If gravity were to suddenly change, would the lift generated by a airfoil also change If gravity were to suddenly change, would the lift generated by a airfoil also change?
I realise that if gravity were to increase, then weight would also increase, leading to a change in the resultant force on the object (say a aeroplane). However, would the lift itself change, or just the resultant force.
I'm assuming it would due to possible change in fluid density, but am not sure.
If it were to change, would the opposite affect happen if gravity suddenly decreased?
 A: If gravity changes, then so will the density of air. Air pressure at the surface is  proportional to the weight of the column of air above it, so if gravity increases, the pressure would, too. That will in turn affect the lift from an airfoil. NASA says lift varies linearly with density.  I suspect the two effects would pretty much balance each other.
A: For this problem we shall strict our calculations to incompressible flow, troposphere in atmosphere and standard ground temperature  is independent of gravity. Please note this is not true, modelling that is  difficult so this assumption is required to close this problem. This may give error in calculations but we shall get an overall picture.
Here we are considering two calculations one is, What will be the value of lift force at 10 $km$ altitude? and another is What would be the value of lift force at 10 $km$ altitude, if $g$ was 20 $N/m^2$? For that we need to calculate free steam at in those cases, so we briefly look about standard atmospheric calculations.
International Standard Atmosphere
Lets see how atmospheric values are calculated in troposphere.
Variation of pressure with height in troposphere is given by
$$P=P_0 \left (1-   \lambda \frac{h}{T_0} \right )^{\frac{g.M}{R.\lambda}}   $$
Here
$P_0$ is standard atmospheric pressure at sea level = 101327 $N/m^2$.
$\lambda$ is lapse rate = $ 6.5*10^{-3} K/m$
$h$ is height of altitude in meters
$g$ is acceleration due to gravity =9.8076 $m/s^2$
$M$ is molecular mass of air = 0.0289644 kg/mol
$R$ is  universal gas constant = 8.31432 $N.m. mol^{-1}·K^{-1}$
and Temperature is given by
$$T=T_0- \lambda h$$
Here,
$T_0$ is sea level temperature = 288.15 K
Temperature, pressure and density at 10 $Km$ is 223.15 $K$, 26437.3 $N.m^{-2}$ and 0.41271 $Kg/m^3$ respectively for real case.
Since its difficult to do real time calculations we are using dry adiabatic laps rate =$9.810^{-3}K/Km$ (this we shall cover in later section) for our calculation then the temperature, pressure and density at 10 $Km$ would be 190.15 $K$, 23779.17 $N.m^{-2}$ and 0.4355 $Kg/m^3$ respectively for real case.
Let see what are the parameters of atmosphere will change if $g$ is a variable:
Let say our $g$ is increased to 20 $m/s^2$ then it will affect $P_0$, $T_0$, $\rho_0$ but its very difficult to estimate or model those thing. We already assumed that $T_0$ is same as actual sea level temperature. But temperature may increase because all heavy and green house gases try to find there place next to earth crust, pressure and density is two times of actual value when gravity is doubled because weight of atmosphere is doubled.
Increasing $g$ value also change lapse rate ($\lambda$). Actual  $\lambda$ is Thermodynamic-based lapse rate. That formula is complicated and tedious to analyze so we strict our self to dry adiabatic lapse rate . Though adiabatic lapse rate value is $ 9.8*10^{-3} K/m $ more than actual one, it can be used to get overall picture about this process.
The adiabatic lapse rate is given approximately by :
$\lambda_w = g/c_p$
$\lambda_w$ = Dry adiabatic lapse rate, K/m
$g$= Earth's Standard gravity, gravitational acceleration= 9.8076$m/s^{2}$
$c_p$=The specific heat of dry air at constant pressure, = 1003.5 $JKg^{-1}K^{-1}$
$\lambda_w$ with $g$ = 20 $m/s^2$, is $19.93*10^-3 K/m$ assuming gas is calorically perfect.
Pressure temperature and density at 10 Km altitude with $g$ =20 $m/s^2$ is 3314.8 $N/m^2$, 88.67 $K$ and 0.1302 $kg/m^3$ receptively.
Definition of lift is:
When a fluid flowing passes the surface of a body exerts a force on it. Lift is the component of this force that is perpendicular to the oncoming flow direction. Please don't confuse this with resulting upward force opposite in direction to weight, that is different, lift is different.
Let see the formula for lift,
$$L=0.5*C_l*\rho*v^2*S$$
where
$L$ is the lift force,
$\rho$ is fluid density
$v$ is true airspeed
$S$ is platform area
$C_l$ is lift coefficient
here $C_l$ depends on Reynolds number and body shape; Reynolds number depends on free stream parameters like temperature, density etc. but variation of $C_l$ w.r.t Reynolds number is negligible. Since our body shape will not change, we shall assume $C_l$ is more or less a constant parameter and $C_l$ is independent of gravitational force. Here I'm assuming lift force acting opposite in direction to weight. Please not that in most of the practical applications this is not true.
Lets calculate lift force for those two cases:
Lets take $C_l$=1,$v$ =1, $S$ =1 and mass=1 $kg$;
For $g$ =9.81 $m/s^2$
Lift force is 0.21775 $N$, weight is 9.81 $N$ and resulting upward force is -9.59225 $N$
For case 2: $g$ =20 $m/s^2$
Lift force is 0.066, weight is 20 $N$ and resulting upward force is -19.934 $N$
Answer:

*

*At ground resulting upward force slightly reduces with increase in gravity but that is negligible.

*If $g$ is increased lift force will decrease with altitude

*lift force also decrease  increase in gravity . Lift force is a
nonlinear function of $g$

*$C_l$ is almost constant; Because of increase in Reynolds number it will increase but that is negligible.

*Resulting upward force in airfoil will decrease if $g$ is increased because of reduction in lift and increase in body weight at some height.

(Please note that I made a big assumption that sea level temperature would not change with change w.r.t gravity)
A: If gravity increases, so does air density and pressure, but also air temperature. Since density decreases with increasing temperature, the increase in density will not be sufficient to fully compensate for the increased weight of the aeroplane.
The aeroplane will begin to sink, which in turn will increase its angle of attack. Consequently, it will settle at a new flow condition where lift equals weight, given that it had enough speed initially so the increased angle of attack does not stall it.
If you look at the long-term equilibrium at the new gravity, the air should cool down again to a temperature close to the initial conditions. Then the density increase is proportional with the gravity increase, and at the same flow conditions, the aeroplane will create just enough lift to support its increased weight.
