (Here W and x are constants.) If you know the solution to one of them you know all of them.
Discuss the hanging spring problem again, as in one of the homework problems.
Gravity. State Newton's law and show how it is related to one of the experimentally derived Kepler laws -- the one relating the period of a planet (time to go around the sun) to its distance to the sun: T2 = k r3 / 2.
29 Nov Newton's gravitational force law applies to point masses. The Earth is not a point mass, so how do I figure out its gravitational pull on a mass near its surface? Answer: imagine treating it as a bunch of point masses and add up the pull by all of them -- an integral. State the result that IF you assume the Earth to be a sphere, THEN its graviational pull on a mass at or above its surface is the same as if the Earth were a point mass at its center. If you apply this to anything other than a sphere it's wrong, or at most only an approximation.
The Earth isn't exactly a sphere; it bulges a bit at the equator because of its rotation. It's also a bit more massive in the Northern hemisphere than in the Southern. Still, for our purposes it is very good approximation to treat it as spherical.
You can weigh the Earth using this. Your weight is mg. The result that I just stated says that you can treat the pull by the Earth as if it were from a point mass at the center, so the pull on you is also G ME m / RE2. Equate these and solve for the mass of the Earth. The result is about 6 x 1024kg.
The planets are in orbit around the sun. Apply F = ma to one of them, say the Earth. The orbits are nearly circular. (Actually they're ellipses, but with small eccentricity.) The acceleration is toward the center of the circle, and the force is toward the sun, so the directions agree. All that's left is the magnitudes.
The speed v = 2 p r / T, so this is
Rearrange this and you get
For the Earth, you know that T = 1 year. The only problem is figuring out the distance from the Earth to the sun. That was first accomplished a couple of hundred years ago, though with modern technology it's much easier and more accurate. r = 150 x 106km, and that yields a solar mass of 2 x 1030kg.
The sun is orbiting the center of our galaxy with a period of about 250,000,000 years and at a distance of about 30,000 light-years. Our galaxy isn't very spherical, so to treat it the same way that we would the sun or the Earth is not a very good approximation. Still, it's a place to start to get a rough estimate of the mass of our galaxy. Use the same equation as I did for the mass of the sun and you get a mass of about 1011 solar masses.
