A dumb question about escape velocity
Anthony Giorgianni
I'm hoping this is the right place to ask this question that I've been
wondering about for a long time (please let me know if there is a better
place to ask).
Why is the escape velocity from earth 25,000 miles an hour? At sea level,
one can throw a rock vertically or fire a model rocket vertically at far
less speed than that. And as one increase their distance from the center of
an object (like the earth), gravity decreases, right? So one would think
that the speed to leave the plant's gravitation pull would decrease as one
climbs, not increase. So where does this 25,000 miles an hour suddenly kick
in?
Similarly, (and I guess this is related) why does it take so much of a
ruckus to get the space shuttle into space. Why can't they just put an
engine on the thing, fly it to the to the edge of the atmosphere (taking
advantage of lift instead of rocket power), then just rocket it the rest of
the way?
I know these are probably really dumb questions. But if anyone can explain
it, I'd appreciate it.
Thanks.
Regards,
Tony Giorgianni
mkhan
Anthony Giorgianni wrote:
> Why is the escape velocity from earth 25,000 miles an hour? At sea level,
> one can throw a rock vertically or fire a model rocket vertically at far
> less speed than that.
Yes. And then it falls back, because the speed was not sufficient.
> And as one increase their distance from the center of
> an object (like the earth), gravity decreases, right? So one would think
> that the speed to leave the plant's gravitation pull would decrease as one
> climbs, not increase. So where does this 25,000 miles an hour suddenly kick
> in?
I never saw this figure expressed in mph and can't be bothered to do the
conversion from km/s, so here goes with metric units:
The escape velocity is a function of altitude. It defines the pericentre
velocity on a trajectory that will allow escape from the Earth. Such a
trajectory (if you allow some simplifications) is called a parabola.
If the pericentre velocity - the velocity at the closest point of the
trajectory - is lower than the escape velocity, the object will be in an
elliptical orbit. It will then not escape from the gravity field of the
central body, but remain in a bound orbit around it.
The semi-major axis of a parabola has the value infinity. The enery on a
parabola is 0.
The equation for velocity on a Keplerian trajectory is:
v = SQRT (mu * (2/r - 1/a) )
mu, for the Earth, is 398600.44 km**3/s**2
Assume that the trajectory starts on the Earth surface. Then, the radius
r at the pericentre is equal to the Earth radius, r = 6378 km.
As a is infinity, 1/a is nil. The equation simplifies to:
v = SQRT (2* mu / r) = 11.18 km/s (and yes, this correspons to slightly
over 25,000 mph)
This means that if you want to launch something from the Earth surface
such that it can leave the Earth gravity field, you have to impart to it
a velocity of 11.18 km/s, minus whatever velocity you alreadyt have at
the launch pad due to Earth rotation.
If you are already in orbit, say at 200 km over the Earth surface, the
escape velocity is slightly less, namely, 11.01 km/s. But of course, a
body on a circular orbit will already have a velocity of 7.78 km/s -
otherwise it would not be in that orbit in the fist place. So you only
have to give it the differential velocity, dv=11.01 km/s - 7.78 km/s =
3.23 km/s.
> Similarly, (and I guess this is related)
It don't really see the relation.
> why does it take so much of a
> ruckus to get the space shuttle into space. Why can't they just put an
> engine on the thing, fly it to the to the edge of the atmosphere (taking
> advantage of lift instead of rocket power), then just rocket it the rest of
> the way?
At the time when the Shuttle was conceived, the technology you described
was far from available and the whole thing would have cost a lot more to
develop.
Henry Spencer
Anthony Giorgianni <giorg...@my-deja.com> wrote:
>Why is the escape velocity from earth 25,000 miles an hour? At sea level,
>one can throw a rock vertically or fire a model rocket vertically at far
>less speed than that.
And it falls back. The trick is to get it high enough that it doesn't
fall back. At sea level, that requires a leaving with a velocity of about
25,000mph (plus any extra needed to overcome air resistance along the
way).
>And as one increase their distance from the center of
>an object (like the earth), gravity decreases, right? So one would think
>that the speed to leave the plant's gravitation pull would decrease as one
>climbs, not increase.
Correct; it does. Escape velocity varies inversely with the square root
of radius (distance from Earth's center).
>Similarly, (and I guess this is related) why does it take so much of a
>ruckus to get the space shuttle into space. Why can't they just put an
>engine on the thing, fly it to the to the edge of the atmosphere (taking
>advantage of lift instead of rocket power), then just rocket it the rest of
>the way?
Not really related, actually. The key thing to understand is that
reaching orbit is mostly a matter of *velocity*, not altitude. You can
reach an altitude of 200km quite easily, but then you have to accelerate
to orbital velocity, which is about 8km/s. That is much the hardest part
of the job.
Notice that although the shuttle launches vertically, it starts to tip
over toward the horizontal almost immediately, and by the time the SRBs
are jettisoned, it is moving nearly horizontally. Almost all of its fuel
is burned accelerating roughly horizontally at very high altitude. Some
initial climb is needed, partly to get clear of air resistance and partly
to avoid falling back while accelerating, but that's very much a secondary
issue.
--
When failure is not an option, success | Henry Spencer he...@spsystems.net
can get expensive. -- Peter Stibrany | (aka he...@zoo.toronto.edu)
L.C.
constantly at a speed lower than escape velocity and escape
gravity that way. Well.. You can. Just keep powering the thing
long enough.
Why, then, do people talk about escape velocity? First, it is the
speed one would have to impart to a vehicle at the planet's
surface that would, in the absense of drag, allow the vehicle to
escape gravity. It's a statistic that can be easily compared to
similar statistics in other environments. Secondly, if a non-trivial
number of upper atmospheric molecules have speeds that exceed
the statistic, then the planet will lose its atmosphere. Third, in the
absence of atmosphere, a quick buildup of speed is the most
efficient way to escape gravity. Dense atmosphere is present only
near the earth's surface, so it's a good approximation, and it was
used particularly by early threorists. The so-called Goddard
problem of altitude on minimum impulse has been explored more
rigorously since.
I believe your second question is why booster stages are not air
breathing. There is no hard-fast reason. Several such schemes
have been seriously proposed. There are pro's and con's. Every
engineering project is a compromise. The problem can be addressed
in many ways. It hasn't been addressed that way yet.
I'm not certain why other replies focused on orbital velocity. I think
you have that covered with the rocket-powered upper stages.
Regards,
-Larry Curcio
Anthony Giorgianni
So let me see if this is right: Theoretically, you take a rocket and power
it so that moments after liftoff it hits 1,000 miles an hour straight up.
And you keep powering it so it maintains 1,000 miles an hour and on a
straight line away from the earth's center. The rocket could or couldn't
leave the earth's gravitational pull? If not, what would happen? Would it
slow down despite the fact that you are maintaining a speed of 1,000 miles
an hour, or 500 miles an hour or even 20 miles an hour? Would it travel in a
curve even though it's engine is aiming it straight up? And if you could do
this, why in the case of the space shuttle aren't things done this way,
especially since the shuttle is a flyable aircraft? Isn't it simpler and
much more efficient to use the wings and lift to move the craft through the
flyable portion of the atmosphere and then, beyond that level, crank up the
engines, point the thing at the moon for example and leave the earth's
gravitational pull at maybe 100 or 1000 miles per hour? Wouldn't this save
lots of fuel and eliminate the real violent liftoff and the stages and all
that? And maybe you could even fly the shuttle piggyback style (like they
did with the X-15,) and save even more fuel. As a matter of fact, wasn't the
shuttle for a while transported on the back of a Jumbo jet? I thought I saw
photos of that.
I guess I'm missing some real critical concept here. But it seems to me like
this should work.
Thanks for any further clarification.
Regards,
Tony Giorgianni
"L.C." <lcu...@bellatlantic.net> wrote in message
news:3A3F50DA...@bellatlantic.net...
Ian Stirling
>Thank you for the responses everyone.
>So let me see if this is right: Theoretically, you take a rocket and power
>it so that moments after liftoff it hits 1,000 miles an hour straight up.
<snip quoted message improperly placed at bottom>
If you had a long enough ladder, you could climb till you could jump
hard enough to reach escape velocity.
(the fact the earth is rotating complicates things a little)
Escape velocity is that velocity at which, if no forces other than gravity
act, will let something moving directly away from some altitude, keep slowing,
but never to such an extent it'll fall back.
(It reaches zero speed at infinity)
In real life, things are slightly more complex, though it's often a good
ballpark number.
--
http://inquisitor.i.am/ | mailto:inqui...@i.am | Ian Stirling.
---------------------------+-------------------------+--------------------------
"An enemy will usually have three courses open to him. Of these he will
select the fourth." -- Helmuth von Moltke
Tom
If you maintain a constant speed straight up you must expend a great
deal more fuel than needed. As you fly upward at 1000 mph you are
fighting about 21 mph per second tendency to slow down. This declines
with altitude at 1 / (R^2) where R starts around 4000 miles. Escape will
decline with altitude until you are about 2,474,750 miles altitude
before it reaches 1000 mph. At 1000 mph this will take you 2,474 hours
or a little over 100 days.
If you expend thrust so you are accelerating like the shuttle does, it
takes only an extra 3,955 mph to overcome the gravity and drag losses.
Doing it your way would expend many times more.
mkhan
Anthony Giorgianni wrote:
> So let me see if this is right: Theoretically, you take a rocket and power
> it so that moments after liftoff it hits 1,000 miles an hour straight up.
