MontBlanc2012
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This is a follow on from dgatroulis' post Elliptical trajectory from 3 points and focus?.
In effect, dgatsoulis posed the question: what is the elliptical arc that connects two points separated by an angle [MATH]\theta[/MATH] for a known apoapsis radius? A number of solutions were given in comments.
This note examines a closely related question: what are the Keplerian arcs that connects two points separated by an angle [MATH]\theta[/MATH] for a known periapsis radius? An example would be: starting from an orbital radius of 30,000 km from Earth what is the Keplerian trajectory needed to achieve a periapsis altitude of, say, 200 km and get back to the starting radius of 30,000 km displaced by and angle of, say, 300 degrees.
Solving problems of this kind efficiently is the bread and butter of orbital mechanics. And largely form my own edification, I decided that I would go through and map the solutions to this problem. Since the solutions aren't immediately obvious, this post captures the solution so as to serve as an online aide de memoir for myself; and, perhaps, as a useful guide for anyone else covering similar ground.
Now, because this forum's ability to typeset LaTeX is rather limited, I've take the liberty of using madoko.net to write the solutions and construct an html page that can be accessed by others. Here is the link to the resulting html page:
https://montblanc2012.github.io
As presented, the solutions are light on theory and just focus on presenting the results. If anyone is interested in the proofs, they can send me a private message.
The html page includes some sample Python code with which to enumerate the results. For example, for the example given above with a vessel at a radius of 30,000 km from and wishing to execute a close periapsis encounter (altitude of 200 km) with Earth while swing around 300 degrees before returning to a radius of 30,000 km the solution given by the same code is:
which identifies that there is one and only one Keplerian arc - an ellipse - with the required properties - namely an ellipse with semi-major axis of 23446586.2626 m; orbital eccentricity 0.719745982; and with time-of-flight 14857.46 seconds.
From this solution, and to test this result in Orbiter, we can set up a scenario with state vectors corresponding to this solution. These state vectors are:
x - 30,000,000.0 m
y - 0.0 m
z - 0.0 m
vx - -2137.323476 m/s
vy - 0.0 m/s
vz - 2237.1521334 m/s
where the velocity components of the velocity vector have been determined using standard expressions based on knowledge of the semi-major axis and orbital eccentricity.
A .scn file with these state vectors is given below:
And one can readily confirm by running this file that the resulting trajectory does indeed have the required properties and that the transfer time is 14857.46 seconds.
In effect, dgatsoulis posed the question: what is the elliptical arc that connects two points separated by an angle [MATH]\theta[/MATH] for a known apoapsis radius? A number of solutions were given in comments.
This note examines a closely related question: what are the Keplerian arcs that connects two points separated by an angle [MATH]\theta[/MATH] for a known periapsis radius? An example would be: starting from an orbital radius of 30,000 km from Earth what is the Keplerian trajectory needed to achieve a periapsis altitude of, say, 200 km and get back to the starting radius of 30,000 km displaced by and angle of, say, 300 degrees.
Solving problems of this kind efficiently is the bread and butter of orbital mechanics. And largely form my own edification, I decided that I would go through and map the solutions to this problem. Since the solutions aren't immediately obvious, this post captures the solution so as to serve as an online aide de memoir for myself; and, perhaps, as a useful guide for anyone else covering similar ground.
Now, because this forum's ability to typeset LaTeX is rather limited, I've take the liberty of using madoko.net to write the solutions and construct an html page that can be accessed by others. Here is the link to the resulting html page:
https://montblanc2012.github.io
As presented, the solutions are light on theory and just focus on presenting the results. If anyone is interested in the proofs, they can send me a private message.
The html page includes some sample Python code with which to enumerate the results. For example, for the example given above with a vessel at a radius of 30,000 km from and wishing to execute a close periapsis encounter (altitude of 200 km) with Earth while swing around 300 degrees before returning to a radius of 30,000 km the solution given by the same code is:
Code:
Elliptical solution of the first kind:
-------------------------------------------
Semi-major axis 23446586.2626
Eccentricity 0.719745982361
Time-of-flight 14857.4666227
Elliptical solution of the second kind:
-------------------------------------------
No solution
Hyperbolic solution:
-------------------------------------------
No solutionwhich identifies that there is one and only one Keplerian arc - an ellipse - with the required properties - namely an ellipse with semi-major axis of 23446586.2626 m; orbital eccentricity 0.719745982; and with time-of-flight 14857.46 seconds.
From this solution, and to test this result in Orbiter, we can set up a scenario with state vectors corresponding to this solution. These state vectors are:
x - 30,000,000.0 m
y - 0.0 m
z - 0.0 m
vx - -2137.323476 m/s
vy - 0.0 m/s
vz - 2237.1521334 m/s
where the velocity components of the velocity vector have been determined using standard expressions based on knowledge of the semi-major axis and orbital eccentricity.
A .scn file with these state vectors is given below:
Code:
BEGIN_DESC
END_DESC
BEGIN_ENVIRONMENT
System Sol
Date MJD 51981.7728006566
Help CurrentState_img
END_ENVIRONMENT
BEGIN_FOCUS
Ship GL-01
END_FOCUS
BEGIN_CAMERA
TARGET GL-01
MODE Cockpit
FOV 50.0
END_CAMERA
BEGIN_MFD Right
TYPE Orbit
PROJ Ship
FRAME Ecliptic
REF Earth
END_MFD
BEGIN_SHIPS
GL-01:DeltaGlider
STATUS Orbiting Earth
RPOS 30000000.0 0.0 0.0
RVEL -2137.323476 0.0 2237.152133
AROT -164.095 17.242 -173.532
VROT -0.0011 0.0099 -0.0023
AFCMODE 7
PRPLEVEL 0:0.200000 1:0.999046
NAVFREQ 402 94 0 0
XPDR 0
HOVERHOLD 0 1 0.0000e+000 0.0000e+000
END
END_SHIPSAnd one can readily confirm by running this file that the resulting trajectory does indeed have the required properties and that the transfer time is 14857.46 seconds.
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