Sometimes satellites loop around a planet to increase velocity.How does orbiting a planet increase velocity?Wouldn't its gravity slow it down when departing?
I have an orbit question: How do you calculate the orbital inclination of a dog-leg launch?Say for some reason you needed to reach the ISS from Vandy. Because inclination = arccos(cos(launch latitude)*sin(launch azimuth)), if we want the inclination to be 51.6, and Vandy's latitude is 34.742, then the required launch azimuth would be 49.1 degrees or 130.9 degrees. That would take it either just over Salt Lake City or just over San Diego. Neither would be good, obviously. And certainly it's outside of Vandy's approved launch azimuths of 158-202 degrees.Bu let's say we got approval to do a "coast hugging" maneuver to do a dogleg. Obviously that's a lot more dynamic, and I bet there's an insane delta-V penalty. But it looks like a coast hugging trajectory is a launch azimuth of about 140 (ignoring going over the islands south of Vandy), resulting in about a 61 degree inclination on direct assent. If we instead traveled that, and then turned more to the east after clearing Baja,what the inclination would end up being? Say you traveled a launch azimuth of 140 at launch, then after traveling 3500 km, you turned to 110 degrees until reaching orbital speed. What would your resultant orbital inclination be? And what would the delta-v penalty be for the maneuver (say, if it was a Falcon 9 doing it).
Quote from: jasperval on 07/22/2020 06:27 pmI have an orbit question: How do you calculate the orbital inclination of a dog-leg launch?Say for some reason you needed to reach the ISS from Vandy. Because inclination = arccos(cos(launch latitude)*sin(launch azimuth)), if we want the inclination to be 51.6, and Vandy's latitude is 34.742, then the required launch azimuth would be 49.1 degrees or 130.9 degrees. That would take it either just over Salt Lake City or just over San Diego. Neither would be good, obviously. And certainly it's outside of Vandy's approved launch azimuths of 158-202 degrees.Bu let's say we got approval to do a "coast hugging" maneuver to do a dogleg. Obviously that's a lot more dynamic, and I bet there's an insane delta-V penalty. But it looks like a coast hugging trajectory is a launch azimuth of about 140 (ignoring going over the islands south of Vandy), resulting in about a 61 degree inclination on direct assent. If we instead traveled that, and then turned more to the east after clearing Baja,what the inclination would end up being? Say you traveled a launch azimuth of 140 at launch, then after traveling 3500 km, you turned to 110 degrees until reaching orbital speed. What would your resultant orbital inclination be? And what would the delta-v penalty be for the maneuver (say, if it was a Falcon 9 doing it). It is not a simple equation or easy to get an accurate answer. The easy approximation is to use the minimum launch azimuth allowed from Vandenberg (158 degrees) to find the lowest launch inclination (72 degrees) and do a plane change once on orbit.
I'm not quite sure how your xyz-coordinate frame is oriented.
Quote from: Proponent on 05/09/2023 09:19 pmI'm not quite sure how your xyz-coordinate frame is oriented.In the space shuttle program, the standard inertial frame was Mean of 1950. +X along mean vernal equinox (intersection of equatorial and ecliptic planes), +Z along mean rotational axis (north pole), +Y completes a right-handed system (and therefore both X and Y are in the mean equatorial plane).Really should use the atan2 function to get an unambiguous, quadrant-correct solution for RA:RA = atan2(y,x)And since it's already established that these are unit vectors, the solution for DEC reduces to:DEC = asin(z)
Quote from: Jorge on 05/09/2023 11:03 pmQuote from: Proponent on 05/09/2023 09:19 pmI'm not quite sure how your xyz-coordinate frame is oriented.In the space shuttle program, the standard inertial frame was Mean of 1950. +X along mean vernal equinox (intersection of equatorial and ecliptic planes), +Z along mean rotational axis (north pole), +Y completes a right-handed system (and therefore both X and Y are in the mean equatorial plane).Really should use the atan2 function to get an unambiguous, quadrant-correct solution for RA:RA = atan2(y,x)And since it's already established that these are unit vectors, the solution for DEC reduces to:DEC = asin(z)Jorge, Thank you for your brilliant explanation and reduction of the situation here. I knew they would have spacified if they were Starball, and that why some anwers seemed right and others were in wonky had something to do with ambiguity. ATAN2 saves the day!
My grasp of orbital mechanics is limited, so I'm trying to understand the purpose of a perigee raise maneuver, in particular when heading to the Moon. When an upper stage or spacecraft initially climbs into a high elliptical orbit, is the perigee of this orbit typically inside the Earth's atmosphere? Thus, you fire the upper stage or spacecraft engine(s) at apogee to raise the perigee above the atmosphere? I assume that launching into a high elliptical orbit where perigee is already above the atmosphere is not as fuel-efficient. I hope I've properly stated the question here.Thx