(21) Kepler's 3rd Law
(21a) Application of Kepler's 3rd Law 
Part of a high school course on astronomy, Newtonian mechanics and spaceflight
by David P. Stern
This lesson plan supplements: "Kepler's 3rd Law," section #21 http://www.phy6.org/stargaze/Skepl3rd.htm "Application of Kepler's 3rd Law," section #21a http://www.phy6.org/stargaze/Sappl3rd.htm
"From Stargazers to Starships" home page: ....stargaze/Sintro.htm 
Goals: The student will learn
Terms: Orbital velocity, escape velocity, synchronous orbit.
Starting the lesson: Today we will apply Newton's calculation to circular Earth orbits. As already noted, all three of Kepler's laws can be derived from an inversesquare gravitational force, assuming only that it acts primarily between only two bodiese.g. Sun and planet, Earth and Moonwhile the pull of all others can be neglected. With our limited math we can handle here only circular orbits, not elliptical ones. Circular orbits are a somewhat special and simple case. The first law for instance is automatically satisfied, because yes, the circle is an ellipseof zero eccentricity and with both its foci combined at the center. The second law also does not add any information about the motion. The speed of a satellite or planet in a circular orbit is constant, and so is its distance from the center. The area swept by the radius per unit time is then automatically constant. The third law&, however, is meaningful even for circular orbits. Newton's calculation allows it to be checked, and today we will show that the inversesquares law indeed gives the expected result. In this day and age we are interested not just in planetary orbits but also in the orbits of artificial satellites. We will therefore use such orbits to illustrate the calculation, and some useful results will also be derived.
We start by calculating the velocity V_{0} required by a spacecraft in a circular low Earth orbitwhat is called in Russia "the first cosmic velocity." If g is the acceleration due to gravity and R_{E} the radius of the Earth, we find Using the energy equation of the Kepler motion, one can also show that the "escape velocity" V_{1} from distance R_{E} ("second cosmic velocity" in Russia) is given by It is larger by a factor equal to the square root of 2, which is 1.4142.. Questions and worked examples This section is again mostly concerned with one calculation, and the questions asked are all related to it.
If you give the above velocity V_{0} to a rocket launched vertically, straight up, how far from Earth will it get before falling back?
Since the energy is the same as the one required for low Earth orbit, the semimajor axis a will also be the same. For the low Earth orbit, a = 1 R_{E}, hence the same value holds here, too, making the ellipse 2R_{E} long. The highest point of the trajectorythe apogee of the orbitwill therefore be at the height of 1 R_{E} above ground.]
Given the orbital velocity for a circular orbit at 1 R_{E} as 7.9 km/sec, what is the escape velocity at the surface of the Earth (r = 1 R_{E})? What sort of "escape" does it provide, and what sort doesn't it?
If an object above the atmosphere has a velocity greater than the escape velocity V_{1} in what upward direction must it move to escape the Earth's gravity?
The Earth orbits the Sun in a near circular orbit of radius 1 AU (Astronomical Unit) at 30 km/sec. How much extra velocity, above and beyond 11.2 km/s, does a spacecraft escaping the Earth's gravity need, to escape the Sun's gravity as well?
V_{1} = 30*SQRT(2) = 30*1.4142 = 42.4 km/s. Thus an additional velocity of (42.4  30) = 12.4 km/s is required.
(The question below is really just for debating fun!) In some weird alternate universe (already met in the lesson on Newton's 2nd law) weight and mass are not proportional. Two materials, astrite and barite, have the same weight per unit volume, but a volume of astrite has twice the mass of a similar volume of barite. Both are strong and light metals, and are a natural choice for spacecraft construction; we can assume that barite behaves the way aluminum does on Earth. Which of the two would be a better choice?
The inertiaof the astrite satellite is twice as large. So the centripetal force needed to hold it in orbit Requires a velocity U_{1} satisfying while the barite satellite, with smaller inertia, needs U_{2} satisfying The astrite satellite needs less velocity by a factor of square root of 2. However... it is also harder to accelerate! Since both materials need the same kinetic energy for the orbit, probably either is equally suitable.
The planet Mars has a radius of R_{M}=3390 km and its satellite Deimos orbits it in a nearcircular orbit with orbital period T_{D} = 1.26244 days and a mean distance of R_{D} = 23,436 kilometers. What is (1) The acceleration g_{M} due to gravity at the surface of Mars and (2) The escape velocity there?
where v is the velocity of Deimos in its orbit. From a previous calculation, if N is the number of orbits per second and T_{D} is also given in seconds v = 2π R_{D}N = 2πR_{D}/T_{D} T_{D} = 1.26244*86400 sec = 109,075 sec. v = 6.2832*(23,436,000 m)/109,075sec = 1350 m/sec In calculating a ratio we can use kilometers, so R_{D}/R_{M} = 23,436/3390 = 6.9133 We then have g_{M} = (v^{2}/R_{D}) (R_{D}/R_{M})^{2} g_{M} = ((1350)^{2}/23,436,000)*(6.9133)^{2} = = 0.077764*47.793 = 3.717 m/s^{2} i.e. slightly more than 1/3 the acceleration of free fall on Earth. The orbital velocity V_{0m} at the surface of Mars, in analogy with V_{0} derived for Earth, is found from V_{0m}^{2} = g_{M} R_{M} = 3.717*3,390,000 = 12,600,000 V_{0m} = 3549.6 m/s = 3.5496 km/s The escape velocity is SQRT(2)=1.4142 times that, or 5.02 km/sec.
According to Kepler's 3rd law, T is proportional to a, where T is the orbital period in seconds and a the semimajor axis in meters. That implies T^{2} = ka^{3} . In a circular orbit around Earth, a=r where r is the orbital radius. Can you derive k for such orbits? The result is derived in "Stargazers" and is k = g R_{E}/4π^{2}
Here the teacher may continue with lesson 21a (which is optional). Some questions related to that lesson (in all that follows, * marks multiplication): From Kepler's 3rd law, the orbital period T around Earth in a circular orbit at distance R is T = Q *R*SQRT(R) If R is given in Earth radii, what is the value of Q?
At what distance R is the orbital period around Earth 24 hours? Why is that orbit important?
R^{3/2} = 17.0633 R = (17.0633)^{2/3} = 6.628 R_{E} The orbit is important because a satellite in an equatorial orbit at this distance stays all the time above the same point on the Earth's equator, and rotates with the Earth as if it were attached to it by a rigid rod. That makes such a"synchronous orbit" desirable for communication satellite, because they can then be tracked from the ground by a fixed antenna.
(Subject for discussion) It has been speculated that if we could manufacture a lightweight cable of nearinfinite strength, such a satellite could be anchored to the spot below it on Earth, and loads could be passed to it by an elevator guided by that cable, making leaving Earth essentially independent of rockets. As payloads were raised, they would need a motor to raise them along the cable, but the energy can be applied gradually. As payloads rise, the rotation of Earth would also be very, very slightly slowed down. What do you think of the idea of such a "space tether"?
Navigational satellites of the GPS (Global Positioning System) move in 12 hour orbits. What is their distance? The calculation is very similar to the preceding one. We have 43200 sec = 5063.5 R*SQRT(R) = 5063.5 R^{3/2} R^{3/2} = 8.53164 R = (8.53164)^{2/3} = 4.175 R_{E}

Back to the Lesson Plan Index Back to the Master Index
Guides to teachers... A newer one An older one Timeline Glossary
Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: audavstern("at" symbol)erols.com .
Last updated: 10232004