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Yes - it is easier (all other things being equal)

If you want the entire derivation of my below summary, please read this previous answer of mine.

Producing Lift
The ease of flying (overcoming weight with lift) = f($\frac{\rho}{g}$) and $\rho$ = f($\frac {p}{T}$).

Overcoming Drag
The drag equations are identical to the lift ones after swapping the correct coefficient so you got the same forms and drag = f($\frac{\rho}{g}$) and $\rho$ = f($\frac {p}{T}$).

However, thrust is produced by sucking in atmosphere and expelling it at higher velocity (or pressure - but this is less efficient). It is most simply approximated as Thrust (T) = f($\dot{m}$). $\dot{m}$ is known as mass flow rate and $ \dot{m}$ = f($\rho \times v$).

Since you'll need T = D the atmospheric density ($\rho$) on each side cancels out. As long as there's enough chemical to burn (fuel or oxidizer) and introduce energy into the working fluid (which is the atmosphere), thrust production isn't an issue.

Your Planet
Assuming your planet has the same temperature as Earth, then it would be about $ \frac {1 \div 2}{1 \div 3} = 1.5x $ as easy (meaning it is easier) to fly on your planet than it is on Earth.

Titan as an Example
One other note, if your world possesses a reducing atmosphere (hydrogen, methane, ethane, etc.), then your "air breathing" aircraft would carry an oxidizer (like oxygen) and use the "fuel" the atmosphere provides. This sort of configuration would work great on a body like Titan.

Titan's properties:
Gravity ~ 1/7 Earth's
Pressure ~ 1.4 Earth's
Temperature (K) ~ 1/3 Earth's
Density ~ $ 3 \times 1.4 = 4.2$ Earth's
Ease of flying = $ \frac{4.2}{1/7} = 4.2 \times 7 = 29.2 $x

Flying would be 29.2 times easier (much MUCH easier) on Titan than on Earth.

Yes - it is easier (all other things being equal)

If you want the entire derivation of my below summary, please read this previous answer of mine.

Producing Lift
The ease of flying (overcoming weight with lift) = f($\frac{\rho}{g}$) and $\rho$ = f($\frac {p}{T}$).

Overcoming Drag
The drag equations are identical to the lift ones after swapping the correct coefficient so you got the same forms and drag = f($\frac{\rho}{g}$) and $\rho$ = f($\frac {p}{T}$).

However, thrust is produced by sucking in atmosphere and expelling it at higher velocity (or pressure - but this is less efficient). It is most simply approximated as Thrust (T) = f($\dot{m}$). $\dot{m}$ is known as mass flow rate and $ \dot{m}$ = f($\rho \times v$).

Since you'll need T = D the atmospheric density ($\rho$) on each side cancels out. As long as there's enough chemical to burn (fuel or oxidizer) and introduce energy into the working fluid (which is the atmosphere), thrust production isn't an issue.

Your Planet
Assuming your planet has the same temperature as Earth, then it would be about $ \frac {1 \div 2}{1 \div 3} = 1.5x $ as easy (meaning it is easier) to fly on your planet than it is on Earth.

Titan as an Example
One other note, if your world possesses a reducing atmosphere (hydrogen, methane, ethane, etc.), then your "air breathing" aircraft would carry an oxidizer (like oxygen) and use the "fuel" the atmosphere provides. This sort of configuration would work great on a body like Titan.

Titan's properties:
Gravity ~ 1/7 Earth's
Pressure ~ 1.4 Earth's
Temperature (K) ~ 1/3 Earth's
Density ~ $ 3 \times 1.4 = 4.2$ Earth's
Ease of flying = $ \frac{4.2}{1/7} = 4.2 \times 7 = 29.2 $x

Flying would be 29.2 times easier (much MUCH easier) on Titan than on Earth.

Yes (all other things being equal)

If you want the entire derivation of my below summary, please read this previous answer of mine.

Producing Lift
The ease of flying (overcoming weight with lift) = f($\frac{\rho}{g}$) and $\rho$ = f($\frac {p}{T}$).

Overcoming Drag
The drag equations are identical to the lift ones after swapping the correct coefficient so you got the same forms and drag = f($\frac{\rho}{g}$) and $\rho$ = f($\frac {p}{T}$).

However, thrust is produced by sucking in atmosphere and expelling it at higher velocity or pressure. Mass flow rate = f($\rho \times v$).

Since you'll need T = D the atmospheric density ($\rho$) on each side cancels out. As long as there's enough chemical to burn (fuel or oxidizer) thrust production isn't an issue.

Your Planet
Assuming your planet has the same temperature as Earth, then it would be about $ \frac {1 \div 2}{1 \div 3} = 1.5x $ as easy (meaning it is easier) to fly on your planet than it is on Earth.

Titan as an Example
One other note, if your world possesses a reducing atmosphere (hydrogen, methane, ethane, etc.), then your "air breathing" aircraft would carry an oxidizer (like oxygen) and use the "fuel" the atmosphere provides. This sort of configuration would work great on a body like Titan.

Titan's properties:
Gravity ~ 1/7 Earth's
Pressure ~ 1.4 Earth's
Temperature (K) ~ 1/3 Earth's
Density ~ $ 3 \times 1.4 = 4.2$ Earth's
Ease of flying = $ \frac{4.2}{1/7} = 4.2 \times 7 = 29.2 $x

Flying would be 29.2 times easier (much MUCH easier) on Titan than on Earth.