> And you keep powering it so it maintains 1,000 miles an hour and on a
> straight line away from the earth's center. The rocket could or couldn't
> leave the earth's gravitational pull?
Yes, it would escape from the Earth's gravitational field. Just as long
as you manage to keep it moving while the distance from the Earth keeps
on growing, you will eventually end up with a very large distance and a
non-zero velocity, i.e., a hyperbolic orbit. It's axiomatic.
However, the total effort, in terms of the total velocity increment
imparted to the body is much higher than if you have one big
slam-bam-thank-you-ma'am rocket that kicks your spacecraft right into an
escape trajectory.
In fact, what you are describing is more or less the way it is done with
low-thrust electric propulsion. Electric propulsion is very efficient in
that it ejects the propellant at enormously high speeds, more than is
possible with chemical propulsion. However, electric propulsion units
available up to now deliver very low thrust. Although they therefore
have to fire for long times - which means that they have to impart a
higher total velocity increment to the spacecraft - the higher exhaust
velocity still means that they use less propellant in the end. But you
could not use electric propulsion, unless you have one
humongous-yet-lightweight source of energy, for launch from the surface.
> Isn't it simpler and
> much more efficient to use the wings and lift to move the craft through the
> flyable portion of the atmosphere and then, beyond that level, crank up the
> engines, point the thing at the moon for example and leave the earth's
> gravitational pull at maybe 100 or 1000 miles per hour?
It might be possible, but probably not with the Shuttle, which is not a
very flyable aircraft, not providing much lift and yet providing loads
of drag, and would therefore need huge engines and a lot of fuel even
for the atmospheric flight - before you can even think of going orbital.
> Wouldn't this save
> lots of fuel
Probably quite the contrary.
As it stands now, for a shuttle launch around 80% or more of the initial
thrust is provided by the solid rocket boosters - not very efficient and
yet simple and powerful devices.
> As a matter of fact, wasn't the
> shuttle for a while transported on the back of a Jumbo jet? I thought I saw
> photos of that.
Yes, an empty one, after landing, to get it back to the launch site in
Cape Canaveral after a landing in Edwards AFB in California.
But you are talking about a fully loaded and tanked shuttle prior to an
orbital launch. No way any 747 could carry that load.
Henry Spencer
Anthony Giorgianni <giorg...@my-deja.com> wrote:
>So let me see if this is right: Theoretically, you take a rocket and power
>it so that moments after liftoff it hits 1,000 miles an hour straight up.
>And you keep powering it so it maintains 1,000 miles an hour and on a
>straight line away from the earth's center. The rocket could or couldn't
>leave the earth's gravitational pull?
Could. Ascending at 1000mph, eventually it will reach an altitude where
the escape velocity is down to 1000mph, at which point it can shut off its
engines and coast out.
>...And if you could do
>this, why in the case of the space shuttle aren't things done this way...
Because the fuel cost is absolutely ruinous, compared to accelerating hard
at the start and then coasting. Let's see, a bit of math indicates that
escape velocity doesn't get down to 1000mph until about 4Mkm out -- which
is physically meaningless because the Sun's gravity dominates the
situation starting about about 0.9Mkm, but disregard that -- and getting
that far out at such a low speed takes, mmm, about three months. Even
reaching the Moon takes about ten days. Real launchers spend 10-15min
under power -- admittedly at a higher throttle setting, but even so...
>especially since the shuttle is a flyable aircraft? Isn't it simpler and
>much more efficient to use the wings and lift to move the craft through the
>flyable portion of the atmosphere...
The shuttle's wings cannot support anything much heavier than the orbiter
itself... and the ET full of fuel greatly outweighs the orbiter. (Without
the ET, there is no fuel for the main engines at all.)
>and then, beyond that level, crank up the
>engines, point the thing at the moon for example and leave the earth's
>gravitational pull at maybe 100 or 1000 miles per hour?
See above -- there's no possible way to carry enough fuel for months of
continuous engine operation.
>Wouldn't this save
>lots of fuel and eliminate the real violent liftoff and the stages and all
>that?
No.
>And maybe you could even fly the shuttle piggyback style (like they
>did with the X-15,) and save even more fuel. As a matter of fact, wasn't the
>shuttle for a while transported on the back of a Jumbo jet? I thought I saw
>photos of that.
The orbiters are routinely transported from place to place on the back of
a modified 747. But you can't launch one that way -- the orbiter itself
has no fuel for its main engines.
Ian Woollard
>
> Thank you for the responses everyone.
>
> So let me see if this is right: Theoretically, you take a rocket and power
> it so that moments after liftoff it hits 1,000 miles an hour straight up.
> And you keep powering it so it maintains 1,000 miles an hour and on a
> straight line away from the earth's center. The rocket could or couldn't
> leave the earth's gravitational pull?
Yes of course it could.
> And if you could do
> this, why in the case of the space shuttle aren't things done this way,
> especially since the shuttle is a flyable aircraft? Isn't it simpler and
> much more efficient to use the wings and lift to move the craft through the
> flyable portion of the atmosphere and then, beyond that level, crank up the
> engines, point the thing at the moon for example and leave the earth's
> gravitational pull at maybe 100 or 1000 miles per hour? Wouldn't this save
> lots of fuel and eliminate the real violent liftoff and the stages and all
> that?
No actually it costs fuel.
If you do the mathematics, then it is possible to prove
(ignoring air resistance) that the least amount of fuel is used if
you do one huge rocket burn lasting a fraction of a second that
gives you escape velocity straight away and then you coast the
rest of the way. That's why escape velocity is important.
(The reason its more efficient is because the longer you take
to reach escape velocity then the more fuel you have to carry
with you upstairs- it costs more fuel just to raise the fuel up,
if you leave it all at ground level then you win.)
Of course a burn that big would pulp the astronauts and flatten
the rocket, but it is more efficient in theory.
Real rocketry is to do with compromises, astronauts prefer 3G or
less and reducing the acceleration allows the rocket to be
lighter, and high velocity within the atmosphere causes too
much drag and heating of the vehicle etc. etc.
> But it seems to me like this should work.
It works. It probably means a heavier rocket, and higher costs.
There is a case that can be made for airbreathing rockets
though, they initially use the air instead of liquid oxygen
to avoid carrying quite so much. However that requires extra
equipment, the rocket spends longer in the atmosphere so
needs more fuel to deal with the drag, (rockets normally
only spend a very short time in the atmosphere anyway),
but you would need less liquid oxygen, so that would
offset the disadvantages. Still it's generally agreed that
its very marginal and a lot more difficult.
> Thanks for any further clarification.
>
> Regards,
> Tony Giorgianni
--
-Ian (wo...@nortelnetworks.nojunkmail.com)
|"Dynamo is ... an interpreter for HP's PA-8000 instruction set that |
|itself runs on a PA-8000 processor. Programs "interpreted" by Dynamo |
|are often faster... than native. Sometimes by 20% or more." |
Tom Clarke
> ... keep powering it so it maintains 1,000 miles an hour and on a
> straight line away from the earth's center. The rocket could or couldn't
> leave the earth's gravitational pull? If not, what would happen?
Yes it would escape. But you would have to have engine thrust equal to
the weight of the vehicle to maintain the 1000 mph. Maintaining this
thrust takes fuel and it works out that it is more efficient to burn all
the fuel quickly to reach escape velocity than to burn it a little at a time
to maintain a constant speed.
The thrust would gradually decrease as you got further from the earth
and the earth's gravity pull decreases, but if you do the math the fuel
to keep constant speed in decreasing gravity is the same as the fuel
needed to keep constant speed for 4000 miles (earth radius) at
constant earth surface gravity. At 1000 mph this means thrusting
for 4 hours at thrust equal to 1 gee. I gee is 32 ft/sec/sec so 4
hours of thrusting could potentially accelerate the vehicle to
4.6 million feet per second which is 85 miles per second.
Burning the fuel quickly to reach escape velocity only requires
a little more than 7 miles/second of acceleration so the quick
method is much more efficient.
> Would it
> slow down despite the fact that you are maintaining a speed of 1,000 miles
> an hour, or 500 miles an hour or even 20 miles an hour?
20 miles/hour would be 50 times worse in terms of fuel consumption.
> Would it travel in a
> curve even though it's engine is aiming it straight up?
No curve required, just a lot of fuel.
> And if you could do
> this, why in the case of the space shuttle aren't things done this way,
> especially since the shuttle is a flyable aircraft?
Things are done in the way thus uses the least fuel, for the most part.
> Isn't it simpler and
> much more efficient to use the wings and lift to move the craft through the
> flyable portion of the atmosphere and then, beyond that level, crank up the
> engines, point the thing at the moon for example and leave the earth's
> gravitational pull at maybe 100 or 1000 miles per hour?
The wings of the shuttle can barely fly the shuttle alone, they would not
have enough lift for full fuel tanks.
> Wouldn't this save
> lots of fuel and eliminate the real violent liftoff and the stages and all
> that?
There is just too much difference between orbital speed and
practical flight speeds. Even an X-15 tops out at 4000 mph,
the SR-71 blackbird at just over 2000 mph. Orbit needs
18000 mph. Saving 2000 mph by flying in the air is outweighed
by the weight and compexity of the extra hardware needed for
flight in the air.
Although see the blackhorse concept.
