Physics of falling objects
Join Date: Nov 2012
Location: Planet Claire
Posts: 581
Likes: 0
Received 0 Likes
on
0 Posts
A RV, (that's a re-entry vehicle, or the bit that goes BANG!) on an ICBM is released in space.
It's merely 'pointed' quite accurately at its target. Not powered.
By the time it reaches the ground, it's doing about Mach 10.
It's merely 'pointed' quite accurately at its target. Not powered.
By the time it reaches the ground, it's doing about Mach 10.
Thank you, Aram.
Good grief! Mass versus drag for the same object is very apparent in weapon development - ballistic co-efficient. It shows up on basic, old bombs that have different times of flight when released from "x" altitude. Fat ones don't fall as fast as "slick" ones.
It is true that Mother Earth attracts the falling object at basic gee. Forget very small gravitational forces due to altitude above the "Mother".
The point is that F=m*a, and drag is relevant to that equation if in the atmosphere. The falling object is subject to the same "a", but the "F" part has to do with drag, which a function of v^2 and Cd and area of the object. Eventually the drag force equals Mother Earth's force and you have "terminal velocity".
A very good science show years ago was called 'Terminal Velocity". It was about determining a falcon's capability to exceed 200 mph. The sky diver could only reach about 120 mph terminal velocity, so the dude trained the bird to follow a bean bag. He would add heavier lead weights to the bean bag, but the bag was the same area when related to the air while going down. He increased the terminal velocity of the bag to over 200 mph, and the doggone bird could keep up. How? Well, the bird changed its area, so the drag decreased/increased according to the equation of Cd* s*1/2*rho*v^2.
From personal experience dive-bombing, I can attest to the fact that deploying "dive brakes" kept me from exceeding the aero limits of the plane in a very steep dive. All else equal, then it was the area that was part of the drag equation that kept me from ripping the wings off, heh heh. Don't try this at home.
The end of my epistle is that parts of the plane with ottsa area, but relatively low weight will drift in the atmosphere on the way down. The wind will move them away from the point of impact of dense objects of lesser area like engines or fuel pumps or.... They come down slower, so the wind moves them, duhhh?
Been a long time since Aero 101, but I think I got most of it right.
Good grief! Mass versus drag for the same object is very apparent in weapon development - ballistic co-efficient. It shows up on basic, old bombs that have different times of flight when released from "x" altitude. Fat ones don't fall as fast as "slick" ones.
It is true that Mother Earth attracts the falling object at basic gee. Forget very small gravitational forces due to altitude above the "Mother".
The point is that F=m*a, and drag is relevant to that equation if in the atmosphere. The falling object is subject to the same "a", but the "F" part has to do with drag, which a function of v^2 and Cd and area of the object. Eventually the drag force equals Mother Earth's force and you have "terminal velocity".
A very good science show years ago was called 'Terminal Velocity". It was about determining a falcon's capability to exceed 200 mph. The sky diver could only reach about 120 mph terminal velocity, so the dude trained the bird to follow a bean bag. He would add heavier lead weights to the bean bag, but the bag was the same area when related to the air while going down. He increased the terminal velocity of the bag to over 200 mph, and the doggone bird could keep up. How? Well, the bird changed its area, so the drag decreased/increased according to the equation of Cd* s*1/2*rho*v^2.
From personal experience dive-bombing, I can attest to the fact that deploying "dive brakes" kept me from exceeding the aero limits of the plane in a very steep dive. All else equal, then it was the area that was part of the drag equation that kept me from ripping the wings off, heh heh. Don't try this at home.
The end of my epistle is that parts of the plane with ottsa area, but relatively low weight will drift in the atmosphere on the way down. The wind will move them away from the point of impact of dense objects of lesser area like engines or fuel pumps or.... They come down slower, so the wind moves them, duhhh?
Been a long time since Aero 101, but I think I got most of it right.
Join Date: Nov 2005
Location: Zulu Time Zone
Posts: 730
Likes: 0
Received 0 Likes
on
0 Posts
Awblain:
Clearly the guy is talking about a model that accounts for variation of the gravity field by location rather than using the bog standard 9.81ms⁻². It isn't even slightly controversial. NASA have a satellite mapping this because "the Earth's gravity field is not uniform". NASA GRACE Mission
That's reassuring, since there isn't any "variable gravity" until you're worried about altitudes that reach tens or hundreds of km high.
Join Date: Feb 2000
Location: Alaska, PNG, etc.
Age: 60
Posts: 1,550
Likes: 0
Received 0 Likes
on
0 Posts
Indeed. What you're missing though, is that you are the one who doesn't know the basics of physics.
It is astonishing that you continue to insist that you're correct when it has been demonstrated quite thoroughly that you are wrong.
