| Topic: |
Science > Physics |
| User: |
"Jean" |
| Date: |
09 Dec 2003 08:08:47 PM |
| Object: |
Escape Velocity |
It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
Suppose you place a ladder to the moon ..
Can you climb it and reach the moon without attaining any escape
velocity??
Waiting for a comment through google..
(Please don't email me)
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| User: "Franz Heymann" |
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| Title: Re: Escape Velocity |
12 Dec 2003 05:32:42 AM |
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"Jean" <alertjean@rediffmail.com> wrote in message
news:4138e33e.0312091808.10c09282@posting.google.com...
It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
Suppose you place a ladder to the moon ..
Can you climb it and reach the moon without attaining any escape
velocity??
Yes. But the energy you will have expended will be equal (assuming 100%
efficiency) to the kinetic energy you will need if you start off with the
escape velocity and no ladder. (Assuming the moon to be infinitely far away
and neglecting its own gravitational field.)
[snip]
Franz Heymann
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| User: "Big Bird" |
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| Title: Re: Escape Velocity |
12 Dec 2003 07:02:38 PM |
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(Jean) wrote in message news:<4138e33e.0312091808.10c09282@posting.google.com>...
It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
Suppose you place a ladder to the moon ..
Can you climb it and reach the moon without attaining any escape
velocity??
Waiting for a comment through google..
(Please don't email me)
One page physics:
http://p.zapto.org/sven/physics/5b598040-a3f7-4373-a858-a711b5ce657f.pdf
(Unfortunately the search engine for this seems to have been closed for now...)
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| User: "S. Enterprize Company" |
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| Title: Re: Escape Velocity |
11 Dec 2003 12:54:23 AM |
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It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
With a theoretical Smart Model Anti-Gravity Saucer, you don't need escape
velocity. It works on the basic principles of the Meissner Effect.
Basic Smart Model Saucer
------- <--- people in saucer
======== magnet
============== SuperConductor
================ Static Electricity
Generator
The Static Electricity Generator generates free electron flow. This is fed
to the Superconductor Field which grows with more free electrons. The attached
magnet tries to move upward pulling the superconductor and static generator
upward. The result theoretically should be anti-gravity and the saucer hovers.
The stronger the superconductor field the higher the altitude reached.
Smart's Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
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| User: "The Ghost In The Machine" |
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| Title: Re: Escape Velocity |
11 Dec 2003 11:00:33 PM |
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In sci.physics, S. Enterprize Company
<smart1234@aol.com>
wrote
on 11 Dec 2003 06:54:23 GMT
<20031211015423.21755.00000431@mb-m26.aol.com>:
It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
With a theoretical Smart Model Anti-Gravity Saucer, you don't need escape
velocity. It works on the basic principles of the Meissner Effect.
You've built one of these, right?
[rest snipped]
--
#191, -- insert random disbelief here
It's still legal to go .sigless.
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| User: "S. Enterprize Company" |
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| Title: Re: Escape Velocity |
11 Dec 2003 11:55:06 PM |
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In sci.physics, S. Enterprize Company
<smart1234@aol.com>
wrote
on 11 Dec 2003 06:54:23 GMT
<20031211015423.21755.00000431@mb-m26.aol.com>:
It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
With a theoretical Smart Model Anti-Gravity Saucer, you don't need
escape
velocity. It works on the basic principles of the Meissner Effect.
You've built one of these, right?
[rest snipped]
The Big Bang occurred didn't it, you built one right?
Smart's Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
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| User: "Gregory L. Hansen" |
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| Title: Re: Escape Velocity |
09 Dec 2003 08:41:29 PM |
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In article <4138e33e.0312091808.10c09282@posting.google.com>,
Jean <alertjean@rediffmail.com> wrote:
It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
Suppose you place a ladder to the moon ..
Can you climb it and reach the moon without attaining any escape
velocity??
Yes. But pack a few Power Bars and a portable urinal, the climb will take
you a while.
Escape velocity is what you need when you want to leave a planet forever
with no additional propulsion. If you have enough propulsion you can
leave at any speed you like.
--
Irony: "Small businesses want relief from the flood of spam clogging their
in-boxes, but they fear a proposed national 'Do Not Spam' registry will
make it impossible to use e-mail as a marketing tool."
http://www.bizjournals.com/houston/stories/2003/11/10/newscolumn6.html
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| User: "Jim Roberts" |
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| Title: Re: Escape Velocity |
09 Dec 2003 11:03:01 PM |
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"Gregory L. Hansen" <glhansen@steel.ucs.indiana.edu> wrote in message
news:br614p$31d$4@hood.uits.indiana.edu...
