You probably need to replace the simple point-source gravitation potential function with a zonal harmonic gravitational potential function. I don't know much about those things yet.
Jerry Abbott
I examined the problem of "twin Earths" once. I assumed a binary planet having each component habitable and of Earth's mass, with their mutual center of mass going around a star in a circular orbit with radius giving a subsolar temperature of 393.6K (same as Earth with respect to the sun)...
I was using a straightforward three-body program that I wrote myself, using a time step of about 1 second and constant acceleration during each time interval. I put in the Sun, Earth, and a test body. I omitted the moon for most of my tests because what I was trying to determine was a general...
The simulations began with circular orbits and remained so while R << Rmax. As R became larger, there began a tendency for the orbits to ellipticize. As R approached Rmax, the eccentricity of the satellite orbit became highly variable with a rapid precession of the major axis probably...
I believe that I was confused about what was implied by the term Hill radius. I thought that this was the maximum distance for the longterm stability of a circular orbit against a perturbing mass.
Apparently, what I determined with my numerical trials was the maximum stable radius for...
I didn't try retrograde orbits. But I did notice that for prograde orbits, with the test particle's orbit in the ecliptic plane, an initially circular orbit of the test particle around Earth having a radius between ~97% and 100% of the Hill radius led to some exceptionally chaotic (and usually...
That's what I used to think the Hill radius was. But I did some simulations on the three-body problem, Earth-Sun-test particle, and it looks like the outer stability radius is
Rmax = (D/3) { (Me+m) / Ms }^(1/3)
Where
Me is the Earth's mass.
Ms is the sun's mass.
m is the mass of the...
Ah ha! I figured that the maximum radial acceleration for a hyperbolic orbit occurred at true anomalies of +/- 90 degrees. The equation for r-double-dot up there give it a zero value at those true anomalies.
I had only inferred this because it worked out that way in three trials between...
For hyperbolas, the semimajor axis is the distance from the intersection of the asymptotes to the perihelion. The product ae is the distance from the intersection of the asymptotes to the sun (i.e., the focus of the hyperbola).
Q = true anomaly (perihelion at Q=0)
u = hyperbolic eccentric anomaly
You choose Q, then...
u' = ArcCosh {(e + cos Q) / (1 + e cos Q)}
Or, equivalently,
B = (e + cos Q) / (1 + e cos Q)
u' = ln { B + ( B^2 - 1 )^0.5 }
Then...
if sin Q < 0 then u = -u' else u = u'
a =...
It's a pretty standard celestial mechanics chore to start with a state vector (a position and a velocity) and derive from it a set of orbital elements. If you can get a job with NASA or JPL or NORAD in between their EEOC-mandated hiring selections , you'll probably be doing orbit...
It has been a while since I had a look at a celestial mechanics textbook other than the one by Dubyago. What does the variable (lower case l) represent? It seems to have units of m^2 sec^-1. What form of energy is that whole third term in that equation supposed to be?
Jerry Abbott
Or, to put it into one long equation...
apogee altitude = 2 / { 2/Rp - [sqrt(GM/Rp) + tF/m]^2 / GM } - Rp - Re
where
Rp is the geocentric distance of the spaceship at perigee (meters)
Re is the Earth's radius (meters)
t is the duration of burn (seconds)
F is the total thrust during...
I'll post his formulas.
Re = 6371000 meters
R = Re + 300000 meters
R = 6671000 meters
GM = 3.98793E+14 m^3 sec^-2
Vo = sqrt(GM/R)
Vo = 7731.8 m/s
F = 2 (25000 kg m sec^-2)
m = 80000 kg
acc = F/m
acc = 0.625 m sec^-2
t = 35.5 sec
dV = (acc) (t)
dV = 22.1875 m/s
Vf...