Tom Clarke
Zaphod Beeblebrox
"Anthony Giorgianni" <giorg...@my-deja.com> wrote:
>
> So let me see if this is right: Theoretically, you take a rocket and
power
> it so that moments after liftoff it hits 1,000 miles an hour straight
up.
> And you keep powering it so it maintains 1,000 miles an hour and on a
> straight line away from the earth's center. The rocket could or
couldn't
> leave the earth's gravitational pull? If not, what would happen?
Would it
> slow down despite the fact that you are maintaining a speed of 1,000
miles
> an hour, or 500 miles an hour or even 20 miles an hour?
If you can "maintain" a speed (any speed at all), you by definition
will not slow down to less than that speed and will eventually get
anywhere you want to go. Away from the Earths gravitaional influence,
the Sun's, the galaxay's, whatever. The trick is in being able to
carry enough fuel. When you first lift off, you have to be carrrying
your entire fuel load, and have enough thrust to lift it, which means
bigger engines, which means more fuel, etc. (The Catch-22 of rockets)
Escape velocity, at any given distance from a gravitaitonal field, is
the speed at which you would escape that gravitaional field without any
more thrust. The trick is getting to that speed before you run out of
fuel (that you have to bring with you from the start).
The shuttle orbits at (about) 225 miles high at over 17,000 miles per
hour. To lift it with an airplane (with anything like today's
technology) you could get it (to be generous) about 11 miles high going
maybe 600 miles per hour. And you would have to have an airplane (or
wings on the shuttle) capable of lifting far more weight than today's
biggest cargo planes... Not to mention the problems of returning
the "airbreathing" part back to the ground (no way you want to pay the
penalty of carrying all that extra weight the rest of the way to orbit).
Keep studying your math and physics (especially the bits about
conservation of energy). The problem most people have with
understanding rockets and space travel is they don't understand the
huge amounts of energy it takes to accelerate even a small mass to the
speeds needed, the distances involved, or just how small the Earth and
it's chemical powered engines are in comparison.
Believe the numbers, don't go by what "seems like" it should be. The
human senses are a relatively lousy tool for measuring the physical
world.
hope this helps
Tracy
Sent via Deja.com
http://www.deja.com/
Christian Michael
> Thank you for the responses everyone.
>
> So let me see if this is right: Theoretically, you take a rocket and power
> it so that moments after liftoff it hits 1,000 miles an hour straight up.
> And you keep powering it so it maintains 1,000 miles an hour and on a
> straight line away from the earth's center. The rocket could or couldn't
> leave the earth's gravitational pull?
In theory it finally would leave our planetary gravity field.
But:
You must consider that, given your numbers, when you have powered your engines
for one hour you are just 1,000 miles away from the surface, thats about 5
centimeters on a typical school globe (Moon is 30 globes away and still in
earth's gravity field!). If you stop your engines now, you will hit earth again
after a quarter of an hour (not considered any drags, losses, spin of the earth,
etc.). For maintaining speed it needs huge ammounts of fuel, have you seen how
big the main engine tanks for the space shuttle are? And they only need it for a
few minutes. The term to calculate how much fuel you need to lift something
against gravity is an euler function, it reaches infinity realy fast.(You have
to lift the fuel too, you know :-) .)
> If not, what would happen? Would it
> slow down despite the fact that you are maintaining a speed of 1,000 miles
> an hour, or 500 miles an hour or even 20 miles an hour? Would it travel in a
> curve even though it's engine is aiming it straight up?
The thing with escape velocity is that you can't fire your engines as long as
you want, it's a mere throw of a stone by firing several seconds and letting the
rocket fly. If your not fast enough the stone will hit your own head again. To
prevent satellites constantly falling on our heads we let them swing around
earth on a circular or elliptic orbit. It a constant falling back to earth but
missing it and by that flying around our planet. To escape earth's gravity you
just have to swing around fast enough, i.e. firing a kick-engine on the nearest
point of an elliptic orbit, the perigaeum, then your rocket hopefully flies away
on a hyperbolic. And very important: You need 11.2 km/s (6.8 miles/second) to
throw it fast enough.
> And if you could do
> this, why in the case of the space shuttle aren't things done this way,
> especially since the shuttle is a flyable aircraft?
The shuttle flies like a brick with paper wings, it's just a controlled reentry
instead of crashing it into the pacific ocean.
> Isn't it simpler and much more efficient to use the wings and lift to move the
> craft through the
> flyable portion of the atmosphere and then, beyond that level, crank up the
> engines, point the thing at the moon for example and leave the earth's
> gravitational pull at maybe 100 or 1000 miles per hour? Wouldn't this save
> lots of fuel and eliminate the real violent liftoff and the stages and all
> that? And maybe you could even fly the shuttle piggyback style (like they
> did with the X-15,) and save even more fuel.
The german "Saenger" and the british "HOTOL" systems tried something like that
but they where too big, too heavy, too inefficient and too expensive and never
reached prototype status.
> As a matter of fact, wasn't the
> shuttle for a while transported on the back of a Jumbo jet? I thought I saw
> photos of that.
Just to transport it from a landing site back to the cape, it's no good
launching the shuttle from the back of a 747, haven't you seen "007: Moonraker".
:-)
> I guess I'm missing some real critical concept here. But it seems to me like
> this should work.
> Thanks for any further clarification.
> Regards,
> Tony Giorgianni
Regards,
Christian Michael
www.er-ig.de
Anthony Giorgianni
response didn't appear. Sorry if this shows up twice)
So let me see if this is right: Theoretically, you take a rocket and power
it so that moments after liftoff it hits 1,000 miles an hour straight up.
And you keep powering it so it maintains 1,000 miles an hour and on a
straight line away from the earth's center. The rocket could or couldn't
leave the earth's gravitational pull? If not, what would happen? Would it
slow down despite the fact that you are maintaining a speed of 1,000 miles
an hour, or 500 miles an hour or even 20 miles an hour? Would it travel in a
curve even though it's engine is aiming it straight up? And if you could do
this, why in the case of the space shuttle aren't things done this way,
especially since the shuttle is a flyable aircraft? Isn't it simpler and
much more efficient to use the wings and lift to move the craft through the
flyable portion of the atmosphere and then, beyond that level, crank up the
engines, point the thing at the moon for example and leave the earth's
gravitational pull at maybe 100 or 1000 miles per hour? Wouldn't this save
lots of fuel and eliminate the real violent liftoff and the stages and all
that? And maybe you could even fly the shuttle piggyback style (like they
did with the X-15,) and save even more fuel. As a matter of fact, wasn't the
shuttle for a while transported on the back of a Jumbo jet? I thought I saw
photos of that.
I guess I'm missing some real critical concept here. But it seems to me like
this should work. Unless the idea is that an object must be going escape
velocity to leave the gravitational pull without further power. I just don't
get it.
whiten...@my-deja.com
"Anthony Giorgianni" <giorg...@my-deja.com> wrote:
>
> So let me see if this is right: Theoretically, you take a rocket and
power
> it so that moments after liftoff it hits 1,000 miles an hour straight
up.
> And you keep powering it so it maintains 1,000 miles an hour and on a
> straight line away from the earth's center.
That means continuously accelerating it just enough to conterbalance the
pull of gravity - at 9.8 m/sec^2 at first, then decreasing the thrust
as distance to Earth's center increases.
> The rocket could or
couldn't
> leave the earth's gravitational pull?
Yes, it could, if you have enough fuel to keep it up long enough.
Earth's gravitational influence effectively ends about 900,000 miles
away, so "long enough" means 900 hours.
> If not, what would happen? Would
it
> slow down despite the fact that you are maintaining a speed of 1,000
miles
> an hour, or 500 miles an hour or even 20 miles an hour? Would it
travel in a
> curve even though it's engine is aiming it straight up?
Unless you launched from either North or South Pole, yes it would travel
in a curve because of residual velocity from Earth's rotation.
> And if you
could do
> this, why in the case of the space shuttle aren't things done this
way,
Because this method would use up far more fuel than a "short and brutal"
acceleration to escape velocity. It is NOT more efficient.
> especially since the shuttle is a flyable aircraft?
> Isn't it simpler
and
> much more efficient to use the wings and lift to move the craft
through the
> flyable portion of the atmosphere and then, beyond that level, crank
up the
> engines, point the thing at the moon for example and leave the earth's
> gravitational pull at maybe 100 or 1000 miles per hour? Wouldn't this
save
> lots of fuel and eliminate the real violent liftoff and the stages and
all
> that?
As I said, continuous gentle acceleration away from Earth is less
efficient, not more. And as for making use of wings and/or airbreathing
engines within atmosphere - yes it can be done, and it would be a more
elegant design than rocket power from the get-go. Unfortunately it is
also more complicated, so 1960's and 70's engineers chose the easier
path. Some people work on such designs now.
Jarmo Korteniemi
> So let me see if this is right: Theoretically, you take a rocket and power
> it so that moments after liftoff it hits 1,000 miles an hour straight up.
> And you keep powering it so it maintains 1,000 miles an hour and on a
> straight line away from the earth's center. The rocket could or couldn't
> leave the earth's gravitational pull? If not, what would happen? Would it
It could. It would. As somebody stated,
the "escape velocity" is just a theorethical
number, to say how much instantaneous
upwards velocity one would need to get just
away from Earth's gravitational pull. A
humorous/philosophical thing though: would
the craft EVER leave Earth's gravity? Doesn't
gravity effect another body until
infinitely.. it's infinately small, but still. :)
But to answer your question: A craft which has
constant upwards propulsion of say 20 km/hour,
or 1 inch/hour, would indeed leave the Earth.
