Orbital Mechanics I

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© 1998-2008 by Glenn Elert -- A Work in Progress
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Discussion

circular orbits

Newton's laws only. Nothing about energy or momentum. Centripetal force and gravitational force.

Fc = Fg ⇒ m2v2  =  Gm1m2 ⇒ v = √
Gm
r r2 r

Kepler's third law. Derive Kepler's third law of planetary motion (the harmonic law) from first principles.

v = √  =   
Gm 2πr
r T  
  Gm  =  4π2r2  
  r T2  
  r3  =  Gm  = constant
  T2 4π2
  r3  ∝  T2  
   

The "constant" depends on the object at the focus. Although formulated from the data for objects orbiting the sun, Newton showed that Kepler's third law can be applied to any family of objects orbiting a common body.

orbit families


[magnify]

A "snapshot" of the earth and about 500 of its artificial satellites generated one summer evening in 2002. Nearly all of them are GEOs or LEOs. Satellites on the ring are in geosynchronous earth orbit (GEO). Those clustered near the earth are in low earth orbits (LEO). Scattered in between are satellites in medium earth orbits (MEO). The moon, earth's only natural satellite, is approximately nine times farther from the earth than the ring of geosynchronous satellites. Source: NASA.

binary systems

circular motion about the center of mass

still just a balance between centripetal and gravitational force, but slightly more complicated

lagrange points

the three body problem, lagrange libration points are the simplest solutions

still just a balance between centripetal and gravitational forces, but now much more complicated


[magnify]

The five Lagrange points of the earth-sun system. Satellites in orbit at these locations remain fixed with respect to the earth and sun. This figure is not drawn to scale.

L1 and L2 are approximately four times farther from the earth than the moon. L3 is a very near the "anti-earth" point.

L4 and L5 are at the vertex of an equilateral triangle formed with the earth and sun. L4 leads the earth and L5 follows.

Objects can settle in an orbit around a Lagrange point. Orbits around the three collinear points, L1, L2, and L3, are unstable. They last but days before the object will break away. L1 and L2 last about 23 days. Objects orbiting around L4 and L5 are stable because the Coriolis force keeps them spinning around the Lagrange point.

noncircular orbits

qualitative description of noncircular orbits

centripetal-gravitational forces don't balance

uses

Summary

Problems

practice

  1. Geosynchronous Satellite
    There is a special class of satellites that orbit the earth with a period of one day.
    1. How will the satellite's motion appear when viewed from the surface of the earth?
    2. What type of satellites use this orbit and why is it important for them to be located in this orbit? (Keep in mind that this is a relatively high orbit. Satellites not occupying this band are normally kept in much lower orbits.)
    3. Determine the orbital radius at which the period of a satellite's orbit will equal one day. State your answer in …
      1. kilometers
      2. multiples of the earth's radius
      3. fractions of the moon's orbital radius
    Solutions …
    1. Start with the basic principle behind all circular orbits. Set the centripetal force equal to the gravitational force.
       
      Fg = Fc
       
      Replace the speed with the circumference divided by the period.
       
      Gm1m2  =  mv2  =  m   ⎛
      ⎜
      ⎝      		 2πr ⎞2
      ⎟
      ⎠      		  =  4π2mr
      r2 r r T T2
       
      Solve for radius to arrive at a general formula
       
      r =  ⎛
      ⎜
      ⎝      		 GmT2 ⎞⅓
      ⎟
      ⎠
      4π2
       
      (This problem can also be solved using Kepler's third law of planetary motion: the square of the period of a satellite in a circular orbit is proportional to the cube of its radius. That solution is presented in an earlier section of this book entitled Heliocentrism.)
      1. Now substitute in the appropriate values. The period of the earth's rotation is approximately equal to the mean solar day (24 x 3600 s = 86,400 s), but for best results use the sidereal day (86,164 s).
         
        r =  ⎛
        ⎜
        ⎝        		 (6.67 × 10−11 Nm2/kg2)(5.974 × 1024 kg)(86,164 s)2 ⎞⅓
        ⎟
        ⎠
        4π2
        r =  4.216 × 107 m ≈ 42,000 km  
         
      2. Convert to earth radii
         
        r =  4.216 × 107 m  = 6.610 ≈ 7 earth radii
        6,378,140 m
         
      3. Convert to earth-moon distances
         
        r =  4.216 × 107 m  = 0.1097 ≈  1  distance from earth to moon
        384,400,000 m 9
         
    2. Answer the remaining parts.
    3. Answer the remaining parts.
  2. Dark Matter
    1. The orbital speed of the planets decreases with distance from the sun. Why does this happen? Derive a formula that shows the relationship.
       
      [magnify]
       
    2. The orbital speed of the stars remains roughly constant with distance from the center of the Milky Way. (This is true for other galaxies as well.) What does this tell us about the distribution of mass in galaxies? Derive a formula that shows the relationship.
       
