impulsive launch vs rocket equation

Sorry if I am belabouring the obvious but rocketry is not my field
(x-ray optics is) but was curious about solar thermal rockets so
looked in my old physics book at the derivation of the rocket
equation.

Using M=Moexp(-v/ve)
where M is payload mass, v is rocket final velocity and ve is exhaust
velocity of fuel, plugging in some numbers of about 2 km/sec for ve
and 8 km/sec for orbital velocity we get a mass ratio of Mo/M of 55.
That is, the payload is only 1/55 of total initial mass.

Do the same for the same velocity ratio where the fuel is all burned
at one time and we get a mass ratio of 1/5, a huge increase.

Mp/Mo=r Mf/Mo=1-r

MfVf=MpVp

gives (1-r)/r=4 gives r=1/5.

This seems to imply that whenever the exhaust velocity is limited in
some fashion, for a given amount of fuel, it makes more sense to
expend as much fuel as fast as you can than to expend it slowly.

This would seem to imply for ion engines where the fuel speed is
limited by the electric field that can be applied, that capacitors be
charged to ionize a lot of fuel at one time rather than run the
ionization continously. The same fuel expenditure results in far more
velocity for the pulsed case than the continously emitted case.

It also seems to imply that for something like a solar thermal engine
that it might be better to charge capacitors and discharge them to
heat a burst of propellant than to heat it continously.

Does this make sense?

Nothing more confusing than an old former physicist trying to
re-understand basic stuff like this.

(Parallax) writes:

Sorry if I am belabouring the obvious but rocketry is not my field
(x-ray optics is) but was curious about solar thermal rockets so
looked in my old physics book at the derivation of the rocket
equation.

Using M=Moexp(-v/ve)
where M is payload mass, v is rocket final velocity and ve is exhaust
velocity of fuel, plugging in some numbers of about 2 km/sec for ve
and 8 km/sec for orbital velocity we get a mass ratio of Mo/M of 55.
That is, the payload is only 1/55 of total initial mass.

Do the same for the same velocity ratio where the fuel is all burned
at one time and we get a mass ratio of 1/5, a huge increase.

Mp/Mo=r Mf/Mo=1-r

MfVf=MpVp

gives (1-r)/r=4 gives r=1/5.

This seems to imply that whenever the exhaust velocity is limited in
some fashion, for a given amount of fuel, it makes more sense to
expend as much fuel as fast as you can than to expend it slowly.

That is correct.

It also makes more sense to expend it as deep in a gravity-well as possible
(Google on the "Oberth Effect").


This would seem to imply for ion engines where the fuel speed is
limited by the electric field that can be applied, that capacitors be
charged to ionize a lot of fuel at one time rather than run the
ionization continously. The same fuel expenditure results in far more
velocity for the pulsed case than the continously emitted case.

First of all, no capacitor or battery ever built or even theoretically
possible can hold a large enough energy density to power an astronautically
significant orbit change.

Second, it's not practical to produce large enough fields in an ion drive;
they are _inherently_ "low thrust" devices. If you want a "large" thrust
out of an electric propulsion system, you will have to go to some sort
of plasma drive (albeit their thrusts _still_ suck) or an electrothermal
drive (in which case their specific impulses suck unless you go to an arcjet).


It also seems to imply that for something like a solar thermal engine
that it might be better to charge capacitors and discharge them to
heat a burst of propellant than to heat it continously.

Again, it is not possible to store enough energy in a capacitor bank of
reasobale size and mass to make a significant orbit change.


-- Gordon D. Pusch

perl -e '$_ = \n"; s/NO\.//; s/SPAM\.//; print;'

Parallax wrote:
Sorry if I am belabouring the obvious but rocketry is not my field
(x-ray optics is) but was curious about solar thermal rockets so
looked in my old physics book at the derivation of the rocket
equation.

Using M=Moexp(-v/ve)
where M is payload mass, v is rocket final velocity and ve is exhaust
velocity of fuel, plugging in some numbers of about 2 km/sec for ve
and 8 km/sec for orbital velocity we get a mass ratio of Mo/M of 55.
That is, the payload is only 1/55 of total initial mass.

all good

Do the same for the same velocity ratio where the fuel is all burned
at one time and we get a mass ratio of 1/5, a huge increase.


