Escape velocity

[hab...@anony.net]

This is a concept I find difficult to understand , why should
something need to be at 30,000mph or more to escape from earth.
What if the rocket was powered by a nuclear reactor. Could it
not climb at a sedate 20mph to reach the shuttle in a day and then
keep going at this rate to the moon and mars. Of course too slow for
humans but ok for satellites at few millions rather than the tens of
millions now?

[Androcles]

<hab...@anony.net> wrote in message
news:4ed194dc....@news.giganews.com...

It can, escape velocity only refers to unpowered ballistics.
Of course if you went to the Moon at 20 mph it will take rather a
long time to get there...
238857/20 = 11942.85 hours
= 497.61 days
= 1.36 years
The Apollo Moon shots took 3 days.

The faster you throw something away from the Earth the higher
it will go before it falls back. The higher it goes the weaker the
Earth's gravity will be to pull it back. So there is a speed (called the
escape velocity) which is fast enough for it never to return, even
though it gets slower and slower. Escaping the Earth does not mean
it escapes the Sun, though.

[ji...@specsol.spam.sux.com]

In sci.physics hab...@anony.net wrote:
> This is a concept I find difficult to understand , why should
> something need to be at 30,000mph or more to escape from earth.

That is because you are an uneducated idiot that is incapable of unerstanding
the simplest of scientific principals and why you have been posting cartoon
science to this newgroup for years.

> What if the rocket was powered by a nuclear reactor. Could it
> not climb at a sedate 20mph to reach the shuttle in a day and then
> keep going at this rate to the moon and mars. Of course too slow for
> humans but ok for satellites at few millions rather than the tens of
> millions now?

This was essentially the premise of the TV show "Salvage 1" starring
Andy Griffith back in the late 70's except they used "monohydrazine", a
fictiticous chemical that allowed a slow ascent to the moon.

Their first stage was a Texaco semitrailer with a cement mixer as the
capsule with controls from a Ford Mustang; exactly the kind of rediculous
stuff you post.

You would be better off if you were to throw away your computer, never post
again, and write plots for TV shows, though Salvage 1 only lasted for
2 seasons.



--
Jim Pennino

Remove .spam.sux to reply.

[hab...@anony.net]

Thanks Androceles , your reply has cleared up something I have been
thinking of for a long time. Jim has evaded the question with insults
and he is wrong . You do not need escape velocity and all the
fireworks and a tiny satellite on top of a monstrous rocket, all you
need is a nuclear reactor or some way to beam power to it , to lift
heavy objects to the space station and beyond.

[ji...@specsol.spam.sux.com]

In sci.physics hab...@anony.net wrote:
> Thanks Androceles , your reply has cleared up something I have been
> thinking of for a long time.

And most children learned before high school and anyone with the slightest
bit of common sense could have googled.

Google escape velocity - 2,540,000 hits.

You are still a babbling idiot.

[Peter Webb]

<hab...@anony.net> wrote in message
news:4ed1a46b....@news.giganews.com...

Or an escalator. Or a set of stairs.

[Peter Webb]

<hab...@anony.net> wrote in message
news:4ed1a46b....@news.giganews.com...

[Androcles]

<hab...@anony.net> wrote in message
news:4ed1a46b....@news.giganews.com...

It's not quite that easy. As Webb said, a flight of stairs will gain height,
but the ISS is continually falling back to Earth at 17,000 mph.

```````¬¬¬¬¬¬--------........ --->17,000 mph
_______________________ ground

However, the Earth is curved, so the ground gets lower.