      [magnify]
       
    3. Calculate the mass of the Milky Way given a typical orbital speed of 220 km/s and a radius of 50,000 light years. Give your answer in solar masses (m☉ = 2 × 1030 kg) and compare it to the approximate number of stars in the Milky Way (1011).
    4. Dwarf galaxies, star clusters, and gas clouds beyond the edge of the visible galaxy have nearly the same orbital speed as the stars within visible galaxy.There is evidence that rotational speeds remain roughly constant at 220 km/s out to distances of 300,000 light years or six times the radius of the Milky Way. What is so amazing about this observation and what does it imply?
    Solutions …
    1. The basic principle behind all circular orbits is that the the centripetal force needed to keep the planet in orbit is supplied by an inverse-square gravitational force. When these two equations are set equal and solved for speed we get …
       
      Fc = Fg ⇒ m2v2  =  Gm1m2 ⇒ v = √
      Gm
      r r2 r
       
      which shows that speed drops off as one over the square root of the distance from whatever it is we're orbiting — the sun in this case. This makes sense since gravity gets weaker as distance increases. At large distances, the outer planets (jupiter, saturn, uranus, and neptune) can drag along and still stay in orbit. Closer in where gravity is strong, the inner planets (mercury, venus, earth, and mars) need much larger speeds to avoid being gobbled up by the sun.
    2. Starting with the end product of our previous solution we can see that speed could be made constant if mass was allowed to increase with distance. This is exactly what happens in a distributed arrangement of mass like a galaxy. Objects that are further from the center are orbiting around more stuff. On such vast scales, kilograms won't do to describe the mass of things. Instead we will use the mass of the sun as our standard unit. Stars in the core are orbiting only a few million solar masses of material, our sun, which is some two thirds of the way to the edge of the galaxy, is orbiting tens of billions of solar masses, and stars at the edge of the Milky Way are basically going around all one hundred billion solar masses of material that make up our galaxy.

      If the observed speed of stars in the Milky way is more or less uniform, then the mass contained within the orbit of any one star must be proportional to the radius of its orbit, but it's really density that we're after — or rather, a density function. Take the equation derived above and solve for mass.
       
      v = √ ⇒ m =  rv2
      Gm
      r g
       
      This shows us that the mass around which a star orbits is directly proportional to its distance from the center of the galaxy. The sun is roughly two thirds of the way to the edge of the Milky Way. Its orbit should therefore encircle two thirds of the mass of the entire disk of the galaxy. Interesting, ut we're not finished. Substitute this expression for mass and the volume of a sphere into the density formula and simplify. This gives us the density function for a galaxy with an observed flat rotation curve.
       
      ρ =  m  =  rv2 / g  =  3v2
      V 4πr3 / 3 4πGr2
       
      In order for the orbital velocity to remain constant in a galaxy its mass must increase linearly with radius. In order for its mass to increase linearly its density must drop off as the inverse square of its radius. I hope that this makes sense. The density of the core of a galaxy, where stars are tightly packed together, should be greater than the density of the whole thing. If the core has half the radius of the disk, then it should have four times the density of the entire galaxy. It's an interesting "conspiracy" of the natural world that this is the way it should work out.

      The equation above is slightly wrong. It's not really a density function, it's an average density function. It doesn't give us the local density at some distance r from the center, it gives us the average density within a sphere of radius r. To fix this, we need to drop the 3 from the numerator.
         
      ρ =  v2
      4πGr2
         
      This last bit should only be read by those who understand calculus. Everyone else can jump ahead to the last part of this problem. Now when this function is integrated over a series of spherical shells with surface area 4πr2 and thickness dr from the center 0 out to a distance r we get back the expression we derived earlier for mass.
            r       r    
      m =  ⌠
      ⌡      		  ρ dV =  ⌠
      ⌡      		 v2  (4πr2 dr) =  v2 ⌠
      ⌡      		  dr =  rv2
      4πGr2 g g
            0       0    
      And all is well again.
    3. Numbers in. Answer out. Here we go …
             
      m =  rv2    
      g    
      m =  (5 × 107 light years)(3 × 108 m/s)(365.25 × 24 × 3600 s)(2.2 × 105 m/s)2  
      (6.67 × 10−11 Nm2 / kg2)(2 × 1030 kg / solar mass)  
      m =  170 billion solar masses  
       
             
      Although a bit high, this is in the ballpark for the number of stars in the Milky Way.
    4. If the orbital speed remains constant for bodies near but outside the milky way, then the mass and density distributions we derived earlier …
               
      m =  rv2 & ρ =  v2
      g 4πGr2
               
      must apply to the apparently empty regions beyond the edge of the galaxy. But when we look at galaxies like the Milky Way we always see a definite edge to them — on one side there are stars and on the other side an empty void populated only occasionally by a small cluster of stars or a cold cloud of radio waves emitting gas. Beyond this distance one would expect an inverse square root drop in orbital speed as is seen with the planets. But this is not the case. Rotational speeds remain roughly constant six times farther than the edge of the Milky Way. Since mass is directly proportional to radius when speed is constant, this means that the total mass of the galaxy is at least six times greater than its visible mass, or equivalently, that five-sixths (roughly 85%) of all the mass in our galaxy is invisible.