But then again, that requires much more fuel
than a faster launch.
Think about it this way: the length of the trip
is the same, but the time is longer the slower
you go, and all the time you have to provide the
"lift" upwards not to fall down.
If you have a two rabbits with the other one's
legs tied together racing a mile-long track, the
untied bunny can run it in a few minutes and doesn't
burn more than a carrot or two of energy. Then
again, the tied bunny has to drag itself for days
with only its teeth, and burns at least 20 carrots
a day... that wasn't a good example, but I hope
you got the picture. :) (no rabbits were harmed
while preparing this opinion)
> this, why in the case of the space shuttle aren't things done this way,
> especially since the shuttle is a flyable aircraft? Isn't it simpler and
> much more efficient to use the wings and lift to move the craft through the
> flyable portion of the atmosphere and then, beyond that level, crank up the
> engines, point the thing at the moon for example and leave the earth's
> gravitational pull at maybe 100 or 1000 miles per hour? Wouldn't this save
> lots of fuel and eliminate the real violent liftoff and the stages and all
Is the shuttle an aircraft which can take
off horizontally, that I don't know..?
It's gliding like a plane when it comes down
but...
At least is has no fuel like a conventional
aircraft - it doesn't "brathe air" to
burn its fuel. So it would have to be seriously
modified in order to act as a liftoffing plane.
But NO, it would NOT save fuel to first
liftoff like a plane and then slowly get up
towards the "edge of space" and then fire
rockets and get into orbit or God knows
where. You still have to go up the same
distance (say to 300km up), and reach the
same horizontal velocity (in order to obtain
orbital speed). But in addition to
that, you would spend time and fuel cruising
in the atmosphere, with time ticking
by. The longer you stay in the atmosphere,
the more fuel you have to burn due to
drag and friction etc, AND the fact that
you have to fight the gravity for a longer
time - remember that eventually the shuttle
will end up in a "frictionless" (okay,
not really..) place, where it just obtains
the orbital speed and then it doesn't
have to worry (too much) about keeping
up the speed.
> that? And maybe you could even fly the shuttle piggyback style (like they
> did with the X-15,) and save even more fuel. As a matter of fact, wasn't the
...That was a really funny one. Think about
it: You wouldn't ONLY have to get the
Space Shuttle up in the air, but you'd have
to get the carrying jumbo up there also!
Of course (?) you meant that there would be
more fuel in the shuttle left for
orbital manouvers etc.. :)
> shuttle for a while transported on the back of a Jumbo jet? I thought I saw
> photos of that.
Yup, it has. But it's on the roof of the
plane, while the X-15 is under the wing, as
far as I know anyway. X-15 can be dropped
and then (as if it was needed) the rocket
engines would kick in, after falling down.
The shuttle can't be dropped off, as it
simply is too big to fit under the wing
(think about liftoff! :-D ). There would
have to be somekind of a runway for the
shuttle on the roof... If it fired
the rockets on the roof, my guess is that
the plane would be seriously damaged, its
pilots would be fried and if everybody was
lucky, the whole thing wouldn't even be
able to liftoff in the first place due to
the heavy load (shuttle+fuel) on the
jumbo!!!!
> Thanks for any further clarification.
> Regards,
> Tony Giorgianni
Jarmo
--
Jarmo Korteniemi | Planetology student at the Division of
tel: Finland-040-7519185 | Astronomy, University of Oulu, Finland
fax: +1-419-7814825 | - http://www.student.oulu.fi/~jkorteni -
"The current state of knowledge summarized: In the beginning, there was
nothing, which exploded." -T.Pratchett
Anthony Q. Bachler
for testing the aerodynamics and its ability to glide to a landing.
Anthony Giorgianni wrote in message
<91oduq$4nr42$1...@ID-62110.news.dfncis.de>...
Branko Badrljica
Christian Michael wrote:
>
<SNIP>
> To escape earth's gravity you
> just have to swing around fast enough, i.e. firing a kick-engine on the nearest
> point of an elliptic orbit, the perigaeum, then your rocket hopefully flies away
> on a hyperbolic. And very important: You need 11.2 km/s (6.8 miles/second) to
> throw it fast enough.
>
Don't some guns achieve considerable part of that ? I mean more or less conventional
artillery.
Last time I cared to read about it, I think 4-8 km/s speeds were mentioned.
Did someone ever succeeded ( besides Jules Verne ;o) with blasting projectile into
the orbit with a gun?
Regards,
Branko
Henry Spencer
Branko Badrljica <bra...@avtomatika.com> wrote:
>> ...And very important: You need 11.2 km/s (6.8 miles/second) to
>> throw it fast enough.
>
>Don't some guns achieve considerable part of that ? I mean more or less
>conventional artillery.
No, conventional guns don't come anywhere near that. Your memory may be
mixing up units: 4000 *feet* per second is not unheard-of, although 2000-
3000 is more typical.
There are fundamental limits on conventional gun performance, set by the
nature of the propelling gases. I don't remember the numbers offhand, but
they can't reach more than a small fraction of orbital velocity.
You can do better with a light-gas gun, which separates the energy source
and the gas source so that the gas can be optimized as a propellant. In
principle, they can reach near-orbital velocities, although this has never
actually been done.
>Did someone ever succeeded ( besides Jules Verne ;o) with blasting projectile
>into the orbit with a gun?
No. Once or twice there have been projects which saw that as a long-term
goal, but they never got that far.
whiten...@my-deja.com
> The shuttle flies like a brick with paper wings, it's just a
controlled reentry
> instead of crashing it into the pacific ocean.
That's an exaggeration. Empty shuttle's glide ratio is 4:1 - more like a
plastic bottle than a brick ;)
> > As a matter of fact, wasn't the
> > shuttle for a while transported on the back of a Jumbo jet? I
thought I saw
> > photos of that.
>
> Just to transport it from a landing site back to the cape, it's no
good
> launching the shuttle from the back of a 747, haven't you seen "007:
Moonraker". :-)
Intersting tidbit - when that 747 flies at cruising speed, shuttle's
wings generate enough lift to support shuttle's weight, so there is no
vertical stress on the 747. Some pilots call it "world's biggest
biplane".
Anthony Giorgianni
Thank you everyone for the great responses. I guess there was a problem with
posting to this group for a bit, but finally I got all the messages in one
clump! Sorry one of mine appeared twice. I've tried to cancel the copy.
Well I think I understand this a lot better. It's all really interesting and
underscores the vast distances of things. I wasn't aware, for example, of
the height of a standard shuttle orbit compared with the standard ceiling of
air breathing aircraft. But what I really needed to understand was this
concept of what escape velocity is, the speed at which, once it's attained,
you can pretty much coast the rest of the way, and how it's MORE efficient
to do it that way.
Anyway, I have to read everything again to make sure I have it all. But
thanks much for answering this old puzzle. I'm MUCH smarter now :o)
Regards,
Tony Giorgianni
"Ian Woollard" <wo...@nortelnetworks.com> wrote in message
news:3A4256AD...@nortelnetworks.com...
John Schilling
>Christian Michael wrote:
>> And very important: You need 11.2 km/s (6.8 miles/second) to
>> throw it fast enough.
>Don't some guns achieve considerable part of that ? I mean more or less
>conventional artillery. Last time I cared to read about it, I think 4-8 km/s
>speeds were mentioned.
The very best conventional "artillery" in this respect would be modern
tank cannons, which approach 2 km/s. They use every trick in the book
to do this; 4-8 km/s is right out.
--
*John Schilling * "Anything worth doing, *
*Member:AIAA,NRA,ACLU,SAS,LP * is worth doing for money" *
*Chief Scientist & General Partner * -13th Rule of Acquisition *
*White Elephant Research, LLC * "There is no substitute *
*schi...@spock.usc.edu * for success" *
*661-951-9107 or 661-275-6795 * -58th Rule of Acquisition *
Navigaiter
At three earth radii, 19,200 km from center, 12,800 km from surface, gravity is
1/9th of surface gravity. Thrust to maintain 1,000 kmh is 1/9th of what it was
near the surface. However, the ship's mass, having burnt most of its fuel near
earth, is probably 1/10 of what is was near the surface, so the thrust required
to maintain velocity is 1/90th of what it was near the surface.
This explains why it does not require humongous fuel loads to go to Luna at
slower-than-orbital speed. If you convert the fuel spent for orbital speed into
vertical distance out of the very shallow gravity well, you have an efficient
Lunar transport.
Help build a near space beanstalk!
http://members.aol.com/beanstalkr/project/
Anthony Q. Bachler
up to 75Kg into orbital speeds. The government saw fit to discontinue the
project though. Incidentally, he also helped IRAQ build its super-gun, the
largest cannon ever built. It could fire a several ton object thousands of
miles. Luckily it's construction was halted before it was finished. There
are no fundamental limits to the speed a cannon can fire a projectile, as
Bohl proved (and was murdered for it). The only limits to a cannon are due
to its design, which must be changed to achieve the performance Bohl got.
Henry Spencer wrote in message ...
Navigaiter
inch/hour, would indeed leave the Earth. But then again, that requires much
more fuel> than a faster launch.>>
done the math on the other method, i.e., the slow and steady burn method. So,
we don't know if the conventional wisdom really still is the more efficient
method of leaving Earth.