```````¬¬¬¬¬¬--------........ ---> 17,000 mph

```````¬¬¬¬¬¬--------........ ground

The net result is the ISS goes all the way around.

The Moon is falling back too, but Earth's gravity at 239,000 miles away
is less, the Moon takes 29 days to go around once instead of the 45
minutes that the ISS takes.

If you launch a spacecraft from Earth it already has Earth's velocity
around the Sun, 18,500 mph, so to reach Mars you'll need to boost
in the Sun's gravity, and to reach Venus you slow it down and let it
fall toward the Sun. As it falls you let it gain speed to match Venus's
speed. Then you have to slow it again to enter orbit around Venus.

One little problem you'll have with your nuclear reactor idea is reaction
mass. The rocket works by Newton's third law, it throws matter
out of the tailpipe. You see that as a flame, but what is really happening
that huge orange tank of fuel attached to the shuttle is being thrown
backwards to make the shuttle go forward. Even if you use nuclear
power you still have to throw mass backwards. Now that could be
water, heated to steam, but eventually you run out of water.
The Harrier jump jet carries 50 gallons of water (500 lbs) for use
in the hover to use as reaction mass.
http://www.youtube.com/watch?v=-kDb99ftPlY

[and/or www.mantra.com/jai]

In article <OsmAq.31286$R36....@newsfe10.ams2>,
"Androcles" <Headm...@Hogwarts.physics.November.2011> posted:

Escape Velocity

http://hyperphysics.phy-astr.gsu.edu/hbase/vesc.html

Jai Maharaj, Jyotishi
Om Shanti

[hab...@anony.net]

Thanks again Androceles. What is the energy needed to get a one
tonne satellite up at 40,000km geosynchronus orbit and as the
satellite falls would the earth turn underneath it so it stays in
place ?.

Am I correct that mgh is the energy needed
ie 1000kg x 10x 40m meters
ie 10exp3 x 10 exp1 x 10exp6 x 4 Joules
4 x 10exp10 joules
or since 1 kwh is 4x 10exp6 joules this is approximately
10exp4 kwh ie 10,000kwh

and at 10 cents a kwh
its just 100,000 cents
or $1,000 !!

So the energy cost of lifting a one tonne satellite into
geosyncronous orbit is just $1000 , as Jim can verify.

So if we can get a nuclear power reactor , which can move aircraft
carriers unfuelled for years , miniaturised, can we not lift a one
tonne satellite into geo orbit for a few million dollars at the most,
and later on solar panels to supply us with free energy for ever!

The kinetic energy costs would be at say 100km/hr or 2km/min or
2000m/min or 40m/sec
m x vexp2
or 10exp3 kg x 40x40 joules
approx 10exp3 x 10exp3 x 2
2 x 10exp6 joules or about 1 unit of electricity ! Even if the
calculations are out by a thousand fold its only about $10 worth.

So instead of sending satellites via expensive chemical rockets ,
lets use nuclear power and some kind of gas or liquid that can be
heated by radioactivity .

[Androcles]

<hab...@anony.net> wrote in message
news:4ed241d0....@news.giganews.com...

| Thanks again Androceles. What is the energy needed to get a one
| tonne satellite up at 40,000km geosynchronus orbit and as the
| satellite falls would the earth turn underneath it so it stays in
| place ?.
|
| Am I correct that mgh is the energy needed
| ie 1000kg x 10x 40m meters

What?
Oh, I see, 40 million... (40m is 40 meters, you need to be careful
with your abbreviations).
Mass times height isn't anything meaningful, energy is 1/2 mv^2
where 'm' means mass and v is velocity. To avoid any ambiguity
never write m to mean million.

| ie 10exp3 x 10 exp1 x 10exp6 x 4 Joules

You are in trouble here, to.
10^3 is 1000 where the exponent is 3, but exp 3 can be read as
exp(3) which is e^3 = 2.71828^3 ~= 20 where e is the base of
the natural logarithm. Better to write 10.0E+03, which is 1.0E+04
If you have Windows calculator set to scientific mode you can play
with the F-E button (floating point to exponential format) and
the Exp button.

[ji...@specsol.spam.sux.com]

In sci.physics hab...@anony.net wrote:
> Thanks again Androceles. What is the energy needed to get a one
> tonne satellite up at 40,000km geosynchronus orbit and as the
