From the article "Which Way To Mars?. Trajectory Analysis. by Kelsey B. Lynn" that I have read, it states that the Hohmann Trajectory at earth,V1=32.76km/s and the earth velocity,VE=29.82km/s,so the hyperbolic excess velocity,VH=[V1-VE]=2.940km/s.All these parameters lead to the calculation of the total velocity needed for launch from earth[know as the delta V].To launch from one of the earth'spoles,the delta V is a combination of VH[hyperbolic excess velocity] and VES[escape velocity from earth],that is delta V=[VH^2+VES^2]^.5 =11.6km/s. The value of VES=11.18KM/S,of course.What I would like to know is how the formula{deltaV=[VH^2+VES^2]^.5} is derived? Any help for explanation would be appreciated. Thanks a lot.
they are derived from the energy equation. The energy of a satellite can be described using the following formula: E = 1/2 V^2 - mu/r For the situation on the surface of the Earth, with a velocity deltaV (on the pole the rocket doesn't get extra velocity due to the earth's rotation), the energy is: E_1 = 1/2 (deltaV)^2 -mu/Re For a parabolic orbit (the satellite is always at the escape velocity), the velocity is zero at infinite distance. When you actually want to go somewhere, you want to have that excess velocity VEH. At r = infinity (you have escaped), the energy is then: E_2 = 1/2 VEH^2 Energy is conserved after giving the delta-V (assuming the delta-V was there immediately). Equating the two equations gives: 1/2 VEH^2 = 1/2 (deltaV)^2 - mu/Re it happens to be the case that escape velocity is written as: VE = sqrt(2*mu/r) so: mu/Re = 1/2 VE^2 at r=Re so, finally: 1/2 VEH^2 = 1/2 (deltaV)^2 -1/2 VE^2 or: VEH^2 + VE^2= (deltaV)^2 et voila
i now see that Hellfire already made the same derivation here: https://www.physicsforums.com/showthread.php?t=16708
Velocity vector components from orbital elements Note that... a : semimajor axis of the orbit e : eccentricity of the orbit 1 astronomical unit = 1.49597870691E+11 meters GMsun = 1.32712440018E+20 m^3 sec^-2 The canonical velocity in a hyperbolic orbit, in general, is found from Vx’’’ = -(a/r) { GMsun / a }^0.5 sinh u Vy’’’ = +(a/r) { GMsun / a }^0.5 (e^2 - 1)^0.5 cosh u Vz’’’ = 0 Where u is the eccentric anomaly and r is the current distance from the sun. The canonical velocity in an elliptical orbit, in general, is found from Vx’’’ = -sin Q { GMsun / [ a (1-e^2) ] }^0.5 Vy’’’ = (e + cos Q) { GMsun / [ a (1-e^2) ] }^0.5 Vz’’’ = 0 Where Q is the true anomaly. The triple-primed vectors would be rotated (negatively) by the angular elements of the orbit (w,i,L) to heliocentric ecliptic coordinates. Rotation by the argument of the perihelion, w. Vx'' = Vx''' cos w - Vy''' sin w Vy'' = Vx''' sin w + Vy''' cos w Vz'' = Vz''' = 0 Rotation by the inclination, i. Vx' = Vx'' Vy' = Vy'' cos i Vz' = Vy'' sin i Rotation by the longitude of ascending node, L. Vx = Vx' cos L - Vy' sin L Vy = Vx' sin L + Vy' cos L Vz = Vz' The unprimed vector [Vx, Vy, Vz] is the velocity in the orbit, referred to heliocentric ecliptic coordinates. Jerry Abbott