Mp/Mo=r Mf/Mo=1-r

MfVf=MpVp

gives (1-r)/r=4 gives r=1/5.

This seems to imply that whenever the exhaust velocity is limited in
some fashion, for a given amount of fuel, it makes more sense to
expend as much fuel as fast as you can than to expend it slowly.

This is incorrect using simplistic open space assumptions
(zero gravity, or burning perpendicular to constant gravity),
where the rocket equation strictly applies. In this very simple
situation, the burn rate doesn't matter.

The rocket equation is independent of the amount of thrust (how fast
(mass flow rate) you burn), as long as there is some limit (as you
correctly point out). In practice, for a "rocket", there is always
a limit.

Your "v" is usually described as "delta V", the change in velocity.


So, the mass flow rate doesn't matter in open space, where the rocket
equation strictly applies. In an inverse square gravity field, however,
mass flow rate does matter because if you are expending energy raising
the propellant to a higher potential energy state before burning it,
you aren't getting all the energy out of it you otherwise could.


This would seem to imply for ion engines where the fuel speed is
limited by the electric field that can be applied, that capacitors be
charged to ionize a lot of fuel at one time rather than run the
ionization continously. The same fuel expenditure results in far more
velocity for the pulsed case than the continously emitted case.

It also seems to imply that for something like a solar thermal engine
that it might be better to charge capacitors and discharge them to
heat a burst of propellant than to heat it continously.

Does this make sense?


Yes, this is correct in practice. But _only_ because of the
second order orbital mechanics effects described above.
In open space, the delta-V at high mass flow rate
and at low rate would be the same. Value as calculated
by the rocket equation.

Integrate to derive the rocket equation yourself for constant mass flow
rate and exhaust velocity and you will see how it works. (The mass flow
rate drops out)


Nothing more confusing than an old former physicist trying to
re-understand basic stuff like this.

As I google'd for " "rocket eqution" and derivation "
I found people confused because their teachers had asked them a question
intending to get the rocket equation as an answer, but the problem
statements included effects that the rocket equation doesn't consider
(e.g.: gravity) This kind of thing is counter productive.
(Kind of like the Galileo rock dropping experiment that is only
correct in a vacuum.)


So my summary is: your conclusion is correct in practice, but the
reasoning is faulty because it used a solution to a simpler problem
where the effect you seem to be interested in isn't considered.


Your wider question of the optimization of the mass flow rate
I'll leave to others. I'll just point out that adding
energy storage means adding mass that you have to accelerate.

Matthew Jessick wrote in message ...
Parallax wrote:
Sorry if I am belabouring the obvious but rocketry is not my field
(x-ray optics is) but was curious about solar thermal rockets so
looked in my old physics book at the derivation of the rocket
equation.

Using M=Moexp(-v/ve)
where M is payload mass, v is rocket final velocity and ve is exhaust
velocity of fuel, plugging in some numbers of about 2 km/sec for ve
and 8 km/sec for orbital velocity we get a mass ratio of Mo/M of 55.
That is, the payload is only 1/55 of total initial mass.

all good

Do the same for the same velocity ratio where the fuel is all burned
at one time and we get a mass ratio of 1/5, a huge increase.


Mp/Mo=r Mf/Mo=1-r

MfVf=MpVp

gives (1-r)/r=4 gives r=1/5.

This seems to imply that whenever the exhaust velocity is limited in
some fashion, for a given amount of fuel, it makes more sense to
expend as much fuel as fast as you can than to expend it slowly.

This is incorrect using simplistic open space assumptions
(zero gravity, or burning perpendicular to constant gravity),
where the rocket equation strictly applies. In this very simple
situation, the burn rate doesn't matter.

The rocket equation is independent of the amount of thrust (how fast
(mass flow rate) you burn), as long as there is some limit (as you
correctly point out). In practice, for a "rocket", there is always
a limit.

