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By now most here are familiar with Keplar's laws of planetary motion. For the most part these are empirical observations, without the mechanical thermodynamics its' relatively difficult to apply theory to. Specific Mechanical Energy (SME) is a subset of all Specific Energy. Specific Energy is the energy per kilogram in its various and sundry forms. SME is the energy within an orbit. To show the relationship a spaceship during a burn converts chemical energy Specific Chemical Energy in fuel entering the reaction chamber into SME in the form of Kinetic Energy. If this along the prograde the specific KE is converted to specific PE such that the SME remains the same until some other form of energy is converted to SME. While Keplarian laws are useful in calculating position over time, understanding SME is very helpful in describing objects that transit in and out of differernt Hill radius, such as a spacecraft traveling between different planets. Critical to understanding SME is the simplified mechanical energy potential Mgh (Mass x agravity x height) where h is the height over which PE is being analyzed. The result in Joules (J) can be deprecated to specific mechanical potential energy by the removal of M and thus the units are J/kg. The equation gh almost never gives a correct answer unless some sort of gravity correction is applied, but for very small heights the answer is generally well within the relative precision of h measurement. Averaging the gravity between the two ends of the potential works well only when plot of gravity versus height is flat, once the scale is such that the plot is visibly non-linear the estimation of velocity will be incorrect. An example of where this might be a problem is during a Suicide burn countdown. The actual equation is the integral of gh which using standard graviational parameter (µ) instead of g and radius replacing height can be converted to SPE = -(µ/ro - u/r1). For a central point mass (CPM) the resulting equation gives and accurate SPE from the surface of the objects hill sphere to 2 Schwarzschild radii . If we makes the basic assumption that SPE between an objects hill sphere (SOI in kSP) radius and infinity is low the calculations can be relatively simplified, in some cases really simple. The reason for this is that in the SPE equation if starting radius (ro) is infinity the SPE is µ/r where r is the radius that one wants to leave orbit from to escape or circularize to on system entry. Upon entering a Hill sphere the SME = SKE + SPE until exit or some other force is applied. Given that a hyperbolic orbit entering a hill sphere at t=0 has a SPE0 At the bottom of the potential all SPE is converted to SKE and speed (Vorbital, Pe) is = √2SPE0 . It really is that simple. In addition the velocity of a circular orbit at Pe is √SPE0 and the dV required to circularize at PE is 0.4142√SPE0 , [ √2 - 1 = 0.4142135]. Therefore for any given circular orbit of radius r to CPM the amount of dV required to escape is 0.4142 √(µ/r). You cannot get simpler than that. In an orbit SPE and SKE are not always interchangable (as in an elliptical orbit) at all magnitudes, this is the fundemental inequity that causes the oberth effect. The limit to which SKE, at any given radius, can be converted to SPE is specified by µ/r. Any SKE acquired at rexceeding u/r resides in SKE. This can result in some rather bizarre physics. For instance if we convert Kerbin into a nuetron star of radius 0.5 km and entering kerbin from its hill sphere, orbital velocity is slightly adjusted to obtain a hyperbolic Pe of 1 km. The equation tells us that at 1 km we have an SKE of at least 3,531,598,400 (v = 84,083 m/s). If exactly at Pe we could add 1 m/s dV (v = 84,084) we can see the effect. In adding 1 dV, SPE (3,531,682,443.3) leaves 84043.327 SKE on system exit. If the space craft leaves into another CPM gravitational influence along the left bodies prograde motion the bodies residual velocity (409 m/s) is added to that bodies prograde motion speed. In this instance, by entering the hill sphere and approaching the dense body closely the addition of 1 m/s of prograde motion resulted in a 84000 fold return of