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

From: dkomo (dkomo871_at_comcast.net)
Date: 06/23/04 

Date: Wed, 23 Jun 2004 23:55:25 +0000 (UTC)

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

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. I'm
neglecting air resistance here and making no attempt to put these
rockets into orbit. They are moving straight up away from the earth.

I claim that by comparing the total energies of the two rockets at
altitude, one should be able to determine which one burned more fuel.
I've worked the change in masses of the rockets into the analysis, as
well as the change in the acceleration of gravity from sea level.

However, the comparison is not possible unless one knows what their
final masses, m and M are, or at least the ratio of these masses. And
this is equivalent to knowing how much fuel they burned, which is what
we're trying to determine.

The fast rocket *probably* has a greater kinetic energy, T = 1/2 M V^2
because of the V^2 term, than the slow rocket (at t = 1/2 m v^2), but
what is its mass M relative to m? If M is small, perhaps the kinetic
energy is actually less than the slow rocket's.

We face the same problem with the potential energies mgh and Mgh. Which
one is greater? It depends on the masses.

What happens if the slow rocket moves at 90% of the final velocity of
the fast rocket? Is it possible to still say unequivocally that the
fast rocket will burn less fuel than the slow one, as all the hand
waving arguments in this thread are saying, using vague criterion such
as "fighting the acceleration of gravity for a shorter period of time"
or "length of time that the fuel was burning"?

        --dkomo@cris.com

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