In order to achieve a stable orbit around Earth, a spacecraft has to travel at velocity not smaller than approximately 7.8 km/s.

If my understanding is correct, this figure refers to component of velocity vector which is perpendicular to the line connecting the spacecraft and the center of the Earth (or as it's called "orthogonal component", in the answer to this question).

  1. Let's consider an example to go "from the opposite": A launcher with a spacecraft is launched from North pole (to exclude influence of the Earth's rotation), trajectory is controlled to be strictly perpendicular to the surface (i.e. in radial direction from the center of the Earth), when the last burn is complete, at 200km altitude the spacecraft reaches velocity of 8.0km/s.

Am I correct to assume that in this case it is not going to orbit the Earth, instead it would continue in radial direction (with reference to local Earth coordinate system) until velocity drops to 0, and then fall back following the same trajectory (if we exclude solar wind, gravity influence of neighbouring planets etc.), i.e. following straight line down to the Earth's center (in local coordinate system, of course; with relation to the Sun it would be a curve)?.

  1. If in the above example the spacecraft achieves velocity slightly larger than Earth escape velocity, say 11.5km/s at 200km altitude, would it be correct to assume that it would leave the Earth gravity influence and continue to orbit the Sun in an orbit slightly inclined to ecliptic?

  2. Now let's consider a spacecraft in a circular orbit around the Earth at the Moon's radius (again, for simplicity let's assume the Moon is not there) with (orthogonal) velocity of approximately 1 km/s, which should be just enough for the orbital velocity at that altitude. Now the spacecraft performs a burn to reach the Earth escape velocity at that orbital radius, which is approximately 1.4km/s.

My main question: is there a requirement for how the actual velocity vector of a spacecraft has to be located with relation to the Earth center in order to escape?

In example 2 the velocity vector is radial, i.e. goes along the line that connects the spacecraft and the Earth's center, in example 3 it is tangential, but in both cases the spacecraft escapes the Earth.

Or maybe there's another option: a fundamental flaw somewhere in my assumption(s)?

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    $\begingroup$ This is a bit (or perhaps a lot) like Given r/a, what are the limits on the direction that an orbiting body could be moving (e.g. solid angle vs r/a)? in the sense that the answer is the same: the direction doesn't matter. (which also surprised me!) $\endgroup$
    – uhoh
    Commented Dec 24, 2019 at 0:27
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    $\begingroup$ @uhoh Thank you for the link. It was interesting to find out that the formula for escape velocity is actually derived from that vis-viva equation $\endgroup$ Commented Dec 24, 2019 at 1:22
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    $\begingroup$ It's amazingly useful. A quick check shows that I've used it in 37 posts! $\endgroup$
    – uhoh
    Commented Dec 24, 2019 at 1:26
  • $\begingroup$ Escape speed is escape speed, direction is irrelevant. It could even be straight down. Just igore/prevent any slowdown due to factors other than gravity. Such as lithobraking. $\endgroup$ Commented Jul 10, 2021 at 19:48

1 Answer 1


According to Wikipedia:

Escape velocity is actually a speed (not a velocity) because it does not specify a direction: no matter what the direction of travel is, the object can escape the gravitational field (provided its path does not intersect the planet).

It may seem nonintuitive that an escape trajectory directed “downward” could escape just as easily as one directed upward, but on the downward leg you’ll accelerate by some amount, then after you make your closest approach to Earth you’ll begin to slow down in the exact reverse of your acceleration, so when you reach the same altitude on your outbound leg you’ll be going exactly the same speed you started with, in a different direction (provided you don’t hit anything along the way), so the “upward at 11km/s” case is equivalent to the “downward at 11km/s” case. (Proving that this equivalency holds for any escape angle is left as an exercise.)

Your suppositions in cases 1 and 2 are correct, by the way. I think the influence of solar gravity may bring you back into Earth’s sphere of influence some time later in case 2, but that’s another story.


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