Thursday, October 09, 2008

Compute Orbits with Radius Vectors

Determine Elliptic Equation

Given two arbitrary radius vectors from elliptic center, determine equation for elliptic orbit.

0
semimajor axis, a >R1 >R2>b , semiminor axis

<θ1<θ2 <

90°

Angle Y=mXX-valueY-value
θSlope (m)r*Cosθr*Sinθ
(Deg)TanθCosθ Sinθ Tan2θCos2θSin2θ
010010
30° 1/√3√3 / 21/21/33/41/4
45° 1√2 / 2 √2 / 2 11/21/2
60° √31/2√3 / 2 31/43/4
90° 0101
010010
26.565°1 / 22/√51/√51/44/51/5
45° 1√2 / 2√2 / 211/21/2
63.435°2 1/√52/√541/54/5
90°0101
Examples of convenient trigonometric ratios.
x2
a2
+
y2
b2
=
1
Ellipse Equation
x2
a2
+
m2x2
b2
=
1
Recall we're looking for intersection of line and ellipse. Thus, substitute line equation (y=mx, slope times x) into term containing "y".
x2* b2
+
m2x2 * a2
=
a2b2
Solve for a
-a2 b2
+
a2m2x2
=
-x2b2
a2 * (-b2
+
m2x2)
=
-x2b2
a2 * (b2
-
m2x2)
=
x2b2
Solve for b
-a2 b2
+
b2x2
=
-m2x2a2
b2 * (-a2
+
x2)
=
-m2x2a2
b2 * (a2
-
x2)
=
m2x2a2
a2 
=
x2b2
(b2 - m2x2)
 
b2 
=
m2x2a2
(a2 - x2)
 
a 
=
xb
√(b2 - m2x2)
 
b 
=
mxa
√(a2 - x2)
 
Assume 1st vector, (r1, θ1):
a 
=
r1cos(θ1)b
√(b2 - tan21)r12cos21))
 
Assume 2nd vector, (r2, θ2):
b 
=
tan(θ2)r2cos(θ2)a
√(a2 - r22cos22))
 
Make following substitutions:
m = tan(θ)
x = rcos(θ)
a 
=
b * r1 * cos(θ1)
√(b2 - r12sin21))
 
b 
=
a * r2 * sin(θ2)
√(a2 - r22cos22))
 

Note following trig identity:
tan(θ) = sin(θ) / cos(θ)
Therefore,
tan(θ) cos(θ) = sin(θ)

a2 
=
b2 * r12 * cos21)
b2 - r12sin21)
 
b2 
=
a2 * r22 * sin22)
a2 - r22cos22)
 

For convenience, use following substitutions:

x1 = r1cos(θ1)

x12= r12cos21)

y1 = r1sin(θ1)

y1 2= r12sin21)

x2 = r2cos(θ2)

x22= r22cos22)

y2 = r2sin(θ2)

y22= r22sin22)

b2 * x12
b2 - y12
 = a2 = 
b2 * x22
b2 - y22
 
 
 
a2 * y12 
a2 - x12
= b2 =
a2 * y22 
a2 - x22
 
Since both radius vectors cross same ellipse, they can be used interchangeably as shown.
Transitive identity property allow two outside expressions to equal b2 as well as each other.
x12
b2 - y12
 = 
 x22
b2 - y22
 
 
 
y12 
a2 - x12
= 
 y22 
a2 - x22
 
Deleting common term from both sides, above equations can be written as shown.
x12(b2 - y22)
 = 
 x22(b2 - y12)
x12b2 - x12y22
 = 
 x22b2 - x22y12
x12b2 - x22b2
 = 
 x12y22 - x22y12
b2(x12 - x22)
 = 
 x12y22 - x22y12
 
 
y12 (a2 - x22)
 = 
y22(a2 - x12)
 
 
 
y12a2 - y12x22
 = 
y22a2 - y22x12
 
 
 
y12a2 - y22a2
 = 
y12x22 - y22x12
 
 
 
a2(y12 - y22)
 = 
y12x22 - y22x12
 
Rearrange as shown.
b2 
=
x12y22 - x22y12
x12 - x22
a2 
=
y12x22 - y22x12
y12 - y22

FINALLY!!! We've isolated terms a2 and b2.
We can now substitute arbitrary values as shown in next row.

b2 
=
0.54 AU2AU2 - 0.18AU2AU2
0.333 AU2
b 
=
1.08 AU2
=
1.039 AU
a2 
=
(0.18AU2 - 0.54 AU2)AU2
-0.107 AU2
a 
=
3.65 AU2
=
1.834 AU

Arbitrary Example:
Recall restriction

r1 = 1.2 AU

θ1 = 30°

r2 = 1.0 AU

θ2=45°

Recall substitutions

x1 =1.04AU

y1=.600AU

x2= .707AU

y2=.707AU

x12=1.08AU2

y12=.36AU2

x22=.5AU2

y22=.5AU2

x12y22 = 0.54 AU2AU2

x22y12= 0.18 AU2AU2

x12 - x22 = 0.333 AU2

y12 - y22= -0.107 AU2

For this entirely arbitrary example of two radius vectors from elliptic center to elliptic perimeter, we have calculated values for semimajor axis, a, and semiminor axis, b, to determine elliptic equation.

x2
1.834AU2
+
y2
1.08AU2
=
1

These vectors were limited only by convenience and a few common sense restrictions.

