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Freshman Seminar FRS 148
From the Earth to the Moon
Part 2. Understanding Orbits
°°°°°°°°°°
Robert Stengel
Princeton University
March 1, 1997
Copyright 1996-1997 by Robert Stengel. All rights reserved.
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
;[s]
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We let the ball fall through the surface just to show its path. How long did it take for the ball to hit the ground?
All well and good, BUT .... The earth is not really flat, and the acceleration due to gravity is not constant. The flat-earth model is not bad for studying baseball trajectories -- in fact, it is useful for studying aircraft motions within the atmosphere at speeds up to several times the speed of sound (about 300 m/s) -- however, it just won't do for much faster motion. We might recognize that the baseball trajectory that we view as a parabola above actually is the tip of an ellipse (i.e., a segment just past the apogee) that would extend to the center of the earth using the "inverse-square" model associated with Newton's Law of Gravitation, discussed below.
;[s]
7:0,0;601,1;608,0;640,1;646,0;742,1;769,0;788,-1;
2:4,14,9,Times,0,12,0,0,0;3,14,9,Times,2,12,0,0,0;
:[font = subsection; inactive; preserveAspect; startGroup]
Analyzing Motion
:[font = text; inactive; preserveAspect]
Continuing a bit longer with our flat-earth model, we can define vectors of the ball's position, velocity, and acceleration. The world is 3-dimensional, so each of these vectors would have 3 components in the most general case; however, we are concerned about only vertical and horizontal motions here (i.e., no lateral motions), and we can use 2-dimensional vectors:
:[font = input; preserveAspect]
HorizontalPosition =.; Time =.; HorizontalPosition = 20 Time;
:[font = input; preserveAspect; startGroup]
BallPosition = {HorizontalPosition, VerticalPosition}
:[font = output; output; inactive; preserveAspect; endGroup]
{20*Time, 200 - 4.900000000000001*Time^2}
;[o]
2
{20 Time, 200 - 4.9 Time }
:[font = input; preserveAspect; startGroup]
BallVelocity = {HorizontalVelocity, VerticalVelocity}
:[font = output; output; inactive; preserveAspect; endGroup]
{20, -9.8*Time}
;[o]
{20, -9.8 Time}
:[font = input; preserveAspect; startGroup]
BallAcceleration = {HorizontalAcceleration, VerticalAcceleration}
:[font = output; output; inactive; preserveAspect; endGroup]
{0, -9.8}
;[o]
{0, -9.8}
:[font = text; inactive; preserveAspect]
The origin of the position coordinate system is the base of the cliff, i.e., the point {0, 0}. At the starting time, the ball is directly overhead, with coordinates {0, 200}. Put another way, the distance (or radius from the origin to the ball's position) is 200 m, and the elevation angle is 90 degrees (or p/2 radians). The distance is the magnitude of the vector from the origin to the ball, or
;[s]
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:[font = input; preserveAspect; startGroup]
Distance = Sqrt[HorizontalPosition^2 + VerticalPosition^2]
:[font = output; output; inactive; preserveAspect; endGroup]
(400*Time^2 + (200 - 4.900000000000001*Time^2)^2)^(1/2)
;[o]
2 2 2
Sqrt[400 Time + (200 - 4.9 Time ) ]
:[font = text; inactive; preserveAspect]
The elevation angle is derived from the position components using trigonometric functions, and there are 3 ways to do it, using either the sine, cosine, or tangent of the angle. Given the position, we actually use the inverses of these functions (or "arc" functions) to find the elevation angle:
:[font = input; preserveAspect; startGroup]
Elevation1 = ArcTan[VerticalPosition / HorizontalPosition]
:[font = output; output; inactive; preserveAspect; endGroup]
ArcTan[(200 - 4.900000000000001*Time^2)/(20*Time)]
;[o]
2
200 - 4.9 Time
ArcTan[---------------]
20 Time
:[font = input; preserveAspect; startGroup]
Elevation2 = ArcSin[VerticalPosition / Distance]
:[font = output; output; inactive; preserveAspect; endGroup]
ArcSin[(200 - 4.900000000000001*Time^2)/
(400*Time^2 + (200 - 4.900000000000001*Time^2)^2)^(1/2)]
;[o]
2
200 - 4.9 Time
ArcSin[------------------------------------]
2 2 2
Sqrt[400 Time + (200 - 4.9 Time ) ]
:[font = input; preserveAspect; startGroup]
Elevation3 = ArcCos[HorizontalPosition / Distance]
:[font = output; output; inactive; preserveAspect; endGroup]
ArcCos[(20*Time)/
(400*Time^2 + (200 - 4.900000000000001*Time^2)^2)^(1/2)]
;[o]
20 Time
ArcCos[------------------------------------]
2 2 2
Sqrt[400 Time + (200 - 4.9 Time ) ]
:[font = text; inactive; preserveAspect]
Here, Mathematica simply returns the formulas, having substituted the formulas for position and distance; had there been specific numerical values assigned to position and distance, it would have returned the numerical value of elevation angle. For example, suppose Time = 3:
