Rocket acceleration question

Hello,
I have a small problem which I initially thought would be fairly easy to
work out.
Given a rocket or mass M ejecting as its exhaust a mass stream at a rate m,
and energy is imparted into this mass at a rate (power) of P, is it possible
to calculate the acceleration of the rocket? Or is it necessary to to have
the ejection velocity of the mass?
Is it also possible to derive the specific impulse of the fuel from this
information?

Thank you ~

In article ,
"Makhno" wrote:

Given a rocket or mass M ejecting as its exhaust a mass stream at a rate m,
and energy is imparted into this mass at a rate (power) of P, is it possible
to calculate the acceleration of the rocket? Or is it necessary to to have
the ejection velocity of the mass?

You need the ejection velocity of the mass, as well as how much mass is
ejected and the (initial or final) mass of the rocket. Here's a page
that puts these together to give you the acceleration (delta-V):

http://www.strout.net/info/science/delta-v/

Is it also possible to derive the specific impulse of the fuel from this
information?

Well, specific impulse is pretty much just another way of presenting the
exhaust velocity.

So, the question is, can you calculate the velocity of the mass from the
power used to accelerate it?

,------------------------------------------------------------------.
| Joseph J. Strout Check out the Mac Web Directory: |
| http://www.macwebdir.com |
`------------------------------------------------------------------'

Makhno wrote:
Hello,
I have a small problem which I initially thought would be fairly easy to
work out.
Given a rocket or mass M ejecting as its exhaust a mass stream at a rate m,
and energy is imparted into this mass at a rate (power) of P, is it possible
to calculate the acceleration of the rocket? Or is it necessary to to have
the ejection velocity of the mass?
Is it also possible to derive the specific impulse of the fuel from this
information?


You know the ejection velocity.
Kinetic energy. Energy=power*time.

Makhno wrote:
Given a rocket or mass M ejecting as its exhaust a mass stream at a rate m,
and energy is imparted into this mass at a rate (power) of P, is it possible
to calculate the acceleration of the rocket?

Thrust, F, is given by

[1] F = m * c ,

where, m is, in your notation, the rate of mass flow and c is the speed
of the flow. Each unit of mass in the exhaust flow has a kinetic
energy of 0.5*c^2, so the power, P, in the flow is

[2] P = 0.5 * m * c^2 .

This can be solved for c:

[3] c = (2 * P / m)^0.5 .

Now we can plug this result into equation [1] to get the thrust:

[4] F = m * (2 * P / m)^0.5 = (2 * P * m)^0.5 .

To get the acceleration, a, of the rocket, whose mass is M, all we need
to do is divide the thrust by the mass:

[5] a = F / M = (2 * P * m)^0.5 / M .

The specific impulse, Isp, is simply the exhaust velocity divided by
the acceleration of gravity, g = 9.81 m/s^2, at the earth's surface:

[6] Isp = c / g = (2 * P / m)^0.5 / g .

Hello,
I have a small problem which I initially thought would be fairly easy to
work out.
Given a rocket or mass M ejecting as its exhaust a mass stream at a rate m,
and energy is imparted into this mass at a rate (power) of P, is it possible
to calculate the acceleration of the rocket? Or is it necessary to to have
the ejection velocity of the mass?
Is it also possible to derive the specific impulse of the fuel from this
information?

Thank you ~

P=(1/2).m.((g0.Isp)^2) g0= 9.81 m/s2
M.acceleration = m.g0.Isp (in vacuum, no gravity)

Given a rocket or mass M ejecting as its exhaust a mass stream at a rate
m, and energy is imparted into this mass at a rate (power) of P, is it
possible to calculate the acceleration of the rocket?

[2] P = 0.5 * m * c^2 .

This is the equation I have problems with. It looks like you have taken the
definition of Kinetic Energy and differenciated both sides with respect to
time.
Is this a valid step here? The increase in kinetic energy when an
infinitesimal mass dm is accelerated out of the engine is clearly
dE=0.5*dm*c^2. This takes place over a time period T, which is an unknown
constant.
What [2] is saying, is that we replace T with dt, and then simply take
dE/dt = 0.5*(dm/dt)*c^2
(where my previous m=dm/dt)
This substitution of the unknown constant T with dt is what I'm having
problems with.

In article ,
Makhno wrote:
Is this a valid step here? The increase in kinetic energy when an
infinitesimal mass dm is accelerated out of the engine is clearly
dE=0.5*dm*c^2. This takes place over a time period T, which is an unknown
constant.

