Class Notes, Fall 2001

Week of   Aug 22   Sep 5 Oct 1 Nov 5 Phy 205 page
  Aug 27 Sep 10 Oct 8 Nov 12
    Sep 17 Oct 15 Nov 19
    Sep 24 Oct 22  Nov 26
      Oct 29    Notes for Fall, 2000

Phy 205, Week 15

26 Nov Is energy (kinetic plus potential) conserved in the motion of a harmonic oscillator? There's one way to find out and that's to evaluate it.

x( t ) = A cos( ω0 t ),   where   ω02 = k / m

The mechanical energy is

E = m v2 / 2 + k x2 / 2

You have to find the velocity, vx = dx / dt, and then evaluate E, substituting both x and vx. When you do it, you have to use one further property of the oscillator to make this work: the value of ω0.

Gravity.

Newton's gravitational force law says that any two point masses will attract each other with a force of magnitude G m1 m2 / r2, where r is the distance between them. G is a numerical constant given by G = 6.67x10 -11 Nm2/kg2.

This gravitational force is very weak. The force pulling you toward the person sitting next to you is less than about a microNewton.

Where does such a law come from? It follows the historical development from Ptolemy (ca 200) to Copernicus (ca 1500) to Kepler (ca 1600) to Newton (ca 1700). The key in this development was Kepler, who performed a heroic feat of data analysis using the observations of Tycho Brahe. After a twenty year project he was able to describe the motions of the planets in our solar system in terms of a small set of laws

1. Planets move around the sun on elliptical orbits with the sun at one focus of the ellipse.
2. The line from the sun to a planet sweeps out equal areas in equal times.
3. The period of the planet (time to orbit the sun) is related to the semi-major axis of its elliptical orbit by T = k a3/2.

What is an ellipse? There are a couple of pertinent definitions. One is in terms of analytic geometry:
 x2
a2
 +  y2
b2
 = 1
Another is more like the definition of a circle. A circle is a set of points whose distance from a fixed point is constant. For an ellipse, start with two points and find the set of points such that the sum of the distances to these two given points is a constant. The two points are the foci (plural of focus) of the ellipse. If the two foci coincide, this is a circle. For planetary orbits, the sun is not at the center of the ellipse, it is at a focus.

Now back to the question: where did Newton's gravitational force law come from? Doing the complete case with the ellipse is too hard, but we can do the special case of circular motion. Most of the planetary orbits are nearly circular; only Pluto is noticeably eccentric.

For a planet going around the sun in a circle, you already know the acceleration: v2/r toward the center of the circle. The force by the sun must then be toward the sun (F=ma).

F = m v2/r = m ( 2 π r / T )2 / r = 4 π2 m r / T2.

Now use Kepler's law that was based on experiment. T = k a3/2. For a circle, the semi-major axis is just the radius: r = a.

F = 4 π2 m r / ( k a3/2 )2 = 4 π2 m / k2 r2

And this is the inverse square law.

28 Nov How is "G" measured? It's a delicate experiment because the forces involved are so small. Cavendish did it in the late 1700's using an apparatus pretty much like the one that I showed in class and that is sketched in chapter 12 of the text.

This lets you measure the mass of the Earth among many other things. The pull of the Earth on you is mg at the surface. It is also equal to GMm/R2, where M is the Earth's mass and R is the Earth's radius. Set these equal and you get

m g = G M m / R2,   so   M = g R2 / G.

The Earth's radius is 6400 km, measured as long ago as 200 BC. Do the arithmetic and you get 6x1024 kg.

It's not quite this simple. Newton's gravitational force law applies to point masses. the Earth is not a point mass. Then what you have to do is to break the Earth into tiny pieces and find the force caused by each. Add them up; take the limit as all the pieces approach zero size and see what you get. This required Newton to invent integral calculus. The result is that IF the Earth is a sphere, THEN the pull on an object above its surface is the same as if the Earth were a point mass at its center.

This result applies only for a spherical object, not for anything else. The Earth is pretty nearly spherical, so this is a good approximation to use.

You can measure the mass of the sun using related ideas. The Earth is in orbit around the sun, so I can write F=ma for its motion. The Earth's acceleration is toward the sun, so I need only write the magnitudes:

G M m / r2 = m a = m v2 / r = m ( 2 π r / T )2 / r

Here M is the sun's mass and m is the Earth's. The distance from the sun to the Earth is r = 150 000 000 km. Solve for the sun's mass.

M = 4 π2 r / G T2

When you put in the numbers, the sun's mass comes out to about 2x1030 kg.

30 Nov The escape speed from the Earth (or from any other planet). The radial component of the gravitational force is

Fr( r ) = - G M m / r2

The minus sign comes because the r-direction is up but gravity is down. Potential energy obeys the defining equation

Fr( r ) = - dU / dr

and that determines the potential energy (up to the usual constant)

U( r ) = - G M m / r

If a ship is fired away from the Earth, how fast will it have to be moving so that it will never fall back? Use conservation of energy to solve this. Once the ship is above the atmosphere you don't have to worry about air resistance.

E = m v2 / 2 + U( r ) = m v2 / 2 - G M m / r

This is a constant.

m v2 / 2 - G M m / r = m v02 / 2 - G M m / R

Here I'm using v0 for the speed it leaves the surface and R for the radius of the planet.

Escaping completely means that I can have r approach infinity. This leaves the equation

m v2 / 2 = m v02 / 2 - G M m / R

for the final kinetic energy as it drifts away. Whatever else kinetic energy is, it is not negative. This means that if the right-hand side is negative the ship will never escape. Escape then means

m v02 / 2 - G M m / R > 0

Solve this for the minimum speed and you get

v0 > ( 2 G M / R )1 / 2

For the Earth this is about 11 km/sec.

Back