Yes - it is easier (all other things being equal)

If you want the entire derivation of my below summary, please read this previous answer of mine.

Producing Lift
The ease of flying (overcoming weight with lift) = f($\frac{\rho}{g}$) and $\rho$ = f($\frac {p}{T}$).

Overcoming Drag
The drag equations are identical to the lift ones after swapping the correct coefficient so you got the same forms and drag = f($\frac{\rho}{g}$) and $\rho$ = f($\frac {p}{T}$).

However, thrust is produced by sucking in atmosphere and expelling it at higher velocity (or pressure - but this is less efficient). It is most simply approximated as Thrust (T) = f($\dot{m}$). $\dot{m}$ is known as mass flow rate and $ \dot{m}$ = f($\rho \times v$).

Since you'll need T = D the atmospheric density ($\rho$) on each side cancels out. As long as there's enough chemical to burn (fuel or oxidizer) and introduce energy into the working fluid (which is the atmosphere), thrust production isn't an issue.

Your Planet
Assuming your planet has the same temperature as Earth, then it would be about $ \frac {1 \div 2}{1 \div 3} = 1.5x $ as easy (meaning it is easier) to fly on your planet than it is on Earth.

Titan as an Example
One other note, if your world possesses a reducing atmosphere (hydrogen, methane, ethane, etc.), then your "air breathing" aircraft would carry an oxidizer (like oxygen) and use the "fuel" the atmosphere provides. This sort of configuration would work great on a body like Titan.

Titan's properties:
Gravity ~ 1/7 Earth's
Pressure ~ 1.4 Earth's
Temperature (K) ~ 1/3 Earth's
Density ~ $ 3 \times 1.4 = 4.2$ Earth's
Ease of flying = $ \frac{4.2}{1/7} = 4.2 \times 7 = 29.2 $x

Flying would be 29.2 times easier (much MUCH easier) on Titan than on Earth.

Yes (all other things being equal)

Producing Lift
The ease of flying (overcoming weight with lift) = f($\frac{\rho}{g}$) and $\rho$ = f($\frac {p}{T}$).

Overcoming Drag
The drag equations are identical to the lift ones after swapping the correct coefficient so you got the same forms and drag = f($\frac{\rho}{g}$) and $\rho$ = f($\frac {p}{T}$).

However, thrust is produced by sucking in atmosphere and expelling it at higher velocity or pressure. Mass flow rate = f($\rho \times v$).

Since you'll need T = D the atmospheric density ($\rho$) on each side cancels out. As long as there's enough chemical to burn (fuel or oxidizer) thrust production isn't an issue.

Your Planet
Assuming your planet has the same temperature as Earth, then it would be about $ \frac {1 \div 2}{1 \div 3} = 1.5x $ as easy (meaning it is easier) to fly on your planet than it is on Earth.

Titan as an Example
One other note, if your world possesses a reducing atmosphere (hydrogen, methane, ethane, etc.), then your "air breathing" aircraft would carry an oxidizer (like oxygen) and use the "fuel" the atmosphere provides. This sort of configuration would work great on a body like Titan.

Titan's properties:
Gravity ~ 1/7 Earth's
Pressure ~ 1.4 Earth's
Temperature (K) ~ 1/3 Earth's
Density ~ $ 3 \times 1.4 = 4.2$ Earth's
Ease of flying = $ \frac{4.2}{1/7} = 4.2 \times 7 = 29.2 $x

Flying would be 29.2 times easier (much MUCH easier) on Titan than on Earth.

Yes (all other things being equal)

If you want the entire derivation of my below summary, please read this previous answer of mine.

Producing Lift
The ease of flying (overcoming weight with lift) = f($\frac{\rho}{g}$) and $\rho$ = f($\frac {p}{T}$).

Overcoming Drag
The drag equations are identical to the lift ones after swapping the correct coefficient so you got the same forms and drag = f($\frac{\rho}{g}$) and $\rho$ = f($\frac {p}{T}$).

However, thrust is produced by sucking in atmosphere and expelling it at higher velocity or pressure. Mass flow rate = f($\rho \times v$).

Since you'll need T = D the atmospheric density ($\rho$) on each side cancels out. As long as there's enough chemical to burn (fuel or oxidizer) thrust production isn't an issue.

Your Planet
Assuming your planet has the same temperature as Earth, then it would be about $ \frac {1 \div 2}{1 \div 3} = 1.5x $ as easy (meaning it is easier) to fly on your planet than it is on Earth.

Titan as an Example
One other note, if your world possesses a reducing atmosphere (hydrogen, methane, ethane, etc.), then your "air breathing" aircraft would carry an oxidizer (like oxygen) and use the "fuel" the atmosphere provides. This sort of configuration would work great on a body like Titan.

Titan's properties:
Gravity ~ 1/7 Earth's
Pressure ~ 1.4 Earth's
Temperature (K) ~ 1/3 Earth's
Density ~ $ 3 \times 1.4 = 4.2$ Earth's
Ease of flying = $ \frac{4.2}{1/7} = 4.2 \times 7 = 29.2 $x

Flying would be 29.2 times easier (much MUCH easier) on Titan than on Earth.