Did Juno start off its course on the ecliptic plane from the earth orbit and then left it at insertion to Jupiter's orbit over its north pole or it was always in that plane?

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    $\begingroup$ Jupiter is not exactly at the ecliptic, so Juno was only approximately in that plane during the transfer. You can get any inclination you want at the orbital insertion though. $\endgroup$ – Hohmannfan Jul 7 '16 at 13:36
  • $\begingroup$ Is it not too costly though to change the plane of orbit? It's not just slowing down the space-craft, it's changing the course into an almost perpendicular plane. $\endgroup$ – kamran Jul 7 '16 at 13:55
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    $\begingroup$ To change the plane is costly, once in orbit, but the initial choice is basically free. $\endgroup$ – Hohmannfan Jul 7 '16 at 14:04
  • $\begingroup$ Are you referring to it being in a polar orbit? Jupiter did the work of putting it in that orbit. The craft only had to be in a spot where the pull of Jupiter's gravity bent it into an orbit over the poles. So it was above it, or below it, when that pull became strong enough. $\endgroup$ – kim holder Jul 7 '16 at 14:40
  • $\begingroup$ That is the question!! because the tangent intercept trajectory is least costly if it remains on the ecliptic plane which goes through the center of Jupiter not above or below its poles. the pull of gravity of Jupiter has been centering Juno for all its trajectory in the planetary orbit of Jupiter not above it! $\endgroup$ – kamran Jul 7 '16 at 15:12

The plane of the orbit around the Sun is not directly related to the planet-relative plane of the hyperbola on approach to Jupiter, or correspondingly the orbit around Jupiter after orbit insertion. The plane and shape of the orbit around the Sun relative to Jupiter's equatorial plane determines its approach declination to Jupiter. That is essentially the negative of the latitude on Jupiter of your atmospheric entry point if you tried to target your spacecraft to hit the center of Jupiter. The magnitude of the approach declination limits how equatorial the orbit can be, but not how polar it can be.

On approach to a body, your target can be anywhere in a plane perpendicular to your approach trajectory, at essentially zero cost. That plane is called the B-plane:


For convenience, you target a point on the B-plane that you would intersect if the planet were not there, or if it had no gravity. You actually cross the B-plane inside of that point due to the planet's gravity. You get to pick how far from the center of the planet your target point is, and the angle of that point anywhere around the planet. The distance from the center determines your closest approach distance, or your atmospheric entry flight path angle. The angle determines the plane of your planet-relative approach trajectory, which will be the plane of your orbit.

Your approach declination defines a line through the center of the planet entering at a latitude equal to the negative approach declination on one side and exiting at a latitude equal to the positive approach declination on the other side. That line is labeled "S" in the diagram. Imagine a plane that contains that line, and you are allowed to rotate that plane around that line. Those are your allowed orbit planes. One such plane is the "Trajectory Plane" in the diagram. Since your plane must go over the latitude of the magnitude of the approach declination, your orbit plane inclination to the planet equator cannot be less than that. If it were less than that, then your orbit could not go over that latitude.

Juno's approach declination was about 8°. Then by choosing how you target the Jupiter closest approach, which can be anywhere in a circle around Jupiter, you can pick any final orbit inclination from 8° (prograde) to 90° (polar) to 172° (retrograde).

After orbit insertion, you can do maneuvers or swingbys of satellites to change the orbit plane, e.g. if you wanted to be more equatorial than the approach geometry permitted. Cassini is the champion of this, having dramatically changed its orbit plane many times over the course of its mission at Saturn, using Titan gravity assists.

  • $\begingroup$ Nice diagram! Again the point is that not all the points around the Jupiter on your B plane are possible and of the possible ones some are much more costly. let's not forget both Earth and Jupiter lay in the ecliptic plane. therefore even though we have a wide choice of shooting targets far scattered horizontally in this plane we have a hard time shooting at targets off this plane, up or down. We take advantage of centripetal force of Juno's sun orbit to keep it from falling and crashing into Jupiter but that luxury is not available for targets off the heliocentric plane meaning up or down! $\endgroup$ – kamran Jul 7 '16 at 17:44
  • $\begingroup$ @kamran Jupiter is not centered in the ecliptic plane. It is inclined 1.31 degrees to the ecliptic. Earth, of course, is in the ecliptic by definition. $\endgroup$ – called2voyage Jul 7 '16 at 18:00
  • $\begingroup$ @kamran No, all the points on the B-plane near the planet are possible and targeting any point you like in the B-plane is essentially zero cost. $\endgroup$ – Mark Adler Jul 7 '16 at 18:09

As noted in comments, by selecting the launch time and initial trajectory, any relative inclination between Jupiter and the spacecraft could be achieved.

