Peter Macinnis replied:
>> 5. Does the falling body
accelerate at a uniform rate until it reaches
>> terminal velocity, or does the rate of acceleration gradually
decrease,
>> so that velocity asymptotically approaches the terminal?
>
>Yes and no. The asymptotically is incorrect, I think.
A little analysis will reveal that
you have (in effect) two forces: acceleration due to gravity,
which remains just about uniform within the range of the atmosphere,
and retardation, which depends on drag, which is dependent on
velocity. My reasoning is that the "force" of air
resistance is zero when the body is at rest and starts to fall, and
steadily increases until the air resistance force cancels out the
acceleration force due to gravity, so that the velocity becomes a
constant. On that basis, it has to be asymptotic -- as I
understand Bruce to be using the term, where the effective acceleration
falls away as the terminal velocity is approached.
Paul Williams answered:
> > 1. Is terminal velocity a
product of air resistance, or would it still
> > exist in a vacuum?
> >
> Yes, and no it would not.
>
<snip>
I think that in Newtonian mechanics a body falling in a gravitational
field in vacuum would have an infinite terminal velocity.
Zero
Sum replied:
> A little
analysis will reveal that you have (in effect) two forces:
> acceleration due to gravity, which remains just about uniform
within the
> range of the atmosphere, and retardation, which depends on drag,
which
> is dependent on velocity. My reasoning is that the
"force" of air
> resistance is zero when the body is at rest and starts to fall, and
> steadily increases until the air resistance force cancels out the
> acceleration force due to gravity, so that the velocity becomes a
> constant. On that basis, it has to be asymptotic -- as I
understand
> Bruce to be using the term, where the effective acceleration falls
away
> as the terminal velocity is approached.
I cannot argue with your
reasoning. I can only offer the counter argument.
Terminal velocity is
reached. If the effect were truly asymptotic, then terminal
valicity could only be approached, not reached. If it were
asymptotic there would always be some acceleration (until the ground
was hit). This does not appear to be so in practice. Hence
it cannot be asymptotic although it may resemble such a curve.
Anthony Morton wrote:
> Terminal velocity is
reached. If the effect were truly asymptotic, then
> terminal valicity could only be approached, not reached. If
it were
> asymptotic there would always be some acceleration (until the
ground was
> hit). This does not appear to be so in practice. Hence
it cannot be
> asymptotic although it may resemble such a curve.
There's a difference between attainment 'for all practical purposes'
and attainment 'in theory'. To a good approximation, the equation
of free fall under gravity with air resistance has the same form as
other familiar 'asymptotic decay' processes, such as radioactive decay
or progress toward thermal equilibrium. So terminal velocity is
'reached' in the same sense that short-lived isotopes 'vanish', or a
body free of heat sources 'attains' ambient temperature.
A good first approximation to the physical situation is that the force
of air resistance on a body is proportional to velocity: Fr = - K *
v(t), where K is a constant depending on the
surface-area-to-volume ratio. (A sheet of paper has large K; a
scrunched-up sheet of paper has small K.) In free fall, the only
other force acting is that of gravity: F
g = M * g where
M is the mass and g is a constant acceleration (9.8 m/s per second at
the Earth's surface). So Newton's Second Law reads
M * dv/dt = M * g - K * v(t)
or
dv/dt + (K / M) * v(t) = g.
This equation for velocity as a function of time is typical of most
equilibrium scenarios in physics. On the assumption that velocity
ultimately attains a constant value (say V
o), this can be
found by setting dv/dt = 0. We obtain V
o = M * g / K,
which is the approximate formula for the terminal velocity of a falling
body. It is equal to
the weight-force divided by the coefficient of air resistance.
However, in practice this velocity V
o is attained only
asymptotically; the rate of change dv/dt can be made as small as we
please if we wait long enough, and eventually no measuring device will
be capable of distinguishing the actual velocity from V
o.
dv/dt will eventually differ from zero only by a quantum fluctuation,
but classically will
never be precisely zero.
A more precise analysis of free fall recognises that the force of air
resistance is nonlinear, depending on higher powers of the velocity.
Thus, a better approximation is F
r = - K
1 * v - K
3
* v
3.
This makes the equation of motion more difficult to solve, but the
essential feature is retained: the terminal velocity is obtained
asymptotically, rather than precisely after a finite time.