In article <4138e33e.0312091808.10c09282@posting.google.com>,
Jean <alertjean@rediffmail.com> wrote:
It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
Suppose you place a ladder to the moon ..
Can you climb it and reach the moon without attaining any escape
velocity??
Yes. But pack a few Power Bars and a portable urinal, the climb will take
you a while.
Escape velocity is what you need when you want to leave a planet forever
with no additional propulsion. If you have enough propulsion you can
leave at any speed you like.
--
Irony: "Small businesses want relief from the flood of spam clogging their
in-boxes, but they fear a proposed national 'Do Not Spam' registry will
make it impossible to use e-mail as a marketing tool."
http://www.bizjournals.com/houston/stories/2003/11/10/newscolumn6.html
Exactly. As you get further away from the Earth's center of mass, the escape
velocity decreases. Ve=(2GM/R)^.5 Of course, it doesn't change
significantly in LEOs as compared to the surface.
Regards,
Jim
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| User: "" |
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| Title: Re: Escape Velocity |
09 Dec 2003 11:04:42 PM |
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In article <4138e33e.0312091808.10c09282@posting.google.com>, (Jean) writes:
It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
Suppose you place a ladder to the moon ..
Can you climb it and reach the moon without attaining any escape
velocity??
Sure.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Sam Wormley" |
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| Title: Re: Escape Velocity |
09 Dec 2003 08:12:27 PM |
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Jean wrote:
It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
Escape Velocity
http://scienceworld.wolfram.com/physics/EscapeVelocity.html
Suppose you place a ladder to the moon ..
Can you climb it and reach the moon without attaining any escape
velocity??
Waiting for a comment through google..
(Please don't email me)
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| User: "Uncle Al" |
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| Title: Re: Escape Velocity |
10 Dec 2003 09:52:16 AM |
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Jean wrote:
It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
Take a stationary test mass at infinity and allow it to drop onto the
Earth (no atmosphere, no rotation), integral of a=MG/r^2. The impact
velocity is the escape velocity by symmetry. The first principles
math is trivial given the surface gravity,
Suppose you place a ladder to the moon ..
Can you climb it and reach the moon without attaining any escape
velocity??
Indian rope trick/"Fountains of Paradise"/"Mouse That Roared, II".
That is not the definition of escape velocity. Escape velocity is the
minimum initial impulse velocity at and normal to the surface (no
planetary atmosphere, no rotation) necessary to launch a test mass to
infinity
Waiting for a comment through google..
Use Google and look up the math - and the definition. If you do not
know what you are talking about, how do you know when you get the
desired answer?
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
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| User: "John Zinni" |
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| Title: Re: Escape Velocity |
10 Dec 2003 10:49:11 AM |
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"Uncle Al" <UncleAl0@hate.spam.net> wrote in message
news:3FD74130.52B03DC5@hate.spam.net...
Indian rope trick/"Fountains of Paradise"/"Mouse That Roared, II".
That is not the definition of escape velocity. Escape velocity is the
minimum initial impulse velocity at and normal to the surface (no
planetary atmosphere, no rotation) necessary to launch a test mass to
infinity
Hi Uncle Al
It does not need to be normal to the surface. Any old direction will do, it
will just take a slightly less direct route to infinity.
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| User: "John Zinni" |
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| Title: Re: Escape Velocity |
10 Dec 2003 12:02:10 PM |
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"John Zinni" <j_zinni@NOCRAP.sympatico.ca> wrote in message
news:g8IBb.2336$aF2.299726@news20.bellglobal.com...
"Uncle Al" <UncleAl0@hate.spam.net> wrote in message
news:3FD74130.52B03DC5@hate.spam.net...
Indian rope trick/"Fountains of Paradise"/"Mouse That Roared, II".
That is not the definition of escape velocity. Escape velocity is the
minimum initial impulse velocity at and normal to the surface (no
planetary atmosphere, no rotation) necessary to launch a test mass to
infinity
Hi Uncle Al
It does not need to be normal to the surface. Any old direction will do,
it
will just take a slightly less direct route to infinity.