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2:2,14,9,Times,0,12,0,0,0;1,14,9,Times,2,12,0,0,0;
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Time = 3; Elevation3
:[font = output; output; inactive; preserveAspect; endGroup]
1.203407596345267
;[o]
1.20341
:[font = text; inactive; preserveAspect]
Which formula is best to use depends on the values of the quantities; note that computing Elevation1 is problematical whenever HorizontalPosition equals zero (Why? How do the Elevation2 and Elevation3 calculations get around the problem?). Mathematica returns the angle in radian measure; if you want the corresponding degree measure, how do you obtain it? Our original definition of BallPosition is expressed in Cartesian coordinates. Alternatively, we could express the position in polar coordinates, as, for example,
;[s]
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:[font = input; preserveAspect; startGroup]
Time =.; AltBallPosition = {Distance, Elevation1}
:[font = output; output; inactive; preserveAspect; endGroup]
{(400*Time^2 + (200 - 4.900000000000001*Time^2)^2)^(1/2),
ArcTan[(200 - 4.900000000000001*Time^2)/(20*Time)]}
;[o]
2
2 2 2 200 - 4.9 Time
{Sqrt[400 Time + (200 - 4.9 Time ) ], ArcTan[---------------]}
20 Time
:[font = text; inactive; preserveAspect]
Similarly, we can express the velocity and acceleration vectors in polar coordinates. The ball's initial speed is 20 m/s, and its flight path angle is zero (deg or rad). More generally,
:[font = input; preserveAspect; startGroup]
Speed = Sqrt[HorizontalVelocity^2 + VerticalVelocity^2]
:[font = output; output; inactive; preserveAspect; endGroup]
(400 + 96.04*Time^2)^(1/2)
;[o]
2
Sqrt[400 + 96.04 Time ]
:[font = input; preserveAspect; startGroup]
FlightPathAngle = ArcSin[VerticalVelocity / Speed]
:[font = output; output; inactive; preserveAspect; endGroup]
-ArcSin[(9.8*Time)/(400 + 96.04*Time^2)^(1/2)]
;[o]
9.8 Time
-ArcSin[-----------------------]
2
Sqrt[400 + 96.04 Time ]
:[font = input; preserveAspect; startGroup]
AltVelocity = {Speed, FlightPathAngle}
:[font = output; output; inactive; preserveAspect; endGroup]
{(400 + 96.04*Time^2)^(1/2),
-ArcSin[(9.8*Time)/(400 + 96.04*Time^2)^(1/2)]}
;[o]
2 9.8 Time
{Sqrt[400 + 96.04 Time ], -ArcSin[-----------------------]}
2
Sqrt[400 + 96.04 Time ]
:[font = text; inactive; preserveAspect; endGroup; endGroup]
With the flat-earth assumption, gravitational acceleration is unchanging and always pointed down; hence, we know that its polar-coordinate form (magnitude, angle) is {9.8, - p/2}. You might find it interesting to use Mathematica to plot some of these quantities, revealing how they evolve over time. In all of the above, it is sufficient for you to remember the input definition (which is simple) and to let Mathematica work with the output (which may be complicated).
;[s]
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:[font = section; inactive; preserveAspect; startGroup]
Newton's Laws of Motion
:[font = text; inactive; preserveAspect]
First Law - If no force acts on a particle, it remains at rest or continues to move in a straight line at constant velocity.
Second Law - A particle acted upon by a force moves with an acceleration proportional to and in the direction of the force; the ratio of force to acceleration is constant for any particle.
Third Law - For every action, there is an equal and opposite reaction.
Law of Gravitation - Two particles attract each other with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them.
;[s]
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Weight, Mass, and Inertia
:[font = text; inactive; preserveAspect]
An object possesses mass, volume, and density, all of which are scalar quantities. The density is the concentration of mass at a point; integrating the density over the volume of the object produces the mass. Alternatively, we can think of the object as being composed of individual particles, each with its own mass, the constant referred to in Newton's second law. The mass of the object is then the sum of the masses of the particles. A cubic centimeter of water has a mass of one gram and a density of one gram per cubic centimeter. The position of the object can be characterized by the position of its center of mass, a point about which the object is "balanced," and its shape is unimportant. To study the orbits and trajectories of spherical objects like the Earth or the Moon, we can treat each as a single particle located at the object's center of mass, ignoring the more complicated effects of component particles affecting each other. For "small" objects like spacecraft, the effect of the object's shape on the trajectory is inconsequential. We might suppose that the mass of the baseball is about 1/10 kilogram (kg):