Zero
Sum wrote:
> > Terminal velocity is
reached. If the effect were truly asymptotic,
> > then terminal valicity could only be approached, not
reached. If it
> > were asymptotic there would always be some acceleration
(until the
> > ground was hit). This does not appear to be so in
practice. Hence it
> > cannot be asymptotic although it may resemble such a curve.
>
> There's a difference between attainment 'for all practical
purposes'
> and attainment 'in theory'. To a good approximation, the
equation of
> free fall under gravity with air resistance has the same form as
other
> familiar 'asymptotic decay' processes, such as radioactive decay or
> progress toward thermal equilibrium. So terminal velocity is
'reached'
> in the same sense that short-lived isotopes 'vanish', or a body
free of
> heat sources 'attains' ambient temperature.
You are correct and I understand
this. However, it seems to me that by changing the ground you can
be on either side of the question.
In practice, terminal velocity
*is* attained. Theoretically, it can never be reached.
Approaching luminal velicity would be asymptotic, reaching teminal
velocity is not. That which can be reached is not and cannot be,
the product of an asymptotic curve. Remember that maths *models* the
real world and the equations you present mandate a very simple model
which only roughly approximates reality.
Any further discussion would, I
think, be mere debate.
Christopher Luke wrote:
It seems to me that you would be more likely to attain and remain at
terminal velocity (or something close to it, either above or below)
while descending through a thickening medium, as the decelerating force
from the medium would always be increasing, so you would always have a
significant mismatch between gravity and air resistance.
Bruce
Harris responded:
Wow. Thank you, gentlemen, asking
a question of this list is indeed a rewarding experience. Without
wishing to sound like a crawler or a pisser-in-pockets, I get immense
pleasure reading the words of people who know what they're talking
about (and that's not everyone on the list, but a significant subset of
them).
All this talk of air resistance
has evoked a memory of another question. I swear this was not an
ulterior motive in the original posting, it's just something I was
reminded of while reading the responses about terminal velocity.
A year or so back I was showing my
sister, her husband and three of their four kids, over from Perth, the
delights of Springbrook National Park. Inspired by watching a waterfall
from the top, and later walking under it, my nephew asked my opinion on
something that had bugged him for a while. Suppose you are standing at
a cliff edge, and you throw a rock horizontally. It's a mile to the
valley floor, and the rock hits a certain horizontal distance from you.
Now suppose that instead of one mile, it's two miles down to the valley
floor. How much further, horizontally, would the rock have flown while
descending the extra mile?
His dad reckoned the extra
horizontal distance would be next to nothing, "no more than a metre or
so" (these people from the west are very liberal in jumping between
mensuration systems, I thought), because by the time it's travelled a
mile downwards its trajectory is so close to vertical that any forward
movement is negligible.
I told my nephew I didn't know the
answer in the real world. All I could offer with any certainty was that
this idea that the downward acceleration somehow "swallowed up" the
horizontal motion was a misconception, that the two were independent of one another. If
there were no air resistance, the horizontal velocity would be
unchanged throughout the drop. Doing a "back of the envelope"
calculation without the aid of the envelope, I reckoned it would take about 18 seconds to fall the
first mile and another 7 seconds to fall the second mile. If the
initial horizontal velocity was, say (wild guess) 40 metres per second,
then that would still be the horizontal velocity during the second mile
of freefall (in the no air resistance scenario) so the rock would move
an extra 280 metres horizontally during that second mile of vertical
drop. However, in the real world where air resistance is a factor, I
simply had no idea.
I
still reckon my answer was essentially correct for the "no air
resistance" model. However I still have no idea what the answer would
be in the real world. Is my brother-in-law right, that it would only
advance an extra metre or
so in the second mile of falling?
Zero Sum replied:
> I still reckon my answer was
essentially correct for the "no air
> resistance" model. However I still have no idea what the answer
would be
> in the real world. Is my brother-in-law right, that it would only
> advance an extra metre or so in the second mile of falling?
Reasonable. The two dimensions can be considered independantly,
but that model fails on one point. The air resistance applies
along the vector of movement (which is constantly changing from the
horizantal towards the
vertical) so there would be a 'rate of change' in air resistance along
each dimension that was not entirely dependent on the velicity in that
dimension.