One thing to remember is that gravity reduces sharply with distance and that
at 3200 km from the earth's surface, gravity is already only half as strong as
at the surface! This means the throttle of a slow and steady craft can be cut
in half. Maybe even cut by 3/4 since most of the fuel mass has been burnt by
then.
And yes, any type of airborne piggy-back launch will improve the efficiency
of any rocket because any reduction in fuel mass to be accelerated means more
altitude and speed with less onboard fuel.
The arguers say the advantage would be small but a 25% savings in fuel mass
is actually a very important savings for ships that would escape our shallow
gravity well.
George William Herbert
Navigaiter <navig...@aol.com> wrote:
>>A craft which has constant upwards propulsion of say 20 km/hour, or 1
>>inch/hour, would indeed leave the Earth. But then again, that requires much
>>more fuel than a faster launch.
>
> This is "conventional wisdom." It is so conventional that no one has ever
>done the math on the other method, i.e., the slow and steady burn method. So,
>we don't know if the conventional wisdom really still is the more efficient
>method of leaving Earth.
That is not true. We've done the math repeatedly, including showing it
to you last time this discussion came up. Slow launches lose fatally
on gravity losses until you reach orbital velocity, and slow burns from
LEO to transfer orbits are less efficient than fast (order of magnitude
10% of orbital period) burns are.
Think about it a bit. Most of the velocity produced by modern rockets
goes into the vehicle's final velocity. A rocket might produce 9,200
meters per second of velocity change, and end up traveling 7,400 meters
per second in low orbit (this is roughly typical). That's about 80%
efficient use of propulsion capability translating to final velocity.
About half the loss (900 m/s or so) is in gravity losses, equivalent
to thrusting upwards at 1 G for 90 seconds, one and a half minutes.
The rocket actually burns for typically near 10 minutes, but the
gravity losses are low because of the early vertical accelleration
and gravity turn type trajectory being efficient uses of the upwards
thrust component. Very little of the time is spent burning downwards.
Any launch method which takes 15 minutes hovering or thrusting upwards
to reach stable orbit will burn as much fuel in transit just on resisting
gravity as a normal rocket needs to burn in total.
Navigaiter, I don't want to be mean, but rocketry and in particular
trajectories are not fields where an innumerate intuition gets you
very far. You have to be willing and able to analyze the physics,
work out a numerical analysis, and analyze the results. It's good
that you're enthusiastic about the field, but if you want to understand
it you really need to get to a point in understanding the relevant
physics and math and being able to analyze it yourself that you don't
make fundamental mistakes like this.
-george william herbert
gher...@retro.com
frank...@delphi.com
>Why is the escape velocity from earth 25,000 miles an hour? At sea level,
>one can throw a rock vertically or fire a model rocket vertically at far
>less speed than that. And as one increase their distance from the center of
>an object (like the earth), gravity decreases, right? So one would think
>that the speed to leave the plant's gravitation pull would decrease as one
>climbs, not increase. So where does this 25,000 miles an hour suddenly kick
>in?
you down. Ignoring atmospheric drag (but see next answer), you'd have to
start as 25,000 mph from Earth's surface before your velocity never fell
to zero, evan at an 'infinite' distance away.
>Similarly, (and I guess this is related) why does it take so much of a
>ruckus to get the space shuttle into space. Why can't they just put an
>engine on the thing, fly it to the to the edge of the atmosphere (taking
>advantage of lift instead of rocket power), then just rocket it the rest of
>the way?
as X-30.
The problem is, as a pure rocket, you want to get out of the denser
lower atmosphere (and eventually all of it) as soon as possible, so as
not to have to fight its drag and aerothermal issues as soon as possible.
An air-breathing spaceplane must get as close to orbital velocity as it
can while still *in* the atmosphere, so as to take on outside oxidizer.
This means you *are* still fighting drag, heat, and the problems of
taking on air at increasingly greater mach numbers and burning fuel
in it so as to expel it even faster. This also means complex inlets
whose shape must change (this already is done to some extent on the SR-71)
as airspeed and altitude do. Then you fire a relatively small rocket
to get all the way to orbital velocity. None of this is easy, and the
price you pay for 'free' oxidizer is an engine that's heavier than a
pure rocket for the same thrust, espically if you include the variable
geometry inlets that a rocket doesn't need.
If you decide to switch to all-rockets at a lower, less challenging
speed, then you have to take the weight of an air-breathing engine to
orbit that's not doing anything to help you over a greater part of the
ascent, now.
And all of this is easier with a vehicle that has a smaller, lighter
payload capacity than the shuttle. (Though others would say, and I'd
agree, that something like that has greater commercial value than the
shuttle, anyway.)
Others may explain it better than I, but that's basically it.
Frank
frank...@delphi.com
>especially since the shuttle is a flyable aircraft? Isn't it simpler and
>much more efficient to use the wings and lift to move the craft through the
>flyable portion of the atmosphere and then, beyond that level, crank up the
>engines, point the thing at the moon for example and leave the earth's
>gravitational pull at maybe 100 or 1000 miles per hour? Wouldn't this save
>lots of fuel and eliminate the real violent liftoff and the stages and all
>that? And maybe you could even fly the shuttle piggyback style (like they
>did with the X-15,) and save even more fuel. As a matter of fact, wasn't the
for most of such a mission.
>shuttle for a while transported on the back of a Jumbo jet? I thought I saw
>photos of that.
Frank
frank...@delphi.com
>> shuttle for a while transported on the back of a Jumbo jet? I thought I saw
>> photos of that.
>
> Yup, it has. But it's on the roof of the
> plane, while the X-15 is under the wing, as
> far as I know anyway. X-15 can be dropped
> and then (as if it was needed) the rocket
> engines would kick in, after falling down.
> The shuttle can't be dropped off, as it
> simply is too big to fit under the wing
> (think about liftoff! :-D ). There would
> have to be somekind of a runway for the
> shuttle on the roof... If it fired
> the rockets on the roof, my guess is that
> the plane would be seriously damaged, its
> pilots would be fried and if everybody was
leaving the back of a 747, C-5 or An-225. One needs to to a pretty hefty
pitch-up maneuver at the time, but I could be wrong. Even the shuttle
orbiter (Unpowered, of course) has been flown off the back of a 747 for
approach and landing tests in 1977. But...
> lucky, the whole thing wouldn't even be
> able to liftoff in the first place due to
> the heavy load (shuttle+fuel) on the
> jumbo!!!!
it just can't be done with orbiter/ET/SRBs as we know them. Again, this
could be done with a much smaller spaceplane, but that's all.
Frank
frank...@delphi.com
>>So let me see if this is right: Theoretically, you take a rocket and power
>>it so that moments after liftoff it hits 1,000 miles an hour straight up.
>
><snip quoted message improperly placed at bottom>
>
>If you had a long enough ladder, you could climb till you could jump
>hard enough to reach escape velocity.
>(the fact the earth is rotating complicates things a little)
Frank
frank...@delphi.com
>Thank you for the responses everyone. (I'm reposting this because my first
>response didn't appear. Sorry if this shows up twice)
Frank
Navigaiter
discussion came up.
I beg to differ with the kind and learned Mr Herbert. The math proof has never
appeared here to support the preferability of orbiting before departing earth.
How could one even manage to post a spreadsheet of differential equations to
compare a direct versus orbital trajectory to Luna? I'm sure I would remember
that!
A LunaDirect flight may spend much fuel on gravity losses but it makes up
for it by not wasting fuel on "velocity losses." 18,000 mph is not required for
a moon trip. [indeed, it's a waste of power because you just have to turn
around and burn it all off to stop on the moon!] 1,000 kmh will get you there
fast enough, ten days. The fuel saved by not accelerating to heroic speeds is
sufficient to steadily thrust directly out of the earth's very shallow gravity
well.
Best regards and thanks for the dialog, Allen
Navigaiter
mechanics. [It means the velocity required to escape from ORBIT, not from
earth] In truth, any upward velocity, if it is maintained, is an "Escape
Velocity."
In other words, the earth can be escaped at any speed.
Henry Spencer
Anthony Q. Bachler <cwhi...@no.spam.mail.socket.net> wrote:
>Actually Dr. Bohl (sp?) worked out the HARP cannon, which launched payloads
>up to 75Kg into orbital speeds.
Uh, no. You might try reading some of the HARP papers sometime; I have.
HARP launched quite substantial payloads to high *altitudes*, but not to
velocities anywhere near orbital. There was no hope of reaching such
velocities with a powder gun, as he knew full well; his orbital designs
used a gun launch, but had three or four rocket stages following that.
It was Dr. Gerald Bull, by the way. I recommend William Lowther's
biography "Arms and the Man" if you want to know more about him.
Tom McWilliams
> WH wrote:
> >We've done the math repeatedly, including showing it to you last time this
> >discussion came up.
>
> I beg to differ with the kind and learned Mr Herbert. The math proof has never
> appeared here to support the preferability of orbiting before departing earth.
I don't think that was the objection; rather, it is that slow vertical
ascents are inefficient, in terms of final height, compared to
higher-acceleration, horizontal ascents.
> How could one even manage to post a spreadsheet of differential equations to
> compare a direct versus orbital trajectory to Luna? I'm sure I would remember
> that!
I don't remember it either, but you can start here:
v = -gt + uln(m0/m) where u is the speed of the exhaust, m0 is the
inital mass of the rockt+fuel, m is the rocket after (m-m0) has been
ejected at speed 'u', t is time, and v is your vertical speed, with 'up'
being positive.
