Physics of falling objects
 

An argument developed off topic in the Malaysian thread about the physics of falling objects. It was off topic so was mostly deleted.

I and one other were taken to task by Skipness One Echo and others for stating that objects of the same size but different densities (eg a basketball and an identical one filled with concrete) would not hit the ground at the same time, dropped from say 35000 feet in an atmosphere.

Discuss.

there was an experiment many years ago where objects were dropped off the Eiffel tower to explore the effect. you are wrong.

the fall is due to gravity (whatever that is)
the gravitational attraction acts on every particle of the two balls at the same rate of influence.
so it doesn't matter whether you have a ball full of nothing or a ball full of concrete, every particle of both balls is under the same influence.
you don't get any more influence exerted on the concrete ball.

It has got to be a wind-up.

In a vacuum, all objects free fall due to gravity at the same rate of acceleration, regardless of their mass or density.

In a vacuum.


Objects dropped on earth, in the atmosphere, behave differently.


Ever see the feather and hammer dropped on the moon?
Feather & Hammer Drop on Moon - YouTube

Actually differing masses will accelerate under the sole force of gravity at exactly the same rate. The more massive object will only accelerate faster if there is reasonable air resistance.

Every kilogram of mass is acted upon by gravity (9.8N/kg on our planet). This is 'f'.

F/m=a

10kg(m) x 10 (9.8 simplified) = 100N/10kg(m) = 10m/s/s

100kg x 10 = 1000N/100kg = 10m/s/s

Factor in enough air resistance and time and the more massive object will hit the ground first. Would it in the OP's case? Dunno. Not enough info.

Is it possible to drop anything off the Eiffel tower without having it bounce off the side a couple of times?

http://aparisguide.com/eiffeltower/eiffel-tower1.jpg

Now the Tower of Pisa I could see working. But that could be too short to measure the effect of drag.

The formula for the terminal velocity of a falling object depends on its density (mass for identically sized objects).

Terminal velocity - Wikipedia, the free encyclopedia

So I would expect that, given sufficient altitude to allow terminal velocity to be approached and enough time for that difference in velocity to manifest itself as a noticeable difference in descent time, that difference would become apparant.

So a balloon filled with air dropped from the Eiffel Tower or even my garage bench would fall at the same rate as a balloon filled with concrete?

The denser ball will always hit the ground faster, in air, as it has a lower ratio of drag to weight.

Galileo is reported to have used different-sized cannonballs dropped from the tower of Pisa to show that overall mass, popularly thought to determine rate of descent, did not affect the outcome. (Although the bigger cannonball will fall a little faster owing to: a lower ratio of drag to weight.)

Provided the both objects are identical except for mass, they will fall at the same rate. Ignoring drag from the air for a moment, eg in a vacuum, gravity acts to accelerate each particle at the same time.

Consider this thought experiment: Three identical masses, dropped at exactly the same moment. Each will accelerate the same as the other two. Now connect two of the masses by a loose piece of string that is so small that its mass & surface area are negligible (or even non-existent if you posit a thread small enough). Do you think the two connected masses will suddenly behave differently to the independent mass? Now imagine the thread replaced by a rod so now the two masses are locked together. Do you think the connected masses will suddenly change behaviour? If it helps, imagine the connector suddenly appears mid-fall. Now imagine, instead of three identical masses with two of them connected together, two identical masses differing only in density. Gravity's effect is the same, regardless of size or shape.


Now add a fluid such as air. The only effect the fluid has is to provide a retarding force - 'drag' - reducing the acceleration of gravity. Drag is not affected by mass. C=Cd 1/2 rho v^2 s remember. So, as long as the objects *only differ in density* they will fall at the same rate.

Drag is affected by mass.

For the same density drag rises as size squared, while mass rises as size cubed.

Added - if shape is the same.

Mass is a good choice of a dependent variable to link that.

Don't forget Tinstaafl that while Cd does have that dependence, the drag force derived from it has an area term too.
If objects only differ in density, and have the same size and shape, they will definitely not fall at the same rate.

Yes, actually. The first elevated floor is a square "doughnut" with a hole in the center overlooking the plaza underneath. Easy to drop something 190 feet from there without it hitting anything before the ground (there is a glass wall, previously a low railing, to prevent amateur gravity experiments. ;) )

File:Sous la Tour Eiffel 1.jpg - Wikipedia, the free encyclopedia

Uh, no.

