How to calculate dv for inclination change?

[guitarxe]

How can I find out how much dv is required to do an inclination change from X degrees to Y degrees?

I'm planning a munar satellite mission and want to know how much dv to pack to put it in munar polar orbit.

[WOODY01]

It cost a lot less dv to get the inclination you want during the transfer burn or a correction burn on the way to the target. Inclination changes are costly inregards to dv.

The formula and what to insert into it can be found with a google search.

[OhioBob]

The formula for be found here:

http://www.braeunig.us/space/orbmech.htm#maneuver

Equation 4.73 for a simple plane change and 4.74 for combined plane change and altitude change.

Edited June 7, 2015 by OhioBob

[Snark]

The thing about inclination changes is that the cost completely depends on where you do them. For example, the worst possible way to get into a polar orbit around the Mun (or anywhere else) is to get into an equatorial orbit first, and then wrestle it around by 90 degrees.

The most efficient way to set your orbital inclination around a body is to do it as far away as possible from that body.

For example, if you're in equatorial LKO and want to go to a polar orbit around the Mun, the best way to do that is to do a slight inclination change long before you even enter the Mun's SoI. Then it takes very little dV.

Even better is to launch directly into a slightly inclined orbit when taking off from Kerbin, much better than launching equatorial and then changing inclination later.

[guitarxe]

Even better is to launch directly into a slightly inclined orbit when taking off from Kerbin, much better than launching equatorial and then changing inclination later.

How would I get an encounter with the mun if my orbital plane doesn't match it?

[Snark]

How would I get an encounter with the mun if my orbital plane doesn't match it?

By picking the right time to launch, so that the Mun is located at your orbit's ascending or descending node when you get there.

For example, let's say you decide to forego circularizing to LKO first, and are launching directly to munar intercept. If you were doing that equatorial all the way, how would you do it? You'd sit on the pad and wait until about the time that the Mun rises over the horizon (give or take a few degrees), then launch eastwards to intercept, right?

So, suppose you do that same thing, and burn about the same amount, but instead of heading due east when you make your gravity turn, you head northeast (or even due north). You're launching from the same place, at the same time, with the same periapsis and apoapsis, just with a different inclination. Which means that you'll reach apoapsis at the same time and place as you would have in the equatorial case, which in turn means you will get a Mun intercept-- just from a different angle.

[cybersol]

Even better is to launch directly into a slightly inclined orbit when taking off from Kerbin, much better than launching equatorial and then changing inclination later.

In the general case this would be tricky to pull off timing wise, and you could easily lose more sidereal rotation boost than the small maneuver at mid-plane would take.

EDIT: Ninja'd, but still relevant. Generally this is complicated enough to easily mess it up enough to hurt your fuel situation more than you help it. It the kind of thing I might try if I tried the normal way and ran out of fuel. However, if you know you are going to run out of fuel at launch time, you also have easier options to fix low fuel

Edited June 7, 2015 by cybersol

[Arkalius]

2 * v * sin(i/2)

v is your orbital velocity while making the change, and i is the inclination change amount.

A 60 degree change will require a delta-v that is equal to your orbital velocity at the time you do it. It kind of makes sense. You're trying to change your velocity vector angle by 60 degrees and maintain the same velocity. Thus, the delta-v vector will form an equilateral triangle with the other two.

The slower you are moving, the less delta-v you need to change inclination.

I've done some math on this. If your need to change your inclination by notably more than 45 degrees, and are in a low-eccentricity orbit, it can actually be cheaper to raise your apoapsis of your orbit, then at apoapsis, do your plane change, then lower your apoapsis again than it would be to simply do the plane change on your existing orbit. Of course, this will take a lot longer, but if you have the time you can save up to 40% of your delta-v depending on how much of a plane change you're doing and how high up you push the apoapsis.

[cybersol]

For >90* you can save even more by aerobraking some of the retrograde component, which is particularly useful in the case where you do a full 180 around a body with an atmosphere.

[Coga19000]

Actually, for the Mun, going equatorial and changing inclination at the highest point during the injection may be a bit cheaper.

Also note that any inclination you have at Kerbin orbit is generally going to be accentuated at Munar SOI, so you'll have to take this into account too.