"From Stargazers to Starships" home page and index: Sintro.htm
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• Get an intuitive understanding for the way orbital velocities vary along each orbit, according to Kepler's second law.

• Get a first exposure to the concepts of "potential energy" and "kinetic energy." Motion in the presence of of gravity, in everyday life, conserves total energy, but can swap one form for another--as in a roller coaster, which speeds up at the low points and slows down at high ones.

• Know that orbital motion conserves also energy, in a somewhat similar way, although the formulas look different.

• (Optional) Learn (very qualitatively--how orbital motion is calculated: the polar angle f, ("true anomaly") varies unevenly around the orbit, but another angle, the "mean anomaly" M, varies steadily, and formulas exist to derive from M yet another angle, the "eccentric anomaly" E, from which f can be obtained by another transformation.

Note: At this stage, knowing that M exists is the most the student can do. An optional section (12a) "How Orbits are Calculated" goes further into the subject, but is not covered in these lesson plans. It introduces the student to the idea of successive approximationa for solving a difficult equation--here, "Kepler's Equation" which relates M and E. It also describes what the other orbital elements are.

Terms: Perigee, apogee, perihelion, aphelion, radius vector; Energy, kinetic energy, potential energy, conservation of energy, escape velocity, [true anomaly, mean anomaly, eccentric anomaly]

Stories and extras: Thomas Jefferson's clock in Monticello, driven by suspended cannonballs.

With suggested answers, brackets for comments by the teacher or "optional")

[Note to the teacher: Enrico Fermi, the Italian physicist--Nobel prize winner, one of the founders of nuclear physics and designer of the first nuclear reactor--once described the way he felt a lesson should be given:
"Tell them what you are going to tell them, tell them, then tell them what you told them."
What follows below follows the first part of Fermi's advice.]

   As we will see, the meaning of Kepler's 2nd law is "planets speed up when closer to the Sun, slow down when further away."

   We already know another motion which does exactly that: a roller coaster speeds up in the "valleys," slows down on the "hilltops." The two motions are related: both involve gravity.

   As we will see, both also involve something known as energy. That energy comes in two forms--energy of speed, or kinetic energy, and energy of position, or potential energy, higher at the hilltops and higher at apogee. The sum of the two is kept constant--which is why, as the roller-coaster carriages rush down the steep slope or the satellite approaches Earth, both gain speed, losing it again as they pull away.

Let us go into the details.

(Here give the material of section (12), except for the optional part about the true and mean anomaly, which (if included) is given separately.

The questions below can be used in the lesson and/or in the review afterwards.

The line between it and the Sun. [strictly speaking, center-to-center]

The closer the planet to the Sun, the faster it moves.

Yes, it does.

[What follows next is a calculation: you may let one of the better students "help you with it" at the board, illustrating the problem with a drawing as you describe it. The rest of the class should copy.
Note: This calculation also appears in "More about Kepler's 2nd Law"]

Here is how. Let us mark the distances the satellite covers in one second-- D₁ at perigee, D₂ at apogee (enter in drawing). The area covered in one second by the radius vector at perigee is a triangle, its base is D₁ and its height is r₁, and the two lines are perpendicular. Its area is therefore...?         (1/2) r₁ D₁

A similar triangle at apogee (draw) has area ...?

                      (1/2) r₂D₂     (shades of "Star Wars"!)

Therefore...? By the second law, r₁ D₁ = r₂D₂. But the velocity is defined as the distance covered in one second, so in place of (D₁,D₂) we may just as well write (v₁,v₂), the velocities at both these points. So...

r₁v₁ = r₂v₂

Divide both sides by r₁v₂ to get

v₁/v₂ = r₂/r₁ .

The second ratio is 10, so the first ratio must be 10 too: the velocity near Earth is 10 times larger.

No. At other points, distances such as r₁ and D₁ are not perpendicular to each other, and the area of the triangle also depends on the angle between them.

The stone slows down as it rises and loses energy, then regains velocity and energy as it falls down again.

Loosely, anything that can make a machine move.

Have the students answer: electricity, calories in food, heat, sunlight, sound, nuclear energy...

The total energy in an isolated body or system of bodies is conserved, though it can change from one kind to another.

