The misery of school

[Exnihiloest]

...
But since by design an airplane in flight at a constant altitude has almost zero weight (otherwise it would climb or descend permanently), this is not dramatic....

The weight is always the same, it's P=mg. What you mean is that the resultant of the vertical forces on it is zero. Sure.

[SebastianL]

It's great, here we are with a formula that is used to calculate almost nothing, without giving any clue as to what is happening under the "physical" hood and with that we manage to criticize the school that educates students to formulas that do not really understand and just as quickly forget their field of application.
You are all received at the national education, mention very well

[sicetaitsimple]

...
But since by design an airplane in flight at a constant altitude has almost zero weight (otherwise it would climb or descend permanently), this is not dramatic....

The weight is always the same, it's P=mg. What you mean is that the resultant of the vertical forces on it is zero. Sure.

You understood!

[Remundo]

All the other answers confirm that the weight of the plane is well transferred to the ground. We have to stop the bad faith.

OK, OK, you're right, of course the atmosphere below distributes the resulting additional weight over a large floor area

absolutely

But since by design an airplane in flight at a constant altitude has almost zero weight (otherwise it would go up or down constantly), it's not dramatic....

The weight of an airplane is never zero.

Surprisingly, however, the weight of an airplane decreases with altitude, and even tends towards zero if it moves infinitely far from the Earth.

However, as a first approximation, the weight is more or less constant in the interval of altitudes usually reached by an airplane.

You develop far too many notions of "apparent weight" and other "zero weight" which are quite scabrous.

Better to come back to a balance of forces on a specific system, and clearly identify all the mechanical systems that interact.

The case of an airplane is already not so simple!

[Exnihiloest]

... the weight of an airplane decreases with altitude ...

I did the calculation with mgh: at 10000m, it is reduced by 0,157%.

At the height of the orbital station, that would be only 6,2%. The error is to believe that weightlessness is due to the spatial position far from the earth, whereas it is due to ballistic flight (centrifugal acceleration compensates that of gravity).

[Remundo]

ah for those who want the variation, here is the exact formula of g (which is not constant, but almost for "little birds")

g = g0 x (RT/(RT+z))²

with g0 the gravitational field 9,81 m/s² at sea level.

RT terrestrial radius (6400 km)
z altitude of the plane (10 km, hardly more).

with these assumptions, the aircraft still weighs 99,688% of its weight on the ground at 10 km altitude.

[Christophe]

... the weight of an airplane decreases with altitude ...

I did the calculation with mgh: at 10000m, it is reduced by 0,157%.

At the height of the orbital station, that would be only 6,2%. The error is to believe that weightlessness is due to the spatial position far from the earth, whereas it is due to ballistic flight (centrifugal acceleration compensates that of gravity).

My God and that gives lessons in physics or other to others...

Run away !

Listen to Remundo instead!

[Exnihiloest]

... the weight of an airplane decreases with altitude ...

I did the calculation with mgh: at 10000m, it is reduced by 0,157%.

At the height of the orbital station, that would be only 6,2%. The error is to believe that weightlessness is due to the spatial position far from the earth, whereas it is due to ballistic flight (centrifugal acceleration compensates that of gravity).

My God and that gives lessons in physics or other to others...

Run away !

Listen to Remundo instead!

I did a quick calculation with an approximation method that I provided, and we can see that the result, even if it is simple to double, still gives the order of magnitude (0,157% instead of 0,314 % by the exact method).

And the radius of the earth is 6371 km, not 6400.

But hey, you've never been in the rational but in denigration.

[Christophe]

Yes yes you are right…100% error in physics is zero points!

[Remundo]

Nihilo reasoned about energy, which gave him an order of magnitude. But physically, there is an error, it is necessary to reason on the field in itself.

In reality, the quadratic character of the gravitational fields multiplies by two the small quantity z/RT

RT/(RT+z) = 1 / (1+z/RT)

dong g/g0 = (1+z/RT)^(-2)

And if you do a little limited development because z/RT is a very low quantity...

g/g0 is close to: 1 - 2 z/RT

g/g0 close to 1 - 2 * 10 / 6400 = 0.996875

we are very close to the exact calculation which gave 99,688%.