# Homework Help: Why do things not float in a airplane like they do on a satellite?

1. Jun 4, 2005

### michaelw

Such as SkyLab? If the airplane got more horizontal velocity would things float?

2. Jun 4, 2005

### whozum

Things float in spacecraft because the spacecraft's velocity is just right (for a given altitude) to the point where they are simply falling aruond the earth. Objects in orbit are in constant freefall, and are weightless. In an airplane however, the airplane isnt bound to orbit by gravitation, it uses the air drag force to keep it up, it would need to travel much faster to maintain an orbit. It would never be able to due to the air resistance.

3. Jun 4, 2005

### OlderDan

whozum has given you a correct answer to this question, for airplanes on normal flight paths, but airplanes have been used to create a "weightless" environment for times approaching a minute. Such flights have been used for astronaut training for years. If you can imagine an object moving under the influence of gravity, with no air resistance, near the surface of the earth or at any nearly consatnt distance from the surface, the path of that object is a parabola. More precisely, if you include the small changes in gravitational force with distance from the center of the earth, the path is a portion of an ellipse near the turning point. All that matters is that there is a smooth path that an object will follow if the only force acting is earth's gravity. If an airplane containing interior objects (including people) flies exactly along that path, maintaining the exact speed of a "free-flying" object, the objects inside the plane wil experience "weightlessness", which really means there are no forces from the walls or seats of the plane pushing on them. It also means the usual forces a body exerts to keep internal organs in place are now gone, and that funny feeling in your stomach appears. Although it only lasts for several seconds, this is the same experience that persists in the space lab.

The same effect can be achieved by a roller coaster by shaping the track to conform exactly to the path an object would travel if it were a projectile. For a roller coaster, the practical limitations of size limit the duration of the experience. That's why somebody invented the "free fall" kinds of rides for ammusement parks, (and bungee jumping for those who like their experience inverted and more in touch with the elements). They crank you up to some height above the ground and let the platform that lifted you essentially free fall for a ways before slowing you to a stop. The initial experience is nearly "weightless" followed by a greater than normal force to bring you to a stop, which makes you feel excessively heavy.

4. Jun 4, 2005

### michaelw

thank you
but im still a little confused..
If there was no air resistance, does that mean that anything traveling at a horizontal velocity will continue at that velocity (orbit) forever? intuitively thats not true, because of simple projectile motion

so what makes something orbit? is there a min or max speed horizontal speed neccessary to get it to orbit? how can this speed be found? if the speed is lower than this magic speed, does that mean the object will fall to the earth, and if its higher then it will escape earths gravitational pull?

thanks :)

5. Jun 4, 2005

### Gokul43201

Staff Emeritus
At a certain specific value of horizontal velocity, yes !!

This is only because there are some underlying assumptions in simple projectile motion (that rarely get mentioned :grumpy:)

One of these is that the distances in a projectile problem are small compared to the radius of the earth. This allows us to assume that gravity is a constant vector (both in magnitude and direction.

Remember that something that is traveling horizontally over large distances is actually traveling along a circular path (the earth is a large sphere), and hence has an acceleration towards the center of the earth. This acceleration is precisely equal to 'g'.

That's almost exactly right !

This magic speed, known as the orbital velocity is calculated from the last statement in my previous paragraph.

$$g = centrip.~ accel.=v^2/r=\frac {v^2}{R+h}$$
$$=>v_{orbital} = \sqrt{(R+h)g}$$
where h is the orbiting height and R is the radius of the earth.

As v increases, the orbiting height will increase. At another magic speed, called the escape velocity (=1.4*orbital velocity), the object will completely be free of the earth's gravity.

Last edited: Jun 4, 2005
6. Jun 5, 2005

### michaelw

thank you very much :)

7. Jun 5, 2005

### jdavel

OlderDan,

So does what you said about the training planes that simulate zero gravity mean that the plane has to be flown at just the right speeds along just the right parabolic path? I assumed they just got it going really fast, pointed it up at some angle, and cut the engines. I always wondered why air resistance and lift didn't screw everythinng up though. A controlled flight could avoid this, but it sure sounds as though it would be hard to do.