> satellite falls would the earth turn underneath it so it stays in
> place ?.

Oh goodie, two mental defectives are going to stroke each other.

> Am I correct that mgh is the energy needed
> ie 1000kg x 10x 40m meters
> ie 10exp3 x 10 exp1 x 10exp6 x 4 Joules
> 4 x 10exp10 joules
> or since 1 kwh is 4x 10exp6 joules this is approximately
> 10exp4 kwh ie 10,000kwh

No, you are never correct about anything as you are a babbling idiot.

> and at 10 cents a kwh
> its just 100,000 cents
> or $1,000 !!
>
> So the energy cost of lifting a one tonne satellite into
> geosyncronous orbit is just $1000 , as Jim can verify.

This is just so moronic it is beyond comment.

> So if we can get a nuclear power reactor , which can move aircraft
> carriers unfuelled for years , miniaturised, can we not lift a one
> tonne satellite into geo orbit for a few million dollars at the most,
> and later on solar panels to supply us with free energy for ever!

So if pigs had wings they could fly.

> The kinetic energy costs would be at say 100km/hr or 2km/min or
> 2000m/min or 40m/sec
> m x vexp2
> or 10exp3 kg x 40x40 joules
> approx 10exp3 x 10exp3 x 2
> 2 x 10exp6 joules or about 1 unit of electricity ! Even if the
> calculations are out by a thousand fold its only about $10 worth.

Yet more babble that is just so moronic it is beyond comment.

> So instead of sending satellites via expensive chemical rockets ,
> lets use nuclear power and some kind of gas or liquid that can be
> heated by radioactivity .

Sort of like Heinlein's torchship of the 50's.

Habshi's sci-fi wet dreams have caught up to the 50's.

[Marvin the Martian]

Try integrating the force over the distance to get the work, and then
setting the work to the kinetic energy at the surface of the earth and
solve for the velocity.

Work = Integral from r_earth to infinity of G*M/r^2
= G*M/r_earth = 6.6E-11 * 6.0E24 /6.3E6 = 6.3E7 Joules/kg
k.e. = 6.3 E7 Joules/Kg = 1/2 V^2
=> V = sqrt(6.3E7 *2 ) = 11200 m/s

Check;
http://en.wikipedia.org/wiki/Escape_velocity

[CWatters]

You could do that but work out how much fuel you need to reach a given
altitude. At 20mph you need to burn fuel just to maintain altitude let
alone increase it.

At >25000mph a spacecraft doesn't need anymore Kinetic energy to escape
from earths gravity.

http://en.wikipedia.org/wiki/Escape_velocity

[hab...@anony.net]

Try integrating the force over the distance to get the work, and then
setting the work to the kinetic energy at the surface of the earth and

solve for the velocity.

Work = Integral from r_earth to infinity of G*M/r^2
>> = G*M/r_earth = 6.6E-11 * 6.0E24 /6.3E6 = 6.3E7 Joules/kg
k.e. = 6.3 E7 Joules/Kg = 1/2 V^2
=> V = sqrt(6.3E7 *2 ) = 11200 m/s

6.3E7 joules per kg ie 6E10 joules per tonne <<

lets says the energy needed from above is 10E10 joules/kg or 10E4 kwh
or 10,000 units of electricity and at a cost of 10cents a unit its
100,000 cents or $1000 , confirming the figures I gave , ie it needs
just $1000 of electric energy to put a tonne of satellite into
space!!!

The second bit is the escape velocity for ballistic missiles but if
the speed was limited to say 100km/hr for a nuclear rocket , which
would still get a one tonne supply to the space station within a day,
and to geosync orbit in about a month then the kinetic energy needed
is half 1000kg 100km/hr squared
or half 10E3 x 2km/min squared
or 10E3 x 1000meters/min squared
or 10E3 x 20meters/second squared
or 10E3 x 400joules
or 10E3 x 10E3 joules
or 10E6 joules , or since one kwh is 3x10E6 joules , its about 3 kwh
!!
or 30 cents worth of electricity. Again somebody please check these
figures.

The reason chemical rockets are so energy inefficient is that the
most of the energy is wasted in lifting the fuel the first ten km or
so and then its burnt away.
So we do need some nuclear reactor, ?what is the smallest size one
can be , or some means of laser transmission of energy to a risking
rocket.

[ji...@specsol.spam.sux.com]

Babbling, comicbook science moron.

You are still an idiot.

[G=EMC^2]

