if load factor equation is lift devided by weight , so in very high altitude for example 1000 km above the earth atmosphere, there is no air to produce lift so load factor is zero but in newton equation we see that even for very high altitudes there is a small gravitational accelereation due to object weight not lift! so how load factor is same as g but in high altitude load factor is zero but object feels gravtiy?
$\begingroup$ In most cases the centripetal force is small enough to be ignored, although in supersonic flight it would be measurable. However, in retrograde (east to west) flight the aircraft is revolving more slowly than the Earth’s surface and so would in fact appear to be heavier than when at rest. Load factor doesn’t generally make allowance for the direction of flight, which would need to be taken into account for flights at near-orbital speed. $\endgroup$– FrogOct 21, 2022 at 18:36
$\begingroup$ At an altitude of 1000 km, gravity is not so weak (g=7,3 m/s2), and the atmosphere is extremely thin, so there can be no lift at all, and the 'load factor' has no meaning. It's true that satellites 'fly' orbiting planet Earth, at very high altitudes, but that's a different thing, those satellites being in continuous free fall, so the Earth's gravitational acceleration, as measured in the satellite, is zero... $\endgroup$– xxavierOct 21, 2022 at 19:25
$\begingroup$ how gravititional acceleration is zero and object is in continuous falling? $\endgroup$– alirezaOct 21, 2022 at 19:28
$\begingroup$ Imagine yourself in an elevator hanging from a rope. Suddenly, the rope breaks and –assuming no air drag and other resistances– the elevator and its contents fall at 9,8 m/s2. As you are falling –and accelerating– with the elevator, you are weightless... The acceleration that you feel or measure with a dynamometer is zero... $\endgroup$– xxavierOct 21, 2022 at 20:52
$\begingroup$ Just to add to the answer of @xxavier over here: the object (satellite) is not only falling like an elevator with the brocken rope but it is falling on a curved path which is just perfectly circular around the earth $\Rightarrow$ it is in continuous free fall. And that's why people on the ISS are weightless and not because they are so far from earth that they do not feel gravity anymore $\endgroup$– sophitOct 21, 2022 at 21:06
Here, like in your other question, we have a typical situation when definitions break down when removed from their context. All definitions, especially engineering ones, have a certain domain (often implied) where they remain valid and meaningful.
What is lift? Do engines in a vertical climb provide lift, or what? Is load factor the thing that the G-meter in the cockpit shows? Does it have to point down? What is "down"? What is weight?
In a purely physical sense, you can define load factor as the ratio between the weight and any reaction force that counteracts this weight. We don't need to call this force "lift". When you are standing on a static floor, the floor provides a reaction force that is exactly equal to your weight, and you have "load factor" 1. Is it enough? Is it useful?
This depends on how you define "weight". Surprisingly, there is no universal definition: weight is not an essential physical quantity. Mass is. In many cases, weight is defined exactly as the reaction force, i.e. what the scales show. If they show 1000 N1 when you are standing on them on the floor, then that's your weight.
Now let's build a 1000 km tower (with all the amenities) and put you up there. At that elevation, gravity is less (but not too much less), and your scales will show about 750 N. Now that's your weight! The load factor is still 1 by our definition: the floor reaction ("lift") matches your weight.
Now let's put you in a centrifuge and spin you up. The scales under you might show 2000 N now. Still, the floor is intact, and now that's your weight - even though in this case not all of it comes from gravity! And the load factor still 1.
Doesn't feel very useful, right? But this comes from the "common" definition of weight: you "actually" feel lighter in the first case and heavier in the second, and the floor is stressed less on the tower and more in the centrifuge. If you buy cheese on the tower "by weight", you'll get more of it for the same money!
So, it seems more useful to define "weight" in the load factor equation as some standard quantity, fixed for the "normal" gravity at the surface. In this case we'll relate the actual reaction force (lift or whatever) to your "normal" weight of 1000 N. And then we'll get load factor G=0.75 on the tower and G=2 in the centrifuge. And this actually reflects your stress, as well as the stress of the floor (or airframe).
Note that the G-meter in an aircraft is essentially a weight scale, calibrated to the ground level and graded in "g" of 9.81 m/s2.
