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Lesson Plan #31         http://www.phy6.org/stargaze/Lkepl3rd.htm

# (21)  Kepler's 3rd Law

### (21a)  Application of Kepler's 3rd Law

In these closely linked units, Newton's calculation is applied to artificial Earth satellites. It is shown that at least for circular orbits, this calculation leads to Kepler's 3rd law. The velocity required for a low Earth orbit is derived, and a practical formula is obtained for the orbital period in a circular Earth orbit of any radius.

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 Lesson plan home page and index:             ....stargaze/Lintro.htm

 Goals: The student will learn To derive the velocity in a circular orbit of any radius. To derive the Earth escape velocity. To prove Kepler's 3rd law for circular orbits A simple formula for the orbital period in a circular Earth orbit of given radius. 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 inverse-square gravitational force, assuming only that it acts primarily between only two bodies--e.g. Sun and planet, Earth and Moon--while 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 ellipse--of 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 inverse-squares 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 V0 required by a spacecraft in a circular low Earth orbit--what is called in Russia "the first cosmic velocity." If g is the acceleration due to gravity and RE the radius of the Earth, we find V02 = g RE (Teacher may go over derivation) Using the energy equation of the Kepler motion, one can also show that the "escape velocity" V1 from distance RE ("second cosmic velocity" in Russia) is given by V12 = 2g RE It is larger by a factor equal to the square root of 2, which is 1.4142..     (This calculation may be included or omitted, depending on the level of the class and the time available). 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 V0 to a rocket launched vertically, straight up, how far from Earth will it get before falling back? Hint: the semi-major axis of an orbit depends only on the energy. [The rocket will rise and then come down along the same radial line, following what is essentially a very narrow and elongated ellipse (eccentricity close to 1). Since the energy is the same as the one required for low Earth orbit, the semi-major axis a will also be the same. For the low Earth orbit, a = 1 RE, hence the same value holds here, too, making the ellipse 2RE long. The highest point of the trajectory--the apogee of the orbit--will therefore be at the height of 1 RE above ground.] Given the orbital velocity for a circular orbit at 1 RE as 7.9 km/sec, what is the escape velocity at the surface of the Earth (r = 1 RE)? What sort of "escape" does it provide, and what sort doesn't it? The escape velocity is V1 = 7.9*SQRT(2) = 7.9*1.4142 = 11.2 km/s. A velocity greater than V1 makes it possible for the launched object to escape Earth and not fall back. However, it will still be bound by the Sun's gravity, and will stay in an orbit around the Sun, similar to the Earth's. If an object above the atmosphere has a velocity greater than the escape velocity V1 in what upward direction must it move to escape the Earth's gravity? It does not matter. An escape orbit is the limiting case of an orbit with very, very large semi-major axis, and therefore a very distant apogee. With a starting speed close to the escape velocity, in whichever upward direction the object is moving, its starting point is always close to the perigee, and it is always headed for the apogee. In the limit (=at escape velocity), this means it is heading for infinity. 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? The calculation is similar to the preceding one. Escape velocity from the Sun at 1 AU is     V1 = 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? Say an Earth-like planet in the alternate universe has a radius RE and surface gravitational acceleration g. In a circular orbit having a radius of R planetary radii, the weight of each satellite is the same, say F   =   mg/[R/RE]2 The inertiaof the astrite satellite is twice as large. So the centripetal force needed to hold it in orbit Requires a velocity U1 satisfying F   =   2m U12/R while the barite satellite, with smaller inertia, needs U2 satisfying F   =   m U22/R 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 RM=3390 km and its satellite Deimos orbits it in a near-circular orbit with orbital period TD = 1.26244 days and a mean distance of RD = 23,436 kilometers. What is (1) The acceleration gM due to gravity at the surface of Mars and (2) The escape velocity there? If the acceleration on the surface is gM, then for a circular orbit at distance RD gM (RM/RD)2 = v2/ RD where v is the velocity of Deimos in its orbit. From a previous calculation, if N is the number of orbits per second and TD is also given in seconds         v   =  2π RDN   =   2πRD/TD         TD  =   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 RD/RM = 23,436/3390 = 6.9133 We then have         gM   =   (v2/RD) (RD/RM)2         gM = ((1350)2/23,436,000)*(6.9133)2 =        = 0.077764*47.793 = 3.717 m/s2 i.e. slightly more than 1/3 the acceleration of free fall on Earth. The orbital velocity V0m at the surface of Mars, in analogy with V0 derived for Earth, is found from         V0m2 = gM RM = 3.717*3,390,000 = 12,600,000         V0m = 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 semi-major axis in meters. That implies T2 = ka3 . 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 RE/4π2 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? The calculation in "Stargazers" gives 5063.5 seconds. --At what distance R is the orbital period around Earth 24 hours? Why is that orbit important? We have 86400 sec = 5063.5 R*SQRT(R) = 5063.5 R3/2 R3/2 = 17.0633         R = (17.0633)2/3 = 6.628 RE 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 near-infinite 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"?     It is not quite so simple, since each part of the tether will try to orbit at its own velocity. If however the satellite could be anchored at a somewhat greater distance, its centrifugal force will tighten the tether (and add stress to it, too). That might work, if such a strong material existed. As far as we know, it's pure science fiction (and was used that way in Arthur Clarke's book "Fountains of Paradise"). No known material is strong enough. --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 R3/2 R3/2 = 8.53164         R = (8.53164)2/3 = 4.175 RE

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Author and Curator:   Dr. David P. Stern
Mail to Dr.Stern:   stargaze("at" symbol)phy6.org .

Last updated: 10-23-2004

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