An article in yesterday's New York Times by Andrew Revkin dealing with rise in sea level from melting glaciers cites Dr. Eric Rignot of NASA as saying that unabated global warming could result in 3 feet of sea level rise from melting ice on Greenland and another 3 feet from Antarctica and about 18 inches from the melting of Alpine glaciers. He quotes Dr. Rignot as stating: " it is too early to reassure that all will stabilize and similarly there is no way to predict a catastrophic collapse," ....." But things are definitely far more serious than anyone would have thought five years ago." (Emphasis mine).
There's an old expression-"Stop the world, I want to get off." Well just in case the worst does happen, we may have think of having to go elsewhere(getting off in a sense). Where? I don't know, but here's the calculation on what upward speed we will need to get off.
We're going to determine the velocity of a particle projected outward in a radial direction from the earth and acted on by only one force, that of gravity.
According to Newton's theory of gravitation , the acceleration of the particle will be inversely proportional to the square of the distance of the particle from the center of the earth.
Let r be that distance and R be equal to the earth's radius, t is time, a represents acceleration and k is the proportionality constant in Newton's law.
Then a = dv/dt=k/r^2
Since the acceleration is negative because the velocity is decreasing as it travels upward, then k is negative. When r=R and a =-g, then g=-k/R^2. Substituting for k above, gives a= -gR^2/r^2.
Since a =dv/dt and v=dr/dt then:
a= dr/dt(dv/dr) =vdv/dr so that our diff equation is : vdv/dr= -gR^2 / r^2
The solution is v^2= 2gR^2 /r +c
When v=v0 and r=R then c = vo^2- 2gR so that a particle projected radially outward will go with a velocity
v^2=2gR^/r +v0^2- 2gR.
In order to escape from the earth the quantity v0^2-2gR > 0 (must be equal to or greater than zero).
So then the vo=the square root of 2gR is the escape velocity required . R=3,960 miles(approx.) and
g= 6.09x10^-3 miles/second^2, so that v0 =6.95 miles per second. Frictional air resistance in the lower atmosphere may not be negligble so that a slightly higher value may be required.