And TINSTAAFL, this is directed at you also, as you too seem determined to continue posting your incorrect understanding of this.
There have been several very detailed explanation of *why* a more dense object will fall faster through the atmosphere than will less a dense dense object of the same dimensions. Your example of an empty container falling at the same terminal velocity is, quite simply, wrong. It has been repeatedly explained in this thread *why* it is wrong, with detailed discussions of the forces involved. Instead of simply repeating the claim that they fall at the same speed, without offering proof, why don't you *explain* why you think the explanations are incorrect?
Serious question? If you wish to be taken seriously, why aren't you explaining your case instead of just contradicting the explanations which have been given? Merely repeating your contradiction makes you appear as John Cleese in Monty Python's argument sketch.
Additionally, a number of posters have cited a very compelling example which makes it abundantly obvious that you are wrong, yet, you pretend that you haven't even noticed. Again, consider the example of two balloons, one inflated with air, and one filled to the same dimensions with water, or concrete, or whatever dense material . Drop them both from the same height. They do not fall at the same rate. Everyone realizes this, except, apparently, you. The air filled balloon is identical to the "empty container" you claim will fall at the same rate as one filled with concrete. In fact it *is* by any definition a container (Not technically empty, as it is filled with air, but so too is the "empty" container in your post actually filled with air)
So how can an "empty" (air filled) container fall at the same speed as the same container filled with concrete (as you claim) when it is obvious to all that an air filled balloon does not fall at the same rate as one filled with concrete?
That, by the way, is not merely a rhetorical question. I actually would like to see your answer. You have a very simple, obvious, very easily understood example that shows without any question that you are wrong, and somehow you rationalize that away. I am truly interested to hear your explanation for how or why you disregard this.
If the fact that it the balloon is filled with air is confusing you let's consider a different form of the same example. A feather. No doubt you have seen feathers falling, and noted that they fall rather slowly. Now lets suppose that a jeweler had very carefully crafted a replica of that same feather from gold. The form of the feather is identical on a microscopic level, it is the same dimensions, the same thickness, the shaft is the same dimensions, as are the barbs, and the barbules between them, everything is duplicated to the same dimensions, except that it is made of solid gold instead of ... well whatever it is feathers are made of.
With me so far? OK, I think that everyone will agree that the real feather will fall decidedly slower than the gold feather, right? Is that not a pretty conclusive demonstration that , all else being equal, the speed of a falling object through air is *not* independent of it's mass ? If you believe that it is not, explain *why* not. If you can't explain *why* not, than that's likely an indication that you're mistaken.
While you are mulling over your explanation of the previous obvious flaws in your claim, you might consider the following:
NASA webpage on Terminal Velocity
and
Wikipedia Article on Terminal velocity
Both of these pages contain an equation for calculating an object's terminal velocity (and both pages describe how that equation is derived) the equations are the same equation in slightly different formats. The NASA page puts the equation in a format that can be shown using only ASCI text and no special formatting (such as fractions, radical signs, or greek letters) but mathematically the equations are identical.
Now here's the point which demonstrates quite conclusively that you're wrong. Both equations contain a term for mass, which can *only* mean that terminal velocity *is* in fact affected by mass. If terminal velocity was (as you erroneously claim) independent of mass, mass would not be included as a variable in the equation for calculating it. But, it is included, so it must affect the velocity*. Now, I suspect that you're getting ready to say; “ But the NASA equation uses “weight” which is not the same as mass” True, in the context of physics, weight is not the same as mass. In physics, “weight” is the force of gravity on an object. And weight is equal to gravity times mass (g*m ) so where the NASA equation has W, it really means g*m so the first term in parentheses (2 *W) becomes (2*g*m) which you will note is the same as the upper term beneath the radical in the Wikipedia Equation: 2gm The asterisks in the NASA equation denote multiplication, while the multiplication is implied in the Wikipedia equation, so 2*g*m is just a different way of writing: 2gm. Bottom line, the equations are identical, and they both contain mass as a variable.
This leaves us with this:
Your (fallacious) opinion:
Originally Posted by NASA
Terminal Velocity = sqrt ( (2 * g* mass ) / (Cd * r * A)
All the preceding can be summarized neatly in two questions:
How do you resolve you claim that mass doesn't affect the velocity of a object falling through air, when a balloon filled with air and a balloon filled with concrete obviously and unquestionably fall at different speeds?
If Mass doesn't affect the speed of a falling body, why does NASA, one of the worlds foremost organizations for the study of aerodynamics and ballistics, say that it does?
Serious questions. Response expected. Lack of response will indicate a tacit admission of being mistaken.
* if anyone is having difficulty following this reasoning, all you have to do is make some simple calculations of terminal velocity using those equations and the same values for all terms, but different values for mass. The calculated terminal velocity will be different.
