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Why is it the most energy efficient to change orbit inclination while crossing the equator? Specifically, it's most efficient to do a plane change at one of the two "nodes" where the origin orbital plane intersects the destination plane. ANASIS-II is destined for geostationary orbit, so its destination plane is the plane of the equator. Any orbit ...


29

A great aid to intuition is to remember one principle about orbit changes: if the engine is off, the orbiter always returns to same point one orbit later. So for any orbit change, if you want to do only a short burn, it has to be at a point that is common for both the current orbit and the destination orbit. This applies to inclination changes, altitude ...


11

To change the inclination by an angle $\alpha$, you need to apply a velocity change of $\Delta v = v \times sin \alpha$, so the higher the velocity, the higher the velocity change that you need to apply, and thus the energy that you need.


10

Your comment on this answer has, I think, led me to understand what you are really asking about. What I am saying is that, presuming the inclination to be ZERO (Plane of orbit parallel to the equatorial plane), the entire plane keeps shifting from pole to pole - parallel to the equatorial plane. You want to move the orbit like this: If that is the right ...


8

If you're going to a higher inclination than your launch latitude (for example, going from Cape Canaveral at 28º to ISS at 51º), you'll want to do it as early as possible, while your horizontal velocity is minimal -- essentially immediately when you start your gravity turn. The velocity you start with from Earth's rotation is less than that of even a very ...


8

Inclination changes are very expensive in terms of delta V/fuel. To do a 90 degree plane change you are thrusting against your current ~7 kilometer a second velocity to bring it to zero and also adding thrust to get a ~7 kilometer per second velocity in the new orbital plane. This is more than it took to get into orbit in the first place, and would require a ...


8

It's not just the most efficient way, it's the only way to achieve this particular target orbit. As the other answers have pointed out, an orbital inclination change must occur at the so-called ascending/descending nodes, which are the two points in the orbit at which the current and target orbital planes intersect. Anytime a spacecraft moves from one orbit ...


7

Although a little vexing, the GIF below and the answer it comes from shows that even fixed orbit will eventually pass over all places on a body that rotates underneath it that are at any latitude smaller than the inclination. That being said you have to choose an altitude that spread them out evenly so that you don't have to wait a very long time. That being ...


7

Let's first look at a typical orbit first for comparison. Note that the object being orbited could be anywhere. Note that it spends a time both above and below the equator. A few things about a Geosynchronous orbit: The inclination varies with time, although this isn't that important for day to day operations. The period is 24 hours, but a geostationary ...


7

I found this kind of counterintuitive as well at first. This is the way I rationalized it to myself: consider a high-thrust maneuver to change inclination. Obviously if it's impulsive, you perform it at the node. If it lasts for, say, one minute, you would burn +/- 30 seconds around the node. Now, take that to the limit, where the burn time is the entire ...


5

It will be a combination of things. Like TonyK mentioned in his comments, most of the change will occur during cis-lunar transfer. This is optimal because inter-planetary trajectories can be designed to result in an optimal parking orbit around your target planetary body. One neat trick astrodynamists use is B-plane targeting (a nice AGI article here: http://...


5

New Horizons used its monopropellant attitude control thrusters for course correction. Because it couldn't make large course changes with the limited fuel available, the choice of targets was severely constrained: Mission planners searched for one or more additional Kuiper belt objects (KBOs) of the order of 50–100 km (31–62 mi) in diameter as targets for ...


5

While delta-V - or change in velocity - remains the same, the effect on the trajectory depends in large part on the absolute velocity. While a small change of speed (a small modification of the velocity vector) at periapsis can convert to large apoapsis change, maneuvers like plane change (normal/antinormal burn) or moving argument of periapsis (rotating ...


5

It's just a matter of some spherical trig. $$ \cos(inclination) = \cos(lat)*\sin(azimuth) $$ So in your example, the inclination would be equal to: $$ \arccos(\cos(30.56)*\sin(123)) = 43.77\deg $$ To simplify the calculation, I assumed that your given azimuth was an inertial one. The calculations that take into account the rotation of the Earth and what ...