      Astronomers have decided to call this stuff dark matter, but I don't particularly like this term since people have a tendency to think "dark" means "black". Dark matter does not interact with light or any other form of electromagnetic radiation. You and I are giving off plenty of infrared. Many communications devices give off microwaves. Both forms of radiation are invisible to our eyes, but we have other means of detecting them. I can feel infrared on my skin as heat and detect microwaves with a cellular phone or a satellite dish. Dark matter will have nothing to do with any of these forms of radiation. Dark matter neither emits, nor absorbs, nor reflects, refracts, diffracts, or interacts electromagnetically in any way with radio waves, microwaves, infrared, visible light, ultraviolet, x rays, or gamma rays. The only way dark matter can be detected is through its gravitational effects — and they are significant. So much so that the dark matter haloes around galaxies will bend spacetime from its normally flat geometry. As we all know light travels in straight lines. But when light encounters the warped spacetime around a galaxy, straight lines have no choice but to bend. The result is a phenomena called gravitational lensing (a more complete discussion of which is best left to another part of this book). What's important to note here is that this phenomena can be used to measure the amount of matter in moderately distant galaxies and that results always show a significantly larger amount of dark matter than ordinary matter (as high as 10:1 in some cases).

      Dark matter exists in other galaxies besides the Milky Way. Flat rotation curves have been plotted for other nearby spiral galaxies and gravitational lensing has been used to measure dark matter distributions of more distant galaxies. Computer simulations of colliding galaxies don't work (that is, they don't agree with observations of actual colliding galaxies) unless they include dark matter as a variable. They need dark matter to give realistic results. In summary, dark matter exists. It exists as much as electrons or radio waves exist despite the fact that they can't be seen. The only remaining question is, unfortunately, a really big one. What is it? Let me know when you find out.

      More on the dark side of the universe in the next section: Gravitational Potential Energy II.
  3. Some sort of binary star problem would be good here.
    • Answer it
  4. Locate the L1, L2, and L3 Lagrange points for the earth-sun system using dynamical principles. State your answers as distances …
    1. from the sun and earth in meters
    2. from the earth as multiples of the moon's orbital radius
    3. from the sun as multiples of the earth's orbital radius

    Solution …

    The first three Lagrange points lie on the line connecting the earth and sun.

    [magnify]

    For this problem let …

    ms be the mass of the sun
    me be the mass of the earth
    re be the radius of earth's orbit
    r be the displacement from the sun to the satellite
    x be the displacement from the earth to the satellite

    so that …

    x = re − r

    All of the Lagrange points move together with the earth about the sun as if they were fixed on a rotating disk. In this situation, where angular velocity is constant, centripetal acceleration is directly proportional to the distance from the center of rotation.

    ac = −ω2r = −  ⎛
    ⎜
    ⎝    		 2π ⎞2
    ⎟
    ⎠    		 r = −  4π2  r
    T T2

    A satellite will travel on a circular orbit wherever the required centripetal acceleration can be provided by the net gravitational field.

    gnet =      gearth      +      gsun    
                 
    gnet =  ⎛
    ⎜
    ⎝    		  −  Gme  ˆi  ⎞
    ⎟
    ⎠    		  +  ⎛
    ⎜
    ⎝    		  −  Gms  ˆr  ⎞
    ⎟
    ⎠
    x2 r2

    Solving this pair of equations is difficult for two reasons.

    1. It's a vector problem, which means we must contend with direction. Since the first three Lagrange points lie on the line connecting the earth and sun, the vector aspects of this problem are not that serious. In a one-dimensional problem like this one, directions are are indicated with plus and minus signs. The nature of this problem requires that we deal with the signs in a piecewise fashion.
    2. It's also fifth order, which means an exact analytical solution is impossible. The way around this is to graph both equations on a calculator and let it find the points of intersection.