This assumes g is constant, which is incorrect, of course, but you can
do the numerical analysis to remove that assumption, or massage the
equation to figure out dg/dt based on u, and the mass of the body
causing 'g'.
What you are looking for is the mass, (m-m0) which is required to
acheive a final velocity=escape or, since you don't seem to like that, a
g-factor sufficiently low for whatever purposes you have in mind.
> A LunaDirect flight may spend much fuel on gravity losses but it makes up
> for it by not wasting fuel on "velocity losses." 18,000 mph is not required for
> a moon trip. [indeed, it's a waste of power because you just have to turn
> around and burn it all off to stop on the moon!] 1,000 kmh will get you there
> fast enough, ten days. The fuel saved by not accelerating to heroic speeds is
> sufficient to steadily thrust directly out of the earth's very shallow gravity
> well.
Very 'shallow,' but with a rather high gradient, which is the important
aspect.
Do the math...you'll see where the difficulty is.
-Tm
--
* . * '^
,.. " . *
,
' Tommy Mac
Jorge R. Frank
<20001226223919...@ng-ff1.aol.com>:
>I beg to differ with the kind and learned Mr Herbert. The math proof has
>never appeared here to support the preferability of orbiting before
>departing earth. How could one even manage to post a spreadsheet of
>differential equations to compare a direct versus orbital trajectory to
>Luna? I'm sure I would remember that!
You're expecting *way* too much. The equations themselves have been given
to you, multiple times. It's not Mr. Herbert's job to build the
spreadsheet for you. That's left as an exercise for the student. :-)
> A LunaDirect flight may spend much fuel on gravity losses but it
> makes up
>for it by not wasting fuel on "velocity losses." 18,000 mph is not
>required for a moon trip. [indeed, it's a waste of power because you
>just have to turn around and burn it all off to stop on the moon!]
Umm, no you don't. First of all, it's 25,000 mph, not 18,000. Second,
just because you achieve a speed of 25,000 mph in low Earth orbit 100-200
miles up doesn't mean you have to "turn around and burn it all off" at the
moon. Once the engines have cut off and you're coasting toward the moon,
the Earth's gravity is still acting on you, slowing you down. By the time
you reach the point where the Earth's and moon's gravities balance, you've
slowed to just a couple of thousand mph. As you get closer to the moon,
the moon's gravity speeds you up. It is *that* extra speed imparted by the
moon that you have to "turn around and burn off" in order to enter lunar
orbit.
--
JRF
Reply-to address spam-proofed - to reply by E-mail,
check "Organization" and think one step ahead of IBM.
Navigaiter
acheive a final velocity=escape or, since you don't seem to like that, a
g-factor sufficiently low for whatever purposes you have in mind.>>
spreadsheet.
The purpose in mind is to reach the Earth/Lunar gravity boundary as
efficiently as possible, in aid of the CATS goal.
Navigaiter
fighting about 21 mph per second tendency to slow down. This declines
with altitude at 1 / (R^2) where R starts around 4000 miles.>>
Most people underestate the rapidity of the inverse square reduction of the
earth's G field with distance. The decline starts immediately and rapidly at
the earth's surface. At an altitude of just 2500 km, G is already only 50% of
sealevel value. At 1 radius high, 6400 km, G is only 25% of sealevel. Plus, at
1radia, the ship has burnt 75% of her fuel and is that much lighter, further
reducing mutual gravitational attraction. All the fuel saved by not racing for
orbital velocity has been constantly throttled-down to provide one long burn
directly toward the target, Luna.
In other words, the gravity well is very shallow and only has steep sides
within half a radia from the earth. Slow and steady is the way to go cheaply.
At one radia, power is a piece of cake -- deploy an STR system and get to Luna
with energy from the sun.
Navigaiter
the Earth's gravity is still acting on you, slowing you down. By the time
you reach the point where the Earth's and moon's gravities balance, you've
slowed to just a couple of thousand mph.>>
So what is the need for an orbit in the first place? [besides letting the
personel catch their breaths.]
Robert Shimmin
On 27 Dec 2000, Navigaiter wrote:
> reducing mutual gravitational attraction. All the fuel saved by not racing for
> orbital velocity has been constantly throttled-down to provide one long burn
> directly toward the target, Luna.
Let's consider a limiting case: the slowest, steadiest rocket I can
imagine begins its burn with exactly its own weight in thrust. It just
barely counteracts gravity, and thus does work to go nowhere, an
astounding zero percent efficiency. Why NASA has not yet funded a
prototype of such a design is left as an exercise to Congress.
Rockets with some higher initial thrust go somewhere, and thus have a
non-zero efficiency (however you measure it). At what point does this
trend reverse, as you insist, and why?
--RS
John Hare
(Navigaiter) writes:
This reply is to people that don't have a grasp of the numbers.
Navigator refuses to believe in numbers in favor of intuition.
><<As you fly upward at 1000 mph you are
>fighting about 21 mph per second tendency to slow down. This declines
>with altitude at 1 / (R^2) where R starts around 4000 miles.>>
> Most people underestate the rapidity of the inverse square reduction of
>the
>earth's G field with distance. The decline starts immediately and rapidly at
>the earth's surface. At an altitude of just 2500 km, G is already only 50% of
>sealevel value. At 1 radius high, 6400 km, G is only 25% of sealevel. Plus,
At the proposed velocity of 1,000 mph=1,600 kph, it takes
over an hour and a half to reach this 2500 km altitude.
That is 5400 seconds of thrusting. Given a half gee of thrust
for this whole time, this is the equivelent fuel consumption
of 3 gees for 900 seconds. With the gee losses of a verticle
assent, the 3 gee climb will leave an excess velocity of 18,000
meters per second. This is 7,000 meters per second in excess
of escape velocity. Compare this to the 1,000 mph climb that
will begin falling back within a couple of minutes of ceasing
thrust.
>at
>1radia, the ship has burnt 75% of her fuel and is that much lighter, further
>reducing mutual gravitational attraction. All the fuel saved by not racing
>for
>orbital velocity has been constantly throttled-down to provide one long burn
>directly toward the target, Luna.
If the numbers had actually been done, that 75% would read 99+%.
> In other words, the gravity well is very shallow and only has steep sides
>within half a radia from the earth. Slow and steady is the way to go
>cheaply.
>At one radia, power is a piece of cake -- deploy an STR system and get to
>Luna
>with energy from the sun.
At one radius, an STR is helpless in that trajectory. It will fall
all the way to the ground in a few minutes.
>Help build a near space beanstalk!
>http://members.aol.com/beanstalkr/project/
>
Not likely. To accept a team leader that makes up facts
as he goes along is to place a very low price on your own worth.
Tom McWilliams
It might be easier to just write a 'rocket simulation' program with
settings for exhaust velocity, rocket-mass, etc etc, and just try
different choices.
Efficiency has many meanings. Presumably, you mean mass-into-space vs.
mass-used-as-fuel.
If you go to the trouble of writing such a simulator, you can add a
setting for fuel costs, and another for radius and density of the world
you are lauching from...
I recommend you write it in JAVA, so you can put it on a web-site for us
to play with :-)
Henry Spencer
Navigaiter <navig...@aol.com> wrote:
>>We've done the math repeatedly, including showing it to you last time this
>discussion came up.
>
>I beg to differ with the kind and learned Mr Herbert. The math proof has never
>appeared here to support the preferability of orbiting before departing earth.
Uh, that's not what we were talking about at all. An efficient ascent
does not have to go via orbit. Via orbit or not, short burns followed by
coasting are vastly more efficient than a slow climb under power.
>How could one even manage to post a spreadsheet of differential equations to
>compare a direct versus orbital trajectory to Luna?
Why do you imagine that this is necessary? Simple approximations, based
on known results from the differential equations, demonstrate that your
slow-climb approach is absurd; precise calculations (which require more
powerful tools than spreadsheets) are unnecessary.
> A LunaDirect flight may spend much fuel on gravity losses but it makes up
>for it by not wasting fuel on "velocity losses."
There's no such thing as "velocity losses". The kinetic energy obtained
in the conventional approach is converted -- 100% efficiently -- into
potential energy during the ascent. Whether you do the ascent by rocket
burns or by witchcraft, you have to supply that potential energy from
somewhere; nothing is being lost.
>18,000 mph is not required for
>a moon trip. [indeed, it's a waste of power because you just have to turn
>around and burn it all off to stop on the moon!]
Uh, no. Entry into lunar orbit does require a rocket burn, but it's a
very small one by comparison.
>...The fuel saved by not accelerating to heroic speeds is
>sufficient to steadily thrust directly out of the earth's very shallow gravity
>well.
No. Wrong. Please learn how to do arithmetic, rather than trying to
arrive at numerical results by blind faith and intuition.
Jorge R. Frank
<20001227193424...@ng-ff1.aol.com>:
> So what is the need for an orbit in the first place? [besides letting
> the
>personel catch their breaths.]
For Apollo, that was certainly part of it - they wanted time to check out
the spacecraft and make sure everything was OK before committing to the
moon. It's a lot easier to abort a mission from low Earth orbit than from
translunar coast, as Apollo 13 learned...
In general though, the bigger reason for a parking orbit is that it expands
your launch window. You can launch at a slightly non-optimal time and
coast to a more optimal point in the parking orbit. For a direct ascent,
you must launch at the right time.