By the way, for anyone who didn't already know, when Ron Howard was shooting his movie Apollo 11, he got to use a "vomit comet" to shoot all the zero gravity scenes. The realism makes other movies about space travel seem pretty lame!

8. Jun 5, 2005

### OlderDan

The plane does have to be guided because of the air resistance and wing lift at any velocity. Just cutting the engines would make the plane glide and it would still have lift. I assume the whole thing is computer controlled or an autopilot is fed from a three axis accelerometer and responds to cancel any acceleration in any direction.

9. Jun 5, 2005

### Andrew Mason

I think you meant Apollo 13. Also, of course, it is not really zero gravity. It is a zero weight environment.

AM

10. Jun 5, 2005

### Integral

Staff Emeritus
I once was a passenger in a small single engine plane that flew a parabolic trajectory for a few seconds. I do not think that it is particularly difficult to do. We need a pilots comment.

11. Jun 5, 2005

### Huckleberry

I'm no pilot, but wouldn't it be as simple as flying in a -1 G turn in a parabola towards the Earth. Where's Danger? He was a pilot and loves these questions.

12. Jun 5, 2005

### OlderDan

It happens that I am a also a licensed single engine pilot, with several hundred hours of flying time, though regrettably it's been a long time since I last handled the controls. Achieving a true zero-g environment might be approximated for a second or two in a small plane, but it is no simple task to get it exact or try to maintain it. It takes a fair amount of skill just to learn to maintain a constant-g turn, and pilots rarely do that to perfection. Part of flight training used to involve stalls (I think they stopped requiring that) where you put the plane into a steep climb attitude until the plane lost lift and the nose dropped (assuming you had paid proper attention the load distribuion). To come out of it you accelerate on the downward path until you have enough lift to pull it out. All that happens pretty fast and though you are nearly weightless for that first drop, it's nothing like the sustained effect of true zero-g. Pushing a small plane into a dive, or better yet doing an outside loop in an aerobatic model might do it for a while, but actually flying a true zero-g path for any length of time is beyond the skill level of most pilots. I seriously doubt anyone can do it just by feel.

13. Jun 5, 2005

### Andre

That's because your speed is pretty low and you need quite some degrees per second pitch change to create the centrifugal force that simulates 0 g. Now in a jet doing 500 knots you could do a parabolic zero g flight that lasts several minutes.

However, nobody is going to try that because you probably have to bail out afterwards since all the oil floats out of the engine and the fuel in the zero g compartment only lasts some 10 seconds, leading to a flame out, since the jet was not designed that way.

But 0,25 g or -0,25g is not a problem.

14. Jun 5, 2005

### Huckleberry

Last edited by a moderator: May 2, 2017
15. Jun 5, 2005

### michaelw

thanks everyone
one more problem :/
I seem to get different values for v if i use conservation of energy vs newtons 1st law

Fg = Fc
mg = mv^2/R
v = sqrt(gR)

but..
W = Fd = Ek
mgR = mV^2/2
v = sqrt(2gR)

why are there 2 answers for v?

16. Jun 5, 2005

### OlderDan

Because these are completely different things. The first is an equation saying that the centripetal force at the top of a circular path is being provided by gravity. It would be appropriate for a problem of a roller coaster momentarily achieving exactly zero-g at a top.

The second equation is appropriate for finding the velocity of an object falling from height R if it inititally started from rest. It is more common to express that height as h, but it could be R or any other letter you want to use. It also applies do any other path of descent where energy is conserved, such as a frictionless inlcine or track.

17. Jun 7, 2005

### Gokul43201

Staff Emeritus
...because you are incorrectly using energy conservation.

$$KE_{initial} + PE _{initial} = KE_{final} + PE _{final}$$ is the correct form. You can use this if you know all the quantities that need to be plugged in. Your other calculation is correct.