Best to keep in mind an object was shot into orbit with just one
single blast out of a cannon. That was done over 35 years ago. If
interested I will tell you what was needed and how it was done.
TreBert

[G=EMC^2]

On Nov 27, 7:52 pm, hab...@anony.net wrote:

Time to have a base on the Moon and use a rail gun. TreBert

[Arindam Banerjee]

On Nov 27, 12:42 pm, hab...@anony.net wrote:
>         This is a concept I find difficult to understand , why should
> something need to be at 30,000mph or more to escape from earth.

When you are orbiting the Earth, you must have a certain tangential
speed with respect to the Earth you are orbiting.
Consider the formula
mg=mMG/R^2=mVV/R
for the forces acting upon such an orbiting body.
mg is the gravitational force, pulling it down.
It is equal to the attraction between the earth and the satellite.
If the orbital radius does not change, then the gravity force has to
be equal to the so-called centrifugal force, which is mVV/R
The centrifugal force is a virtual force - it really does not exist,
but appears to. It is the force apparently keeping the satellite from
falling into the earth.
Why it is not falling into the earth? For the same reason that the
earth is not falling into the sun.
Meaning,
the satellite is always falling into the earth, just as the earth is
always falling into the sun.
But the radial distance is always remaining the same because the
satellite has a tangential velocity, just as the Earth.
Meaning, that over the time the earth/satellite has fallen into the
Sun/earth, the earth/satellite with its velocity has also moved
sideways.
The net effect is that moving both radially and sideways at the same
time, it has moved at an angle such that the radius always remains
constant.
Of course, the orbit need not be circular. Then what happens is that
the speed V is not constant over the entire path. It increases as R
decreases. Comets like Halley;s Comet are like that. Thus the angle
of attack is very important in orbiting. If it is 90 deg to the
surface it will orbit at uniform speed. If it is less, then the speed
has to be increased if we want a fast low trajectory orbit over any
space on earth. The Soviets used such low attitude satellites for
electronic communication, as opposed to geostat satellites.

>         What if the rocket was powered by a nuclear reactor. Could it
> not climb at a sedate 20mph to reach the shuttle in a day and then
> keep going at this rate to the moon and mars. Of course too slow for
> humans but ok for satellites at few millions rather than the tens of
> millions now?

Ridiculous. But it does show the quality of mind at the top money-
granting levels, preventing new technologies.

Cheers,
Arindam Banerjee

[Richard Tobin]

In article <4ed194dc....@news.giganews.com>, <hab...@anony.net> wrote:

> This is a concept I find difficult to understand , why should
>something need to be at 30,000mph or more to escape from earth.
> What if the rocket was powered by a nuclear reactor. Could it
>not climb at a sedate 20mph to reach the shuttle in a day and then
>keep going at this rate to the moon and mars.

As you now know, it is not essential to reach such a speed. Why then
do rockets work like that? It's just the most efficient way to do it
with a chemical fuel; if the rocket went steadily it would have to
transport more of the fuel up against the force of gravity. It's
better to use up the heavy fuel as quickly as possible.

A nuclear reactor does indeed change the calculation, in that it
doesn't use up a large mass of fuel as it accelerates.

-- Richard

[Marvin the Martian]

You guys are not familiar with rocket science 101 are you?

Going up slowly will require a LOT more energy to get to escape velocity.

[PD]

On 11/26/2011 7:42 PM, hab...@anony.net wrote:
> This is a concept I find difficult to understand , why should
> something need to be at 30,000mph or more to escape from earth.

Habshi, first look at what it means. It means the velocity required to
escape WITHOUT continuous firing of an engine. It means the speed it
