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Let me post my take on orbital mechanics:
I make frequent use of the simple formula
v*v = 2/r - 1/a
where
r = distance to center (Sun or planet)
v = velocity in units of the orbital velocity at r=1
a = the semi-major axis of the orbit
(This is just the formula for energy conservation after cancelling out the mass, kinetic energy being v*v/2, potential gravitational energy being 1/r, and the total energy being 1/(2a).)
For orbits around the Sun I take Earth distance to the Sun as unit for r and a (i.e. use Astronomical Units) and Earth's orbital velocity (~30km/s) as unit for v. (And indeed, plugging in r = a = 1 then gives v = 1.)
For planetary orbits I express r and a in planetary radii, and v in units of the orbital velocity near the surface (which is escape velocity divided by square root of 2).
E.g. for mars at the perihelion v*v = 2/1.381 - 2/(1.666 + 1.381) = 0.792, or v = 0.890, or (multiplying with 30) 26.7 km/s. An orbit stretching from Earth at perihelion to Mars at perihelion in ithe orbit's aphelion has v*v = 2/1.381 - 2/(1 + 1.381) = 0.608, or v = 0.780 (23.4km/s). So arrival at the Mars orbit occurs with a velocity difference of 3.3km/s.
The second requirement is to use such 'asymptotic' relative velocities to calculate the velocity after falling into the planet's gravitational well. For this you have to add or subtract the velocities 'according to Pythagoras', i.e. square, add/subtract, and then take the square root. Falling from infinity to the surface gives you by definition the escape velocity. For Mars the latter is 5km/s, so you would arrive near the surface at square root of 3.3*3.3 + 5*5 = 5.99 km/s. That means you have to slow at least 990m/s to get into orbit rather than just having a single fly-by.
Another example of the orbital formula:
To get in a higly elliptic Mars orbit stretching from near the surface to 19 Mars radii, you need a velocity at periapsis for which v*v = 2/1 - 2/(1 + 19) = 2 - 2/20 = 1.9, or v = 1.379. This is in units of the Mars orbital velocity, which is 5km/s / sqrt 2 = 3.55km/s. This gives 4,87 km/s. So you only have to slow down to 130m/s below escape to get into such an orbit.
Such an orbit is a good starting for aerobraking, which will lower the apoapsis, circularizing the orbit. But it is also a good place to park the mothership for the return journey; it would only have to accelerate in the by 130 + 990 m/s to travel back to Earth. (If that journey would also start from Mars' perihelion.)
Another useful tool is the conservation of angular momentum. Which is the distance to the orbited object times the velocity component perpendicular to that (the 'tangential' velocity). The orbital formula can give you the magnitude of the velocity at any point of the orbit. But only in the tips of the elliptical orbit is the full velocity perpendicular to the line to the center. At other points there also is a radial component to the velocity.
The tangential velocity can be calculated by first calculating the velocity V in one of the extreme points (e.g. perihelion), and multiply this with the radius there to get the angular momentum. This can then be divided by the radius at the desired point, to get the tangential velocity there. The radial velocity then is the square root of the total velocity squared minus the tangential velocity squared.
For example, if we have an orbit with semi-major axis 1 AU (so that its orbit time is 1 year), which stretches to Mars its perihelion. Then its velocity where it intersects Earth orbit (the point from which you can launch it) would also be 1 (= 30km/s). For the velocity at aphelion you would have v*v = 2/1.381 - 2/(1+1) = 0.448, or v = 0.669 (20.1 km/s). The angular momentum is thus 1.381*0.669 = 0.924. At r = 1 we would thus have tangential velocity 0.924/1 = 0.924. Then the radial velocity is square-root 1*1 - 0.924*0,924 = 0.382 (11.4km/s). Since Earth has hardly any radial velocity of itself (near-circular orbit) all that velocity would have to be acquired with rockets, requiring a starting speed of more than 15km/s in LEO.
Last edited by HG Muller (2026-10-18 14:01:29)
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