# Orbital mechanics problem

by MonstersFromTheId
Tags: mechanics, orbital
 Emeritus Sci Advisor PF Gold P: 2,361 First I'd turn the recovery craft so that the recoverery ring rotates in the same plane as the craft's orbit. This simplifies things greatly. As far as the actual burns go, the simpliest, if not most direct, method goes like this: Perform a burn that lifts your craft into a transfer orbit with a higher apogee of the proper value (to calculated below). Do a second burn that puts you into a circular orbit at the same altitude as the new apogee. Wait until you reach the correct point in your new orbit and do a braking burn which puts you into another tranfer orbit, one with a perigee equal to the altitude of the orbit of the recoverery craft. cone right, this gives you the correct relative speed to the recovery craft as you pass. The equations. To determine the altitude for the new orbit for the deployable craft: given that $V_r$ is the relative velocity the two craft must have to dock. $M$ is the mass of the planet being orbited $R_1$ is the radius of the recovery's craft's orbit. (as measured from the center of the planet) $R_2$ is the radius of the initial orbit of the deployable craft. $G$ is the gravitational constant. $m$ is the mass of the deployable of the craft. Then we use the conservation of energy to find the new orbit's radius. First we find the orbital velocity of the recovery craft by $$v_2 = \sqrt{\frac{GM}{R_1}}$$ The total energy of the deployable craft is as it passes the recovery craft is found by $$E_t =\frac{2(V_1+V_2)m}{2}- \frac{GMm}{R_1}$$ It is also found by $$E_t = -\frac{GMm}{2a}$$ a is the semimajor axis of the second transfer orbit. we equate the two equations and solve for a We find the apogee of this orbit by: $$R_{a1} = 2a-R_1$$ This is the same as the apogee for the first transfer orbit. Now you can find the delta v needed for the first burn. $$\Delta v1 = \sqrt{\frac{Gm}{R_2}} \left [ \sqrt{\frac{2R_{a1}}{R_2+R_{a1}}}-1 \right ]$$ The second delta v burn needed to move into a circular orbit is: $$\Delta v2 = \sqrt{\frac{GM}{R_{a1}}} - \sqrt{\frac{2GMR_2}{R_{a1}(R_2+R_{a1})}}$$ The delta v needed for the last burn is: $$\Delta v3 = \sqrt{\frac{2GMR_{a1}}{R_1(R_1+R_{a1})}}$$