Anthony
Morton responded:
>
Reasonable. The two dimensions can be considered independantly,
but that
> model fails on one point. The air resistance applies along
the vector of
> movement (which is constantly changing from the horizantal towards
the
> vertical) so there would be a 'rate of change' in air resistance
olong
> exch dimension that was not entirely dpendeant on the velicity in
that
> dimension.
This is the main
complication. Nonetheless it's not hard to see that the main
conclusion is the same as for free fall: the terminal velocity is
always directed vertically, and is independent of the initial
conditions (in particular, it doesn't matter how hard you throw the
stone, it'll always wind up falling vertically).
But if one is prepared to
approximate the drag force as being proportional to speed (that is, Fr = K * v, directed opposite to the
velocity vector), it is still possible to decouple the horizontal and
vertical motion, just as when air resistance is not present. In
this case it is possible to define a 'half-life' for the horizontal
component of velocity. Say this is 10 seconds, and you threw the
stone horizontally at 20 m/s: then after 10 seconds the horizontal
speed is 10 m/s, after 20 seconds it's 5 m/s, after 30 seconds it's 2.5
m/s and so on.
The approximate half-life can be
determined from the ratio of the terminal velocity to the gravitational
acceleration, with a factor (ln 2) to convert from natural to base-2
exponents. That is:
Th = ln 2 * Vt / 9.8 = Vt / 14.14
where Vt is the observed terminal
velocity. So a rock with a terminal velocity of 10 m/s will have
a half-life of about 0.7 seconds for decay of the horizontal
motion. If it takes 18 seconds to fall the first mile down the
cliff and starts with a horizontal velocity of 40 m/s, its velocity
after falling the first mile will be about one-thousandth
of a millimetre per second!
So your brother-in-law just might be right in thinking that the extra
horizontal distance travelled after that is next to nothing.
But the decay is *very* sensitive
to the terminal velocity of your rock. A boulder with a terminal
velocity of 100 m/s (or 360 kph) will have a half-life closer to 7
seconds, and if you shoot that out of a cannon with an initial
horizontal velocity of 40 m/s, it'll still be travelling at nearly 7
m/s (or 25 kph) in the horizontal direction after falling for 18
seconds. So it'll easily cover a few metres in the following 7
seconds.
In reality the magnitude of the
drag force is a nonlinear function of the speed, so the horizontal and
vertical motions don't nicely decouple like this. But the
half-life still gives a fairly good approximation to the exact
behaviour.
Bruce Harris wrote:
Zero wrote:
> Reasonable. The two
dimensions can be considered independantly, but that
> model fails on one point. The air resistance applies along
the vector of
> movement (which is constantly changing from the horizantal towards
the
> vertical) so there would be a 'rate of change' in air resistance
along
> each dimension that was not entirely dpendeant on the velicity in
that
> dimension.
Conceded, excellent point, and it does make a difference. However,
alas, it gets me no closer, or rather not much closer, to an answer to
the question.
Anthony Morton answered:
>
Although you may reach terminal velocity with or without the parachute
> opening, it seems to me that it is much more terminal in the latter
> case. :=)
According to my first year physics
textbook, the terminal velocity of a skydiver is 200kph without a
parachute, and 18kph with a parachute. The numbers tell the story :-)
Paul Williams posted:
Just playing around:
I realise that this is not predicted
but could we postulate a super massive 'dark star' in Newtonian
mechanics containing all the mass of the universe?
If we can do this, and if we further
postulate infinite mass for our 'dark star', it seems that we end up
with an infinite escape velocity and therefore an infinite terminal
velocity...??
Playing around some more:
Superman comes from the planet
Krypton (a noble gas of course - but he is a noble man :-)).
If we ignore the silliness of his
'Superstrength' and just look at the physics of the deep gravity well
which was originally postulated to produce this superstrength:
- If Superman has the strength of, say, 1,000 men, would this not
denote a mass for the planet Krypton 1,000 times that of the Earth?
- If this is so, would not the escape velocity of Krypton be about
11,000 km/s?
(Actually the simplistic 1,000 times
the mass of Earth cannot work - ordinary matter cannot attain the
density needed for a suitable surface gravity)
My understanding is that the escape
velocity of the Sun is about 618 km/s.
A white dwarf of the Sun's mass has
an escape velocity of about 3,380 km/s.
It seems that if we add enough mass
to increase the (now white dwarf) Sun's mass to attain a Krypton-like
gravity, we end up with a neutron star.
