Re: Question on rocket physics & mechanics [redirected from talk.origins]

From: Don Cates (catHORMELes_at_ms.umanitoba.ca)
Date: 06/25/04 

Date: Fri, 25 Jun 2004 02:54:57 +0000 (UTC)

On Thu, 24 Jun 2004 22:45:26 +0000 (UTC), dkomo <dkomo871@comcast.net>
wrote:

>Don Cates wrote:
>
>> On Thu, 24 Jun 2004 14:16:32 +0000 (UTC), dkomo <dkomo871@comcast.net>
>> wrote:
>>
>>
>>>Don Cates wrote:
>>>
>>>
>>>>On Wed, 23 Jun 2004 23:55:25 +0000 (UTC), dkomo <dkomo871@comcast.net>
>>>>wrote:
>>>>
>>>>
>>>>
>>>>>Old Man wrote:
>>>>>
>>>>>
>>>>>
>>>>>>"Cyde Weys" <vze23tnc@verizon.net> wrote in message
>>>>>>news:az4Cc.22578$Yb1.14431@nwrddc02.gnilink.net...
>>>>>>
>>>>>>
>>>>>>
>>>>>>>Ken Shaw wrote:
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>>dkomo wrote:
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>>Ken Shaw wrote:
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>>dkomo wrote:
>>>>>>>>>
>>>>>>>>>[snip for brevity]
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>>>Good grief, haven't we heard about conservation of energy? It takes
>>>>>>
>>>>>>the
>>>>>>
>>>>>>
>>>>>>
>>>>>>>>>>>same amount of energy to escape the earth's gravity whether you do it
>>>>>>
>>>>>>at
>>>>>>
>>>>>>
>>>>>>
>>>>>>>>>>>1 m/s or 25,000 mph. That means you use the same amount of fuel
>>>>>>
>>>>>>either way.
>>>>>>
>>>>>>
>>>>>>
>>>>>>>>>>>Look at this way. If you go up slowly you burn fuel more slowly over
>>>>>>
>>>>>>a
>>>>>>
>>>>>>
>>>>>>
>>>>>>>>>>>longer time period. If you go up fast you burn fuel rapidly over a
>>>>>>>>>>>shorter time period. Either way, you end burning the same amount of
>>>>>>
>>>>>>fuel.
>>>>>>
>>>>>>
>>>>>>
>>>>>>>>>>You're neglecting the amount of energy expended counteracting Earth's
>>>>>>>>>>gravity.
>>>>>>>>>
>>>>>>>>>No, I haven't. That's the energy I posted about.
>>>>>>>>>
>>>>>>>>>If you attempt to reach orbit at 1m/s you must also accelerate
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>>at 10m/s at first decreasing with time to nearly zero simply to
>>>>>>
>>>>>>maintain
>>>>>>
>>>>>>
>>>>>>
>>>>>>>>>>your present altitude. If you travel at a much higher speed gravity
>>>>>>>>>>acts on you for a shorter period of time therefore using less fuel to
>>>>>>>>>>reach the same objective.
>>>>>>>>>
>>>>>>>>>The beauty of an energy argument is that you don't need to worry about
>>>>>>>>>the details of the force, in this case gravity. You simply take the
>>>>>>>>>difference between the potential energy of the rocket at the earth's
>>>>>>>>>surface and the potential energy when it reaches earth orbit, and that's
>>>>>>>>>the work the rocket does in counteracting Earth's gravity. And that
>>>>>>>>>amount of energy is independent of how the force varies, and what speed
>>>>>>>>>the rocket had. That's a basic law of physics.
>>>>>>>>>
>>>>>>>>>You could do this problem the hard way as well. You could put the
>>>>>>>>>formula for gravity (with its inverse square dependence) under the
>>>>>>>>>integral for computing work (force times distance) and integrate from r
>>>>>>>>>= earth's surface to r = orbit, and you'll end with exactly the same
>>>>>>>>>amount of energy. Speed of the rocket simply is irrelevant.
>>>>>>>>
>>>>>>>>You are simply wrong. The longer you take to do the work the more
>>>>>>>>acceleration due to gravity you have to overcome. Remember that Earth's
>>>>>>>>surface gravity can be expressed as 10m/s^2. This is the very reason
>>>>>>>>that rockets burn their fuel so fast. If you were correct the shuttle
>>>>>>>>could burn it's liquid fuel much more slowly and still reach orbit which