;[s]
13:0,0;20,1;24,0;26,1;32,0;38,1;45,0;545,1;553,0;612,1;626,0;682,1;687,0;1139,-1;
2:7,14,9,Times,0,12,0,0,0;6,14,9,Times,2,12,0,0,0;
:[font = input; preserveAspect]
BallMass = 0.1;
:[font = text; inactive; preserveAspect]
On the other hand, if we want to know about the orientation of the object and the distributed effects of forces on many particles, we must consider the object's shape and the effect that it has on the distribution of mass within the object. The distribution of mass produces a quantity called inertia, and it must be represented not by a scalar, not by a vector, but by a (3 x 3) inertia matrix. There are three moments of inertia (Ixx, Iyy, Izz), which represent the object's primary rotational properties about the three axes of the coordinate frame used to define its shape, and six products of inertia (Ixy, Ixz, Iyx, Iyz, Izx, Izy), which represent the coupling of rotational effects between axes. In the worst case, there are only three independent products of inertia, i.e., Ixy = Iyx and so on, and the inertia matrix is symmetric. By convention, the products of inertia are defined so that their negatives appear in the matrix:
;[s]
33:0,0;294,1;301,0;381,1;395,0;414,1;432,0;435,2;437,0;440,2;442,0;445,2;447,0;588,1;607,0;610,2;612,0;615,2;617,0;620,2;622,0;625,2;627,0;630,2;632,0;635,2;637,0;786,2;788,0;792,2;794,0;832,1;841,0;941,-1;
3:17,14,9,Times,0,12,0,0,0;5,14,9,Times,2,12,0,0,0;11,22,13,Times,64,12,0,0,0;
:[font = input; preserveAspect]
InertiaMatrix = {{Ixx, -Ixy, -Ixz}, {-Ixy, Iyy, -Iyz}, {-Ixz, -Iyz, Izz}};
:[font = input; preserveAspect; startGroup]
MatrixForm[%]
:[font = output; output; inactive; preserveAspect; endGroup]
MatrixForm[{{Ixx, -Ixy, -Ixz}, {-Ixy, Iyy, -Iyz},
{-Ixz, -Iyz, Izz}}]
;[o]
Ixx -Ixy -Ixz
-Ixy Iyy -Iyz
-Ixz -Iyz Izz
:[font = text; inactive; preserveAspect]
Moments of inertia are always positive, while products of inertia may have either sign. If the products of inertia are zero, then the (x, y, z) coordinates are called principal axes, and the inertia matrix is diagonal. because all of the off-diagonal elements are zero. The perfect baseball (forget the seams!) has spherical symmetry, so not only is its inertia matrix diagonal, all three moments of inertia are the same. In such case, we could write the inertia matrix as an identity matrix (a diagonal matrix of ones) multiplied by a scalar constant representing the common moment of inertia. By comparison, a football, a sounding rocket, or a compact disc has axial symmetry; thus, two moments of inertia are the same, and one is different. The moment of inertia of a sphere is (2/3) mr2, so the moment of inertia of the baseball is about (2/3)(0.1)(0.04)2 = 1.067 x 10-4 = 0.0001067 kg-m2; therefore,
;[s]
17:0,0;168,1;182,0;210,1;218,0;479,1;494,0;667,1;681,0;794,2;795,0;864,2;865,0;878,2;880,0;897,2;898,0;912,-1;
3:9,14,9,Times,0,12,0,0,0;4,14,9,Times,2,12,0,0,0;4,22,13,Times,32,12,0,0,0;
:[font = input; preserveAspect; startGroup]
BallInertia = 0.0001067 IdentityMatrix[3]
:[font = output; output; inactive; preserveAspect; endGroup]
{{0.0001067, 0, 0}, {0, 0.0001067, 0}, {0, 0, 0.0001067}}
;[o]
{{0.0001067, 0, 0}, {0, 0.0001067, 0}, {0, 0, 0.0001067}}
:[font = input; preserveAspect; startGroup]
MatrixForm[%]
:[font = output; output; inactive; preserveAspect; endGroup; endGroup]
MatrixForm[{{0.0001067, 0, 0}, {0, 0.0001067, 0},
{0, 0, 0.0001067}}]
;[o]
0.0001067 0 0
0 0.0001067 0
0 0 0.0001067
:[font = subsection; inactive; preserveAspect; startGroup]
Momentum
:[font = text; inactive; preserveAspect]
The linear momentum of an object is simply its mass times its velocity, and the baseball's momentum (expressed in 2 dimensions) is
;[s]
3:0,0;3,1;19,0;131,-1;
2:2,14,9,Times,0,12,0,0,0;1,14,9,Times,2,12,0,0,0;
:[font = input; preserveAspect; startGroup]
LinearMomentum = BallMass BallVelocity
:[font = output; output; inactive; preserveAspect; endGroup]
{2., -0.98*Time}
;[o]
{2., -0.98 Time}
:[font = text; inactive; preserveAspect]
The velocity is a vector, as is the linear momentum. If the baseball is rotating at 50 rad/s about its lateral (y) axis, its angular momentum, measured with respect to the rotational center defined by its center of mass, is
;[s]
3:0,0;173,1;190,0;225,-1;
2:2,14,9,Times,0,12,0,0,0;1,14,9,Times,2,12,0,0,0;
:[font = input; preserveAspect]
AngularVelocity = {0, 50, 0};
:[font = input; preserveAspect; startGroup]
AngularMomentum = BallInertia . AngularVelocity
:[font = output; output; inactive; preserveAspect; endGroup]
{0., 0.005335, 0.}
;[o]
{0., 0.005335, 0.}
:[font = text; inactive; preserveAspect]
The angular momentum also is a vector, the product of the (3 x 3) inertia matrix and the (3 x 1) angular velocity vector.