Of course, if you're willing to live without both of the above, you *can*
skip the parking orbit and launch directly into a translunar coast
trajectory.
Joseph Nebus
>Let's consider a limiting case: the slowest, steadiest rocket I can
>imagine begins its burn with exactly its own weight in thrust. It just
>barely counteracts gravity, and thus does work to go nowhere, an
>astounding zero percent efficiency. Why NASA has not yet funded a
>prototype of such a design is left as an exercise to Congress.
Well, some of the early Atlas and Vanguard test flights did that,
but that wasn't technically their design.
Joseph Nebus
------------------------------------------------------------------------------
L.C.
broadens the launch window, for example. For
deeper space missions, it will eventually allow the
use of ion rockets, which would accelerate from
orbital velocity to escape velocity very slowly.
Henry Spencer
Navigaiter <navig...@aol.com> wrote:
> So what is the need for an orbit in the first place? [besides letting the
>personel catch their breaths.]
There have been launches to the Moon (and beyond) which have not gone via
parking orbit, and it does improve efficiency *slightly*. The price you
pay for it is very narrow and infrequent launch windows, because Earth's
rotation puts your launch site in precisely the right place only once a
day. With a stop in parking orbit, it suffices to arrange that the
parking orbit passes through the right place, which is much easier.
Note that a lunar launch without parking orbit still looks nothing like
Navigaiter's slow vertical ascent. The rocket tips over to horizontal as
quickly as possible, given the need to clear the atmosphere first, and
accelerates as hard as it can; it's just that it doesn't stop accelerating
as it reaches orbital velocity. The trajectory differs only slightly from
what you'd get with a very brief stop in parking orbit.
Don Stokes
Henry Spencer <he...@spsystems.net> wrote:
>In article <20001227193424...@ng-ff1.aol.com>,
>Navigaiter <navig...@aol.com> wrote:
>> So what is the need for an orbit in the first place? [besides letting the
>>personel catch their breaths.]
>
>There have been launches to the Moon (and beyond) which have not gone via
>parking orbit, and it does improve efficiency *slightly*. The price you
>pay for it is very narrow and infrequent launch windows, because Earth's
>rotation puts your launch site in precisely the right place only once a
>day. With a stop in parking orbit, it suffices to arrange that the
>parking orbit passes through the right place, which is much easier.
It also decouples the trans-lunar injection from the vagaries of weather
and booster performance in the atmosphere. Launch and low altitude
flight is pretty rough on the vehicle, and it's useful to park in LEO,
check that nothing important fell off and correct any discrepancies.
An (eg) Apollo-Saturn stack is a lot smaller at the time the TLI
decision is made from the relative safety of a parking orbit than it is
at launch. That represents rather less to go wrong ...
-- don
frank...@delphi.com
>Of course, if you're willing to live without both of the above, you *can*
>skip the parking orbit and launch directly into a translunar coast
>trajectory.
would've put the entire Apollo CM (with appropriately larger descent/
ascent stages) on the Moon, that's how it would've been done. The mode
was even called Direct Ascent. (As opposed to the Lunar Orbit Rendezvous
[LOR] approach we did use, or Earth Orbit Rendezvous [EOR] which would've
required Earth orbit refueling [and another Saturn to do it] to do a
similar CM on the surface mission.)
I didn't know Direct Ascent involved a more constrained launch window,
though. Thanks.
Frank
frank...@delphi.com
>>Let's consider a limiting case: the slowest, steadiest rocket I can
>>imagine begins its burn with exactly its own weight in thrust. It just
>>barely counteracts gravity, and thus does work to go nowhere, an
>>astounding zero percent efficiency. Why NASA has not yet funded a
>>prototype of such a design is left as an exercise to Congress.
>
> Well, some of the early Atlas and Vanguard test flights did that,
>but that wasn't technically their design.
Of course, hovering, lateral translation and powered descent was exactly
what was wanted of it in the first flights. The LM could do this too, but
you didn't want to waste one second more than necessary in that state...
Frank
Henry Spencer
Don Stokes <d...@news.daedalus.co.nz> wrote:
>>...very narrow and infrequent launch windows, because Earth's
>>rotation puts your launch site in precisely the right place only once a
>>day. With a stop in parking orbit, it suffices to arrange that the
>>parking orbit passes through the right place, which is much easier.
>
>It also decouples the trans-lunar injection from the vagaries of weather
>and booster performance in the atmosphere. Launch and low altitude
>flight is pretty rough on the vehicle, and it's useful to park in LEO,
>check that nothing important fell off and correct any discrepancies.
While this is undoubtedly useful, it's given too much weight in many
discussions of the matter, because the launch-window issue is harder to
explain. The dominant issue is wider launch windows; having some time for
orbital checkout etc. is usually a helpful side effect rather than a
primary motive. Unmanned deep-space launches via parking orbit, in
particular, tend to run on tight preprogrammed schedules with little
opportunity for tinkering in parking orbit, because of limited
communications and inflexible control systems. The parking orbit is
primarily a way of reaching a tightly constrained departure path without
severely restricting the launch window.
Greg D. Moore
Navigaiter wrote:
>
> >So where does this 25,000 miles an hour suddenly kick in?>>
> It doesn't. It's a mis-named mathematical term pertaining to orbital
> mechanics. [It means the velocity required to escape from ORBIT, not from
> earth]
No, it mean the velocity required at that point with NO additional
acceleration.
You can escape from orbit using an ion drive. You won't be traveling at
25,000 mph any time soon.
> In truth, any upward velocity, if it is maintained, is an "Escape
> Velocity."
> In other words, the earth can be escaped at any speed.
Yes, provided a constant acceleration. As you approach an infinite
distance from Earth the escape velocity approaches 0. No one here is
arguing with that.
However, this is not necessarily the most efficient use of your energy.
That is what people are trying to point out to you.
Navigaiter
acceleration.>>
You're reading more into the term "escape velocity" than is there. It is
often used as a short form of "escape velocity at X distance." And it is left
to the hearer to know the speaker was possibly, maybe, perhaps, referring to
the velocity necessary to escape from LEO. Hence the proper confusion stated
in the original post of this thread.
Thankfully, this board has clarified the fact that the earth can be escaped
at any speed. Mankind is NOT marooned here by a high-speed velocity trap as may
be inferred by lazily saying "escape velocity" without any modifier.
cheers for CATS!
Navigaiter
primarily a way of reaching a tightly constrained departure path without
severely restricting the launch window. >>
The remaining question is called the Goddard Problem, what throttle setting to
use to gain the most distance from a given quantity of fuel and mass ratio.
Consensus says that the least exposure time to near-earth gravity is the
only way to get out of here. That necessitates heroic speed and expense to make
a capable engine which will not bring about CATS.
A pellet-fed solid rocket could be the answer. Cheap and easy to handle
fuel, lightweight firing chamber, stageable pellet tanks and even a primitive
throttleability offer a CATS solution.
Naysayers can only say it's hard to make a pellet-feeder but so what, if
it's the only workable method? It could offer a >50% (?) weight reduction over
conventional solid motor casings.
The reduction in weight would allow slower speed to succeed in the shallow
gravity well.
Don Stokes
Navigaiter <navig...@aol.com> wrote:
> Thankfully, this board has clarified the fact that the earth can be escaped
>at any speed. Mankind is NOT marooned here by a high-speed velocity trap as may
>be inferred by lazily saying "escape velocity" without any modifier.
Velocity has never been considered the trap. Energy is. Regardless of
how you do it, you still have to liberate a certain amount of energy
to lift a given mass to a given altitude.
If you do it by thrusting all the way you have to carry not only your
payload but your propellant for the higher altitude parts of your trip
through all the lower altitudes. This costs more propellant -- a *lot*
more -- than is required to lift the payload alone.
On the other hand, of you burn all your propellant at low altitude as
quickly as you can, you don't have to carry any propellant to higher
altitude. This is a Very Big Win. This is why folks worry about escape
velocity -- it's the speed you have to reach _at_low_atitude_ that lets
you escape from a gravity well without having to waste energy lifting
additional propellant into higher altitude.
-- don
Henry Spencer
Navigaiter <navig...@aol.com> wrote:
> So, the consensus is that orbit is nice but not necessary to get to Luna.
>The remaining question is called the Goddard Problem, what throttle setting to
>use to gain the most distance from a given quantity of fuel and mass ratio.
Who, exactly, calls it the "Goddard Problem"? The math of the situation
is crystal-clear and can be found in any introductory rocket book. The
way to minimize gravity losses, and hence fuel required, is rapid
acceleration (in a direction as close to horizontal as possible) followed
by coasting outward. There is no "problem" except in the minds of the
innumerate.
Bruce Dunn
Navigaiter wrote:
> A pellet-fed solid rocket could be the answer. Cheap and easy to handle
> fuel, lightweight firing chamber, stageable pellet tanks and even a primitive
> throttleability offer a CATS solution.
> Naysayers can only say it's hard to make a pellet-feeder but so what, if
> it's the only workable method? It could offer a >50% (?) weight reduction over
> conventional solid motor casings.
> The reduction in weight would allow slower speed to succeed in the shallow
> gravity well.
