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The escape velocity of earth is about 11.2 km/s. Let's say it was, for the sake of argument, exactly 11.2 km/s. If I go 11.199999 km/s, I'm going to fall back down to earth. But how can I determine how far I will get before I do so? What pieces of data do I need for this?

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  • $\begingroup$ Suggest you look at space.stackexchange.com/questions/4727/… which is a possible duplicate. $\endgroup$ – Erik Feb 13 '15 at 20:23
  • $\begingroup$ I did, but it doesn't seem to be a duplicate. It might be related in some way, but if it is a duplicate it only underscores how little I know about the topic! $\endgroup$ – corsiKa Feb 13 '15 at 20:41
  • $\begingroup$ Both questions ask for relationships between speeds and apogees. But Corsika seems interested in perigees speeds just below escape. A very interesting set of ellipses! In my opinion this question's more specific interest make it different enough it's not a duplicate. $\endgroup$ – HopDavid Feb 13 '15 at 21:21
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You are talking about an elliptical orbit about the earth. Two points on this ellipse are of interest.

Perigee - point on the orbit closest to earth. Apogee - point on the orbit farthest from earth

If you're launching from earth's surface, the perigee would be 6378 km from earth's center. (6378 is earth's radius). For this problem I will pretend there's no earth atmosphere.

The apogee is what you're interested in, the farthest point from the earth.

The vis viva equation is a nice tool for finding velocity:

$$v=\sqrt{Gm(2/r-1/a)}$$

What is a? That's the semi-major axis. We can find a by taking the average of perigee distance from center and apogee distance from earth's center.

Here is a pic of the stuff I've described so far:

enter image description here

We're interested in the speed when the object leaves earth so r would be 6378 km.

Let's take a look at the vis viva question again:

$$v=\sqrt{Gm(2/r-\mathbf{1/a})}$$

Noticed I bolded the term that uses the semi-major axis quantity. What happens when a gets real big? 1/a gets closer and closer to zero.

So for high apogees you have something very close to

$$v=\sqrt{Gm(2/r)}$$

Which is escape velocity.

"Just a hair under escape" describes a multitude of ellipses!

For example let's imagine an ellipse whose apogee reaches EML1, about 5/6 of the way to the moon. Now let's compare it to an ellipse whose apogee extends all the way to Sun-Earth L2, about 1.5 million kilometers away. The perigee speeds of these two ellipses differ by only .11 km/s!

But when you get apogees that high, 2 body mechanics is no longer a good model. The gravitational influences of the moon and sun are enough to bend the object out of an elliptical path. Launch something as high as SEL2 and it is likely the sun will wrest the object from earth's sphere of influence.

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    $\begingroup$ I see what you're saying - if there was nothing else in the universe, I could go forever before getting pulled back down. But realistically, something else will snag me if I get too far away. $\endgroup$ – corsiKa Feb 13 '15 at 21:33
  • $\begingroup$ That's pretty much it. The big something else is the sun. You can achieve escape velocity from earth but still not get much farther from the sun than 1 A.U. (earth's distance from the sun). To get to Mars you'd need a hyperbolic orbit with a Vinfinity of about 3 km/s. A hyperbola's speed is sqrt(Vinf^2 + Vesc^2). $\endgroup$ – HopDavid Feb 14 '15 at 0:56
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You wouldn't get far, since that wouldn't quite get you escaped from the Earth-Moon system. Your escape velocity is what you would need at the surface of the Earth just to escape Earth. (I get $11.18\,\mathrm{km/s}$.) To escape the Earth-Moon system, you would need $11.25\,\mathrm{km/s}$.

For an escape from a body, if you depart radius $r$ with $1-\epsilon$ times the escape velocity, and you are going directly away from the center of the body with that velocity, then for small $\epsilon$ you will get to $r\over 2\epsilon$ before falling back.