The concepts of momentum are relative because velocity (linear or angular) must be measured with respect to some frame of reference, and the reference frame itself could be moving. We can, for example, also think of the angular momentum of a non-rotating baseball about some rotational center other than its center of mass, such as the base of the cliff. In such case, the angular momentum is the cross-product of the radius vector (from the origin to the baseball's center of mass) and its angular momentum. Here, we have to recognize that the world really is 3-D; calling our original axes x and z,
;[s]
3:0,0;524,1;537,0;729,-1;
2:2,14,9,Times,0,12,0,0,0;1,14,9,Times,2,12,0,0,0;
:[font = input; preserveAspect]
Position3D = {HorizontalPosition, 0, VerticalPosition};
:[font = input; preserveAspect]
Momentum3D = {2, 0, -0.98 Time};
:[font = input; preserveAspect; startGroup]
AngularMomentum = CrossProduct[Position3D, Momentum3D]
:[font = output; output; inactive; preserveAspect; endGroup]
{0, 400 + 9.8*Time^2, 0}
;[o]
2
{0, 400 + 9.8 Time , 0}
:[font = text; inactive; preserveAspect; endGroup]
Only the middle (y) term is non-zero, because the axis of rotation is "out of" (i.e., perpendicular to) the vertical plane of motion.
:[font = subsection; inactive; preserveAspect; startGroup]
Changing Momentum
:[font = text; inactive; preserveAspect]
Forces change linear momentum, and moments (or torques), which are forces applied perpendicular to a moment arm (i.e., a radius from some point of reference like the center of mass) change angular momentum. Both statements are consequences of Newton's Second Law. A 3-D force (Fx, Fy, Fz) on the baseball produces a 3-D linear acceleration:
;[s]
14:0,1;6,0;35,1;42,0;47,1;54,0;101,1;111,0;280,2;281,0;284,2;285,0;288,2;289,0;343,-1;
3:7,14,9,Times,0,12,0,0,0;4,14,9,Times,2,12,0,0,0;3,22,13,Times,64,12,0,0,0;
:[font = input; preserveAspect; startGroup]
LinearAcceleration3D = {Fx, Fy, Fz} / BallMass
:[font = output; output; inactive; preserveAspect; endGroup]
{10.*Fx, 10.*Fy, 10.*Fz}
;[o]
{10. Fx, 10. Fy, 10. Fz}
:[font = text; inactive; preserveAspect]
A 3-D moment (Mx, My, Mz) on the baseball produces a 3-D rotational acceleration:
;[s]
7:0,0;15,1;16,0;19,1;20,0;23,1;24,0;82,-1;
2:4,14,9,Times,0,12,0,0,0;3,22,13,Times,64,12,0,0,0;
:[font = input; preserveAspect; startGroup]
RotationalAcceleration3D = Inverse[BallInertia] . {Mx, My, Mz}
:[font = output; output; inactive; preserveAspect; endGroup]
{9372.07122774133*Mx + 0.*My + 0.*Mz,
0.*Mx + 9372.07122774133*My + 0.*Mz,
0.*Mx + 0.*My + 9372.07122774133*Mz}
;[o]
{9372.07 Mx + 0. My + 0. Mz, 0. Mx + 9372.07 My + 0. Mz,
0. Mx + 0. My + 9372.07 Mz}
:[font = text; inactive; preserveAspect; endGroup]
This latter results reveals one of my pet peeves about Mathematica: it doesn't know that the symbolic dot product of zero with anything else is zero. There is an involved way to teach it this rule each time you run into the problem, but it is not worth the trouble. The result shown above should simply be {9372.07 Mx, 9372.07 My, 9372.07 Mz}
;[s]
9:0,0;55,1;66,0;318,2;319,0;330,2;331,0;342,2;343,0;345,-1;
3:5,14,9,Times,0,12,0,0,0;1,14,9,Times,2,12,0,0,0;3,22,13,Times,64,12,0,0,0;
:[font = subsection; inactive; preserveAspect; startGroup]
Action and Reaction
:[font = text; inactive; preserveAspect; endGroup]
The Earth exerts a force on the Moon, and the Moon exerts an equal but opposite force on Earth. The accelerations produced by these equal but opposite vector forces
(Fx, Fy, Fz)Earth = - (Fx, Fy, Fz)Moon
are different because the masses of the two bodies are different.
