Re: Konstantin Tsiolokvsky was wrong?!


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In sci.physics, Spaceman
<Realspace@xxxxxxxxxxx>
wrote
So if e = 500 meters per second exhaust speed
The rockets will be crazy fast and will go much
faster than 500 meters per second.
(care to put that e into place and come up with rocket speed?)
:)

I get about 1150 m/s, given a 10:1 fuel/rocket ratio. Is
there a problem here?

So you say, that a 500 meter per second exhaust speed
would produce a 1150 meter per second rocket speed
in outerspace, as long as the fuel had 10 times more total
mass than the rocket?
I do think there is a problem there.
:)

I haven't been following all these follow-ups, but FWIW I get the
maximum (final) speed of the rocket to be e * Log(1 + m1 / m2), where
m1 is the initial mass of fuel, and m2 is the mass of the rest of the
rocket. So a 10:1 ratio of fuel to rest of rocket, with e = 500, would
give a final speed of 1199 m/s, while a 9:1 ratio (possibly what was
meant) would give a final speed of 1151 m/s.

So how long would it take for such to get to the
speed predicted?
Instantly I guess huh?
Since you have no timing factor?.

The time taken depends on the flow rate of the fuel (number of
kilograms shot out per second). However, unless I made a mistake, the
final maximum speed attained by the rocket is *independent* of the flow
rate. It just takes different times to reach it.

But as I pointed out,
The time you did your calculation was at time 0,
after half of the fuel burn you would not longer
get the same result.

No, the answer is not calculated "at" or "for" any particular time. The
answer encompasses the *whole* time that the motor burns, as well as
the fact that the fuel mass is constantly decreasing and the rocket
speed is constantly increasing over that time. It may sound like
alchemy, but that really is the way it works.

With such (instant end result) type of thinking
in the equation, it shows it lacks "reality"
and just gives a beginning with no time involved
calculation.
Excluding time in any equation will not be
a reality based equation.
:)


The equations take full account of the continuously varying mass of
fuel, as well as the continuously varying speed of the rocket. This is
what the dm's and dv's are doing, but you need to be familiar with
calculus and differential equations to fully appreciate how the maths
works.

It does not do such if it is using the 10:1 that is only
present at time 0 and lacks a time function itself.

As I said, the answer takes into account the fact that the ratio
changes continuously with time.

Is e changing over time?

No, e is presumed to be constant over time. This is an assumption that
is made in the statement of the problem. As such it is not a debatable
point - it just *is* constant because it is stated to be so.

Supposedly and I guess logically, No,
Is mass changing over time,
Yes. and logically yes.

Yes, the total mass of the rocket decreases over time as the fuel is
expelled.

So..
Where is the "timing" function for the mass change
in that equation?

I assume you are referring to the equation:

Max. rocket speed = e * Log(1 + m1 / m2)

There is no "timing function" for mass change here because it is
irrelevant to the answer. You don't need to know how quickly the fuel
is burnt, only that all the fuel eventually shoots out of the back of
the rocket at speed e. The fuel could burn out in one millisecond or
burn over one million years, and the final speed of the rocket would be
the same. According to my calculations anyway.

.

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