>>>>>>>>it can't.
>>>>>>>>
>>>>>>>>Ken
>>>>>>>
>>>>>>>Let's ask the sci.physics people. We're not gonna get this resolved here.
>>>>>>
>>>>>>
>>>>>>Consider a rocket of mass, m, and thrust, F, moving
>>>>>>vertically in a uniform gravitational field, g. Thrust is
>>>>>>applied for the first portion of the ascent. The engine
>>>>>>is then shut-off, and the rocket ascends ballistically to
>>>>>>a maximum altitude, h. The energy, E, expended by
>>>>>>the rocket engine is then given by
>>>>>>
>>>>>>E = mgh / sqrt[ 1 - mg / F ]
>>>>>>
>>>>>>Thus, for a given apogee, greater thrust yields less fuel
>>>>>>energy consumption. For low thrust, as F -> mg (hover),
>>>>>>total fuel consumption increases without limit. For large
>>>>>>thrust, the rocket burn time approaches zero, and total
>>>>>>energy consumption approaches mgh, which is just the
>>>>>>change in gravitational potential energy between the launch
>>>>>>pad and apogee.
>>>>>>
>>>>>>[Old Man]
>>>>>>
>>>>>>
>>>>>
>>>>>Ok, I'll make one last attempt to present an energy argument for the
>>>>>fast and slow rockets. I realize now that the previous argument I made
>>>>>using only the potential energies of the rockets was incorrect because
>>>>>it didn't include the kinetic energies of the rockets when they reached
>>>>>their final altitude.
>>>>>
>>>>>I am now concluding that the question of which rocket uses less fuel is
>>>>>indeterminate without more information. If there is another mistake in
>>>>>the following analysis, point it out.
>>>>>
>>>>>When the slow rocket reaches altitude, its energy will be
>>>>>
>>>>>e = t + u where t = 1/2 m v^2 is its kinetic energy
>>>>> and u = mgh is its potential energy
>>>>>
>>>>>here m is the mass of the rocket at altitude h. This m is less than the
>>>>>mass the rocket had on the launch pad because it burned up some fuel.
>>>>>For convenience I've set the potential energy of both rockets to 0 at h
>>>>>= 0 (on the launching pad). In the u = mgh equation, the g here is the
>>>>>acceleration of gravity at altitude h, which for h = 200 miles is about
>>>>>10% less than it is at sea level.
>>>>>
>>>>>I'll use small letters for the variables for the slow rocket, and
>>>>>capital letters for the variables for the fast rocket.
>>>>>
>>>>>When the fast rocket reaches altitude, its energy will be
>>>>>
>>>>>E = T + U where T = 1/2 M V^2 is its kinetic energy
>>>>> and U = Mgh is its potential energy
>>>>
>>>>
>>>>You are making this much too complex. Let's take a simpler scenario.
>>>>The fast rocket runs out of fuel and begins to slow down. At some
>>>>altitude h it has velocity v upwards. The slow rocket maintains a
>>>>constant velocity v and runs out of fuel at altitude h.
>>>
>>>This isn't just a simple scenario, it's a virtually impossible scenario.
>>>How do you guarantee that the slow rocket runs out of fuel at *exactly*
>>>the altitude h going exactly at velocity v?
>>
>>
>> By firing the fast rocket and measuring the altitude at which its
>> velocity is v. According to your argument, the slow rocket with the
>> same amount of fuel burned so as to give a constant velocity of v will
>> run out of fuel at that altitude. It's a thought experiment; so
>> *think*.
>>
>
>Are you beating a dead horse here? I'm no longer claiming that the fast
>and slow rockets burn the same amount of fuel when they reach the *same*
>altitude h. Look through the energy equations I posted above more
>carefully. All four energy terms, the kinetic and potential energies of
>the fast rocket, and the kinetic and potential energies of the slow
>rocket could be different.