Derivation, starting with the velocity $v$ at radius $r$, just shy of the escape velocity:

$$v_e^2={2\mu\over r}$$ $$v=v_e\left(1-\epsilon\right)$$ $$v^2={2\mu\over r}\left(1-\epsilon\right)^2$$

Relating the velocity to periapsis and apoapsis using the conservation of energy:

$$v^2=2\mu\left({1\over r}-{1\over r_p+r_a}\right)$$

For a trajectory directly away from the center of the body, use a periapsis of zero:

$$v^2=2\mu\left({1\over r}-{1\over r_a}\right)$$

Then:

$${2\mu\over r}\left(1-\epsilon\right)^2=2\mu\left({1\over r}-{1\over r_a}\right)$$ $${1\over r}\left(1-\epsilon\right)^2={1\over r}-{1\over r_a}$$ $$\left(1-\epsilon\right)^2=1-{r\over r_a}$$

For small $\epsilon$:

$$1-2\epsilon\approx 1-{r\over r_a}$$ $$2\epsilon\approx {r\over r_a}$$ $$r_a\approx {r\over 2\epsilon}$$

So in the case where you make the mistake of using the Earth escape velocity without taking into account the Moon, then $\epsilon$ is about $0.006$. So you would get about $500,\!000\,\mathrm{km}$ from Earth before falling back. A little farther than the Moon. That assumes however that the Moon is not close to your path. If it is, then it might either a) give you enough energy to completely escape, or b) send you back to the Earth with less energy.

In the spirit of your example, e.g. if you assume that $11.2\,\mathrm{km/s}$ is exactly the escape velocity, then your $\epsilon$ is about $10^{-7}$. So you would get about 36 billion kilometers away (240 AU!). However other bodies such as the Sun, Venus, Jupiter, and re-encountering the Earth-Moon system would alter your course well before that.

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  • $\begingroup$ So umm... how far is not far? If I was going jjuusstt under whatever value we use for the escape velocity, how far do I get? $\endgroup$ – corsiKa Feb 13 '15 at 21:12
  • $\begingroup$ Wouldn't you be perturbed out of the system (or, less likely, into one of the bodies) quite quickly due to lunar or solar influence? $\endgroup$ – pericynthion Feb 13 '15 at 22:50
  • $\begingroup$ Yes. Perturbed out of the system or back into it. $\endgroup$ – Mark Adler Feb 13 '15 at 23:31
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The easiest way how to compute this is to use specific orbital energy. The following quantity is preserved over time:

$$\frac{v^2}{2} - \frac{G(M+m)}{r}$$

since the mass of a spaceship is negligible compared to the mass of the Earth, we can omit $m$.

The furthest point where we can get is when we consume all kinetic energy, so $v=0$. Knowing the velocity $v$ at the Earth surface $R$, we get

\begin{equation} \frac{v^2}{2} - \frac{GM}{R} = - \frac{GM}{r_{\mbox{max}}} \end{equation}

For the escape velocity $v_e$ we have $r_{\mbox{max}}=\infty$ so

$$\frac{v_{e}^2}{2} - \frac{GM}{R} = 0$$

which leads to

$$\frac{GM}{R} = \frac{v_{e}^2}{2}$$

where $R$ is the radius of the Earth.

Substituting this into the equation for $r_{\mbox{max}}$, we obtain

$$ \frac{v^2}{2} - \frac{v_{e}^2}{2} = - \frac{GM}{r_{\mbox{max}}} $$ and $$ r_{\mbox{max}} = \frac{2GM}{v_{e}^2 - v^2} $$

Substituting the velocity you asked for and the Earth's standard gravitational parameter we get (in km)

$$ r = \frac{2 \times 398600.4419}{11.2^2-11.199999^2} \doteq 3.56\times 10^{10} $$

which is approximately $238 \mbox{AU}$.

(Of course we completely disregarded the effect of all the other celestial bodies.)

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  • $\begingroup$ @MarkAdler Yes, you're right, correcting. $\endgroup$ – Petr Pudlák Feb 14 '15 at 20:35

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