would require to *coast* and still escape the the Earth.

[Richard Tobin]

In article <p-idnZE4wO3fJE7T...@giganews.com>,

Marvin the Martian <mar...@ontomars.org> wrote:

>Going up slowly will require a LOT more energy to get to escape velocity.

Youe sentence is unclear. Obviously if you go up slowly you don't
have to get to the speed that would be escape velocity at the surface.

The question is not whether going up slowly requires more energy, but
whether it is practical to do it if you have a motor which doesn't
work by having a huge chemical mass that gets burned up.

-- Richard

[Chris Richardson]

On Sun, 27 Nov 2011 01:42:29 +0000, habshi wrote:

> What if the rocket was powered by a nuclear reactor. Could it
> not climb at a sedate 20mph
>

20 mph with respect to what?

For describing space flight, a coordinate system relative to the
fixed stars is ordinarily used. This coordinate system is usually
made to coincide with the center of the earth at t=0.

With this coordinate system in place, at t=0, the spacecraft
already has an initial velocity due to both the earth's rotation
and the earth's revolution about the sun. It would therefore
require tremendous energy to counteract these initial velocities
and to cruise at a sedate 20 mph through this coordinate system.

I am too lazy to do calculations, but even nuclear propulsion
may not provide sufficient energy to navigate the gravitational
fields involved at the rates and trajectories which you propose.

The way things are done now is perhaps the most efficient.
Rockets are not designed to achieve escape velocity; they
are designed to achieve low earth orbit (LEO). This is done by
staging. The first stage, or stages, are meant to reach
the proper altitude and attitude. Then, a subsequent stage
is used to attain the proper velocity for earth orbit at that
particular altitude. From the initial LEO, other orbits can be
attained by Hoffman orbit transfer maneuvers, and these other
orbits can be either around the earth or around the sun.

One cannot just "cruise" at will through the solar system.
There are powerful gravitational fields and initial velocities
to consider. Energy is the key, but it is impossible, even
with nuclear propulsion, to construct a small space craft
(of human dimensions) to provide that energy.

[Arindam Banerjee]

On Nov 29, 7:27 am, Chris Richardson <r...@localhost.localdomain>
wrote:

So far so good, with old technology. Rockets are for fireworks. The
conquest of space will require much better internal force engines than
the first generation of same, namely rockets.

An internal force engine operates on the principle of having no
backward reaction to forward action. Continuous action leads to
continous acceleration. A modified electromagnetic rail gun is the
basis for this new engine. It will operate on an upgraded physics,
based upon the formula e=0.5mVVN(N-k) that I have been talking about
for the past few weeks on the ngs.

Details were presented by myself in my book "To the Stars!" published
in Internet in 2000. I wish it had been taken more seriously then.
Still, better late than never.

Cheers,
Arindam Banerjee

[Marvin the Martian]

On Mon, 28 Nov 2011 19:13:25 +0000, Richard Tobin wrote:

> In article <p-idnZE4wO3fJE7T...@giganews.com>, Marvin the
> Martian <mar...@ontomars.org> wrote:
>
>>Going up slowly will require a LOT more energy to get to escape
>>velocity.
>
> Youe sentence is unclear. Obviously if you go up slowly you don't have
> to get to the speed that would be escape velocity at the surface.

The work done is the same, sure enough.

But the power of the rocket is 1/2* dm/dt * v^2, while the thrust is
T=v*dm/dt. Your net upward acceleration T/M_rocket - g (for small
displacements from the surface of the earth).

Even if you're standing still, you're using a considerable amount of
power just to fight gravity.

> The question is not whether going up slowly requires more energy, but
> whether it is practical to do it if you have a motor which doesn't work