Unfortunately, the escape velocity
of a neutron star of about 1.5 solar masses ends up being a lot more
than we need - approaching 200,000 km/s.
It appears that the problem now is
to create a Krypton with a neutron star centre but with a
'non-neutronian' surface surround.
Clutching desperately at straws to
create this homeworld for Superman, we may perhaps posit a neutron star
which is spinning rapidly and has a halo of material suspended
(rapidly orbiting) around at a distance which
gives us the requisite
gravity/escape velocity.
We now appear to have a further
problem in that gravitational heating of this material would produce
temperatures of millions of degrees. That the whole lot would be bathed
in x-rays we may assume would not be
health promoting for our Kryptonians
in any way. (This material would also likely be spiralling into our
Neuton star
core)
I think we must sadly discount a red
giant sun for our Krypton, for if one had existed, it would have
actually been orbiting our Krypton - but not for long - it seems that a
black hole 'Sun' may be more likely.
This of course complicates the
possibility of escaping this hellish 'solar system' somewhat (well
actually, I believe it makes it *impossible*)
If we accept all this as being
remotely possible, one may cease to wonder that Krypton was so unstable
(not to say uncomfortable) that finding a safer abode for Kal-el
(Superman) to live would have been an immediate
concern for his parents.
One may say that none of this
matters as our Kryptonians can certainly not be made of ordinary matter
- even extraordinary matter... but that will be another story...
David Martin responded:
Lots of questions here, and lots of
good physics too. This will just be a
brief reply, I'm busy right now.
> Just
playing around:
> I realise that this is not predicted but could we postulate a super
> massive 'dark star' in Newtonian mechanics containing all the mass
of
> the universe?
> If we can do this, and if we further postulate infinite mass for
our
> 'dark star', it seems that we end up with an infinite escape
velocity
> and therefore an infinite terminal velocity...??
I'm not game to touch concepts
like infinite mass :-)
> Playing
around some more:
>
> Superman comes from the planet Krypton (a noble gas of course -
but he
> is a noble man :-)).
> If we ignore the silliness of his 'Superstrength' and just look at
the
> physics of the deep gravity well which was originally postulated to
> produce this superstrength:
>
> 1) If Superman has the strength of, say, 1,000 men, would this not
> denote a mass for the planet Krypton 1,000 times that of the Earth?
> 2) If this is so, would not the escape velocity of Krypton be about
> 11,000 km/s?
> (Actually the simplistic 1,000 times the mass of Earth cannot work
-
> ordinary matter cannot attain the density needed for a suitable
surface
> gravity)
Not necessarily. The surface
gravity, g, of a (spherical) object of mass M and radius R is given by
GM divided by R squared (G is the gravitational constant). If the mass
of Krypton was 1000 times that of the Earth then the surface gravity would be as well,
*provided* that it had the same radius as Earth. As you say,
however, that would mean that the density of Krypton was also a
thousand times that of Earth.
Any satisfactory answer now
requires a knowledge of the relationship between pressure, density and
temperature for matter under these extreme conditions. Such a
relationship is called the "equation of state (EOS)". I don't know much
about this subject I'm afraid. There are a number of theoretically
derived EOS's based on quantum effects, such as electron degeneracy,
but, to my limited knowledge, these haven't been tested experimentally
so I don't know how much faith can be put in them.
My feeling is that ordinary matter
increases its density only slowly as the pressure is increased, and
then undergoes catastrophic collapse; so densities of around 1000 times
Earth's may not be possible at ordinary temperatures.
The (Newtonian) escape velocity
for any (spherical) object, of mass M and radius R, is sqrt (2GM / R);
escape velocity doesn't scale linearly with mass. Interestingly enough,
putting this expression equal to c, the speed of light, gives the Schwartzchild radius
for a simple black hole of mass M as well; i.e. it's the same
expression in general relativity. If the hole is spinning and / or
charged, Newtonian physics can't handle the problem.
Anyway, back to Krypton; a cold,
rapidly spinning neutron star would seem to be the best bet. Krypton's
inhabitants would have to live near the equator, however, and cope with
some interesting electrical / magnetic effects. The magnetic fields of
neutron stars are trillions of times stronger than Earth's and spinning
magnetic fields of that magnitude would give enormous voltages all over
the place.
