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That's the figure cited for "escape velocity", (or a hair less than that perhaps), which would be valid if the Earth had no atmosphere.
Saint Exupery
reply to post by Mads1987
The short answer is: Yes - It is possible for material to be blasted off of Earth at escape velocity (>12 km/sec) so that it goes into solar orbit, then eventually re-impact the Earth.
Here is an article about Meteorites from Other Planets. It includes a post-script about meteorites from Earth.
and thanks for the link!
For the lower atmosphere, where most of the air is, the temp, pressure, and density of air is given by:
$T=T_0-Lh$
$p=p_0left(frac[T][T_0]right)^[frac[gM][RL]]$
$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/(molÂ·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 constant) 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 $A approx 4times 10^[-4]$ m$^2$. This leads to a initial velocity of $13.5$ km/s. While not much higher than the vacuum 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 $C_d=0.4$, $A=4times 10^[-3]$ 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!
Right. That's why I said "I've never seen a rigorous calculation of the escape velocity from Earth taking the atmosphere into account."
eriktheawful
reply to post by Arbitrageur
A few of things to point out:
1) The air pressure at sea level is at 101.325 kPa. However, as we rise in altitude that pressure drops.....dramatically. Go up 10 km and the pressure drops to 24.28 kPa. At 20km the pressure has dropped to only 5.8 kPa. At 30km to only 1.3 kPa.
I saw a video where scientists tried to re-create impacts using high speed projectiles into the ground, in a lab. I don't recall seeing anything ejected vertically, in those test videos, regardless of the angle of impact.
2) Velocity of the rock ejected from a massive impact will depend upon the energy imparted to it from the impact event. We know that that escape velocity for Earth (not taking into account the air drag as you mentioned) is about 12km per second. Assuming a vertical path for the rock, the first second of flight it will be at 12km, where the air pressure has already dropped from 101.325 kPa to only 18.24 kPa. I do not see where this is taken into account in what you presented.
3) Density and size of the ejecta: Obviously the larger and more dense the object, the more energy needed to impart that escape velocity. However, it is not very hard to believe that a 7km wide object hitting the earth (ala KT event) would not be able to get some super sized boulders into space, the size of houses or even bigger.