[This part is best handled by the teacher without much elaboration. Energy will come up again in section 15 and will be discussed there in a more complete fashion. Here students are told what the energy of a moving satellite is, but remembering the formula may be optional].

Kinetic energy and potential energy.

Energy of motion.

The amount of matter being moved (the mass m) and its velocity v. The formula is E_K=(1/2)mv².

Energy due to position.

m g h, with g close to 10 (meters/second²), the acceleration due to gravity.

E = (1/2)mv² + m g h = constant

As one part grows, the other decreases to keep the sum the same.

At the bottom of its swing, its speed is largest, and therefore, so is its kinetic energy. At the top of its swing, it comes to a brief stop: its kinetic energy is zero, but its height and therefore its potential energy are largest. It therefore seems reasonable that the sum of the two energies stays the same.

...exactly the same as the swing.

The constant which gives the value of E will depend on it, but we may use any reference height we wish.

[Note: the energy equation of an orbiting body may be too much to memorize, and deriving it is beyond the scope of "Stargazers." Therefore, in the question below, the student is given the formula and is only asked to explain its meaning.]

E   =   1/2 mv2 – k m/r   =   constant

  where k = gR_E². The part representing kinetic energy is the same, but a different expression gives the potential energy.   With a stone, the higher we lift it, the greater is its potential energy: is this also true here?

  Yes, though the fact the potential energy is negative may be confusing. The higher the satellite goes, the larger r and the less negative the potential energy is, which means it grows: at infinity it is zero, a number larger than any negative value.

  This again illustrates that the value of the potential energy E_P depends on the reference level at which we choose E_P = 0. Any reference level is OK: what matters are differences in potential energy, which dictate the gain or loss of kinetic energy E_K.

The velocity with which an object would fly off the surface of the Earth and never return.
    (Note however, we are talking about motion relative to the Earth alone. In practice, such an object would still be in orbit around the Sun. To also escape the Sun's gravity would take considerably more energy!)

[Give details on the blackboard as you go along, and let students copy.]

The energy of an object with velocity V, at the surface of the Earth, is

        E = 1/2 mV² – k m/R_E

When the object is far enough from Earth to be considered "escaped", its distance r is so big that its potential energy k m/r is virtually zero. Also, it has used up all its kinetic energy to get so far, hence v=0   too.

This suggests that for such a motion E=0. Then

        1/2 mV² – k m/R_E = 0
so
        V² = 2k/R_E = 2gR_E     (k=gR_E²)

If we calculate in meters and seconds, take g = 10, RE = 6 371 315 meters... quick, the calculator!

The polar angle f in its orbiting motion, measured from the position of perigee (or perihelion).

One calculates the mean anomaly M, an angle which is also zero at perigee, but which grows at a constant rate--unlike f.

Transformations exist from M to an intermediate angle E ("Eccentric anomaly") and from E to f.

We can obtain its position in the orbital plane, because we then can calculate
r = a(1 – e²)/(1 + e cos f), giving its location in polar coordinates.

But we still need to know how the orbital plane is oriented in 3-D space, which requires 3 more inputs:

   (1) the inclination of that plane to the equator

   (2) in which direction on that plane does apogee point.

   (3) The orbital plane can still be rotated around the Earth's axis without changing its inclination. Imagine a slanted board on a record turn-table; one more angle is needed to give the amount of rotation of that turntable.

In 2003, fall equinox occurred 184 days after spring equinox, but the next spring equinox occurred only 181 days later. Other years show the same imbalance. Knowing that Earth is closest to the Sun around January 4, how do you explain this?

(Added hint, if necessary): Draw a sketch of the orbit of the Earth. The directions to the Sun at spring equinox and at fall equinox are in the plane of the orbit, and since they are in opposite directions, they form one continuous straight line, dividing the orbit into two parts. Draw it in your sketch too!

The closest approach to the Sun ("perihelion"), around January 4, is about halfway between fall equinox and spring equinox. That should therefore be "the smaller half" of the orbit. (redraw the orbit, distorting it so that it is notably elliptic, and you will see why).

Also, by Kepler's 2nd law, Earth moves a bit faster when closer to the Sun. Both these factors make it spend a little less then half of its time in the section of its orbit closer to the Sun.