Cyde Weys wrote:

dkomo wrote in message ...

Dale wrote:


"dkomo" wrote in message
...


Dale wrote:



"dkomo" wrote in message
...



Thomas H. Faller wrote:

[...]



So trying to escape the earth's gravity at less than escape velocity

requires



very large power sources.


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.


No, if you go up slowly, you burn fuel at pretty much the same rate as

if


you go up rapidly, but it takes more time, therefore you burn more fuel.

What? If you burn fuel at a slow rate, you generate less thrust,
produce less acceleration, and therefore you go up more slowly. If you
burn fuel at a fast rate, you generate more thrust, produce more
acceleration, and therefore you go up faster.


As long as you're fighting gravity, you have to burn fuel at a certain rate
just to hover. It's pretty complicated to explain fully, but just take a
look at the limits. If you have just enough thrust to hover, you could be
burning fuel at some given rate forever and never get anywhere. If you don't
have enough thrust even to hover, you'll come to a rest on the Earth's
surface.

If you burned a little extra, say enough to accelerate at 1 m/s^2 you could
achieve orbital velocity in 123 seconds. Well, actually, that's against a
constant gravity, but let's just leave it at that for now. It's also
assuming a constant mass, when of course, the mass would decrease as fuel
was expended, and thus the acceleration would increase over time.

But anyway, if you burned enough to accelerate at 10 m/s^2, you'd achieve
orbital velocity in 39 seconds. Now let's say that accelerating at 1 m/s^2
against Earth's gravity costs 1.1 times as much fuel as hovering, and that
accelerating at 10 m/s^2 takes 2 times as much fuel as hovering.

Where do you get this stuff? Accelerating at 10 m/s^2 will take 10
times the thrust as accelerating at 1 m/s^2. Therefore you'll burn fuel
at ten times the rate.

Look at it in terms of energy required. From high school physics, the
work done by the fuel on the rocket is force times distance. Let's look
at a small distance the rocket covers and assume the mass of the rocket
stays approximately constant in that distance. Then since force equals
mass times acceleration, the work done by the fuel on the rocket when it
is accelerating at 10 m/s^2 is ten times that when it is accelerating at
1 m/s^2 across that distance. Where does this work(energy) come from?
From the burning of the fuel. The fuel consumed will be ten times as
great.


Yeah but there's a big difference. When your engine is accelerating
at 1 m/s^2 you're falling towards the Earth at the rate of 8.8 m/s^2.
That's not enough thrust to even counteract gravity. And when your
engine is accelerating at 10 m/s^2, you're pulling away from the Earth
at the rate of 2 m/2^2. So although you're right, the difference
between fuel expenditures in the two situations is ten times
different, in one situation you're burning fuel sitting on the
launchpad, and in the other situation, you're actually getting
somewhere.

Lets take another example: two rockets, one with the acceleration of
10 m/s^2 and another with the acceleration of 20 m/s^2. The second
one burns twice the amount of fuel. But its acceleration away from
the Earth is 10.2 m/s^2, which is much higher than the first rocket's
acceleration of the paltry 0.2 m/s^2. So you see, although you're
using twice the fuel, your acceleration rate is over fifty times
higher - and thus you won't have your engine on for nearly as long.


Except that this argument doesn't jive with the argument from energy
considerations, which everybody so far just ignores.

Let's take a very similar example using elevators. Imagine you raise an
elevator to the top of a 100 story building slowly. Then do the same
thing very rapidly. Using the exact reasoning you give above you would
claim that in the second case you use less energy than in the first case.

However, let's look at the two cases in terms of the energy the elevator
has when it reaches the top of the building. Its potential energy in
both cases is the same -- mgh where h = 100 stories. I've assigned a
potential energy of 0 at h = 0 for convenience.

But there is big difference in the kinetic energy -- 1/2 m v^2. The
fast moving elevator has a far greater kinetic energy than the slow
moving elevator because of the v^2 dependence.

Now, the sum of the potential and kinetic energies the elevator has
gained could only have been gotten from the motor, which means the motor
had to deliver substantially more energy in the second case of the
rapidly moving elevator. This contradicts the conclusion you'd reach
using your arguments.

dkomo, at some point you have to accept that, if you don't get it, but
everyone still says that you're wrong, including the physicists over
at sci.physics ... then you are wrong, the reason why is just beyond
you.


What the f---? The only person who responded from sci.physics that I
could tell was Old Man. Everyone else in this thread is from
talk.origins. And Old Man didn't by any means say that I was wrong.

In fact I did make a mistake in the energy argument I made originally in
that I neglected the kinetic energies of the fast and slow moving
rockets when they reach their final height above the earth. I will
correct this when I make a reply to Old Man later today.

I'll take into account the changing mass of the rocket and the change in
acceleration of gravity and conclude that the question of which rocket
uses less fuel is *indeterminate* without further details about the fuel.

And that will be my final statement. There's no point in trying to
argue with everybody when they just totally ignore the energy argument
anyway.