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Including air resistance, what is the escape velocity from the surface of the earth for a free-flying trajectile?

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For the lower atmosphere, where most of the air is, the temp, pressure, and density of air is given by:

$T=T_0-Lh$

$p=p_0\left(\frac{T}{T_0}\right)^{\frac{gM}{RL}}$

$\rho = \frac{pM}{RT} $

using the following constants:

  • sea level standard atmospheric pressure p0 = 101325 Pa

  • sea level standard temperature T0 = 288.15 K

  • Earth-surface gravitational acceleration g = 9.80665 m/s2.

  • temperature lapse rate L = 0.0065 K/m

  • universal gas constant R = 8.31447 J/(mol·K)

  • molar mass of dry air M = 0.0289644 kg/mol

The force due to air resistance can be written:

$F = -\rho v^2 C_d A$

where $C_d$ is the coefficient of drag, $v$ is the velocity, and $A$ is the surface area of the projectile. The goal is to get out of the atmosphere (where force of gravity is roughly constante) with the Earth's escape velocity, $11.2$ km/s. For a bullet-shaped 1-kg projectile of steel, $C_d \approx 0.04$ and $A \approx 4\times 10^{-4}$ m$^2$. This leads to a initial velocity of $13.5$ km/s. While not much higher than the vaccuum value, it is still high enough that the bullet would probably vaporize in the atmosphere.

Interestingly, if one used a sphere instead of a bullet, then $C_d=0.4$, $A=4\times 10^{-3}$ m$^2$, the air resistance is 100 times higher, and thus a much greater velocity is needed. But since the drag scales like $v^2$, this leads to much higher drag. So much so, that even if one could launch the ball at the speed of light (Newtonians only, please!), it still could not make it out of the atmosphere!

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I doubt that a $v^2$ dependence is the whole story for objects with speeds on order of km/s. It is certainly not the whole story for objects at or above Mach 1, which means that achieving those kinds of velocities in the lower atmosphere renders these kinds of estimates very rough indeed. –  dmckee Dec 28 '10 at 16:01
    
Is the drag force is still the same when the bullet speed is much higher than the sound speed? –  hwlau Dec 28 '10 at 16:04
    
That is what I thought too. I'm no expert, but from looking around the web, it seems the $v^2$ scaling is pretty universal, even for supersonic flight. But the drag coefficient may change, as "wave drag" becomes important around the sound barrier. In hypersonic flight, the problem actually reverts to a simpler one, where the air molecules are treated as elastic scattering events, which explains the $\rho v^2$ scaling. –  Jeremy Dec 28 '10 at 16:27
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why you are asking and answering the same question at the same time? –  Graviton Dec 29 '10 at 15:56

It depends on the altitude, but let's assume sea level. The base escape velocity, assuming a vacuum, is 11.2 km/s. From there, it depends on the aerodynamical properties of the object. From Hitchhiker's Guide to Model Rocketry, we get this formula:

$$rag = \frac{1}{2}V ^ 2 \times (Air Density) \times CD \times (Projected Area)$$

The CD, or Coefficient of Drag, is a dimensionless number dependent on the shape of an object. A typical model rocket has a CD of about 0.75, while a high performance, highly streamlined rocket may be as low as 0.4. Smaller fins will reduce the CD, but also cause the stability to go down.

So I'm going to assume for simplicities sake that we've got a 0.4 CD rocket. I'll assume a 1 m diameter circular projected area, or pi/4 projected area. That leaves us still needing an adequate Air Density model. The air density model I'm using is pulled from Wikipedia, and is:

T term

p term

Density Term

I might try and model this sometime, but let me just give you the force at sea level, given a normal escape velocity. The drag is 2414 kN. Assuming this object was roughly cylndrical, having a density of 1, and is 10 m long, that would give it a weight of roughly 8000 kg. That would be somewhat less then the affect of gravity on the same object. I'd guess from these numbers that for this object, if you were going around 14-15 km/s, you'd be able to escape the Earth in one shot, but I'd have to crunch the numbers more carefully.

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I had been using drag coeff's from en.wikipedia.org/wiki/Drag_coefficient, which are a bit lower. –  Jeremy Dec 28 '10 at 15:52
    
Only the Streamlined body is lower, but, well, thanks for that source. Very interesting. –  PearsonArtPhoto Dec 28 '10 at 15:54

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