Drag is affected by size (among other things) and Mass is affected by size (among other things).

But there is no direct correllation between drag and mass (or vice versa) (with the exception where one introduces a wing into the equation, in which case one can say that induced drag (from the wing's lifting effort) is affected by mass.

That has nothing to do with form drag and air resistance in a free fall, however.

You can have high-mass objects with low drag (Saturn/Apollo moon rocket, for example) and low-mass objects with high drag (the average mattress, for example). And objects of equal mass and even equal density, with very different drag (a flat plate of aluminum vs. a needle-shaped piece of aluminum, both massing 5 kilos).

Induced drag depends on lift, not mass. If you put two wings in a wind-tunnel at a fixed AoA and filled one with concrete the induced drag would be the same in each. For aircraft though more mass means more lift is required and induced drag increases.

Exactly...

Unless you're misrepresenting the discussion (and I have n reason to believe that you are, just covering that possibility) Skippiness 1 E hasn't a clue what he's talking about.

Well, no, he's not wrong. I don't know about the Eiffel tower experiment, but Galileo is famously (and perhaps apocryphally) credited with doing the same thing at the Tower of Pisa. Some accounts have the balls falling at the same speed, other accounts have the heavier ball falling perceptibly faster.

In either case, it doesn't matter, the experiment was to test the (then) current theory of "Natural Motion" concocted by Aristotle which claimed that the speed of a falling object is proportional to it's mass. And even if the denser ball fell slightly faster, it was still obvious that the difference between the two was not proportional to the respective mass differences.



well, yeah, actually you do. The force of gravity on an object is proportional to it's mass. The basketball filled with air is about .6kg while the one filled with concrete will have a mass of 17 kg. So the force of gravity (in round numbers) near the earth's surface will be 6 newtons on the air ball, and 170 newtons on the concrete ball.

In the absence of any other force, the acceleration will be identical (because the force is proportional to the mass) but the balls are falling thru the atmosphere, not a vacuum. There is also drag to consider.

OK, forget weight for a moment. you have two objects moving through the same air. they have identical drag coefficients and identical frontal areas, and identical wetted area. Aerodynamically, they are identical. One has 6 newtons of force propelling it thru the air, and the other has 170 newtons of force propelling it through the air.

Which will travel faster thru the air?

It should be obvious that the object with 170 newtons propulsive force will travel faster than the one with 6 newtons.

It's the same with falling objects. As the balls fall faster, the drag begins to predominate over inertia, and more and more of the force of gravity is counteracted by the drag and less of it accelerates the balls. And the ball that has the least force will accelerate less than the ball with more force.

Ultimately both balls will reach their terminal velocity*, and the terminal velocity of the concrete ball will be much higher than the terminal velocity of the air filled ball.


*Dangerous territory when dealing with a poor understanding of physics. There is a popular misconception that there is some single "terminal velocity" for any and all falling objects This ain't true. Not even close. Every object has it's own terminal velocity determined by it's density and aerodynamic properties. A feather has a very low terminal velocity. Doesn't matter how high you let it fall, it ain't falling any faster. A high penetration aerial bomb on the other hand has a very high terminal velocity. There are bombs which will exceed the speed of sound while in a free fall. This is because they have a lot of mass and have a very low drag coefficient. (lots of gravitational force vs. not much drag force) The terminal velocity of a skydiver in free fall is somewhere between a feather and a bunker buster bomb. In fact, different individual skydivers will have different terminal velocities depending on their technique.

That's a good illustration of the complete absurdity of the claim. Frankly I'm astonished that we're even having this discussion on a board of professional pilots.

This discussion started on the ML thread when there was disagreement about the duration of free fall wreckage from 35000ft, with one maths man saying it would only take 40 seconds!

Real world examples differ from pure maths..
The Space Shuttle Challenger cockpit detached at 45,000 feet continued to 65,000 feet then free fell, slowly spinning for 2 min 25 seconds before it hit the water.
Challenger 1

Halo skydivers jump from 35000ft, it takes around 2 minutes to deploy their chute at between 3000 to 5000 feet.

Gopros weight around 100 grams and when not clad in a waterproof housing, are the size of a matchbox.

Unverified go pro footage, falling 2 minutes from 12500 feet, it flutters and rotates at around 7 revs per second.