Your "v" is usually described as "delta V", the change in velocity.


So, the mass flow rate doesn't matter in open space, where the rocket
equation strictly applies. In an inverse square gravity field, however,
mass flow rate does matter because if you are expending energy raising
the propellant to a higher potential energy state before burning it,
you aren't getting all the energy out of it you otherwise could.


This would seem to imply for ion engines where the fuel speed is
limited by the electric field that can be applied, that capacitors be
charged to ionize a lot of fuel at one time rather than run the
ionization continously. The same fuel expenditure results in far more
velocity for the pulsed case than the continously emitted case.

It also seems to imply that for something like a solar thermal engine
that it might be better to charge capacitors and discharge them to
heat a burst of propellant than to heat it continously.

Does this make sense?


Yes, this is correct in practice. But _only_ because of the
second order orbital mechanics effects described above.
In open space, the delta-V at high mass flow rate
and at low rate would be the same. Value as calculated
by the rocket equation.

Integrate to derive the rocket equation yourself for constant mass flow
rate and exhaust velocity and you will see how it works. (The mass flow
rate drops out)


Nothing more confusing than an old former physicist trying to
re-understand basic stuff like this.

As I google'd for " "rocket eqution" and derivation "
I found people confused because their teachers had asked them a question
intending to get the rocket equation as an answer, but the problem
statements included effects that the rocket equation doesn't consider
(e.g.: gravity) This kind of thing is counter productive.
(Kind of like the Galileo rock dropping experiment that is only
correct in a vacuum.)


So my summary is: your conclusion is correct in practice, but the
reasoning is faulty because it used a solution to a simpler problem
where the effect you seem to be interested in isn't considered.


Your wider question of the optimization of the mass flow rate
I'll leave to others. I'll just point out that adding
energy storage means adding mass that you have to accelerate.

Thanks everyone.


Actually the derivation in my 20 yr old physics text neglected
gravity. However, the point about raising the fuel to a higher
potential seems like good reasoning. Is there some way to optimize
the use of fuel by means of repeated "detonations" to get more v out
than you would get by expending fuel constantly? I assume that the
limiting factor is the acceleration the payload can handle.

For Gordon, dont be so literal. I was using capacitor in a generic
way to mean some hypothetic storage device that can release it
suddenly.

Rather than "burning" the H2 and O2 as in conventional LH2/LO2
engines, might it be better to allow it to accumulate and then
"detonate" it. The flow rate between detonations might be lower than
with turbopumps although the integrated flow would be the same.

For the rocket equation, your model is that the rocket is throwing
mass out the back at some finite rate, and you get the right answer:

Mp=Mo * exp(-v/ve)

Mp = rocket dry mass (payload plus engines, etc)
Mo = rocket fully fuelled mass
Mf = rocket fuel mass. Mf + Mp = Mo
v = change in velocity of rocket payload
ve = velocity of exhaust RELATIVE TO ROCKET

Your other case, all fuel burned at one time, assumes that the fuel
velocity relative to the mass centroid is Vf == ve above. You use
conservation of momentum to get your payload velocity.

You violated conservation of energy.

How much energy is in the rocket fuel? Imagine an infinitely heavy
rocket, that gets no delta-V from burning Mo-Mp fuel. After burning
that fuel, the kinetic energy of the rocket and fuel in the initial
frame of reference of the rocket is Ke == 0 + 0.5*Mf*ve.

The derivation looks tough, so I skipped it, but I'm sure if you
integrate for a rocket of finite mass, you'll find that the final
kinetic energy of the rocket plus the exhaust is also = Ke above. In
the initial frame of reference, the initial little bit of exhaust is
going ve in the opposite direction. Quite a bit of the exhaust is
going slower than that, in the opposite direction. Some is actually
stopped. A bit is going in the same direction as the rocket, and some
is even going faster than ve! But the root-mean-squared velocity is
less than ve.

But consider your instantaneous case. The kinetic energy of the
two is

0.5*Mf*ve^2 + 0.5*Mp*v^2

Whoa! Where'd that extra energy come from?