Try it with the following terms to keep it simple:
Drag Coefficient (Cd in NASA's equation) = 1
Area(A in NASA's equation) = 1 (square meter)
Gravity (g) = 9.8 (meters/sec/sec)
Rho ("r" in NASA's equation) = 1.3 (kg per cubic meter sea level standard temp at pressure )
Mass = 1 (kg)
Don't worry about the units, just plug the numbers into the equation and see what you come up with.
Now do it again, using all the same numbers except you use 10 kg for mass instead of 1kg. Notice that the calculated terminal velocity is different? the point here is that you can't include a quantity in an equation and not have it affect the result. That means the terminal velocity of an object is dependent on mass. QED.
Join Date: Dec 2007
Location: Pasadena
Posts: 633
Likes: 0
Received 0 Likes
on
0 Posts
Oggers,
Look at those variations though, and consider their nature. At constant height above sea level / geoid, they are very modest, requiring days of data from satellite orbit measurements to measure. The differences are typically at the 10 ppm level, and up to about 60 ppm.
See GRACE Gravity Model- Gravity Recovery and Climate Experiment Gravity Model for a description of the GRACE mission and its products. GOCE produced more too.
If the author is including GRACE maps is his fluttering debris data, then he's missing the wood for the trees.
Is he using pressure altitude to start his simulation? If so, then he should have a further think about the "variable gravity" bit and where isobars lie in 3D.
Look at those variations though, and consider their nature. At constant height above sea level / geoid, they are very modest, requiring days of data from satellite orbit measurements to measure. The differences are typically at the 10 ppm level, and up to about 60 ppm.
See GRACE Gravity Model- Gravity Recovery and Climate Experiment Gravity Model for a description of the GRACE mission and its products. GOCE produced more too.
If the author is including GRACE maps is his fluttering debris data, then he's missing the wood for the trees.
Is he using pressure altitude to start his simulation? If so, then he should have a further think about the "variable gravity" bit and where isobars lie in 3D.
Join Date: Feb 2000
Location: Alaska, PNG, etc.
Age: 60
Posts: 1,550
Likes: 0
Received 0 Likes
on
0 Posts
I suspect that is a little too subtle for some. Can you be more specific on the difference of the effect of gravity? Is the heavier glider pulled to earth faster or at the same rate as the glider which has dumped it's ballast?
Join Date: Dec 2007
Location: Pasadena
Posts: 633
Likes: 0
Received 0 Likes
on
0 Posts
It would be an important check for the thoughtful reader to work it out first and then check the answers; that way the thoughtful reader might be able to tell whether they'd been thoughtful enough.
It might be easier to consider the flight of a lead zeppelin before adding the complexities of lift.
It might be easier to consider the flight of a lead zeppelin before adding the complexities of lift.
Join Date: Nov 2005
Location: Zulu Time Zone
Posts: 730
Likes: 0
Received 0 Likes
on
0 Posts
A squared:
Indeed. What you're missing though, is that you are the one who doesn't know the basics of physics.
It is astonishing that you continue to insist that you're correct when it has been demonstrated quite thoroughly that you are wrong.
If something is worth saying it is worth saying twice.
Will some of the numpties posting here please wake up.
For an object in freefall at terminal velocity: drag = mass x g. That is not up for debate. 'g' being fixed (notwithstanding the fact it does vary very slightly
) - if you increase the mass you must increase the drag that would result at terminal velocity. And that means an increase in terminal velocity, because the OP is predicated on a fixed shape and hence a fixed coefficient of drag.
This result has already been mentioned and proved in other ways in this thread multiple times now. I am bemused - though no longer the least surprised - to see some posters still wading in to contradict this fact with their half baked physics.
Originally Posted by ANCPER View Post
What's the world coming to
when pilots don't know the basics of physics?
What's the world coming to
when pilots don't know the basics of physics?
It is astonishing that you continue to insist that you're correct when it has been demonstrated quite thoroughly that you are wrong.
Will some of the numpties posting here please wake up.
For an object in freefall at terminal velocity: drag = mass x g. That is not up for debate. 'g' being fixed (notwithstanding the fact it does vary very slightly
) - if you increase the mass you must increase the drag that would result at terminal velocity. And that means an increase in terminal velocity, because the OP is predicated on a fixed shape and hence a fixed coefficient of drag.This result has already been mentioned and proved in other ways in this thread multiple times now. I am bemused - though no longer the least surprised - to see some posters still wading in to contradict this fact with their half baked physics.
Last edited by oggers; 13th Mar 2014 at 12:06.
