Short answer for air travel
The practicality does not depend upon the mass or gravity of a body. It is the ratio of surface gravity to air density that makes it practical or impractical.
As long as the ratio of gravitational acceleration to air density remains constant, air travel remains practical.
Increase surface gravity while keeping air density constant will eventually make air travel impractical. When it becomes impractical depends upon how efficient your engines are and what you deem to be the minimum sized payload worth your while.
If you doubled planetary gravitation while keeping air density constant, commercial air travel would be impractical (though some special planes might still be possible). For example, take the numbers from a Boeing 747. If you doubled gravitational acceleration, the aircraft could take off if it was empty but it could carry no cargo.
Take off gross weight: 333,390 kg
1 g empty weight: 162,400 kg
2 g empty weight: 324,800 kg ~ 333,390 kg
In order to double the Earth's gravity, you'd have to double the mass of the planet while keeping the radius the same OR keep the Earth's mass the same and decrease its radius to 70% of current.
It would take the collision of two bodies of mass of Earth or larger to do it and that would liquify the Earth - No survivors. Plus the atmosphere and hydrosphere would be permanently lost.
There is no scenario that I can envision that could do this and leave any survivors.
Short answer for space travel
IMO, chemical rockets are on the verge of being impractical now. Even with staging (which makes the performance better) they aren't widely used now except as a specialty transportation mechanism for very high value transportation.
So if you doubled surface gravity, your only practical method of space launch might be one of these that I documented on The Case for Space section of my blog:
- Nuclear Pulse Propulsion
- Laser Launch/Light Craft
- Ram Accelerator
- Light Gas Gun
- Coilgun
Basically only engines with very high specific power (e.g. nuclear bombs) or don't have to carry their propellant would work for space launch for a 2g planet - air density doesn't affect this much except to make it more difficult.
Lift - Weight
From a first order analysis, lift is the force required to lift the aircraft off the ground. Lift must equal the mass of the aircraft in order to lift off.
$$ L = \frac{m_aM_pG}{r^2} \rightarrow L = m_a a_p $$
$m_a$ - mass air vehicle
$M_p$ - Mass of planet
$a_p$ - Planet's gravitational acceleration
G - Universal gravitational constant
r - radius of the surface of the planet
The lift equation is:
$$ L = \frac {1}{2} C_L \rho V^2 $$
L - Lift force
$C_L$ - Coefficient of lift (dependent upon aircraft & wing shape)
$\rho $ - Density of air
$ V^2 $ - Velocity of vehicle squared
So putting them together we get:
$$ m_a a_p = \frac {1}{2} C_L \rho V^2 \rightarrow a_p = \rho \frac {C_LV^2}{2m_a} $$
Simplifying we get
$$ \frac{a_p}{\rho} = \frac {C_LV^2}{2m_a} $$
This equation shows that $ C_L $, V, $ m_a $ remain constant if the ratio of $\frac{a_p}{\rho}$ remains constant.
Drag - Thrust
In addition to weight issues, you must also pay a drag penalty.
The drag equation is identical to the lift equation but uses a different constant. You can approximate the drag coefficient as 1/10 of the lift equation.
$ C_D $ ~ $ \frac{C_L}{10} $
So
$$ D = \frac{1}{20}C_L \rho V^2 $$
The turbine engine thrust equation is:
$$ D = T = \left(\dot{m_a} + \dot{m_f} \right)v_e - \dot{m_a}v_i $$
$\dot{m_a}$ - Mass flow rate of air, which can also be expressed as $\dot{m_a} = \rho A v$
$\dot{m_f}$ - Mass flow rate of fuel
$v_e$ - Engine exhaust velocity
$v_i$ - Velocity of air at the inlet (when multiplied by $\dot{m_a}$, this is also known as ram pressure
A - Area at the inlet or exhaust (depending upon where you're doing the calculation)
But it is usually approximated with the following (the fuel's contribution to thrust is mostly through heating):
$$ D = \dot{m_a} \left(v_e - v_i \right) $$
I am not going to go through all the gyrations to do this exactly. I'm assuming the inlet and exhaust are the same size (they almost never are) but I simply want the feel of the equations and for this purpose it works.
$$ D = \rho A v \left(v_e - v_i \right) $$
Combining with the Drag equation and we get
$$ \rho A \left(v_e^2 - v_i^2 \right) = \frac{1}{10}\frac{1}{2}C_L \rho V^2 \rightarrow \frac{1}{10} m_a a_p = \rho A \left(v_e^2 - v_i^2 \right) $$
Substituting in the Lift equation equivalence to aircraft mass times surface gravity, I get:
$$ \frac{a_p}{\rho} = 10 \frac{A \left(v_e^2 - v_i^2 \right)}{m_a} $$
Anyway long story short, it looks like its Drag remains the same as long as the ratio of surface gravity to atmospheric density remains constant.
becomes impractical for commercial use
is a quite undefined "definition". As a result of the decreasing utility laws, you will find that increasing prices will progressively move away more users, but up until the very end some applications may be worth enough to keep them until the very end. Which is the point? When local distance flights are replaced by roads/rail? When trasantlantic flight is only affordable to half its current users? or a tenth? $\endgroup$