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Let's say a rocket launches from Iceland at 65oN headed for LEO, but instead of using an inclination that works from that latitude, it heads due East. It attains orbital altitude and acceleration, and cuts its engines.

If its path continued due East it would be defying gravity because the orbit wouldn't be around the Earth's center of mass, it would be off to one side of it so it would essentially be constantly climbing. So the pull of gravity has to torque the orbit to a different inclination. How would that work? If it had used the kind of acceleration profile that would be used to launch a rocket from the equator due East, would the pull of gravity just drain its speed until it falls back to Earth? If that had been calculated for and extra acceleration had been done to compensate for it, what would its path be while it attains orbit and how would it change the energy needed?

orbit trajectories from Iceland

Red trajectory can't work, the path would be something like the green?

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The gravity pulling the rocket toward the center of the Earth doesn't wait for the rocket to achieve orbit, so if the rocket is pointing due East for the entire ascent, the course will be somewhat south of East from the beginning, and at the moment of orbital insertion/engine cutoff, the rocket will be somewhere on the southbound leg of the inclined orbit.

If instead the rocket points in whatever direction is required to keep the net velocity vector of the rocket Eastward, it will be pointed slightly northerly, and at insertion it will be at the peak latitude of the inclined orbit and just about to start the south leg.

I'm not certain how much additional fuel each of these trajectories would require.

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For the sake of simplicity, I'll assume the rocket starts off due east, and quickly climbs and accelerates into orbit. It then turns off it's engines and coasts along it's orbit. We'll ignore that first bit, and think about what happens once it's 150km above Iceland and heading due east.

The first thing to realise is that, as far as something in orbit is concerned, there is nothing special about Iceland. Or the north pole, or the equator or anything else. The only thing that makes these points special is that the earth rotates around it's axis, but once the ship is in orbit it doesn't matter how the planet rotates below it. So lets forget about the rotation of the earth for a minute. Now all points on the surface are the same, and you can launch from any one, in any direction, and you'll end up in an orbit around the planet. If you launch east from Iceland, you'll end up in this orbit (viewed edge on):

A picture of the earth, centred on Iceland, with a red line drawn across it from east to west

Image adapted from www.webglearth.com

If we turn the globe to look straight at Iceland, you can see how an orbit can pass through Iceland going east and still be centred on Earth's centre of mass. If we take that same orbit and draw it on a map of the Earth, it would have the familiar sinusoidal shape we associate with inclined orbits. That's just what happens when you draw a circle on a globe, then unwrap it in the (actually slightly odd) way we traditionally unwrap a sphere to make a flat map.

So we were ignoring the planet's rotation there. What happens if we put that back in? Very little changes. Because the planet is rotating, Iceland is moving eastwards fairly fast, we can take advantage of that to make the rocket launch easier. Let's assume that we launch in the same direction, achieve the same speed, but use a little less fuel doing so. The second change is that once the ship is up in orbit, the planet rotates underneath it. If you look back at the picture above, you can imagine the planet rotating about it's axis, and the orbit staying put. When we flatten out the globe to make a map, this means the sinusoidal shape seems to move across the earth.

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    $\begingroup$ To put it another way, OP's red path is a fallacy - to follow that trajectory would require constanly turning the rocket to follow geographic or magnetic east rather than simply proceeding in a straight great circle path parallel to the east vector at ground zero in Iceland (which points due east only at ground zero in Iceland). $\endgroup$
    – J...
    Sep 15, 2015 at 10:18
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First of all, keep in mind that most of the acceleration happen essentially in a single point. The actual trajectory will be somewhat of a typical satellite pass, although it will be slightly irregular at the very beginning. The green line you have drawn is close, but not quite, it will look more sinusoidal.

If you think about it, for any positive inclination rocket, there will exist a point in it's orbit where it is moving due East, at the top and bottom of the orbit. That point is the point where the orbit was achieved.

The inclination will be approximately the same as the latitude for a due East launch. So in the case you provided, the inclination will be about 65 degrees.

If you think about this, the satellite moving straight East will be pulled towards the center of the Earth. As the center of the Earth isn't straight East, then it will start moving to the south, and continue until it passes the equator, in which case the southward motion will slow down.

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  • $\begingroup$ I tried to state the question better now. I understood it has to end up at the inclination corresponding to the launch site's latitude, but it was the process of getting to that equilibrium that i wondered about. The acceleration happens essentially at a single point - are you referring to it being over a relatively small portion of the first orbit? $\endgroup$
    – kim holder
    Sep 14, 2015 at 17:13

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