energy and 409 fold return on velocity. While the oberth effect is limiited to objects entering and leaving hill sphere's, the underlying physics are not so limited. An example is burning from Kerbin to Moho. The primary issue here is that an object in Kerbins SOI is traveling through kerbins space-time warp. A key dimension is time, the longer time required to exit, the more time space-time has to act on it. By applying a tremendous amount of dV from a minimal circular orbit the less time the space craft will be influenced. If simply 0.41422 √(µ/r) the resulting craft will exit kerbin with dV slightly different than kerbin. The limit orbit around Kerbin is 70km alt (r =670,000). √(µ/r) = 2295.68 if one applies 2000 dV rapidly along the tangent close to angle to prograde 300 the resulting SKE = 9.226MJ/Kg. The residual SKE on kerbin system exit is 3955987.99 J/kg. This results in a residual velocity of 2812.82. IOW we applied 2000 dV to a stable circular orbit, we overcame the energy required to escape (dv = 950) and majically had 2812.82 'dV' to play with on Kerbin exit. Since the Kerbin orbital velocity is 9284.5 and the exit vector is nearly retrograde the resulting crafts circumKerbol velocity is 6741.81 and allows the resulting orbit to cross the elliptical cylinder that moho resides (noting to intersect the orbit itself requires a plane change in kerbin). So the question is how to practically use SME. So lets take moho as an example because it is one of the most challenging objects to intersect. For the hohmann transfer we want an orbit with a Pe very close to the Moho periapsis (r = 4,210,510,628; 4.2Gm). Kerbin has a orbital radius of 13.599840256 Gm. The semi-major axis is the average of the two, 8905175422. Using vis-viva equation (kerbol µ = 1.17023 x 1018] the velocity at kerbin orbit needs to be √(µ(2/r - 1/a) ) = 6378 m/s. With this number in hand if we need to make a plane change of 7 degrees (assuming that one node is radial to its periapsis and we align 180' but along the same axis through kerbol, then we can calculated antipolar dV required. 6378 * sin 7' = 777.28 (south polar). From Kerbin escape minimum of 9284.5 - 6378 = 2906.3 m/s velocity in the retrograde and 777.28 m/s polar. The speed is thus √(2906.32 + 777.282) = 3008.45 m/s. So now we convert back into Kerbin. SKE on exit is 4,525,385.7 J/Kg. We have to add this to KPE of the orbit we are rising from 5270477 requiring an SKE of 9795833 and requiring a velocity at alt = 70k of 4426.3. Since the orbit requires 2295.7 this means dV added is 2130.3. The savings from making these two burns in low kerbin orbit are 1553.3 dV. However, there is one caveot, because kerbin is rotational axis is perpendicular to the plane of its rotation with a equitorial surface velocity of 174.94 and the moho pre-plane matching orbit about kerbin is not perpendicular to the same axis the launch looses some of that rotational velocity, resulting in a slower acquisition of circular orbit. Exit angle relative to the equitorial plane at angle to prograde 270' can be calculated. Its the arcsin of 777.28/3008.45 = -14.97' OR a bearing of 104.97 relative to kerbins axis of rotation. The amount of dV loss due to non-optimal launch and circularization angle is 174.94 - 174.94 * cosine 14.97 is greater than 5.97 m/s dV. Therefore the amount of savings is less than 1547.3 m/s dV. The remaining question is where to launch. There are two places one can launch from, but in either place the launch either needs to cross the Angle to prograde plane of 90/270 degrees that halves kerbin at 36,000 meters, at which point HSI switches from surface to prograde motion (assume you are using a automated navigation system like MechJeb). Because of Kerbins equitorial surface velocity is not zero, the initial launch angle at when crossing Angle to Prograde of 90' below 36,000m is less than 75' or at AtP of 270 crossing is more than 105 degrees. We can also calculate this but the calculations are inexact because speed reaching 36,000 can vary from 400 m/s to 800 m/s a good guess is to use a handicap of 2 degree (73 or 107 degrees, respective to AtP). Any inaccuracies in the tilt can be corrected at Atp90 or 270 using polar burns in LKO.