Stop! But what about the "lift = weight" formula for the level flight? It doesn't work with such "fixed" definition of weight. Yes, there is different weight here! The one we used in the first definition.
This is the crux of the problem. In our daily life, these weights are "practically" the same, and we conflate them. But once you step out of the "normal" conditions, like putting the aircraft in outer space, the implicit conditions in each definitions may no longer be valid, and you need to watch carefully what you mean by each term.
What is lift? Is it the force that "lifts" the body away from Earth and holds it from falling? Or is it the "useful" force that the wing generates? This term starts to break down when the attitude is such that the two directions don't align (like in a vertical climb), or when there is no aerodynamics to speak of.
From the point of view of the aircraft (and its engineer), lift is the aerodynamic force normal to the incoming stream flow.2 Most of the attention is given to this force, and generally to the forces in this direction which is "vertical" for the airplane: they are by far the greatest, and the most stressful for the airplane. The typical G-meter also measures the load factor in this plane. But nothing prevents us from measuring the forces (and defining the load factor) in other planes: after all, pilot's body (and the airframe) experiences stress in all directions.
In a steady vertical climb, aerodynamic lift is zero, but its role is taken up by the engines. The load factor is 1 (when close to ground), but it's in a different direction, and it's not the direction in which the normal load factors are certified. Still, you (and the airframe) will feel it; you just might not care that much from the point of view of stress (or might care more, if it was a side force). But from the point of view of physics, there is no fundamental difference.
In outer space, we have no aerodynamics. But we can still provide "lift" by other means. When a jump jet takes off vertically, its engine arguably provides lift, while its wing is useless. So, get a rocket engine, direct it "down" (towards gravity), and throttle it to provide just enough thrust to keep the altitude. And you'll have a non-zero load factor. Which one? At 1000 km, a G-sensor will show about 0.75, as we discussed earlier. The direction depends on attitude.
When you run out of fuel (which will happen in minutes, if not seconds), you'll start falling down. There is no other force you can use to counteract gravity. So you'll have zero load factor, like in any free-fall situation. Ironically, you don't "feel" gravity when gravity is the only force acting on you.
To avoid crashing into Earth, you can move forward such that you "miss" Earth due to its curvature. That's how satellites fly. But you are still falling constantly. This is why they don't feel weight in orbit.
1 Weight is, of course, force. But common scales are usually (and mistakenly) graded in the units of mass (e.g. kg). Even if we consider it be the related force unit such as "kg-f", this has an implicit definition of the gravity acceleration "g", which is important for our discussion. So let's stick to the proper (even if uncommon) units.
2 And in the plane of symmetry, to separate it from side force.
$\begingroup$ thanks a lot bro. so in vertical climb g is 1 not zero because in this case lift is produced by engines not wings? based on newtons second law if an aircraft goes vertically up without any accelereation the pilot and aiframe feels 1g . based on load factor point of view , in vertical climb there is also lift but not produced by wings and it is duty of thrust to produced the same lift. is my conclusion correct? $\endgroup$– alirezaOct 25, 2022 at 10:35
$\begingroup$ @alireza: in a vertical climb at constant speed, G would be 1 longitudinally to the aircraft (from nose to tail) and G=0 vertically (from belly to canopy). Also lateral G (from left to right) would be 0. Load factor is by definition lift/weight = 0/weight = 0. Engine produces a thrust to win weight plus aerodynamic drag. This should be the complete view about this case. $\endgroup$– sophitOct 25, 2022 at 11:44
$\begingroup$ @sophit, right, except that here again, it's better to qualify the "definition" of load factor. In this "normal" definition, it is implied that this is the "body vertical" load factor. Since it is usually the main point of concern, it is the "default" one. But the longitudinal load factor (which is 1 here) also exists; it's just rarely a concern and is generally well within ±1 G even for very powerful jets, so it's rarely mentioned or certified. Yet when we are trying to work out the confusion in non-usual situations, it's better to qualify things explicitly. $\endgroup$– ZeusOct 25, 2022 at 23:46
$\begingroup$ @Zeus: yes sure, I meant the standard definition of load factor, the one that you find in the books i.e. lift/weight. Very good answer. $\endgroup$– sophitOct 26, 2022 at 4:37