(12) Kepler's Second Law |
Part of a high school course on astronomy, Newtonian mechanics and spaceflight
by David P. Stern
This lesson plan supplements:
#12. "Kepler's Second Law Skepl2nd.htm on the web http://www.phy6.org/stargaze/Skepl2nd.htm #12a. "More about Kepler's Second Law Skepl2A.htm on the web http://www.phy6.org/stargaze/Skepl2A.htm
"From Stargazers to Starships" home page and index: Sintro.htm |
Goals: The student will
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.
Guiding questions and additional tidbits With suggested answers, brackets for comments by the teacher or "optional")
"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.
--Kepler's 2nd law is sometimes stated "The radius vector sweeps equal areas in equal times" What is the "radius vector" of a planet?
--What does Kepler's 2nd law imply, about the way orbital velocity of a planet varies with distance?
--Does Kepler's 2nd law also hold for artificial satellites, orbiting Earth?
[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. --An artificial satellite moves in an elongated orbit with perigee r1= 2 Earth radii (2 RE) and apogee 10 times more distant, at r2= 20 RE. Show that the same ratio also holds for velocities at those points--that at apogee the satellite moves 10 times slower.
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.
--Is the velocity also inversely proportional to distance at other points of the orbit?
--How is this similar to what happens to a stone thrown upwards?
--What is energy?
--Can you give examples of types of energy?
--What is the law of conservation of energy?
[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].
--What types of energy are involved in the throwing of a stone?
--What is kinetic energy?
--What does it depend on?
--What is potential energy?
--If a stone of mass m is lifted to a height h meters, how much is added to its potential energy?
--And what does conservation of energy say?
As one part grows, the other decreases to keep the sum the same.
--How does a playground swing demonstrate conservation of energy?
--How about the pendulum of a grandfather clock?
--Does it matter where we choose h=0 to be, the reference point to measure h from?
[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.]
--The equation giving the energy of an orbiting planet or satellite is
where k = gRE2. 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 EP depends on the reference level at which we choose EP = 0. Any reference level is OK: what matters are differences in potential energy, which dictate the gain or loss of kinetic energy EK.
--What is "escape velocity"?
(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!)
--How can the above equation give you the escape velocity V from the surface of the Earth, r = RE?
The energy of an object with velocity V, at the surface of the Earth, is E = 1/2 mV2 – k m/RE 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/RE = 0 If we calculate in meters and seconds, take g = 10, RE = 6 371 315 meters... quick, the calculator!
--What is the "true anomaly" of an orbiting satellite or planet?
--How does one predicts the value of f at some given time?
Transformations exist from M to an intermediate angle E ("Eccentric anomaly") and from E to f. --If we calculate f of an artificial satellite for some time, can we tell where its position will be?
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.
(This question is discussed in "More about Kepler's 2nd Law") 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!
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. |
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Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last updated: 10-25-2004