> by having a huge chemical mass that gets burned up.

Have fun doing your pretend physics.

[Bill Snyder]

On Mon, 28 Nov 2011 21:09:53 -0600, Marvin the Martian

<mar...@ontomars.org> wrote:

>On Mon, 28 Nov 2011 19:13:25 +0000, Richard Tobin wrote:
>
>> In article <p-idnZE4wO3fJE7T...@giganews.com>, Marvin the
>> Martian <mar...@ontomars.org> wrote:
>>
>>>Going up slowly will require a LOT more energy to get to escape
>>>velocity.
>>
>> Youe sentence is unclear. Obviously if you go up slowly you don't have
>> to get to the speed that would be escape velocity at the surface.
>
>The work done is the same, sure enough.
>
>But the power of the rocket is 1/2* dm/dt * v^2, while the thrust is
>T=v*dm/dt. Your net upward acceleration T/M_rocket - g (for small
>displacements from the surface of the earth).
>
>Even if you're standing still, you're using a considerable amount of
>power just to fight gravity.

How much power does a lump of lead sitting on the ground use to
"fight gravity," retardo? How much power does a helium balloon
use to "fight gravity?" Where does all this power come from? How
do lead weights and balloons magically repeal the laws of
thermodynamics?

Moron.

--
Bill Snyder [This space unintentionally left blank]

[Richard Tobin]

In article <aZOdnZW4nfuc0EnT...@giganews.com>,

Marvin the Martian <mar...@ontomars.org> wrote:

>Even if you're standing still, you're using a considerable amount of
>power just to fight gravity.

Yes, obviously. There's no need to pretend that other people don't
realise this.

But a chemical rocket has the *additional* consideration that you use
up most of the mass you are lifting, and can therefore save energy by
using it early. You seem to want to ignore this point and just go on
stating the obvious.

-- Richard

[Richard Tobin]

In article <0321e9f0-b96c-44fe...@20g2000prn.googlegroups.com>,

Arindam Banerjee <adda...@bigpond.com> wrote:

>It will operate on an upgraded physics,

ITYM "imaginary".

-- Richard

[Richard Tobin]

In article <jb2g43$23gf$1...@matchbox.inf.ed.ac.uk>, I wrote:

>But a chemical rocket has the *additional* consideration that you use
>up most of the mass you are lifting, and can therefore save energy by
>using it early.

On reflection, this may not be very significant, because a non-chemical
rocket will need something else to use as reaction mass, which will
have the same consideration. I suppose it depends whether it can
use significantly less mass than a chemical rocket (by ejecting it
at a faster average speed).

-- Richard

[jbriggs444]

On Nov 29, 6:36 am, rich...@cogsci.ed.ac.uk (Richard Tobin) wrote:
> In article <aZOdnZW4nfuc0EnTnZ2dnUVZ5qedn...@giganews.com>,

The delta-v available from a rocket is fixed regardless of how quickly
or how slowly you expend the fuel store. The rocket equation does
not have a term for time.

delta-v = exhaustvelocity * ( ln(initialmass/finalmass))

A simplistic understanding based on this equation, is that you save
energy by not dawdling deep in a gravity well spending delta-v to
fight gravity.

A slightly more sophisticated analysis would note that the energy
provided to the payload is highest if the delta-v is applied when
the craft is moving fastest and when the delta-v is applied in the
direction of the craft's existing motion. That is the effect that
is used in a gravitational slingshot. It is also the effect that
motivates an eastward launch angle and a trajectory that
does, in fact, "dawdle" deep in a gravity well.

If it were not for the solidity of the earth and the viscosity
of the pesky atmosphere, the optimum launch angle for
a rocket-powered craft would be below the horizontal.

[John Polasek]

On Sun, 27 Nov 2011 01:42:29 GMT, hab...@anony.net wrote:

> This is a concept I find difficult to understand , why should
>something need to be at 30,000mph or more to escape from earth.

> What if the rocket was powered by a nuclear reactor. Could it

>not climb at a sedate 20mph to reach the shuttle in a day and then
>keep going at this rate to the moon and mars. Of course too slow for