A: It came from the nonphysical instantaneous burn.

Parallax wrote:
Actually the derivation in my 20 yr old physics text neglected
gravity. However, the point about raising the fuel to a higher
potential seems like good reasoning. Is there some way to optimize
the use of fuel by means of repeated "detonations" to get more v out
than you would get by expending fuel constantly? I assume that the
limiting factor is the acceleration the payload can handle.

If you burn in short bursts each time you swing around at periapsis
then you will get the most energy out of a fixed delta velocity.
(because you are burning at a higher average speed, and the
derivative of energy gain with delta velocity is proportional
to velocity.) Of course it takes a lot longer.

Coasts between the solid rocket motor burns on the IUS
sending Magellan, Galileo and Ulysses away from the Earth
were made as short as possible because of this effect.
(The longer the coast, the slower you are when the second
burn ignites as your radius increases as you coast upward
after the first burn.)

(Parallax) wrote in message . com...
Sorry if I am belabouring the obvious but rocketry is not my field
(x-ray optics is) but was curious about solar thermal rockets so
looked in my old physics book at the derivation of the rocket
equation.

Using M=Moexp(-v/ve)
where M is payload mass, v is rocket final velocity and ve is exhaust
velocity of fuel, plugging in some numbers of about 2 km/sec for ve
and 8 km/sec for orbital velocity we get a mass ratio of Mo/M of 55.
That is, the payload is only 1/55 of total initial mass.

Do the same for the same velocity ratio where the fuel is all burned
at one time and we get a mass ratio of 1/5, a huge increase.


This is a classic mistake; I've done it myself.

When I teach undergraduate mechanics, I give my students the problem
of propelling the cart you are standing on by throwing bricks off the
back at speed "ve" relative to the cart. First, do the calculation
throwing one brick at a time. Then, repeat the calculation throwing
the *entire* load of bricks at the same velocity "ve".

You will find the same result you did above. However, think about
what is involved in throwing the entire load of bricks at once at
velocity "ve". Now, thick about rocket propellant and the statement
"the fuel is all burned at one time." Get it?
--
Andrew J. Higgins Mechanical Engineering Dept.
Assistant Professor McGill University
Shock Wave Physics Group Montreal, Quebec CANADA
http://www.mcgill.ca/mecheng/staff/academic/higgins/

(Iain McClatchie) wrote in message . com...
For the rocket equation, your model is that the rocket is throwing
mass out the back at some finite rate, and you get the right answer:

Mp=Mo * exp(-v/ve)

Mp = rocket dry mass (payload plus engines, etc)
Mo = rocket fully fuelled mass
Mf = rocket fuel mass. Mf + Mp = Mo
v = change in velocity of rocket payload
ve = velocity of exhaust RELATIVE TO ROCKET

Your other case, all fuel burned at one time, assumes that the fuel
velocity relative to the mass centroid is Vf == ve above. You use
conservation of momentum to get your payload velocity.

You violated conservation of energy.

How much energy is in the rocket fuel? Imagine an infinitely heavy
rocket, that gets no delta-V from burning Mo-Mp fuel. After burning
that fuel, the kinetic energy of the rocket and fuel in the initial
frame of reference of the rocket is Ke == 0 + 0.5*Mf*ve.

The derivation looks tough, so I skipped it, but I'm sure if you
integrate for a rocket of finite mass, you'll find that the final
kinetic energy of the rocket plus the exhaust is also = Ke above. In
the initial frame of reference, the initial little bit of exhaust is
going ve in the opposite direction. Quite a bit of the exhaust is
going slower than that, in the opposite direction. Some is actually
stopped. A bit is going in the same direction as the rocket, and some
is even going faster than ve! But the root-mean-squared velocity is
less than ve.

But consider your instantaneous case. The kinetic energy of the
two is

0.5*Mf*ve^2 + 0.5*Mp*v^2

Whoa! Where'd that extra energy come from?

A: It came from the nonphysical instantaneous burn.

Thanks:

I'll go back to x-rays now.

DBO