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I understand that the escape velocity of Earth is 11 km/s. However, Earth's gravitational sphere of influence is not infinite, so it is possible to go slower than that and still escape the sphere of influence (Because of the sun.) If a rocket starts accelerating from 0 on Earth's surface, what speed would it have to be going at, say, 100 km above the Earth's surface for it to escape Earth's gravity? How would you calculate this?

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  • $\begingroup$ Where the rocket starts from is irrelevant. All you need is the gravitational force at the altitude of interest to calculate escape velocity at that location. $\endgroup$
    – Carl Witthoft
    Nov 19, 2019 at 12:41
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    $\begingroup$ Escape velocity is a function of altitude above a body and the mass of that body, it is not a single fixed value. Escape velocity at 1000k is different to escape velocity on the surface of the earth. $\endgroup$
    – PeteBlackerThe3rd
    Nov 20, 2019 at 15:25

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Earth's sphere of influence has a radius of about 924000km. A highly eccentric orbit with a perigee at 100km altitude and an apogee at the SOI radius has a semimajor axis of 465239km. Throwing that into the vis-viva equation $v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right)$ where $G$ is the gravitational constant, $M$ the mass of earth, $r$ the orbital radius (not altitude) at perigee and $a$ the semimajor axis of the orbit gives a velocity at perigee of 11.05km/s. Using the escape velocity equation $V_e = \sqrt{2GM \over r}$ you'd get an escape velocity at 100km altitude of 11.09km/s, so there's a small saving to be had but really, the edge of the sphere of influence is quite a long way away.

Whilst this isn't quite what you asked for, I'd be startled if your direct ascent trajectory involved a speed dramatically different from this figure.

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    $\begingroup$ You're kind of going at a simple problem in a difficult way :-) $\endgroup$
    – Carl Witthoft
    Nov 19, 2019 at 12:43
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    $\begingroup$ @CarlWitthoft eh, it works. Does something really count as difficult if it doesn't involve solving anything, or even engaging in any integration? $\endgroup$
    – Starfish Prime
    Nov 19, 2019 at 21:28
  • $\begingroup$ This doesn't actually answer the question. $\endgroup$
    – Russell Borogove
    Nov 20, 2019 at 2:27
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Earth's gravitational sphere of influence is not infinite

That's your problem. The force of gravity does have an infinite range. There is no place in the universe where earth's gravity is not felt.

As a result, it does not matter where you start, in order to escape earth, you need 11 km/s relative to the earth. If you first get into orbit at 100 km, then you already need a speed of 7 km/s for that. From that orbit, you only need an additional 4 km/s to escape earth altogether - but only because you are already going at 7 km/s.

EDIT

As pointed out by @uhoh the escape velocity does vary with altitude. However, the difference at 100 km is so small that I ignored it. Like the OP I approximated the escape velocity as 11 km/s. In actual fact, at ground level it is 11.186 km/s, and at 100 km it reduces to 11.099 km/s.

The same approximation also ignores the fact that, if you are far enough from earth (924,000 km), the sun's gravity is stronger than earth's, and you leave the earth's Sphere of Influence - which is not an exact sphere. This is further complicated by the fact that it ignores the influence of the other planets and the moon. For example, when travelling to the moon, the gravitational field of the moon exceeds that of the earth (and the sun) when you're about 40,000 km from the moon.

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    $\begingroup$ Earth's gravitational sphere of influence is certainly finite and has a radius of 945,000 km. I recommend that you have a look at that link and then revise your answer accordingly. It is true that there is no limit to a $1/r^2$ force (other than the speed of light), but "sphere of influence" is a well-recognized technical term. $\endgroup$
    – uhoh
    Nov 19, 2019 at 1:03
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    $\begingroup$ Also, Earth's escape velocity is certainly lower at 100 km, about 0.8% lower in fact! Ignoring atmospheric drag and rotation, if it is 11.186 km/s at the surface (the ballistic speed necessary to asymptotically escape to infinity), you'd only need 11.099 km/s starting from 100 km altitude. $\endgroup$
    – uhoh
    Nov 19, 2019 at 1:10
  • $\begingroup$ Escape velocity from a body does depend on starting altitude, as is clear when you remember that no matter what altitude you start from, that body's gravity will be decelerating you the whole way. $\endgroup$
    – Russell Borogove
    Nov 19, 2019 at 2:47
  • $\begingroup$ @uhoh Your comment looks like a pretty decent answer to me. May I suggest you expand it a bit into one? $\endgroup$
    – Diego Sánchez
    Nov 19, 2019 at 10:37
  • $\begingroup$ @DiegoSánchez my comments are meant to be coaching; hopefully the author will revise the answer and use some of this. If that happens I'll reverse my temporary down vote to an up vote. When people revise a post in response to a comment, they usually will leave a short message like @uhoh I've made an edit, how does that look? I'll get a notification then and can adjust my vote. In the mean time if someone else posts an answer that would be great too! $\endgroup$
    – uhoh
    Nov 19, 2019 at 12:13

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