>humans but ok for satellites at few millions rather than the tens of
>millions now?

If you dropped something from a great distance to the earth it would
arrive with what is called escape velocity, about 17000 mph. If it
then somehow reflected, it would bounce and leave at escape velocity
and go back to its original position, all without further
intervention.
A moon rocket operating for a few minutes and rising a few miles from
the earth (4000mi radius) would still have to reach nearly escape
velocity to overcome the earths gravity and coast to the moon. That
few minutes approximates a step function of velocity on a 250,000 mile
trip.

[Marvin the Martian]

On Tue, 29 Nov 2011 11:36:35 +0000, Richard Tobin wrote:

> In article <aZOdnZW4nfuc0EnT...@giganews.com>, Marvin the
> Martian <mar...@ontomars.org> wrote:
>
>>Even if you're standing still, you're using a considerable amount of
>>power just to fight gravity.
>
> Yes, obviously. There's no need to pretend that other people don't
> realise this.

Yet, "other people" are saying they can escape from the earth gravity at
20 mph and that there is no need to reach escape velocity.

Apparently, it does need to be said. Unfortunately, you don't understand.

That's okay. I was just thinking of some of the stupid things I said when
I was a physic undergrad.

> But a chemical rocket has the *additional* consideration that you use up
> most of the mass you are lifting, and can therefore save energy by using
> it early. You seem to want to ignore this point and just go on stating
> the obvious.

I'm sorry you're an idiot, and even more sorry that you post here.

[Marvin the Martian]

On Tue, 29 Nov 2011 05:02:10 -0800, jbriggs444 wrote:

> On Nov 29, 6:36 am, rich...@cogsci.ed.ac.uk (Richard Tobin) wrote:
>> In article <aZOdnZW4nfuc0EnTnZ2dnUVZ5qedn...@giganews.com>, Marvin the
>> Martian  <mar...@ontomars.org> wrote:
>>
>> >Even if you're standing still, you're using a considerable amount of
>> >power just to fight gravity.
>>
>> Yes, obviously.  There's no need to pretend that other people don't
>> realise this.
>>
>> But a chemical rocket has the *additional* consideration that you use
>> up most of the mass you are lifting, and can therefore save energy by
>> using it early.  You seem to want to ignore this point and just go on
>> stating the obvious.
>
> The delta-v available from a rocket is fixed regardless of how quickly
> or how slowly you expend the fuel store. The rocket equation does not
> have a term for time.
>
> delta-v = exhaustvelocity * ( ln(initialmass/finalmass))

That's assuming no gravity. Which is where you're going wrong.

But you know, I don't give a shit anymore. Prattle on!

[jbriggs444]

On Nov 29, 11:03 am, Marvin the Martian <mar...@ontomars.org> wrote:
> On Tue, 29 Nov 2011 05:02:10 -0800, jbriggs444 wrote:
> > On Nov 29, 6:36 am, rich...@cogsci.ed.ac.uk (Richard Tobin) wrote:
> >> In article <aZOdnZW4nfuc0EnTnZ2dnUVZ5qedn...@giganews.com>, Marvin the
> >> Martian  <mar...@ontomars.org> wrote:
>
> >> >Even if you're standing still, you're using a considerable amount of
> >> >power just to fight gravity.
>
> >> Yes, obviously.  There's no need to pretend that other people don't
> >> realise this.
>
> >> But a chemical rocket has the *additional* consideration that you use
> >> up most of the mass you are lifting, and can therefore save energy by
> >> using it early.  You seem to want to ignore this point and just go on
> >> stating the obvious.
>
> > The delta-v available from a rocket is fixed regardless of how quickly
> > or how slowly you expend the fuel store.  The rocket equation does not
> > have a term for time.
>
> >     delta-v = exhaustvelocity * ( ln(initialmass/finalmass))
>
> That's assuming no gravity.

Yes.

> Which is where you're going wrong.

No.

> But you know, I don't give a shit anymore. Prattle on!

Ok, don't bother reading what you respond to.

[Matthew Lybanon]

In article <4ed194dc....@news.giganews.com>, hab...@anony.net

wrote:

> This is a concept I find difficult to understand , why should
