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Would a lighter-than-air craft in the mid atmosphere at 80,000 feet altitude need to achieve the same velocity to escape earth gravity as the space shuttle?

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Yes. Escape velocity does not depend on the flying object's mass, but only on that of the earth. The precise formula is given by


where $G$ is the gravitational constant, $M$ the mass of the earth and $R$ the object's distance from the center of the latter.

This is of course only true if one ignores the interaction between the atmosphere and the vehicle.

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So if the shuttle is on the launchpad, then it would only require more momentum because of the smaller distance to the earth's center of gravity? – David Wilkins Jun 24 '14 at 20:37
The interaction, meaning aerodynamics? Or bouyancy? – David Wilkins Jun 24 '14 at 20:39
@DavidWilkins: Regarding your second question: yes. I do not understand your first one, though. – Frederic Brünner Jun 24 '14 at 20:41
@DavidWilkins for your first question ("So if the shuttle is on the launchpad...") the answer would be "yes" if you substitute "momentum" for "velocity" in your question. The space shuttle would only need a larger velocity to escape than the plane at 80,000 feet because it is closer to the center of the earth. – Joshua Jun 24 '14 at 20:59
Technically, they need the same momentum to escape the Earth's gravitational pull. If one object is pushing another object, and neither one reaches escape velocity, they would escape Earth if they have enough momentum between the 2 of them. Same thing with pulling, objects inside of another object, etc. – trysis Jun 25 '14 at 2:14

Assuming no air resistance, friction, etc., the answer is usually. @FredericBrunner is pretty much correct in his analysis, if we are assuming the object being talked about is the only object in the equations. However, if there are 2 or more objects cooperating or competing in getting away from Earth's gravitational pull, things get slightly more complicated. As noted in answers to this question (similar but not exactly a duplicate), if one object is going at a very low speed, but another object is pushing it along with a force sufficient to accelerate both objects to escape velocity and keep it there, both objects will escape Earth's gravitational pull. Therefore, there is technically no such thing as "escape velocity"; it's more like "escape momentum" or "escape force" (or work, or any other combination of speed & mass).

The main reason either object would not be near escape velocity but would still escape the Earth's pull, however, is usually because the forces pushing the object upward are not much greater in total than the sum of the forces pushing the object down. In absolute terms, if the object is of mass m and the acceleration required to maintain escape velocity assuming no external forces is a, the difference between the forces causing the object to move upward and the forces causing the object to move downward would be between 0 and ma. Thus we normally don't see this very much.

To answer your question, any two objects at the same distance from Earth need the same velocity to escape the Earth's pull if the net forces acting on the respective objects are identical in magnitude and direction. Technically, they need the same velocity to escape at each moment, as once the difference between the forces gets to be less than the force required to escape gravity, they of course start moving towards the Earth again.

I haven't gone into everything involved, but I hope this was a good start. As I said, the previous answer was close enough in most circumstances, and of course my answer also fails in some situations.

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Given that the spacecraft of mass $m$ is moving initially with escape speed $v_e$ away from Earth, and given that the speed of Earth (of mass $M$) is initially $0$, therefore the center of mass of these two objects moves always at speed $$u = \frac{m}{M + m} v_e$$ (with respect to the system in which Earth had initially speed $0$).

In an idealized "escape experiment" both the spacecraft and Earth should eventually be moving almost at (and ever closer to) the same speed $u$ as the center-of-mass system; that's when the (non-relativistically approximated) initial kinetic energies of the spacecraft and of Earth (with respect to the center-of-mass system) are almost (and ever more fully) converted to (also approximate) gravitative potential energy.

Equating these initial kinetic energies and gravitative potential energy:

$$\begin{eqnarray}\frac{m}{2} (v_e - u)^2 + \frac{M}{2} u^2 & = & \frac{G~M~m}{R} \\ \frac{m}{2} \left(\frac{M}{M + m} v_e\right)^2 + \frac{M}{2} \left(\frac{m}{M + m} v_e\right)^2 & = & \frac{G~M~m}{R} \\ \frac{v_e^2}{2}~m~M~\left( \frac{M}{(M + m)^2} + \frac{m}{(M + m)^2}\right) & = & \frac{G~M~m}{R} \\ \frac{v_e^2}{2}~m~M~\left( \frac{1}{M + m}\right) & = & \frac{G~M~m}{R} \end{eqnarray},$$

and consequently

$$v_e = \sqrt{ \frac{2~G~(M + m)}{R} }.$$

In practical cases the mass of the spacecraft may of course be negligible compared to the mass of Earth; so

$$v_e \simeq \sqrt{ \frac{2~G~M}{R} }.$$

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