No, it takes place over an infinitesimal time dt. So just divide both
sides by dt, and you get dE/dt = 0.5*(dm/dt)*c^2 as suggested.

Any mathematicians in the audience are probably having apoplexy at the
"just divide by dt" part, and will tell you at great length that something
like dm/dt is not really a quotient and you can't just casually treat dt
as if it were an ordinary variable.

To which I say: if it was good enough for Leibniz, it's good enough for
me. :-) The dirty little secret of calculus is that just casually treating
dt like an ordinary variable gets you the right answer, every time... and
since the development of "nonstandard analysis", we know why. You can
build a mathematically-rigorous non-standard number system in which dm/dt
*is* a quotient and dt *is* an ordinary variable, and it's provably
equivalent to the vastly more complicated epsilon-delta circumlocutions
that later mathematicians invented to explain why Leibniz's calculus
perversely insisted on working so well. The engineers' deplorable habit
of just dividing dm by dt and getting dm/dt turns out to be rigorously
justifiable after all.
--
spsystems.net is temporarily off the air; | Henry Spencer
mail to henry at zoo.utoronto.ca instead. |

In article ,
Henry Spencer wrote:
In article ,
Makhno wrote:
Is this a valid step here? The increase in kinetic energy when an
infinitesimal mass dm is accelerated out of the engine is clearly
dE=0.5*dm*c^2. This takes place over a time period T, which is an unknown
constant.

No, it takes place over an infinitesimal time dt. So just divide both
sides by dt, and you get dE/dt = 0.5*(dm/dt)*c^2 as suggested.

Any mathematicians in the audience are probably having apoplexy at the
"just divide by dt" part, and will tell you at great length that something
like dm/dt is not really a quotient and you can't just casually treat dt
as if it were an ordinary variable.

To which I say: if it was good enough for Leibniz, it's good enough for
me. :-) The dirty little secret of calculus is that just casually treating
dt like an ordinary variable gets you the right answer, every time... and
since the development of "nonstandard analysis", we know why. You can
build a mathematically-rigorous non-standard number system in which dm/dt
*is* a quotient and dt *is* an ordinary variable, and it's provably
equivalent to the vastly more complicated epsilon-delta circumlocutions
that later mathematicians invented to explain why Leibniz's calculus
perversely insisted on working so well. The engineers' deplorable habit
of just dividing dm by dt and getting dm/dt turns out to be rigorously
justifiable after all.

In the writings of Leibniz, it may seem to have been treated
that way, and if you know what you are doing and when to do
it, it will give you the right answer in those cases. Even
in non-standard analysis, it does not always work. If one
has a function of more than one variable, one can still work
with differentials, but not by dividing.

One can make "epsilon-delta" more understandable, and it is
needed to show why non-standard analysis works when it does,
and why. There are other ways of working correctly with
differentials, and I do this often.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
Phone: (765)494-6054 FAX: (765)494-0558

(Henry Spencer) writes:

No, it takes place over an infinitesimal time dt. So just divide both
sides by dt, and you get dE/dt = 0.5*(dm/dt)*c^2 as suggested.

Any mathematicians in the audience are probably having apoplexy at the
"just divide by dt" part, and will tell you at great length that something
like dm/dt is not really a quotient and you can't just casually treat dt
as if it were an ordinary variable.

Henry, I'm afraid I don't know any mathematicians who'd
protest the manipulation of dt like you've used in this context.
Granted I got the bulk of my training at RPI, where the mathematics
program offers the ``Applied Mathematics'' and the ``More Applied
Mathematics'' tracks, but poking through the references I have on
my desk shows that sort of move done without complaint, past a rare
note that proving this rigorously requires analysis not included.

The point of emphasizing to students that dE/dt is not a
fraction is that if students believe it *is* a fraction, they'll
quite happily cancel the d's out of both sides and get an answer
that's not even wrong. Even with the emphasis to not think about
this like a fraction that happens disappointingly often. Making
the point that it's not a fraction reduces the risk -- and it has
the benefit of getting students to think of the derivative as this
thing done to a function, not just a quotient, which is quite a
useful point of view.

Also the sharper (or more troublesome) students will point
out that if you let dt go to zero, then you get a zero over zero form
and how do you get any answer from that ... which requires appealing
to all sorts of explanation that won't fit in any non-major's courses.
Best to try avoiding the issue altogether.

--
Joseph Nebus
------------------------------------------------------------------------------