    The next step is to set up a coordinate system. For no apparent reason, I've chosen to place the origin at the sun and use r as the independent variable. (The earth would have worked equally well as an origin and x as the independent variable.) This places L1, the earth, and L2 on the positive side of the axis and leaves L3 by itself on the negative side. As is usually done, all vectors pointing to the right will be positive and those to the left will be negative.

    ac  =  gnet      
         
    −  4π2  r  =  ⎛
    ⎜
    ⎝    		  −  Gme  ˆi  ⎞
    ⎟
    ⎠    		  +  ⎛
    ⎜
    ⎝    		  −  Gms  ˆr  ⎞
    ⎟
    ⎠    		  
    T2 x2 r2
        ⎧
    ⎪
    ⎪
    ⎪
    ⎨
    ⎪
    ⎪
    ⎪
    ⎩    		  +  me  +  ms   L3 behind the sun  
    (re − r)2 r2  
    −  4π2  r  =   +  me  −  ms   L1 between the sun and earth  
    GT2 (re − r)2 r2  
         −  me  −  ms   L2 behind the earth  
    (re − r)2 r2  

    Which when graphed looks like this.

    [magnify]

    Well, not exactly. Since the sun is so much more massive than the earth, the L1 and L2 points would lie so close together as to be indistinguishable at this scale. Using the following values …

    symbol value name
    G 6.67259 × 10−11 N·m2/kg2 universal gravitational constant
    T 365.25 x 24 x 3600 s period of the earth's rotation about the sun
    re 1.4959787 × 1011 m radius of the earth's orbit
    me 5.9742 × 1024 kg mass of the earth
    ms 1.9891 × 1030 kg mass of the sun

    Yields these solutions …

    lagrange point distance from sun, r distance from earth, x
    L1 1.481 × 1011 m 1.49 × 109 m
    L2 1.511 × 1011 m 1.50 × 109 m
    L3 1.496 × 1011 m 2.98 × 1011 m

    or in terms of "natural" units …

    lagrange point distance from sun, r distance from earth, x
    L1 0.99 AU 3.88 re-m
    L2 1.01 AU 3.90 re-m
    L3 1.00001 AU 775 re-m

    where an AU (astronomical unit) is the distance from the sun to the earth (1.4959787 × 1011 m) and re-m is the distance from the earth to the moon (3.844 × 108 m).

conceptual

  1. If objects in earth orbit are weightless, why can't astronauts throw objects like baseballs or screwdrivers into outer space? Since they're weightless, it should be possible to heave them to the moon, planets, or distant stars. What's wrong with this thinking? In addition, why would casually discarding junk overboard from a space station or space shuttle be a bad idea?
  2. One way to send a spaceship to the planet Mars would be to point it in the general direction of the Red Planet, ignite the rocket engines, and let it go. This method won't work, however. Give two reasons why this procedure would never result in a successful mission, no matter how precisely the spacecraft was aimed.
  3. Spacecraft in extreme near earth orbit are subject to small but (in the long run) non-negligible amounts of aerodynamic drag from the upper regions of the earth's atmosphere.
    1. What happens to the altitude and speed of such a satellite over time?
    2. Sketch the path of a satellite in such an orbit.
  4. Pluto was discovered in 1930, but it's mass wasn't known with any accuracy until 1978 when Pluto's moon Charon was discovered. What was it about Charon's discovery that enabled astronomers to finally determine the mass of Pluto?

numerical

  1. Satellite Motion
    1. Calculate the speed needed for the space shuttle to travel around the earth in a circular orbit at an altitude of 350 km above the earth's surface.
    2. Calculate the period of the space shuttle at this same orbit.
  2. Black holes are formed when massive stars exhaust their nuclear fuel and collapse. The gravitational field near a black hole is extremely intense. Within a radius known as the event horizon nothing can escape, not even the speediest thing known — light. (We will discuss the event horizon on another day.) Inside the event horizon there is another special radius called the photon sphere. A beam of light directed at a tangent to the photon sphere will be trapped in a circular orbit around the black hole. A black hole may be black on the outside, but inside it is filled with light — light that is locked forever in orbit about the black hole.
    1. Determine the radius of the photon sphere of …
      1. a small black hole with a mass about three times the mass of the sun
      2. a supermassive black hole (like the one at the center of the Milky Way galaxy) with a mass about three million times the mass of the sun
    2. Complete the following table where you compare your answers to the radius of the sun (r☉ = 695,500 km) and the radius of mercury's orbit (r☿ = 58,000,000 km).

      black hole r (km) r☉ r☿
      small      
      supermassive      

statistical

  1. The table below gives the orbital period in days and orbital radius in millions of meters for Jupiter's four largest satellites (named the Galilean moons in honor of their discoverer, Galileo Galilei). Use this data to determine the mass of Jupiter.

    moon period (days) distance (106 m)
    Io 1.769137786 422
    Europa 3.551181041 671
    Ganymede 7.154552960 1070
    Callisto 16.68901840 1883

worksheets

  1. trajectories-satellite.pdf
    The accompanying pdf file shows a satellite in a circular orbit about the earth. Sketch the new path that the satellite would take if its speed were changed abruptly in the ways described.

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