A minor, somewhat interesting side note:
In my copy of Halliday & Resnick there is an entry in the index for "escape
velocity" only, however in the text that it points to, the authors never use
the term "escape velocity" and only use the term "escape speed".
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| User: "" |
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| Title: Re: Escape Velocity |
10 Dec 2003 11:53:51 AM |
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In article <3FD74130.52B03DC5@hate.spam.net>, Uncle Al <UncleAl0@hate.spam.net> writes:
Indian rope trick/"Fountains of Paradise"/"Mouse That Roared, II".
That is not the definition of escape velocity. Escape velocity is the
minimum initial impulse velocity at and normal to the surface (no
planetary atmosphere, no rotation) necessary to launch a test mass to
infinity
The direction is irrelevant. Normal to the surface, tangent to
the surface or anywhere in between will do. For sufficiently
permeable "surfaces", you can even launch straight down. The
required speed to escape is the same.
John Briggs
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| User: "Uncle Al" |
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| Title: Re: Escape Velocity |
10 Dec 2003 12:15:56 PM |
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wrote:
In article <3FD74130.52B03DC5@hate.spam.net>, Uncle Al <UncleAl0@hate.spam.net> writes:
Indian rope trick/"Fountains of Paradise"/"Mouse That Roared, II".
That is not the definition of escape velocity. Escape velocity is the
minimum initial impulse velocity at and normal to the surface (no
planetary atmosphere, no rotation) necessary to launch a test mass to
infinity
The direction is irrelevant. Normal to the surface, tangent to
the surface or anywhere in between will do. For sufficiently
permeable "surfaces", you can even launch straight down. The
required speed to escape is the same.
Point taken. Absent rotation the straight through approach is
especially elegant. Uncle Al will immediately write for funding to
stop the Earth's rotation and bore through the core. Studies msut be
funded before the Chinese steal our thunder!
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
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| User: "G=EMC^2 Glazier" |
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| Title: Re: Escape Velocity |
10 Dec 2003 01:33:00 PM |
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If the initial upward velocity is greater than a critical value called
the "escape" velocity,gravity will never be strong enough to stop the
partical,and it will go off in space. The escape velocity is about 12
kilometers per second for the Earth,and about 618 kiliometers per second
for the sun. Even with the speed of 300,000 kilometers per second light
can't get out of the BH. Bert
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| User: "Rene Tschaggelar" |
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| Title: Re: Escape Velocity |
10 Dec 2003 04:41:39 PM |
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Uncle Al wrote:
Jean wrote:
It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
Take a stationary test mass at infinity and allow it to drop onto the
Earth (no atmosphere, no rotation), integral of a=MG/r^2. The impact
velocity is the escape velocity by symmetry. The first principles
math is trivial given the surface gravity,
Suppose you place a ladder to the moon ..
Can you climb it and reach the moon without attaining any escape
velocity??
Indian rope trick/"Fountains of Paradise"/"Mouse That Roared, II".
That is not the definition of escape velocity. Escape velocity is the
minimum initial impulse velocity at and normal to the surface (no
planetary atmosphere, no rotation) necessary to launch a test mass to
infinity
AFAIK, it needn't be the surface either.
Any orbiting body moves below the escape speed.
Therefore look at the escape speed as limes of the tangential speed
for orbiting objects.
Rene
--
Ing.Buero R.Tschaggelar - http://www.ibrtses.com
& commercial newsgroups - http://www.talkto.net
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| User: "Greg Neill" |
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| Title: Re: Escape Velocity |
11 Dec 2003 08:28:51 AM |
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"Rene Tschaggelar" <none@none.none> wrote in message
news:3FD7A123.8060507@none.none...
AFAIK, it needn't be the surface either.
Any orbiting body moves below the escape speed.
Therefore look at the escape speed as limes of the tangential speed
for orbiting objects.
In particular, if Vcs is the circular orbit speed at a
given location, then sqrt(2)*Vcs is equal to the
escape velocity at that location.
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| User: "" |
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| Title: Re: Escape Velocity |
10 Dec 2003 04:51:10 PM |
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In article <3FD7A123.8060507@none.none>, Rene Tschaggelar <none@none.none> writes:
Uncle Al wrote:
Jean wrote:
It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
Take a stationary test mass at infinity and allow it to drop onto the
Earth (no atmosphere, no rotation), integral of a=MG/r^2. The impact
velocity is the escape velocity by symmetry. The first principles
math is trivial given the surface gravity,
Suppose you place a ladder to the moon ..