Hey! I have got a great idea. Lets improve the pellet fed rocket by
sutstituting kerosene and liquid oxygen for the solid fuel. The new
improved liquid fuel is much cheaper than the solid fuel. It has the
added bonus that it has a higher exhaust velocity. It's exhaust is also
more environmentally friendly than that of solid fuels. Finally,
handling procedures are extremely simple - look to your local airport
for the handling of large amounts of kerosene, and to your local
hospital for the handling of large amounts of liquid oxygen.
Instead of having to devise some sort of high capacity high power pellet
handling mechanism (the world's most sophisticated screw auger or ??) to
mechanically force pellets from the feed hopper into the high pressure
combustion chamber, we can use simple liquid pumps. The pumps can be
powered by turbines running on liquid rocket propellant, avoiding the
problem that pellet fed rockets face of having no convenient way to
generate the large amounts of power needed to force the pellets into the
combustion chamber. As an added bonus, kerosene can be used as a
regenerative coolant to make the engines remarkably more lightweight
than pellet-fed engines, which require ablative chamber and nozzle
linings. The liquid tanks won't be any larger than the pellet hoppers
they are replacing, as although the liquids are less dense than the
pellets, they can be stored without the penalty of empty space between
the pellets. As an added bonus, we can make the liquid propellant tanks
an integral part of the rocket structure by pressurizing them so that
they become rigid.
Yes indeed! Liquid propellant rockets are the wave of the future!
Navigaiter
be overcome more efficiently than the conventional wisdom has achieved. CATS
has been at a standstill because too few think inside the same old box.
<<The
way to minimize gravity losses, and hence fuel required, is rapid
acceleration (in a direction as close to horizontal as possible) followed
by coasting outward.>>
conventional solid rockets, a PF rocket has a small combustion chamber and its
fuel is contained in detachable lightweight tanks, not in a huge motor casing.
Thus, the rocket would have a great mass ratio and might be able to exit the
steep gravity gradient within one radius of earth more slowly than conventional
wisdom has heretofore dictated.
This method could reduce the instantaneous thrust requirements and allows a
more modestly powered engine which should be cheaper.
Navigaiter
how you do it, you still have to liberate a certain amount of energy
to lift a given mass to a given altitude.>>
*speed* of raising that mass.
High acceleration is not free. It costs *a lot* more energy than raising
something slowly. That's the advantage of a slow and steady escape from earth.
Gordon D. Pusch
> <<Velocity has never been considered the trap. Energy is. Regardless of
> how you do it, you still have to liberate a certain amount of energy
> to lift a given mass to a given altitude.>>
>
> The good old work equation. Note that energy required is a function of the
> *speed* of raising that mass.
No. As usual, your notion of physics is wrong. The change in potential
energy depends ONLY on the initial and final _radius_, and NOT AT ALL on
``speed.'' For a body in a free trajectory, the change in kinetic energy
is equal to the change in potential energy, because energy is conserved.
> High acceleration is not free. It costs *a lot* more energy than
> raising something slowly.
High acceleration may not be ``free,'' but those who have WORKED OUT THE MATH
(rather than using your armchair logic based on flawed notions of physics)
have proved beyond any doubt that it is _CHEAPER_ than low acceleration.
Crudely speaking, that is because you do not have to carry the fuel to
carry the fuel to carry the fuel to carry the fuel while you are _WASTING
FUEL_ by bloody taking your sweet time climbing out of the gravity well.
I know --- I've _personally_ set up and solved the optimal control problem
for a ``minimum fuel'' trajectory as an excercise. Like most optimal controls,
the solution turns out to be a ``Bang/Bang'' function: Burn your propellant
as fast as the structural limits of your engines, vehicle, and payload allow,
coast until you reach your target, burn as fast as you can again until you've
matched velocities with it, and end up with dry tanks.
Look --- I'll try one more time to show you the folly of your pseudo-logic.
Consider the following ``thought-experiment:'' Blindly pursue your strategy
of ``slow and steady'' to the ultimate degree, by thrusting at exactly the
strength of local gravity to maintain a constant, _infinitesimally small_
velocity. You will therefore burn your engines at 1g thrust for an INFINITE
amount of time, consume an INFINITE amount of fuel, and not move ONE MICRON
off the surface of the Earth. Can't you see just how stupid that policy is ???
> That's the advantage of a slow and steady escape from earth.
There is _NO SUCH ADVANTAGE_. Work out the math for yourself, instead of
attempting to browbeat us with sheer volume of flawed rhetoric. You will
NOT convince anyone who actually understands the physics involved, because
we already know you are wrong.
-- Gordon D. Pusch
perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;'
Henry Spencer
Navigaiter <navig...@aol.com> wrote:
><<The
>way to minimize gravity losses, and hence fuel required, is rapid
>acceleration (in a direction as close to horizontal as possible) followed
>by coasting outward.>>
>
> We were talking about a pellet fed rocket, not a conventional rocket.
This is independent of the type of rocket -- solid, liquid, nuclear,
electric, you name it.
>Unlike
>conventional solid rockets, a PF rocket has a small combustion chamber and its
>fuel is contained in detachable lightweight tanks, not in a huge motor casing.
Yes, if you set aside some awkward engineering problems, it actually has
some (not all) of the virtues of a liquid-fuel rocket. It's unlikely to
equal liquid-fuel performance, though.
>Thus, the rocket would have a great mass ratio and might be able to exit the
>steep gravity gradient within one radius of earth more slowly than conventional
>wisdom has heretofore dictated.
No.
Henry Spencer
Navigaiter <navig...@aol.com> wrote:
><<Velocity has never been considered the trap. Energy is. Regardless of
>how you do it, you still have to liberate a certain amount of energy
>to lift a given mass to a given altitude.>>
>
> The good old work equation. Note that energy required is a function of the
>*speed* of raising that mass.
Where did you get that ridiculous idea? The energy required (disregarding
some complications with sign conventions) is GMm/x - GMm/y, where x and y
are the starting and ending radii. That assumes no losses; losses are
minimal for rapid acceleration plus coasting, and quite high for slow
powered ascent.
> High acceleration is not free.
True, although not in the way you think -- it runs up the dry mass somewhat,
but the tradeoff has to be accepted, given the ghastly gravity losses of
low acceleration.
>It costs *a lot* more energy than raising something slowly.
No. See above. Please stop babbling and start calculating.
Bill Bonde
>
> >Unlike
> >conventional solid rockets, a PF rocket has a small combustion chamber and its
> >fuel is contained in detachable lightweight tanks, not in a huge motor casing.
>
> Yes, if you set aside some awkward engineering problems, it actually has
> some (not all) of the virtues of a liquid-fuel rocket. It's unlikely to
> equal liquid-fuel performance, though.
>
pellet-LOX rocket, you said you thought it would blow up. This was in a
thread about how to make fuel-oxidizer from Mar's atmosphere. Do you
still think this is really, really hard to do without blowing up?
Henry Spencer
>> Yes, if you set aside some awkward engineering problems, it actually has
>> some (not all) of the virtues of a liquid-fuel rocket. It's unlikely to
>> equal liquid-fuel performance, though.
>
>When I asked this about a carbon pellet-carbon monoxide or carbon
>pellet-LOX rocket, you said you thought it would blow up. This was in a
>thread about how to make fuel-oxidizer from Mar's atmosphere. Do you
>still think this is really, really hard to do without blowing up?
There are two separate issues here: pellet feed, and the particular solid
propellant used.
Pellet feed is a bit weird, and to me it looks seriously inferior to plain
old liquid fuel, but it's not inherently dangerous that I can see.
Almost any propellant which involves premixed LOX and fuel is a very
dangerous explosive.
Navigaiter
some complications with sign conventions) is GMm/x - GMm/y, where x and y are
the starting and ending radii>>
more *power* to raise a mass quickly than slowly.
So, if we halve the speed, we halve the power required to reach the same
altitude, in this case, one earth radius, 6400 km.
We also double our exposure time to a *rapidly* declining gravity field.
Since that field is declining as an inverse square function, we are getting
further away from the earth with less fuel than it would've taken to
over-accelerate inside the steepest gradient of the gravity well, [that
distance within 1/2 radius, 3200 km]
Thanks for helping the innumerate Navigaiter to think this idea through!
Gordon D. Pusch
> <<Where did you get that ridiculous idea? The energy required (disregarding
> some complications with sign conventions) is GMm/x - GMm/y, where x and y are
> the starting and ending radii>>
> I'm sorry. I meant to refer to the power equation. P=Work/time It takes
> more *power* to raise a mass quickly than slowly.
> So, if we halve the speed, we halve the power required to reach the same
> altitude, in this case, one earth radius, 6400 km.
And greatly increased the amount of propellant wasted in climbing to that
altitude, due to the increased burn time.
Since it is propellant capacity that is currently the limiting factor of
rockets, not engine power, you have just wasted something expensive to save
something cheap.
> We also double our exposure time to a *rapidly* declining gravity field.
> Since that field is declining as an inverse square function, we are getting
> further away from the earth with less fuel than it would've taken to
> over-accelerate inside the steepest gradient of the gravity well, [that
> distance within 1/2 radius, 3200 km]
No, your pseudo-logic is still wrong. If you do the math, you will find
that you have spent _MORE_ propellant, not less as you foolishly continue
to insist.
You clearly do not yet understand the difference between velocity and
acceleration. You _REALLY_ should learn to do the math.
> Thanks for helping the innumerate Navigaiter to think this idea
> through!
Since you are still getting the wrong answer, you have still failed to
think the problem through.
Learn to do the math, rather than deliberately wallowing in your ignorance.