Well, I did ask you to think. I'll try to plot it out for you again
and you try to think about it this time. Okay?

This is what *your* hypothesis claims will happen. Why? Because you
claim that all the energy from burning the fuel is transfered to the
rocket (ignoring friction, etc). Right?

Start with two identical rockets with an identical fuel load.

Rocket 1 burns its fuel rapidly and rises rapidly until all its fuel
is gone. It is at some altitude H and travelling at some velocity V.
It has an energy T+U. Now that its fuel is gone, its energy cannot
change, and according to you that energy equals the energy given to it
by the fuel. It continues to rise and slow down, trading kinetic
energy for potential energy. It eventually reaches some altitude h' at
some velocity v' such that its total energy t'+u'=T+U. This is true
for *any* altitude that it can reach.

Rocket 2 burns its fuel slowly, such that it maintains a constant
velocity v, until its fuel runs out. It now has the identical mass as
Rocket 1 when all its fuel is gone. It will be at some altitude h and
have some velocity v such that its energy (obtained from the burning
of the fuel and therefor the same as the energy of rocket 1) is
t+u=T+U=t'+u' (same amount of fuel burned).

Now let h'=h
Since the rockets now have the same mass and are at the same altitude
they must have the same potential energy (u'=u)
Since t'+u'=t+u and u'=u; t' must =t and therefor v'=v
Since they are identical rockets at the same altitude with the same
velocity they will continue to have equal altitudes and velocities at
all times after the slow rocket runs out of fuel.

This is *required* by your hypothesis. The *only* difference for these
two rockets is the length of time that the fuel is burning.

There is one catch to this. The slow rocket must be able to reach
altitude h no matter how slowly it burns its fuel. We know that it
must burn it fast enough to generate a force > mg. If it burns any
slower than this, the rocket will not take off at all before all its
fuel is burned and h=0. By your hypothesis, this is impossible but we
know that it can happen. Therefor something is wrong with your
hypothesis. You have been told why it is wrong. Now all you have to do
is accept it.

If you disagree, show me where my mistake is. Don't just claim that
I've made one.

>> Especially since both
>>
>>>rockets start out with the same amount of fuel. Or putting it
>>>differently, how do you ensure that both rockets are going at velocity v
>>>at altitude h.
>>>
>>>This unrealistic condition makes the rest of your analysis moot.
>>>
>>> At this point
>>>
>>>>they have the same mass, altitude, and velocity so they have the same
>>>>kinetic and potentail energy. Now the fast rocket has burnout at a
>>>>higher velocity and lower altitude but there is no change in total
>>>>energy after burnout.
>>>
>>> So the total energy of both ships is the same at
>>>
>>>>the burnout burnout time for each. So the amount of fuel burned to
>>>>give them this energy is the same for each. So the only difference is
>>>>the time of burn for each.
>>>>
>>>>
>>>>
>>>>>Now if the rockets had 0 total energy on the launch pad, their energy at
>>>>>altitude must have come from the burning of the fuel they had.
>>>>
>>>>
>>>>This is correct. The problem is that there is fuel that *must* be
>>>>burned whose energy is *not* transfered to the rocket.
>>>
>>>As of this morning I added chemical energy terms to the energy analysis,
>>>i.e. I considered the change in energy due to the mass of fuel consumed
>>>times the energy content of that fuel. See my replies to Old Man and
>>>mmeron.
>>
>>
>> Which has absolutely nothing to do with my argument. In the scenario
>> above, using your speculation that all the energy from the fuel goes
>> into the rocket as either kinetic or potential energy, two identical
>> rockets with an identical amount of fuel end up at an identical
>> altitude with an identical velocity with an identical amount of fuel
>> remaining (zero).
>
>This is simply not true. Look at my last energy analysis again. The
>fast and slow rockets will have different masses at h

That is because you picked an h where at least one of the still has
some fuel. I avoided that complication by chosing an h where they had
an identical amount of fuel (zero). That is why I said you were making
it more complicated than needed.

, and will have
>different velocities as well

But in my scenario (h is at or above the altitude where both rockets
have no fuel left) the velocities will be the same. At any altitude
below that, they will have different masses and velocities, making any
calculation more difficult.