5

I suggest building a 3D model to understand it. Start with a nice disk, such as a paper plate or the cardboard packaging of a frozen pizza. Mark the center and draw regularly spaced radial lines (“slices”) across the face of the plate. Number the lines at the rim and duplicate the labels farther inward too. Cut a ring off the plate, about 1 inch wide. ...


5

In simple terms, gravity pulls an object directly towards another object. As an analogy, if you are running down the street and grab hold of a lamp post with your left hand you will swing around it to the left, if you grab it with your right you will swing to the right. You can use this to do a u-turn, letting go whenever you or going the right direction, or ...


4

In the case of ANASIS-II, the situation is far more complicated than can be explained in ten seconds of livestream. Some general rules regarding plane changes: You can only make a plane change at the point where your current orbit's plane intersects the target orbit's plane. The faster you're going, the more fuel it takes to perform a plane change. Because ...


4

First, 0.1° latitude is near zero. Second, latitude should be expressed in Earth-fixed frame, which uses the true equator, you might be using a Mean-Equator, Mean-Equinox frame (such as J2000), which also creates a difference between frames. Third, it seems like you are not using a two-body propagator, such that perturbations are present and do cause ...


3

I will give my current best shot at this problem, and others should feel free to strengthen the argument with additional mathematics. (Or poke holes!) You ask two questions, I will answer the first as the second has been partially answered by the update. Are there other inclination change strategies that are more efficient for some values of $\alpha$? ...


3

Barely, but it involving shenanigans even less practical than a Jupiter flyby. The lowest delta-v cost trajectory without any planetary flybys from Earth to a 1.35AU-5.4AU 79.11° orbit is the following: Do a burn in LEO, reaching a solar system escape trajectory. Cost: 8750m/s At the edge of the solar system, do an inclination change and set the periapsis ...


3

This is a well-known phenomenon in helicopter dynamics and in control systems. In a second order control system (which roughly describes your low-thrust scheme), the phase angle change is 90° when the input control frequency is the same as the system's natural frequency. (See any elementary control text and look at the second order system frequency response.)...


3

The trans lunar injection of the Apollo missions took a Δv of about 3000 - 3200 m/s. Let us assume all inclination change is for free (quite reasonable with careful planning and possibly multiple fly-bys) and lowering the orbit to LEO again is free as well due to the possibility of aero-braking. Compared to that, the direct inclination change of 23 degrees ...


2

I may be rusty in my understanding of orbital mechanics, but here's a try: At apoapsis, an orbiting object will be travelling at the slowest velocity of its orbit. Hence, if you make a plane change at this point, assuming you are not changing any other orbital parameters, the velocity at your target orbit will also be the slowest velocity along that orbit. ...


2

I think I've answered my own question. As I mentioned, acceleration to alter inclination would be perpendicular to the orbital velocity, so any effect that is a function of that velocity should not apply. ...right?


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There's two ways that are typically done. The first is to use a "Super Synchronous" orbit to reduce the delta-v overall. Essentially, the initial orbit is around twice geostationary orbit, which makes the inclination change cheaper. The second is to just do it directly. They are always done as far out in the orbit as they can. There's a number of orbital ...


2

Here are some effective ring parameters. To simplify I'm just using the monopole term GM and not including the "extra gravity" near the planet from the oblateness which is okay since I'm rounding. (see this answer to Equation for orbital period around oblate bodies, based on J2?) ring a T ω a_z/1km Δv/day ...


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If your propulsion system has a better specific impulse, then you can travel farther from your initial orbit. So as CubeSat/smallsat propulsion systems improve their efficiency, the range of orbits that can be reached while being launched as a secondary payload will be increased. Theoretically, a solar sail has infinite specific impulse and therefore could ...


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I used JGM-2 and lunisolar perturbation. After maneuver, the satellite is starting to drift. Evidently, the inclination drift caused by lunisolar is not secular. The drift is secular if I just include the lunar perturbation.


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You can get the same efficiency in the ascending or descending node, it's just that you would need to apply the force in the opposite position if you wanted to switch from ascending to descending and vise versa.


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Although it has been already answered briefly, I would like to add some extra information (increasing the inclination) about on the N/S maneuvers. Well the location of the thruster firing is highly depends on the maneuver type. If the maneuver is: North, then you should fire your thrusters while the satellite passes from ascending node in order to do more ...


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