Join Date: Nov 2005
Location: Zulu Time Zone
Posts: 730
Likes: 0
Received 0 Likes
on
0 Posts
Awblain
There is no need. You said:
Which is an opinion shown to be false by that link to NASA I provided. I am not going to enter into a game of semantics with you because I find that sort of debate really lame.
If you wish to prove the point you have now shifted to:
Then please do get on with it, it might be interesting 
Look at those variations though, and consider their nature.
there isn't any "variable gravity" until you're worried about altitudes that reach tens or hundreds of km high.
If you wish to prove the point you have now shifted to:
If the author is including GRACE maps is his fluttering debris data, then he's missing the wood for the trees.
Is he using pressure altitude to start his simulation? If so, then he should have a further think about the "variable gravity" bit and where isobars lie in 3D.
Is he using pressure altitude to start his simulation? If so, then he should have a further think about the "variable gravity" bit and where isobars lie in 3D.
Join Date: Feb 2000
Location: Alaska, PNG, etc.
Age: 60
Posts: 1,550
Likes: 0
Received 0 Likes
on
0 Posts
Comparison of falling spheres.
Join Date: Dec 2007
Location: Pasadena
Posts: 633
Likes: 0
Received 0 Likes
on
0 Posts
Oggers,
You're not entering into a game of semantics, you're missing the point. You get horizontal gravity variations of about 1 part in 100,000. They don't matter. Period. If anyone really sticks them into a model of falling debris, then they're casting doubt upon the quality of other bits in their model.
It is possible that the vertical changes in gravity, by about 3 parts in 1000 from 10km up to sea level might have a subtle effect on where heavy dense debris lands, since that slows less and spends less time lower down than lighter debris.
You're not entering into a game of semantics, you're missing the point. You get horizontal gravity variations of about 1 part in 100,000. They don't matter. Period. If anyone really sticks them into a model of falling debris, then they're casting doubt upon the quality of other bits in their model.
It is possible that the vertical changes in gravity, by about 3 parts in 1000 from 10km up to sea level might have a subtle effect on where heavy dense debris lands, since that slows less and spends less time lower down than lighter debris.
Join Date: Nov 2005
Location: Zulu Time Zone
Posts: 730
Likes: 0
Received 0 Likes
on
0 Posts
The video contains a lot if irrelevant and annoying dialog
Join Date: Feb 2000
Location: Alaska, PNG, etc.
Age: 60
Posts: 1,550
Likes: 0
Received 0 Likes
on
0 Posts
Join Date: Nov 2005
Location: Zulu Time Zone
Posts: 730
Likes: 0
Received 0 Likes
on
0 Posts
Awblain:
Well, you're half right. And I'm open minded on the other half. Just for the record you said:
...whereas you now say 'there is variation but it isn't enough to make any difference'. I am asking you to put some flesh on this by actually showing us some of this trajectory analysis you claim will not be affected. Because, frankly, compared with the expert opinion of Dr Matthew Greaves of Cranfield University, I find the lack of substance to your contra opinion to be most unconvincing
You're not entering into a game of semantics, you're missing the point. You get horizontal gravity variations of about 1 part in 100,000. They don't matter. Period.
there isn't any "variable gravity" until you're worried about altitudes that reach tens or hundreds of km high.
Join Date: Dec 2007
Location: Pasadena
Posts: 633
Likes: 0
Received 0 Likes
on
0 Posts
Oggers,
What is about "1 part in 100,000" that suggest it matters?
Drag changes by 1 part in 50,000 would have the same effect. How much is that? It's associated with the same density changes as climbing about 10cm. QED.
Perhaps Dr Greaves should worry about density changes from the top to the bottom of the wreckage too? I don't think so, and I assume that in fact Dr Greaves meant something else in his abstract.
What is about "1 part in 100,000" that suggest it matters?
Drag changes by 1 part in 50,000 would have the same effect. How much is that? It's associated with the same density changes as climbing about 10cm. QED.
Perhaps Dr Greaves should worry about density changes from the top to the bottom of the wreckage too? I don't think so, and I assume that in fact Dr Greaves meant something else in his abstract.
Join Date: Oct 2003
Location: n/a
Posts: 1,425
Likes: 0
Received 0 Likes
on
0 Posts
Having read through the paper, the tool is designed to have the ability to deal with very high altitude break ups of re-entering space vehicles where gravitational variation with altitude would have an effect.
Join Date: Dec 2007
Location: Pasadena
Posts: 633
Likes: 0
Received 0 Likes
on
0 Posts
Even then, it's hardly worth the worry.
At 100km, the difference in gravity from sea level is only 3%.
If you get have an exosphere density profile correct to 3% then you're already doing well.
At 100km, the difference in gravity from sea level is only 3%.
If you get have an exosphere density profile correct to 3% then you're already doing well.