That said, while plane changes measured in degrees are indeed expensive, for an orbit that goes out to 5 AU, one degree of relative inclination corresponds to a deflection of something like 12 million km.

If, after Earth flyby, mission control had realized that Juno was on a direct collision course with Jupiter, they could have done a 1-meter-per-second correction "northward"; Juno would then pass over Jupiter's North Pole at a distance of about 60,000km. In the solar frame of reference it's still a low inclination orbit; in Jupiter's reference it's polar.

  • $\begingroup$ You'r assuming the trajectory of Juno to be a straight line from the point of possible correction which it is not! It's a very complex 3d curvature with many inflection points depending on the part of solar system it is passing. Very simplified it could be assumed of piecewise sections of parabolas. let's consider only earth and Jupiter. the path has to start just slightly tilted up from horizon to north of eclectic plane anticipating the gravity of Jupiter, somewhere along the way the plane of its path will twist to Jupiter accelerating! very complex 3d path under different force fields. $\endgroup$ – kamran Jul 8 '16 at 21:34
  • $\begingroup$ As it's not going far out of the ecliptic, the angle of the force produced by Jupiter and the other massive bodies of the solar system will be nearly in-plane over the majority of the flight. For rough approximation like this we can just about ignore it. $\endgroup$ – Russell Borogove Jul 8 '16 at 22:30

Basically, Juno came in towards Jupiter from slightly below it. This allowed it to slow down into an orbit which takes it around the poles of the planet. Over the course of its journey to Jupiter, it was at a very slight angle, relative to the plane of Jupiter's orbit. This angle was determined by the flight engineers at its launch, and refined by two deep space maneuvers and an Earth gravitational assist flyby. This slight angle was enough to put the craft below Jupiter, because its journey was very long. More information about Juno's trajectory can be found in Spaceflight101's article, which includes a Juno trajectory animation.

  • $\begingroup$ @kamran there is no conflict between Mark's answer and this one. Mark's just gives a more technical sense of how it is set up. $\endgroup$ – kim holder Jul 7 '16 at 21:06

I estimate the launch time and coordinates must have been optimally designed so that the gravity assist would already position Juno in an inclined plane much more than what was needed ultimately but adjusted over the long trajectory by the pull of Jupiter into correct entry point.

NASA would have two tools to angle the shot: earth's orbit and its axial rotation.

I'd assume they designed the trajectory purposely not in the plane of solar system at all but in an inclined plane winding up and down the apogee of its orbit as needed riding the great roller coaster. Hats off to Isaac Newton.
quote from Wikipedia page on Ulysses probe

to change the orbital inclination of a spacecraft a large change in heliocentric velocity is needed. However the necessary amount of velocity change to achieve a high inclination orbit of about 80° far exceeded the capabilities of any launch vehicle.


Some of the information contained in this post requires additional references. Please edit to add citations to reliable sources that support the assertions made here. Unsourced material may be disputed or deleted.

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    $\begingroup$ That is a word salad that makes no sense. $\endgroup$ – Mark Adler Jul 7 '16 at 20:21
  • $\begingroup$ dropbox.com/s/4i5b7tpp8vmp0to/juno%20orbit_1.jpg?dl=0 i hope the sketch i did is easy to see. 2 alternative orbits are shown, polar and equatorial. My question is how did they set the orbit entry that it took the polar orbit. I have a hard time believing all entry points are equal energy. A comment about the fact that Jupiter is not exactly on the ecliptic plane doesn't apply because Jupiter will pull Juno into its orbit; it does not recognize ecliptic orbit. I would like to here the detail of how they positioned Juno in the final vertical parabola of pre-entry for polar orbit. $\endgroup$ – kamran Jul 8 '16 at 0:26
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    $\begingroup$ You are still confusing Jupiter orbit inclination with Solar orbit inclination. They are two different things. In either final orbit around Jupiter that you drew, the inclination of Juno's path around the Sun has not substantially changed inclination, since it is going around the Sun with Jupiter. $\endgroup$ – Mark Adler Jul 8 '16 at 0:56
  • $\begingroup$ What your drawing is missing is scale. The ratio of Jupiter's distance from the sun to the apojove of Juno's orbit is over 5000 to 1. A tiny deflection at Earth departure is leveraged over that great distance to make a much larger deflection at B-plane. $\endgroup$ – Russell Borogove Jul 9 '16 at 0:01

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