;[s]
17:0,0;169,1;170,0;173,1;174,0;177,1;178,0;179,1;184,0;191,1;192,0;195,1;196,0;199,1;200,0;201,1;205,0;274,-1;
2:9,14,9,Times,0,12,0,0,0;8,22,13,Times,64,12,0,0,0;
:[font = subsection; inactive; preserveAspect; startGroup]
Gravity
:[font = text; inactive; preserveAspect]
The forces that the Earth and the Moon exert on each other follow Newton's Law of Gravitation. Alas, the flat-earth model was only an approximation! The force magnitude is expressed as Gm1m2/r2, where G is the universal gravitation constant, m1 and m2 are the masses of the Earth and Moon, and r is the distance between them. The force magnitude at their average distance is about 2 x 1020 Newtons, and the forces are aligned with the radius vector between the Earth and the Moon. The specific forces (forces per unit of mass, or the accelerations in the absence of other effects) in this condition are
;[s]
15:0,0;189,2;190,0;191,2;192,0;194,1;195,0;245,2;246,0;252,2;253,0;391,1;393,0;490,3;505,0;608,-1;
4:8,14,9,Times,0,12,0,0,0;2,22,13,Times,32,12,0,0,0;4,22,13,Times,64,12,0,0,0;1,14,9,Times,2,12,0,0,0;
:[font = input; preserveAspect; startGroup]
EarthSpecificForce = (1.98 10^20) / (5.98 10^24)
:[font = output; output; inactive; preserveAspect; endGroup]
0.00003311036789297659
;[o]
0.0000331104
:[font = input; preserveAspect; startGroup]
MoonSpecificForce = (1.98 10^20) / (7.35 10^22)
:[font = output; output; inactive; preserveAspect; endGroup]
0.002693877551020408
;[o]
0.00269388
:[font = text; inactive; preserveAspect; endGroup; endGroup]
More generally, any two bodies exert gravitational forces on each other according to this law. Thus, if we were to consider the Sun as well as the Earth and the Moon, there would be three forces, each proportional to a pair of masses and the inverse-square of the distance between the mass centers, and each directed along the radius separating the pair.
:[font = section; inactive; preserveAspect; startGroup]
Laws of Conservation
:[font = text; inactive; preserveAspect]
Here, we focus on a pair of bodies acted upon only by their mutual gravitational effects. It is a two-particle system, and there are no external forces acting on this two-particle system.
:[font = subsection; inactive; preserveAspect; startGroup]
Momentum
:[font = text; inactive; preserveAspect; endGroup]
In the absence of external forces, the momentum of the system, that is, the sum of the momenta of the individual particles, does not change. There can be changes in the momentum of one particle, but they are precisely balanced by the momentum changes in the other. Remember that momentum (either linear or rotational) is a vector with a three components.
:[font = subsection; inactive; preserveAspect; startGroup]
Energy
:[font = text; inactive; preserveAspect]
Energy is a scalar property that reflects the activity or the potential for activity in a system. Before it is thrown, the baseball has potential energy because it is on a 200-m-high cliff. As it is thrown, kinetic energy is imparted to it. The total mechanical energy of the ball is the sum of the potential and kinetic energies, and unless external forces are applied, the total energy of the system is constant.
Potential energy can be considered to be the work done in moving an object over some distance, which is equal to the integral of the force required over the distance travelled. Given Newton's Laws, the force must act against the gravity, so the energy expended in moving m1 under the influence of m2 is
;[s]
11:0,0;137,1;153,0;209,1;223,0;464,1;468,0;692,2;693,0;718,2;719,0;723,-1;
3:6,14,9,Times,0,12,0,0,0;3,14,9,Times,2,12,0,0,0;2,22,13,Times,64,12,0,0,0;
:[font = input; preserveAspect; startGroup]
PotentialEnergy = Integrate[Gm1m2 / Radius^2, Radius]
:[font = output; output; inactive; preserveAspect; endGroup]
-(Gm1m2/Radius)
;[o]
Gm1m2
-(------)
Radius
:[font = text; inactive; preserveAspect]
(Recall that the first term in the square brackets of the Mathematica function Integrate[¥] is the integrand, and the term following the comma is the variable over which that quantity is integrated.) Kinetic energy can be viewed as the integral of linear momentum with respect to velocity (the precise definition is a little more complicated). For m1,
;[s]
5:0,0;58,2;69,0;351,1;352,0;354,-1;
3:3,14,9,Times,0,12,0,0,0;1,22,13,Times,64,12,0,0,0;1,14,9,Times,2,12,0,0,0;
:[font = input; preserveAspect; startGroup]
KineticEnergy = Integrate[m1 Velocity, Velocity]
:[font = output; output; inactive; preserveAspect; endGroup]
(m1*Velocity^2)/2
;[o]
2
m1 Velocity
------------
2
:[font = text; inactive; preserveAspect; endGroup; endGroup]
and there is a similar relation for m2. We could, of course, divide both of m1's energies by m1 to define energy per unit of mass, or specific energy.
;[s]
9:0,0;37,1;38,0;78,1;79,0;95,1;96,0;135,2;150,0;152,-1;
3:5,14,9,Times,0,12,0,0,0;3,22,13,Times,64,12,0,0,0;1,14,9,Times,2,12,0,0,0;
:[font = section; inactive; preserveAspect; startGroup]
The Restricted Two-Body Problem
:[font = text; inactive; preserveAspect]
We will follow Understanding Space's simplifying assumptions in arriving at the equations for the restricted 2-body problem; however, you can find the same results with less-restrictive assumptions elsewhere, as in Modern Spacecraft Dynamics & Control, M. Kaplan, J. Wiley & Sons, New York, 1976.
