I have been reading on orbital mechanics and its strewn everywhere that beyond the escape velocity from any orbit around the earth, the trajectory is a hyperbola that approaches V_infinity and with this velocity we approach other planets (or solar system objects).

But in reality, isn't the spacecraft leaving the earth's sphere of influence but not yet the Sun's? Giving enough energy to a craft to get out of earth orbit should put it into a heliocentric orbit, which is still elliptic.

I was convinced of this when I read about the Hohmann transfer, which is an elliptic transfer orbit with Sun at the prime focus. But what still confuses me is the hyperbolic approach (V_infinity approach) is still considered for the spacecraft's rendezvous with the planet's atmosphere or local orbit.

Is this an approximation? Or am I missing something in my understanding?

  • 2
    $\begingroup$ Related: space.stackexchange.com/q/40560/38535 $\endgroup$
    – PM 2Ring
    Commented Apr 17, 2022 at 10:12
  • 5
    $\begingroup$ Perfect circles don't exist in nature. Perfect ellipses don't exist in nature. Perfect hyperbolas don't exist in nature. Mathematics is reductionist, exact, and smooth. Nature is complex and fuzzy. All science is an approximation. $\endgroup$
    – J...
    Commented Apr 17, 2022 at 12:01

3 Answers 3


Sure, it's an approximation. Perfect conic trajectories only happen in 1 or 2 body systems, a 2 body system can be reduced to an equivalent 1 body system. When $n>2$ in an n-body system, the trajectories can get rather messy, and you mostly need numerical integration to compute them properly, except in special cases.

The whole point of using spheres of influence is to avoid those messy n-body calculations as much as possible. We break up the path of the spacecraft into sections, and in each section we only look at the gravity relative to the current sphere of influence. We combine everything together using the patched conic approximation.

When we put a spacecraft into orbit around the Earth, its trajectory relative to the Earth is (almost) an ellipse. Its trajectory relative to the Sun is a slightly perturbed version of Earth's orbit around the Sun. The Earth's orbital speed is ~30 km/s and LEO orbital speed is around 7.8 km/s, but the Earth's orbital radius around the Sun (1 AU) is around 22,700 times the craft's orbital radius around the Earth. If you plot the Earth & craft orbits relative to the Sun, they'll look identical, unless your screen is >22,700 pixels wide. ;)

So to compute a Hohmann transfer for the craft from Earth orbit to (say) Mars orbit, we can get a pretty good approximation by pretending that the craft starts on a standard conic solar orbit, with some extra velocity.

Escape speed is $\sqrt2$ times the circular orbit speed at that radius. So when the craft leaves Earth orbit at the start of the Hohmann transfer its speed relative to Earth is hyperbolic, but its speed relative to the Sun is certainly elliptic, since its travelling on a Hohmann ellipse.


This is the difference between two-body approximations and the multibody reality.

"Spheres of influence" don't really exist. You can draw a region (which is not a sphere) where one large, isolated body such as a planet has a larger gravitational influence than any other, but actually every object in the universe has at least a little influence on every other object, and one smoothly shifts from "things other than the Earth barely matter" to "things other than the Sun barely matter".

It changes gradually

If you focus near the Earth, the escape trajectory will look very much like a hyperbola (slightly perturbed by the Moon and by the fact that the Earth is not perfectly spherical). As you zoom out, the Sun will be more and more important and the long, nearly straight tail of the Earth-centric hyperbola will curve into a Sun-centric ellipse.

It's a useful approximation.

Hyperbolas and ellipses are very useful for approximations because they are a way to get pretty close, and because the math is vastly easier, especially in the era before cheap, powerful computers on every desk. This is why Kerbal Space Program uses them in its "patched conics" simplification of orbital dynamics.


First off, it's important to note that there's a very important point about reference frame here. Right now, just standing on the Earth's surface, you are moving on a path that is, to a first order, a (slightly) elliptical orbit around the Sun, because the Earth is carrying you along in its orbit.

If you then climb atop a suitably good rocket and launch yourself to an escape trajectory from Earth, you'll now be moving on what amounts to a modest perturbation of this orbit whose exact shape will depend on the direction of launch and any excess speed in the case where there was more fuel than the minimum needed and the engine was allowed to burn past that point.

However, from the reference frame on the Earth, then of course, you are at rest (presumably, unless moving in some other way while reading this). When you launch off to your escape course, this is the reference frame where a hyperbolic trajectory is seen - up until a point.

I bring up this point about reference frame because it's necessary for the next part, which is in ascertaining where that "point" actually is (it's not actually an exact location, but rather a transition zone), as it has to do with the balance of forces affecting your motion.

Since Earth's core (not surface; we don't want to consider the rotation around the axis for this purpose!) is not an inertial reference frame, the force description must include a "fictitious" (i.e. not generated by an interaction) centrifugal term, along with the gravitational forces due to both the Earth and the Sun (let's ignore the Moon for now, that will just complicate [if not 'terminate with extreme prejudice' :D] the trajectory even more if you have an encounter - you see what's going on here? - and assume the coast is clear under the launch circumstances).

The description of the escape as following the hyperbolic course, then, is valid so long as the assumption under which it's made - which is basically that the sole source of gravity is the Earth, and it is a perfectly symmetric gravitator - is valid. In this case, the latter part is not so important as the former: the model will fail once the gravitational force from Earth is no longer large compared to the sum of these other two forces.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.