@astro.cal let GM=Gn*Msun !Mars orbit: set a 1.523679 e .0934 !Earth circular orbit velocity let vEarth=sqrt(GM/au) ? vEarth 29784.72616569657 !actual Mars orbit: let rmax=a*(1+e) let rmin=a*(1-e) let voMars=sqrt(GM/(a*au*(1-e^2))) let vmin=(1-e)*voMars * ? vmin 21971.77754130247 let vmax=(1+e)*voMars * ? vmax 26498.94282336215 !join at aphelion: let a2=(1+rmax)/2 let e2=1-1/a2 !because: 1=a2*(1-e2) ! .2498098132654847 let vo2=sqrt(GM/(a2*au*(1-e2^2))) let v2max=(1+e2)*vo2 ? v2max 33297.80278960929 ? v2max-vEarth 3513.076623912722 ! above delta V to start Hohmann transfer !find spacecraft velocity when it meets Mars: let v2min=(1-e2)*vo2 ? v2min 19986.78889175906 ? vmin 21971.77754130247 ? vmin-v2min 1984.988649543408 ! above to join Mars orbit from Hohmann ! total delta V=5498 ! trip time in years: ? .5*sqrt(a2^3) .7695076400738805 !years !join perihelion: let a2=(1+rmin)/2 let e2=1-1/a2 ! .1601463866427006 let vo2=sqrt(GM/(a2*au*(1-e2^2))) let v2max=(1+e2)*vo2 ? v2max 32081.15588281668 ? v2max-vEarth 2296.429717120114 ! above delta V to start hohmann transfer let v2min=(1-e2)*vo2 ? v2min 23224.20256536158 ? vmax-v2min 3274.740258000569 ! above to join Mars orbit from Hohmann ! total delta V = 5571. ! time of trip in years: ? .5*sqrt(a2^3) .6496278150325970 * ! about 100 m/s better to go to aphelion, but takes longer !Remark: the total delta V available on the OMS Space Suttle was about 300 m/s Halley's comet: set ecc=.96714 a=17.834 !r=a(1-e^2)/(1+e cos(phi)), if cos(phi)=-e then r=a let phi=acos(-ecc) let u=2*atan(sqrt((1-ecc)/(1+ecc))*tan(phi/2)) !Remark: expect u=90 deg as r=a(1-e cos(u)) !t is the time since perihelion to when it achieves r=a !w t=2 pi(t/T) = u-e sin(u)=pi/2-e !(t/T)=1/4-e/(2 pi) ? 1/4-ecc/(2*pi) .9607488833810435E-01 ? (1-2*res) .8078502233237913 !the fraction closer than a is 2t/T, so seek 1-2t/T Oumuamua: when far: E=KE=1/2 m v^2=GMm/(2 a) Vinfinity=sqrt(GM/a) set a=1.2798 ecc=1.19951 ? sqrt(GM/(a*au)) 26328.28428699930 let wt=sqrt(GM/(a*au)^3)*100*year set xl=7 xr=5 root ecc*sinh(x)-x-wt Root between 6.59924316406250 and 6.59930419921875 ? a*(ecc*cosh(6.59924)-1) 562.5269938127005 Mercury: The orbit plane is fixed in space; the Earth rotates in a orbit period of 88.91 minutes, the earth rotates: ? 360*88.9/(24*60) 22.2250 the nubmers I provided suggest: 35-10=25...close enough launch parallel to the equator to get full benefit of spin velocity ? 2*pi*Rearth*cos(32.5*pi/180)/(24*3600) 391.1912761970551 !nearly 900 mpg let a=(2*Rearth+156e3+285e3)/2 !6598637 ! R+156=a*(1-e) !e=1-(R+156)/a ? 1-(Rearth+156e3)/a .9774745905859028E-02 rendezvous Space Station recall: with phasing we catch up by stepping on the brakes, producing an orbit with a shorter period. One orbit later we'd find ourselves back at the altitude of ISS, but a bit before the ISS could complete its full orbit. !period of space station orbit: let T=2*pi/sqrt(Gn*Mearth/(400e3+Rearth)^3) !we seek an orbit that is 10 minutes less let T=T-10*60 let a=(Gn*Mearth/(2*pi/T)^2)^(1/3) !as this a is smaller that our current R, we will !be at apogee after we fire rockets to slow down: !a*(1+e)=Rearth+400e3 let ecc=(Rearth+400e3)/a-1 ! circular orbit velocity: sqrt(GM/R) ! apogee velocity: (1-e)*v0 let vSS=sqrt(Gn*Mearth/(Rearth+400e3)) let v0=sqrt(Gn*Mearth/(a*(1-ecc^2))) ? VSS-(1-ecc)*v0 309.9436491056822 !by itself this is barely within the space shuttles abilities !and of course we would need to do it again when, one orbit later, !we re-join the Space Station's orbit...solution: makeup 5 minutes !per orbit and orbit twice, etc !This orbit in fact does not work as the perigee is inside the Earth: ? a*(1-ecc)-Rearth -594876.7953126021 !so a better strategy would be, say, to catch up 1 minute per orbit let T=T+9*60 !remember we went back 10 minutes let a=(Gn*Mearth/(2*pi/T)^2)^(1/3) let ecc=(Rearth+400e3)/a-1 let v0=sqrt(Gn*Mearth/(a*(1-ecc^2))) ? VSS-(1-ecc)*v0 27.91843817025710 ? a*(1-ecc)-Rearth 302183.9395177895