# Is higher delta-V required to enter polar orbit than equatorial orbit around another planet?

I suppose gravity assist on a flyby only works in the orbital plane of the assisting mass. But I wouldn't think that matters for orbit insertion. Is it as easy to enter into a polar orbit, as into an equatorial orbit around another planet? Could one simply aim at a certain latitude as first periapsis at arrival in order to get a capture orbit with any inclination one wants?

• As an aside, regarding gravity assist, if your next destination is on an inclined orbit relative to the body you're doing the assist around, you will want to be slightly out-of-plane on the flyby, so the assist helps with the plane change as well as the gross velocity change. – Russell Borogove May 12 '15 at 15:48

With vanishingly small $\Delta V$ long before you arrive, you can select any orbital insertion point on a circle around the planet. However your approach declination relative to the destination body equatorial plane limits what orbital inclinations you can get to, where that declination is defined by your interplanetary trajectory and is very expensive to change. The orbital inclination after a single insertion burn cannot be less than the approach declination. So it will always be more expensive to do a plane change to an exactly equatorial orbit, unless your approach declination is exactly zero (which never happens). It will always be free to go into a polar orbit.

The plane change to get to an equatorial orbit does not have to be large, if you first insert into a very loose orbit. Then the cost of the plane change at apoapsis will be small. Once the orbital plane is set to what you want, you can continue with the lowering of the apoapsis to the desired destination orbit.

Assuming both bodies have no axis tilt and an initial orbital inclination of zero degrees, it will require more delta-V to enter a polar orbit around the second body. However, since inclination changes are more efficient the farther away you are, this increase becomes very small at interplanetary distances.

This depends on the target body's axial tilt and its inclination to transfer orbit. For example, it would require less $\Delta v$ to inject into a polar orbit around Uranus with axial tilt of nearly 98° and orbital inclination of 0.8° to ecliptic, than it would into its equatorial orbit. You'd have to compute for exact injection epoch what this inclination would be to your transfer orbit at the time of orbital injection though.

One other note, that gravity assisted $\Delta v$ changes still count towards total $\Delta v$ required to achieve a target orbit, some of it just might not be achieved propulsively. It also speaks nothing of propulsive efficiency, for example during Oberth burns. So, it's good to know how much of it comes free (can be achieved by exchanging momentum with gravity assisting bodies, by means of aerocapture,... say to reduce $v_\infty$ during injection or to slowly crank your orbital inclination) when reading $\Delta v$ charts, because that will define how much of propulsive $\Delta v$ your mission needs to be capable to achieve, but it still counts towards total $\Delta v$.

Bottom line, how easy will be to achieve a specific target orbital inclination around some other planet (as per your question's title) depends on many factors, and would be specific to each planet. Take for example bodies with planetary rings (Jupiter, Saturn, Uranus, Neptune and possibly Pluto), where there might even be rings, moonlets,... in the way of achieving equatorial orbits of certain altitudes. Large portion of this problem will also be reducing your transfer trajectory's $v_\infty$, or hyperbolic excess velocity. Inclination change, once you've reduced $v_\infty$ and are captured in orbit, might be relatively easy. Perhaps, partially, by even slowly cranking your orbital inclination with propulsion methods that don't consume any reaction mass, like magnetorquers.

You can think of it this way - you can approach a planet from any direction and enter orbit in a plane intersecting with the poles. You are aligning yourself with two points, not a specific plane. But when you want to take up orbit aligned with the equator, you have to enter from a specific direction, or adjust inclination once in orbit. The more the planet's axial tilt deviates from the ecliptic, the more you will have to adjust to do that. The first goal only requires you aim at a certain point, the second requires that you also adjust the angle of the craft's motion. And of course, your craft is moving at high speed so changing direction takes a lot of energy.

• I just thought maybe it would be useful to post a version that is more plain English :P Let me know if there are any mistakes, please. I know that is rather likely, but i try... – kim holder May 12 '15 at 16:58