We can use now use values for a and b to calculate:

  • focus [c = √(a2 - b2) = √(3.65 - 1.08) = 1.6 AU]
  • eccentricity [e = c/a = 1.6 / 1.834 = 0.872]







Problem: Can't Count on Convenient Vectors

Previous table used "arbitrary" vectors deliberately chosen for angles with convenient trig ratios

This table assumes that choosing random values from a useable range would be much more realistic of typical observed values.

Thus, this table takes two random radius vectors from an elliptic orbit and determines equation for elliptic orbit.

0
a >R1 >R2>b

<θ1<θ2 <

90°

Example of two random vectors:

r1 = 1.45 AU

θ1 = 13.5°

r2 = 1.424 AU

θ2=21°

x1 = r1cosθ1

y1= r1sinθ1

x2= r2cosθ2

y2= r2sinθ2

x1 =1.41AU

y1=0.34AU

x2= 1.33AU

y2=0.51AU

x12=1.99AU2

y12= .11AU2

x22=1.77AU2

y22= .26AU2

x12y22 = 0.52 AU2AU2

x22y12= 0.20 AU2AU2

x12 - x22 = 0.22 AU2

y12 - y22= -0.15 AU2

b2
=
x12y22 - x22y12
x12 - x22

a2 
=
x22y12 - x12y22
y12 - y22

b2
=
0.52AU2AU2 - 0.20AU2AU2
0.22 AU2

a2
=
0.20AU2AU2 - 0.52AU2AU2
-0.15AU2

b = √(2.364AU2 - 0.909AU2)

a = √[(-1.333+ 3.47)AU2]

b = √1.455AU2

a = √2.16 AU2

b = 1.206 AU

a = 1.47 AU

x2
a2
+
y2
b2
 =
1 =
x2
(1.47 AU)2
+
y2
(1.20 AU)2
 
 
Ellipse Equation
√[ a2
 -
b2
] =
c = √[ (1.47 AU)2
 -
(1.20 AU)2
] =
0.86 AU
Determine focus.
Q = a + c = 2.33 AU and q = a - c = 0.61 AUDetermine aphelion, Q, and perihelion, q.
e = (Q - q) / (Q + q) = 1.72/2.94 = 0.58 AUDetermine eccentricity, e.
l = 2qQ/(Q + q) = 2.84/2.94 = .97 AUDetermine semilatus rectum, l.
R(ν) =
l
1 + e * Cos(ν)
Familiar Polar Equation for ellipses. Recall the independent variable, ν, is the angle of the radius vector from the relevant focus (Sol, our sun) to the object's position in the orbit. This differs from above angle, θ, angle from orbit's center.





Problem: Orbits Don't Share Common Center

By definition all Solar objects orbit Sol, our sun. Thus, all ecliptic orbits share the Sun as a common focus.

Thus, it makes a lot of sense to analyze R vectors from Sun and not the Ecliptic center, because every orbit's ecliptic centers will vary widely, but the common focus will be very close to same position. Therefore, this table will examine two random positional vectors from Sol.

q

<R1 <R2<l

<ν1<ν2 <

90°

For discussion purposes, restrict current considerations to vectors in above ranges. However, following methods should work with vectors from any quadrant in an orbit.

Example of two random vectors follow:
(data for Apollo orbit from
JPL Horizon's web site.)

r1 = 0.694 AU

ν1 = 36°

r2 = 0.847 AU

ν2=70°

x1 = r1cosν1

x1= 0.561 AU

x2= r2cosν2

x2= 0.290 AU

R(ν) =
l
1 + e * Cos(ν)

Polar Equation for Ellipse:
Generates vectors from focus.

l =
R(1 + e * Cos(ν))

Rearrange to solve for l, semilatus rectum.

 
R1(1 + e * Cos(ν1))
= l =
R2(1 + e * Cos(ν2))
 
R1 + R1*e * Cos(ν1))
= l =
R2+ R2*e * Cos(ν2))

Since both radius vectors belong to same ellipse, they can be used independently to compute same l value as shown. Transitive identity property allows two outside expressions to equal l as well as each other.

R1 + e * x1
=
R2+ e * x2
e (x1 - x2)
=
R2 - R1

Let x1=R1*cos(ν1)
Let x2=R2*cos(ν2)

Group like terms.

e
=
R2 - R1
x1 - x2

Solve for e, eccentricity.

e = 0.565
e
=
(0.847 - 0.694) AU
(0.561 - 0.290) AU

Substitute values and determine e.

R1 + e * x1 = l = R2+ e * x2
0.694+0.565*0.561=l=0.847+0.565*0.290
1.01 = l = 1.01

Solve for l in both vectors independently.

R(0°) =
l
1+e*Cos(0°)
 =
1.01 AU
1+0.565*1.0
=
0.645 AU = q
By definition, perihelion is determined at ν = 0°.
R(180°) =
l
1+e*Cos(180°)
 =
1.01 AU
1+0.565*(-1.0)
=
2.322 AU = Q
By definition, aphelion is determined at ν = 180°.
c = (Q - q) / 2 = (2.322-0.645)AU/2 = 0.8385 AUDetermine focus.
a = Q - c = 2.322 AU - 0.8385 AU = 1.4835 AUDetermine a, semimajor axis.
b = (a2-c2) = (1.4835 AU2-0.8385 AU2) = 1.224 AUDetermine b, semiminor axis.




Problem: Observations are from Earth, not Sun. This, Earthly vectors must be translated to Solar vectors. This solution could use special case of midnight observations at zenith from Equator.

Problem, many problems when observing from Earth, atmospheric refraction, diff lat/lon. If we choose to pursue this, we'd need to translate all observations to special case of midnight observations at zenith from Equator.

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