;[s]
5:0,0;15,1;34,0;215,1;251,0;297,-1;
2:3,14,9,Times,0,12,0,0,0;2,14,9,Times,2,12,0,0,0;
:[font = subsection; inactive; preserveAspect; startGroup]
Coordinate Systems
:[font = text; inactive; preserveAspect; endGroup]
An inertial frame of reference (or Newtonian frame) is "unaccelerated," that is, it is moving at constant velocity (or not moving at all) with respect to the universe, and it is not rotating. Of course, everything IS moving with respect to something else, so approximations are in order. In the current situation, it is acceptable to assume that the center of mass (or barycenter) of the 2-body system is not moving, and if one body is much larger than the other, the barycenter is essentially the same as the center of mass of the larger body. The reference (x, y, z) axes are orthogonal to each other and are not rotating. To study Earth satellites, we can choose the North-South polar axis as our z axis, with x and y in the plane of the equator (the Fundamental Plane). We cannot fix x and y to specific longitudes because the Earth itself is rotating once per day. Instead, we choose theVernal Equinox Direction for our x axis.
;[s]
9:0,0;371,1;381,0;674,1;696,0;758,1;775,0;898,1;922,0;939,-1;
2:5,14,9,Times,0,12,0,0,0;4,14,9,Times,2,12,0,0,0;
:[font = subsection; inactive; preserveAspect; startGroup]
Equation of Motion
:[font = text; inactive; preserveAspect; endGroup]
Newton's Second Law provides the fundamental force-acceleration balance, and in the real world, there are other forces besides gravity to consider. Integrating this dynamic vector equation provides solutions for the three linear velocity components. We need another three kinematic equations to solve for the three position components. In the restricted 2-body problem, we can use the conservation of momentum and energy to provide analytical solutions to these equations rather than having to resort to numerical integration.
;[s]
5:0,0;166,1;173,0;274,1;283,0;530,-1;
2:3,14,9,Times,0,12,0,0,0;2,14,9,Times,2,12,0,0,0;
:[font = subsection; inactive; preserveAspect; startGroup]
Simplifying Assumptions
:[font = text; inactive; preserveAspect]
Within the assumptions, the restricted 2-body equation of motion describes the accelerations on the smaller body, the larger body being fixed in space. The gravitational constant m is Gm2, where m2 is the mass of the larger body, and Rhat is the unit vector in the direction of R. This unit vector may be a little confusing; let's define the radius vector as
;[s]
7:0,0;180,1;181,0;187,2;188,0;197,2;198,0;361,-1;
3:4,14,9,Times,0,12,0,0,0;1,19,13,Symbol,0,12,0,0,0;2,22,13,Times,64,12,0,0,0;
:[font = input; preserveAspect; startGroup]
RadiusVector = {x, y, z}
:[font = output; output; inactive; preserveAspect; endGroup]
{x, y, z}
;[o]
{x, y, z}
:[font = text; inactive; preserveAspect]
Then the unit vector is
:[font = input; preserveAspect; startGroup]
Rhat = RadiusVector / Radius
:[font = output; output; inactive; preserveAspect; endGroup]
{x/Radius, y/Radius, z/Radius}
;[o]
x y z
{------, ------, ------}
Radius Radius Radius
:[font = text; inactive; preserveAspect]
where the radius magnitude is
:[font = input; preserveAspect; startGroup]
Radius = Sqrt[RadiusVector . RadiusVector]
:[font = output; output; inactive; preserveAspect; endGroup]
(x^2 + y^2 + z^2)^(1/2)
;[o]
2 2 2
Sqrt[x + y + z ]
:[font = text; inactive; preserveAspect]
Consequently, the gravity specific force vector (i.e., divided by m1) is given by
;[s]
3:0,0;67,1;68,0;82,-1;
2:2,14,9,Times,0,12,0,0,0;1,22,13,Times,64,12,0,0,0;
:[font = input; preserveAspect; startGroup]
Radius =.;
GravityVector = (Mu / Radius^2) Rhat
:[font = output; output; inactive; preserveAspect; endGroup]
{(Mu*x)/Radius^3, (Mu*y)/Radius^3, (Mu*z)/Radius^3}
;[o]
Mu x Mu y Mu z
{-------, -------, -------}
3 3 3
Radius Radius Radius
:[font = text; inactive; preserveAspect]
The restricted 2-body dynamic equation could be written (defining second derivatives of the radius-vector components as xdd, ydd, and zdd) as
:[font = input; preserveAspect; startGroup]
{xdd, ydd, zdd} = -GravityVector
:[font = output; output; inactive; preserveAspect; endGroup]
{-((Mu*x)/Radius^3), -((Mu*y)/Radius^3), -((Mu*z)/Radius^3)}
;[o]
Mu x Mu y Mu z
{-(-------), -(-------), -(-------)}
3 3 3
Radius Radius Radius
:[font = text; inactive; preserveAspect]
and we would integrate this once again to calculate the three components of position. The conic-section radius magnitude takes the general form
:[font = input; preserveAspect; startGroup]
OrbitalRadius = k1 / (1 + k2 Cos[Nu])
:[font = output; output; inactive; preserveAspect; endGroup; endGroup]
k1/(1 + k2*Cos[Nu])
;[o]
k1
--------------
1 + k2 Cos[Nu]
:[font = subsection; inactive; preserveAspect; startGroup]
Orbit Geometry
:[font = text; inactive; preserveAspect]
Three derivations included in Kaplan's book but skipped here manipulate the restricted 2-body equation to produce the following results:
1) The angular momentum of the system is constant; therefore, the orbit of the smaller body about the larger lies in a plane.