Unverified go pro footage falling for 1 minute 40 seconds from around 10,500 feet,

Unverified go pro footage, falling for 22 seconds from 2200 feet.

Unverified go pro footage, 27 seconds to fall 3000 feet.

It's true as far as it goes. Drag is not affected by mass, but the *force* of gravity certainly is; in fact it's proportional to mass.

And for objects of different masses, which are otherwise externally identical, the *force of gravity is higher for the object with greater mass, so the *force* of gravity will reach equilibrium with the *force* of drag, at a higher airspeed. And when the *force* of gravity reaches equilibrium with the *force* of drag, the object will no longer accelerate. And like I said that, occurs at a higher airspeed for the more dense object (assuming same volume) .

You are correct. When they reach terminal velocity Drag = weight. If the shape is the same the only way for drag to be different is if the velocity is different.

They would be correct on the moon..
https://www.youtube.com/watch?v=5C5_dOEyAfk

EDIT: should be easy to prove by dropping a table tennis ball and something the same size like an egg...or perhaps not an egg as that's streamlined :-)

When you're faking the footage on a soundstage in Burbank, you can make anything fall at any speed you want. :}

pattern…

OK, but we were talking about objects of the same shape with the concrete and air basketballs.

To be more correct:

For the same density, speed and shape, drag rises as size squared, while mass rises as size cubed.

I believe the mathematical models which the OP may be trying to grasp are the 'wind-drift' models developed by the AAIB, initially developed to try and locate a missing Comet in the mid-1950's. Assuming an in-flight break-up, this methodology attempts to predict where the wreckage will land.

PM

Isn't building an evacuated sound stage about as much trouble as flying to the Moon?

When we are skydiving it is regular practise to wear lead weight vests to increase your freefall speed vs. your standard speed, particularly for lighter weight jumpers jumping with their heavier weight colleagues.

Must be something in it cos I can guarantee it works!

Mick,

It never free fell, in the technical sense, as it was suffering drag throughout.
A free-fall parachutist is also only briefly in physical (rather than semantic) free fall as they leave the aircraft door; as soon as drag builds, the weightless and acceleration go.

Vomit comets also don't free fall, they are flown precisely so that thrust balances drag so the zero g is imposed by the aerodynamic forces.

The test of the free-falliness of a gopro is that a real dragless free fall covers a height that's the square of time spent falling, while at a terminal velocity it's a linear relationship. 120s from 12000 feet, 100s from 10500 feet, 22s from 2000 feet and 27s from 3000 feet, are all consistent to within 10% of a 100 ft/s terminal velocity being quickly attained.

The freefall time from 35000ft is indeed about 45s, without air, or for an object so dense/streamlined that it doesn't suffer from any effective air resistance.
In contrast, dust and smoke floats, and never makes it the ground.

I saw much misunderstanding, in the discussion on the Malaysian Airlines thread about the shortest time needed for a transport jet to reach the surface from cruising altitude.

Formulas from elementary physics class are not useful. In all cases, and at every point in its trajectory, the jet's vertical velocity will be limited by aerodynamic forces.

Estimating the lower bound of descent time turns out to be quite simple: the airframe as a whole, or any detached part thereof, cannot attain a Mach number that exceeds one by more than a minute margin. Drag increases to such a great magnitude by Mach 1, that it easily matches aircraft weight plus engine thrust. Even to reach 1.0 Mach is very unlikely.

We know this on two grounds:

(1) In the early days of high-speed flight research, aerodynamicists designed shapes that were carefully optimized for minimum drag in the transsonic region, fabricated these shapes in solid steel (many times denser than any actual aircraft), and dropped them from high altitudes (20 to 30 thousand feet). It was very difficult for them to reach Mach 1 even in this condition.

(2) Some hair-raising incidents with transport jets in which gross upsets resulted in uncontrolled dives, showing maximum Mach numbers near 1 even with engine thrust pushing the jet into its dive. For example, in the 727 dive near Detroit (of Hoot Gibson fame), the maximum Mach number achieved was 0.96.

To find a lower bound for descent time, assume purely vertical motion (no "forward speed") at Mach 1. For the example of initial altitude of 35,000 ft, this comes to roughly 34 seconds.