, even if the rockets were identical on the
>launching pad. And I never made any assumption that either rocket would
>use up all its fuel to reach h.

Yes, that is what is adding unneeded complications. So don't do that,
instead chose an h where both rockets *have* used up all their fuel.
Make it simple.

> In fact, if your speculation is true, this is
>> inevitable. If they both start with the same amount of fuel, they
>> *must* be at the same altitude with the same velocity at any time
>> after the the last one has burned all its fuel.
>>
>>
>>> The energy
>>>
>>>>released to create thrust developed by the burning fuel less than the
>>>>local force due to gravity is not transfered to the rocket. I'm not
>>>>sure, but I suspect that it is ultimately transfered to the earth
>>>>through the gravitational coupling of the earth to the rocket.
>>>
>>>This sounds mystical. I think the energy of the fuel burned is entirely
>>>transferred into the potential and kinetic energy of the rocket at
>>>altitude h (and probably a lot of energy into overcoming air resistance,
>>>but I'm trying to neglect that for the time being).
>>
>>
>> What's "mystical" about it? The rocket and the earth are attracting
>> each other with a force mg. This is balanced by the repulsive forces
>> of the atoms at the point of contact of the rocket and the earth when
>> the rocket is at rest. When the fuel starts to burn, the contact force
>> decreases but the gravitational force is the same, so the energy from
>> the fuel is going into moving the earth/rocket system. Due to the huge
>> mass of this system, the acceleration is extremely tiny, but it is
>> there. You could calculate it by dividing the force generated by the
>> fuel by the mass of the system. Once the force from the fuel exceeds
>> mg the rocket will leave the surface but it will continue to attract
>> the earth with a force of mg. As long as the rocket is hovering or
>> moving away from the earth, this force must be supplied by the fuel.
>> The energy required can be calculated as the integral of force (mg)
>> over time. It is only the energy produced by the fuel greater than
>> this that goes into the kinetic and potential energy of the rocket.
>>
>
>This sounds just as mystical. Look, I checked several college level
>physics texts this morning. They work through the equations of lifting
>a weight of mass m to a height h against the force of gravity.

Did any of them discuss the total energy expended by the lifter? Or
did they just discuss the amount of work done on the mass lifted?
Yes, mgh is the amount of work done to the rocket, but the fuel must
do other work as well.

  It is in
>fact the integral of mg across the distance h. In a constant
>gravitational field, the work required is mgh, which is simply the
>gravitational potential energy. None of these books mention any other
>factors involved.

Then you should try and find a discussion that includes the total
amount of energy expended by the 'lifter', not just what is required
to be transfered to the 'lifted'.

  If this simple analysis doesn't apply to the rocket,
>you'd better come with a really good explanation as to why not. If I
>were you I'd check a physics book or two before you continue with the
>claims you are making in the paragraph above.

And you might want to find a more thorough discussion of the subject
in one of those books. Not just the introductory "what is work?" part.

>Speaking of hovering, now I have a question for you. A helicopter of
>weight 5 tons hovers 100 feet above the ground for one hour. How much
>energy is required to do this? A simple calculation is what I'm asking for.

Okay. A helicopter hovering in a 1g gravitational field for 1 hour is
equivalent to a helicopter being accelerated at 1g for 1 hour. The
equivalent distance travelled is 1/2 gt^2 so the energy is 1/2
m(gt)^2. Just plug in m, g, and t.
1/2 * 4545kg * 96.04m^2/s^4 * 12960000s^2
2828531664000 kgm^2/s^2

>Hint: it is not, as you state above "The energy required can be
>calculated as the integral of force (mg) over time." This integral
>doesn't have units of energy. In fact, this integral is defined as
>*impulse*, which is equal to the change in momentum of the mass the
>force is acting on. If you don't believe me, check a physics book or two.

Yep, you are correct. I made an error there.

>Anyway, for a hovering helicopter, which doesn't move, there is no
>change in momentum. Likewise, the usual equation of integrating force
>over distance can't be used since the distance moved is zero. So how do
>you calculate the energy required to keep that chopper up there?

See above.
--
Don Cates ("he's a cunning rascal" - PN)

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