2) There is a constant vector, the eccentricity vector, lying in the orbital plane that defines a reference direction.
3) The radius vector between the bodies traces out a conic section whose characteristics depend on the angular momentum, the magnitude of the eccentricity vector, and m, the gravitational constant. Depending on these parameters, the orbit is either a circle, an ellipse, a parabola, or a hyperbola.
For an elliptical orbit, the points of greatest and least separation of the two bodies are denoted by the prefixes "apo-" and "peri-". If the principal center of attraction is the Earth, the point of closest approach is the perigee; for the Sun, it is the perihelion. The semi-major axis of an Earth orbit is
;[s]
25:0,0;145,2;187,0;299,2;318,0;352,2;381,0;436,2;449,0;550,1;551,0;635,2;641,0;646,2;653,0;657,2;665,0;672,2;681,0;909,2;916,0;941,2;951,0;958,2;973,0;997,-1;
3:13,14,9,Times,0,12,0,0,0;1,19,13,Symbol,0,12,0,0,0;11,14,9,Times,2,12,0,0,0;
:[font = input; preserveAspect; startGroup]
SemiMajorAxis = (ApogeeRadius + PerigeeRadius) / 2
:[font = output; output; inactive; preserveAspect; endGroup]
(ApogeeRadius + PerigeeRadius)/2
;[o]
ApogeeRadius + PerigeeRadius
----------------------------
2
:[font = text; inactive; preserveAspect]
and the orbital radius can be expressed as
;[s]
3:0,0;8,1;22,0;43,-1;
2:2,14,9,Times,0,12,0,0,0;1,14,9,Times,2,12,0,0,0;
:[font = input; preserveAspect; startGroup]
OrbitalRadius =
SemiMajorAxis (1 - Eccentricity^2) / (1 + Eccentricity Cos[TrueAnomaly])
:[font = output; output; inactive; preserveAspect; endGroup]
((1 - Eccentricity^2)*(ApogeeRadius + PerigeeRadius))/
(2*(1 + Eccentricity*Cos[TrueAnomaly]))
;[o]
2
(1 - Eccentricity ) (ApogeeRadius + PerigeeRadius)
--------------------------------------------------
2 (1 + Eccentricity Cos[TrueAnomaly])
:[font = text; inactive; preserveAspect]
The scalar eccentricity can be calculated as
;[s]
3:0,0;11,1;23,0;45,-1;
2:2,14,9,Times,0,12,0,0,0;1,14,9,Times,2,12,0,0,0;
:[font = input; preserveAspect; startGroup]
Eccentricity =
(ApogeeRadius - PerigeeRadius) / (ApogeeRadius + PerigeeRadius)
:[font = output; output; inactive; preserveAspect; endGroup]
(ApogeeRadius - PerigeeRadius)/(ApogeeRadius + PerigeeRadius)
;[o]
ApogeeRadius - PerigeeRadius
----------------------------
ApogeeRadius + PerigeeRadius
:[font = text; inactive; preserveAspect]
and the true anomaly is the angle between the body and perigee in the orbital plane. By way of example, let's plot the orbital radius as a function of the true anomaly (between 0 and p), assuming that the eccentricity is 0.5, the apogee radius is 4 times the Earth's radius, and the perigee radius is two times the Earth's radius:
;[s]
5:0,0;8,1;20,0;184,2;185,0;332,-1;
3:3,14,9,Times,0,12,0,0,0;1,14,9,Times,2,12,0,0,0;1,19,13,Symbol,0,12,0,0,0;
:[font = input; preserveAspect; startGroup]
ApogeeRadius = 4;
PerigeeRadius = 2;
OrbitalRadius
:[font = output; output; inactive; preserveAspect; endGroup]
8/(3*(1 + Cos[TrueAnomaly]/3))
;[o]
8
------------------------
Cos[TrueAnomaly]
3 (1 + ----------------)
3
:[font = input; preserveAspect; startGroup]
Plot[OrbitalRadius, {TrueAnomaly, 0, Pi}]
:[font = postscript; PostScript; formatAsPostScript; output; inactive; preserveAspect; pictureLeft = 34; pictureWidth = 282; pictureHeight = 174]
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[(0.5)] .17539 .01472 0 2 Msboxa
[(1)] .32696 .01472 0 2 Msboxa
[(1.5)] .47854 .01472 0 2 Msboxa
[(2)] .63011 .01472 0 2 Msboxa
[(2.5)] .78169 .01472 0 2 Msboxa
[(3)] .93327 .01472 0 2 Msboxa
[(2.5)] .01131 .16187 1 0 Msboxa
[(3)] .01131 .30902 1 0 Msboxa
[(3.5)] .01131 .45617 1 0 Msboxa
[(4)] .01131 .60332 1 0 Msboxa
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p
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.17539 .01472 m
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P
[(0.5)] .17539 .01472 0 2 Mshowa
p
.002 w
.32696 .01472 m
.32696 .02097 L
s
P
[(1)] .32696 .01472 0 2 Mshowa
p
.002 w
.47854 .01472 m
.47854 .02097 L
s
P
[(1.5)] .47854 .01472 0 2 Mshowa
p
.002 w
.63011 .01472 m
.63011 .02097 L
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P
[(2)] .63011 .01472 0 2 Mshowa
p
.002 w
.78169 .01472 m
.78169 .02097 L
s
P
[(2.5)] .78169 .01472 0 2 Mshowa
p
.002 w
