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As mentioned briefly in this answer and this Everyday Astronaut video:

screenshot of video

And just as a reference, if you launched straight east out of Kennedy Space Center, you'd be on a 28.6° inclination which you may notice is the exact latitude of the Space Center.

I can't seem to formulate a reasonable explanation in my head for why this is the case. I'm well aware of how orbital mechanics often defies intuition, but it would make sense to me that launching east would result in a 0° inclination with the orbital plane raised so it's parallel to the equator but above or below it.

Maybe there's an intuitive graphical way to visualize this behavior?

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    $\begingroup$ Gravity pulls toward the center of the Earth, not towards the nearest point on the polar axis, so the latitude-parallel orbits you’re envisioning aren’t possible. $\endgroup$ – Russell Borogove Jan 12 at 19:49
  • $\begingroup$ Makes sense when put into those words, thanks @RussellBorogove $\endgroup$ – ifconfig Jan 12 at 19:54
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    $\begingroup$ I still can't read "KSC" as anything other than "Kerbal Space Center" $\endgroup$ – Starfish Prime Jan 12 at 20:07
  • $\begingroup$ Imagine if, under your rules, a rocket launched from the north pole. Would it just hover in place? Lines of latitude other than the equator tend to curve to one side, but gravity only pulls objects directly towards the earth. $\endgroup$ – mm201 Jan 14 at 18:05
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The center of the Earth is, for any reasonable approximation, in one of the focus points of an elliptical orbit. For a circular orbit, there is only one focus point, so the center of the Earth is in the center of the orbit.

The plane of the orbit thus would intersect both the center of the Earth as well as the launching site.

If the launch site was on the equator, it's not difficult to imagine that if you launch exactly due East, i.e. tangential to the target orbit, the orbital plane is coincident with the equatorial plane, so your orbital inclination equals the latitude of your launch location: 0 degrees. But as your launch location moves away from the equation, your orbital plane starts to tilt with the same amount of degrees as your latitude, because you must keep the intersection:

Schematic of orbital inclination

(Imagine the grey plane is the equator and the green one the orbital plane - source: Wikipedia)

This is only true if you launch due East. If you launch in any other direction, your orbit inclination is different.

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    $\begingroup$ Thank you very much, the extra bits of explanation which must seem extraneous helped me a lot with understanding this answer. $\endgroup$ – ifconfig Jan 12 at 19:48
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    $\begingroup$ "thus must intersect" is a little strong - launches routinely don't, but they're not optimal. Perhaps "thus would intersect" $\endgroup$ – JCRM Jan 12 at 22:00
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    $\begingroup$ "The center of the Earth is, by definition, in one of the focus points of an elliptical orbit." – By which definition, exactly? (I suggest removing "by definition" from this sentence.) $\endgroup$ – Tanner Swett Jan 13 at 4:19
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    $\begingroup$ I was thinking of doglegs, which mean the orbital plane doesn't pass through the launch site (corrected for the rotation of the earth) $\endgroup$ – JCRM Jan 13 at 8:02
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    $\begingroup$ @JCRM I know. But since the question was specifically about launching east, I didn't want to make it more complicated than necessary and only put the final sentence as disclaimer. $\endgroup$ – Ludo Jan 13 at 9:22
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The short answer is that a spacecraft is attracted to the center point of the earth, not to the earth's rotational axis.

[I]t would make sense to me that launching east would result in a 0° inclination with the orbital plane raised so it's parallel to the equator but above or below it.

Here's one explanation of why that wouldn't happen that you might find intuitive:

Once a spacecraft has been launched into space, it doesn't "know" anything about the earth's rotation. In particular, it doesn't know where the equator is, so there's no way for it to know that it's supposed to stay parallel to the equator. Likewise, it doesn't know which way is east, so there's no way for it to know that it's supposed to keep going east.

So what does happen?

After the spacecraft is launched, there is exactly one plane which contains all of the following:

  • the spacecraft
  • the spacecraft's velocity (or, alternatively, the place where the spacecraft will be one second from now)
  • the center of the earth

This plane will be one that slices through the equator at an angle (unless the spacecraft is flying straight east or west over the equator). It can't be a plane that lies parallel to the equator, because such a plane would not contain the center of the earth.

The earth's gravity will never cause any of these three things to exit this plane, so all three will stay on this plane, and the spacecraft's path will cross over the equator twice per orbit.

One final point. Suppose you launch a spacecraft due east from a location very close to the south pole. Would you expect the spacecraft to orbit in tight little circles, remaining above the south pole the whole time?

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  • $\begingroup$ Great analogy/example $\endgroup$ – Lightness Races with Monica Jan 14 at 15:05
  • $\begingroup$ centre of mass isn't quite the same as the centre of the earth $\endgroup$ – JCRM Jan 15 at 11:47
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Earth's gravity pulls you towards the centre of the Earth, so if you're above Kennedy, that pull has a Southwards component, as well as the component towards the Earth's axis. So your path curves South, so that in the end the orbit spends equal amounts of time North and South of the equator, and the pulls in that direction balance out over time. All orbits lie on planes through the centre of the Earth, for basically this reason.