> something need to be at 30,000mph or more to escape from earth.
> What if the rocket was powered by a nuclear reactor. Could it
> not climb at a sedate 20mph to reach the shuttle in a day and then
> keep going at this rate to the moon and mars. Of course too slow for
> humans but ok for satellites at few millions rather than the tens of
> millions now?

Escape velocity is that velocity such that an object moving WITHOUT
power (that is, falling) can get infinitely far from Earth. If you
throw something up, gravity causes it to slow down and fall back to the
ground. If you throw it upward with a higher speed, it can go higher.
If you could throw it with escape (or higher) velocity, it would never
come back.

But if a rocket motor keeps operating, that's an entirely different
situation.

[hab...@anony.net]

My point is that the amount of energy needed to escape earth
gravity is very small , less than $1000 worth for a one tonne
satellite.
Could it be that MOST of the energy is used to fight friction of the
air , that is why they punch through vertically as fast as possible?
and that this friction increases with speed, so that a nuclear powered
rocket rising at a sedate 100km/hr would use much much less?
Is that why the rocket to get off the moon is so small, because
there is no air on the moon?

[Androcles]

<hab...@anony.net> wrote in message
news:4ed6c7bf....@news.giganews.com...

Gravity : 9.8 metres per second /sec at the Earth's surface,
R = radius of Earth = 6 378.1 kilometres.
What is the gravity at
A: the top of Everest?
B: 300 km, the altitude of the ISS?
C: 384403 km, the altitude of the Moon?
Use the inverse square law, 1/R^2.

[ji...@specsol.spam.sux.com]

In sci.physics hab...@anony.net wrote:
> My point is that the amount of energy needed to escape earth
> gravity is very small , less than $1000 worth for a one tonne
> satellite.

The only point you have is the one on top of your head.

<snip ignorant nonsense>

[Arindam Banerjee]

On Nov 29, 10:40 pm, rich...@cogsci.ed.ac.uk (Richard Tobin) wrote:
> In article <0321e9f0-b96c-44fe-9fd6-d84fd3134...@20g2000prn.googlegroups.com>,

> Arindam Banerjee  <adda1...@bigpond.com> wrote:
>
> >It will operate on an upgraded physics,
>
> ITYM "imaginary".

Nothing could be more wrong than einsteinian physics, so dearly and
wrongly believed in.
>
> -- Richard

[Arindam Banerjee]

Main point with a rocket is that the centre of mass of the whole thing
stays exactly in the same place. So, the further you want a craft to
go, the more mass has to be there to make it go.

[PD]

Fascinating that you think a small nuclear reactor costing less than
$1000 could raise something slowly but inexorably upwards. What
PRECISELY is the lift mechanism that you propose?

[hab...@anony.net]

Inspired air while in the atmosphere for an hour if rising at
100km/hr.
Then stored air heated by the nuclear power. What is the
mechanism on curiosity rover launched this week and powered by nuclear
on mars?

[PD]

On 12/1/2011 6:01 PM, hab...@anony.net wrote:
> Inspired air while in the atmosphere for an hour if rising at
> 100km/hr.

Good. Now do a calculation of the mass of the air that could be inspired
in that trip. This isn't hard: Remember that atmospheric pressure at
ground level, multiplied by the cross-sectional area of a column, is
equal to the weight of the entire atmospheric air column sitting on that
area.

> Then stored air heated by the nuclear power.

Now calculate how long that inspired air will last ejected as propellant
once you get to the top of the atmosphere.

> What is the
> mechanism on curiosity rover launched this week and powered by nuclear
> on mars?

Not inspired air. Also note that the Curiosity rover only has to
overcome (low) surface friction on the surface of Mars, and doesn't have
to fight gravity vertically.

If you think this shouldn't matter, then I invite you to take a look at
how V-8 powered cars that have no problem moving around on horizontal
ground have great difficulty climbing steep hills:
http://www.youtube.com/watch?v=cVAGQsEFjL0