.93327 .01472 m
.93327 .02097 L
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P
[(3)] .93327 .01472 0 2 Mshowa
p
.001 w
.05412 .01472 m
.05412 .01847 L
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p
.001 w
.08444 .01472 m
.08444 .01847 L
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P
p
.001 w
.11476 .01472 m
.11476 .01847 L
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P
p
.001 w
.14507 .01472 m
.14507 .01847 L
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P
p
.001 w
.2057 .01472 m
.2057 .01847 L
s
P
p
.001 w
.23602 .01472 m
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That does not look like an elliptical orbit because Plot[.] simply puts the output on one axis and the input on the other. Incidentally, we could tabulate this result in increments of 0.1 p as an alternative:
;[s]
3:0,0;189,1;190,0;210,-1;
2:2,14,9,Times,0,12,0,0,0;1,19,13,Symbol,0,12,0,0,0;
:[font = input; preserveAspect; startGroup]
N[Table[OrbitalRadius, {TrueAnomaly, 0, Pi, 0.1 Pi}]]
:[font = output; output; inactive; preserveAspect; endGroup]
{2., 2.024774884136934, 2.100279419024431, 2.229787860042145,
2.417636420000058, 2.666666666666666, 2.972891312682886,
3.316454311376249, 3.651329097241321, 3.904451276291236,
3.999999999999999}
;[o]
{2., 2.02477, 2.10028, 2.22979, 2.41764, 2.66667, 2.97289,
3.31645, 3.65133, 3.90445, 4.}
:[font = text; inactive; preserveAspect]
To see an elliptical orbit through an entire revolution, we plot the Cartesian components of the radius and true anomaly:
:[font = input; preserveAspect; startGroup]
xRadius = OrbitalRadius Cos[TrueAnomaly];
yRadius = OrbitalRadius Sin[TrueAnomaly];
ParametricPlot[{xRadius, yRadius}, {TrueAnomaly, 0, 2 Pi}]
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Graphics["<<>>"]
;[o]
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:[font = section; inactive; preserveAspect; startGroup]
Constants of Orbital Motion
:[font = subsection; inactive; preserveAspect; startGroup]
Specific Mechanical Energy
:[font = text; inactive; preserveAspect]
The prior section showed us how to characterize the shapes of orbits. Using the fact that total specific energy is constant, we can compute the velocity at any point on the orbit as
;[s]
3:0,0;91,1;124,0;183,-1;
2:2,14,9,Times,0,12,0,0,0;1,14,9,Times,2,12,0,0,0;
:[font = input; preserveAspect; startGroup]
Velocity = Sqrt[2 (Mu / Radius + SpecificEnergy)]
:[font = output; output; inactive; preserveAspect; endGroup]
2^(1/2)*(Mu/Radius + SpecificEnergy)^(1/2)
;[o]
Mu
Sqrt[2] Sqrt[------ + SpecificEnergy]
Radius
:[font = text; inactive; preserveAspect]
The specific energy is
:[font = input; preserveAspect]
SemiMajorAxis =.;
SpecificEnergy = -Mu / (2 SemiMajorAxis)
:[font = text; inactive; preserveAspect]
and the period of the orbit is found to follow Kepler's Laws:
:[font = input; preserveAspect; startGroup]
Period = 2 Pi Sqrt[SemiMajorAxis^3 / Mu]
:[font = output; output; inactive; preserveAspect; endGroup]
2*Pi*(SemiMajorAxis^3/Mu)^(1/2)
;[o]
3
SemiMajorAxis
2 Pi Sqrt[--------------]
Mu
:[font = text; inactive; preserveAspect; endGroup]
How might you plot the change in velocity over an entire orbit? How would you plot the orbital period as a function of the semi-major axis?
========================================================
Kepler's Laws
First Law - The orbits of the planets are ellipses with the Sun at one focus.
Second Law - The line joining a planet to the Sun sweeps out equal areas in equal times.
Third Law - The square of the orbital period is directly proportional to the cube of the mean distance between the Sun and the planet.
================================================================
;[s]
15:0,0;140,1;221,0;254,2;262,0;291,1;301,0;352,2;378,0;381,1;390,0;397,2;426,0;458,2;514,0;581,-1;
3:8,14,9,Times,0,12,0,0,0;3,14,9,Times,1,12,0,0,0;4,14,9,Times,2,12,0,0,0;
:[font = subsection; inactive; preserveAspect; startGroup]
Specific Angular Momentum
:[font = text; inactive; preserveAspect; endGroup; endGroup; endGroup]
The specific angular momentum magnitude squared, h2, is seen to equal ma(1 - e2).
;[s]
9:0,0;4,1;47,0;50,3;51,0;70,2;71,0;78,3;79,0;82,-1;
4:5,14,9,Times,0,12,0,0,0;1,14,9,Times,2,12,0,0,0;1,19,13,Symbol,0,12,0,0,0;2,22,13,Times,32,12,0,0,0;
^*)