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  • $\begingroup$ I feel like this could be more effectively explained with the help of a visual aid of some sort... as it is I think I can just barely follow the logic here. $\endgroup$ – ifconfig Jan 12 at 19:37
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    $\begingroup$ But it is possible to launch to a polar orbit from a launch pad far away from the poles. $\endgroup$ – Uwe Jan 12 at 19:37
  • $\begingroup$ @ifconfig looks like the accepted answer has covered that. $\endgroup$ – Steve Linton Jan 12 at 20:04
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    $\begingroup$ @Uwe, sure, you just don't launch due East. There is a great circle through Kennedy and the poles and that's what you follow (the rotation of the Earth complicates this a bit). $\endgroup$ – Steve Linton Jan 12 at 20:04
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Reductio ad absurdum

If you could choose freely on which circle to orbit, the most convenient place to take off from would be the North pole. That would set the circle diameter to zero. You would then climb to whichever altitude you pleased and remain there, immobile in space, for as long as you wanted. How cool would that be?

An attempt at analogy

In reality, the trajectory of a circular orbit is like that of a stone spinning in a sling: Earth pulls you toward its centre and the centrifugal force pulls you away in the opposite direction. Since the forces balance each other out, all you can do is follow the circle.

Gravity acts like the rope preventing the stone from flying off. If the planet suddenly disappeared, you would shoot straight into outer space as if released from that sling.

Just like the stone circles around your hand (where the strings of the sling are held), the satellite circles around the centre of the Earth (where the "gravitational rope" is tied).

Now why would we want to launch a rocket due east?

Simply standing on Earth's surface, you spin with the planet (one full circle per day), and this gives you a rotational speed that translates to an initial flight speed in the referential of the orbit (i.e. the Earth immobile in space).

This time, the speed is proportional to the distance to earth's rotation axis.

If you stand on a pole, you just spin around on a zero radius circle, and your speed is zero. If you stand on the equator, your speed is maximal and amounts to about 1,700 km/h or 0.5 km/s.

As it happens, the speed that must be reached to go into a low orbit is roughly 8 km/s. In practice you will consume more fuel due to air drag and other technical issues, for a total equivalent of about 10 km/s. That means the rotation of Earth gives you an initial impulse of about 5% of the speed you need.

This initial speed clearly points due east (the direction toward which the Earth spins). So if you follow that course, you simply add to this initial speed until you reach orbit, spending 9.5 km/s worth of fuel.

If you went due West, you would first have to cancel that initial speed, and then start accelerating in the opposite direction, going from -0.5 to 10 km/s for a total of 10.5 km/s.

This 1km/s difference might not look like much, but as it happens, the fuel cost of each extra km/h is exponential, which means the 10% or so speed difference might translate into a much more costly increase in mass and needed fuel.

Why does the inclination of an orbit launched due east match the launch site latitude?

The rocket is flying straight East until it reaches its orbit. Going East means remaining constantly at the latitude of the launch site.

During this part of the flight, the rocket follows roughly the trajectory you describe: a circle in a plane perpendicular to Earth's rotation axis. In reality it's rather a flat spiral, due to increasing altitude, but it would be a circle if the rocket continued applying thrust after reaching its orbit.

However, this is only possible during the powered part of the flight, because the rocket is actively fighting against gravity.
While the rocket's speed vector points due east, the engine nozzles (controlling the thrust vector) are tilted slightly northwards, to counteract the gravity trying to pull the rocket southwards into this earth-centred circle (or rather ellipse slowly turning into a circle as the rocket comes closer to the orbital insertion point).

As a side note, the guy in the video is slightly wrong when he plays around with Kerbal Space Program. Pointing the engine nozzles of a rocket due east will produce a trajectory that veers slightly southwards.
The drift is not huge because the ascension time is fairly short in comparison with the duration of an orbit, but still...

Anyway, soon as the engine cuts, the trajectory becomes only subject to orbital mechanics laws and starts following this circle centred around Earth.

There is no way gravity could pull the rocket further North. The orbital insertion point is the highest latitude the rocket will ever reach (unless the engine is re-ignited). The orbit must then be a circle centred around Earth that reaches its highest point at the launch latitude.

This is the very definition of an orbit whose inclination matches the launch site latitude.

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Another way to look at it:

Consider an object orbiting Earth at some inclination relative to the equator. Now consider its path projected onto Earth's surface. If you plot that path as latitude vs longitude, you'll get a sinusoid. You can see this in views of mission control rooms where an orbiting spacecraft is displayed on an equirectangular world map. There are only two points where the path is precisely due East (assuming a prograde orbit): at its northernmost point, and its southernmost point.

So, if a rocket were launched due East from some point in the Northern hemisphere, and assuming it achieved orbital velocity promptly, it would have to be at the Northernmost point in its orbit.

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