However, starting from cruise condition, the jet would have an initial vertical speed of zero and forward speed in excess of Mach 0.8, so at least some seconds would be consumed in the acceleration of the velocity vector from horizontal to vertical -- this acceleration (again) controlled by aerodynamic forces.

Note 1: Even in a very severe accident, the actual descent time would likely be substantially longer than estimated by these simple assumptions.

Note 2: A transport jet has much lower drag than any broken part thereof, so the terminal velocities of components after an in-flight breakup can be expected to much less than Mach 1.

Yes, the aircraft could be dived into the sea from cruise in less than two minutes, or if it was vaporized, it would never have come down.

The detailed seabed debris fields from Swiss 111 or AF447 for point impacts on the ocean surface, or from Lockerbie and TWA for the places where high-altitude disintegration put things at sea level mean that there isn't going to much problem reconstructing which took place once the wreckage is found.

Thanks to the replies.

I can't remember the exact posts as they were deleted, but someone had commented about objects of differing density falling at different speeds reaching the ground at different times. That poster was told by someone to go back to school as every object falls at the same rate.

I used the concrete and air basketball analogy to defend that small part of the first poster's premise (as it gets very complicated very quickly with inflight breakup and 470kts fwd velocity) and was then told I needed to go back too school too!

Never one to take a false accusation lying down (to my detriment at times!) thank-you!

Ironic to have someone who is *that* freakin' clueless about physics telling others they need to go back to school, isn't it? That's life on an internet forum though, I guess.

Post #8:

I certainly don't remember that. I do remember that Drag = Cd x ½ρV²S though; as I would expect any pilot, or aerospace engineer to do.

You're quite right. That was my typo. Corrected it now. Don't do longish posts using a phone...

Try this paper by Dr Matthew Greaves, Snr Lecturer in Accident Investigation at
Cranfield Safety and Accident Investigation Centre

Trajectory Analysis

That's reassuring, since there isn't any "variable gravity" until you're worried about altitudes that reach tens or hundreds of km high.

I wonder what he meant?

What's the world coming to
 

when pilots don't know the basics of physics?

s=ut+1/2 at^





^ = squared

s = displacement
u =initial velocity
t = time
a = acceleration


This is the basic physics equation for falling objects where the object size is not considered. There is no mass to take into account, so a mass of 10kg will fall at the same speed as a mass of 40kg, IN EFFECT IF IN A VACUUM! If you take the real world into account and both objects have the same profiles, i.e. two containers of the same dimensions, but one is filled with concrete and the other is empty and both will fall same way down, meaning the empty one won't tumble, the resistance thus terminal velocity will be the same and both will hit the ground at the same time regardless of mass.



When you are considering an object falling in the atmosphere the only determinate when it comes to how fast an object will fall is the area of the object and therefore the drag(resistance) it will produce, mass does not come into it.

Yep, mass does not come into it. Which is why a helium-filled balloon falls to the ground in exactly the same manner as the same ballon filled with water, as everyday experience confirms.

This thread is a telling example of how confused educated people can get - on firm ground.

Not to be a stickler, but isn't that just a bit of an oxymoron?

Well, I've never done it but I'd guess when designing a model it's generally a good idea to cover the variables so that the output is as accurate as possible.

Gravity does vary by altitude, latitude, underlying rock density and so on. 30,000 ft is about 0.3% lower. A rough wiki driven guess suggests total variation may as much as 1%. Whether this is significant or not I don't know.

At 10km up, about 0.15% of the radius of the Earth, gravity is indeed weaker by about 0.3%, but that's completely irrelevant.

By definition, the sea level has a constant gravity. Mentioning there might be changes in gravity just casts doubt on the sensibility of the rest of the discussion.

Well that's not really true, see latitude and density effects.
Anyhow the model seems designed to deal with items which start high up at over 30,000 ft and end low down, ergo they will see a variation in gravitational pull. Now I think that this is likely to make little or no difference compared to wind effects varying with altitude, but I don't know.
Surely the beauty of a model which accounts for varying gravity is that you can find out. If after enough runs of the model to account for statistical significance it turns out that it is irrelevant then you can remove it. If it does make a difference leave it in.

Go measure it. It's not quite an ellipsoid, but it is an equipotential.

What would happen to the water if there was a potential gradient on the surface?

Not quite. Gravity varies slightly across the Earth's surface. Sea level in one place can have a different gravity to somewhere else.