Including air resistance, what is the escape velocity from Earth?

Question

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

Answer 1

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!

Answer 2

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:

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.

Answer 3

Assume the initial velocity is v, assume a function exists v(t) that tells the velocity at time t. The derivative of this function is acceleration. Air resistance is

1/2pv(t)^2CdA

(where Cd and A is the drag coefficient and are respectively) and like in the previous models p, air density is dependent on hight. A function can be made representing the hight at time t, h(t), it is the integral of v(t) from time 0 to t. Since both effect each other it is a differential equation. Acceleration is force divided by mass, the force of attraction of two bodies is

Gm1m2/r^2

(where G is the gravitational constant m1 is earths mass, m2 is the object mass and r is the distance) since Gm1 is just g, the gravitational acceleration constant. Since r is from the centre of the earth Re, the radius of the earth must be added to the hight since hight is measured from the surface.

gm2/(h(t)+Re)^2

(where g is 9.807) The second force is air resistance whose formula has already been discussed. assuming Cd and A are known constants p is the only variable left. p can be expressed as:

p=PM/RT

Where M (0.02896968 kg/mol) and R (8.314462618 J/(mol·K)) are constants, P (pressure) and T (temperature) can be expressed as

P=P0(1-(g/(h(t)+Re)^2)h(t)/(CpT0))^(CpM/R), T=T0-Lh

(where P0 is pressure at sea level (101.3 KPa) Cp is constant pressure specific heat (1004.68506 J/(kg·K)) and L is temperature lapse rate (0.00976 K/m)), this makes the full formula for Force

Force=-gm2/(h(t)+Re)^2)-1/2MP0(1-(g/(h(t)+Re)^2)h(t)/(CpT0))^(CpM/R)/(R*(T0-Lh(t)))v(t)^2CdA,

and as previously stated Acceleration is Force/mass, here mass is m2. By evaluating at small time steps, v=v0+at and h=h0+vt, a model can be made. I did mine in excel. by changing the initial values of velocity and looking if the velocity ever gets negative within some long period of time a velocity similar to what they have previously calculated is found. This is a nice example where a complex model might get the same results as a simple model.