Can you climb it and reach the moon without attaining any escape
velocity??
Indian rope trick/"Fountains of Paradise"/"Mouse That Roared, II".
That is not the definition of escape velocity. Escape velocity is the
minimum initial impulse velocity at and normal to the surface (no
planetary atmosphere, no rotation) necessary to launch a test mass to
infinity
AFAIK, it needn't be the surface either.
Any orbiting body moves below the escape speed.
Below the escape speed at whatever its location is, to be exact.
Escape speed is location dependent.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Pyriform" |
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| Title: Re: Escape Velocity |
10 Dec 2003 06:25:12 AM |
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Jean wrote:
Suppose you place a ladder to the moon ..
Can you climb it and reach the moon without attaining any escape
velocity??
No. It would get all twisted, and break.
--
Pyriform
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| User: "William Oertell" |
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| Title: Re: Escape Velocity |
09 Dec 2003 08:24:45 PM |
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Jean,
Escape velocity is that speed at which an object would have to be
launched, without further propulsion, and not ever fall back. Escape
velocity for earth is about 25,000 mph. If you launch something at 24,999
mph, it will reach some altitude (sorry, don't know what that would be) and
start falling back again. If you launch it at 25,001 mph, it will reach
some altitude and continue leaving the earth's gravitational well at about
one mph. At 25,000 mph it will reach some altitude and stay there. It
won't fall back and it won't travel any further away, either.
"Jean" <alertjean@rediffmail.com> wrote in message
news:4138e33e.0312091808.10c09282@posting.google.com...
It is said that An Escape velocity of 11km/s is needed to get out the
gravitational pull of earth..
Why is it so??
Suppose you place a ladder to the moon ..
Can you climb it and reach the moon without attaining any escape
velocity??
Waiting for a comment through google..
(Please don't email me)
.
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| User: "Richard Henry" |
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| Title: Re: Escape Velocity |
09 Dec 2003 10:38:56 PM |
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"William Oertell" <oertell@pacbell.net> wrote in message
news:vtd0vf1mv6s801@news.supernews.com...
Jean,
Escape velocity is that speed at which an object would have to be
launched, without further propulsion, and not ever fall back. Escape
velocity for earth is about 25,000 mph. If you launch something at 24,999
mph, it will reach some altitude (sorry, don't know what that would be)
and
start falling back again. If you launch it at 25,001 mph, it will reach
some altitude and continue leaving the earth's gravitational well at about
one mph. At 25,000 mph it will reach some altitude and stay there. It
won't fall back and it won't travel any further away, either.
I did not know that.
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| User: "" |
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| Title: Re: Escape Velocity |
10 Dec 2003 12:00:41 AM |
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In article <FrxBb.74621$m24.59430@fed1read02>, "Richard Henry" <rphenry@home.com> writes:
"William Oertell" <oertell@pacbell.net> wrote in message
news:vtd0vf1mv6s801@news.supernews.com...
Jean,
Escape velocity is that speed at which an object would have to be
launched, without further propulsion, and not ever fall back. Escape
velocity for earth is about 25,000 mph. If you launch something at 24,999
mph, it will reach some altitude (sorry, don't know what that would be)
and
start falling back again. If you launch it at 25,001 mph, it will reach
some altitude and continue leaving the earth's gravitational well at about
one mph. At 25,000 mph it will reach some altitude and stay there. It
won't fall back and it won't travel any further away, either.
I did not know that.
Well, you couldn't, since it ain't so. There is no "some altitude at
which you stop".
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Franz Heymann" |
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| Title: Re: Escape Velocity |
12 Dec 2003 05:32:42 AM |
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"William Oertell" <oertell@pacbell.net> wrote in message
news:vtd0vf1mv6s801@news.supernews.com...
Jean,
Escape velocity is that speed at which an object would have to be
launched, without further propulsion, and not ever fall back. Escape
velocity for earth is about 25,000 mph. If you launch something at 24,999
mph, it will reach some altitude (sorry, don't know what that would be)
and
start falling back again. If you launch it at 25,001 mph, it will reach
some altitude and continue leaving the earth's gravitational well at about
one mph. At 25,000 mph it will reach some altitude and stay there. It
won't fall back and it won't travel any further away, either.
No. If it starts with a speed equal to the numerical value of the escape
velocity, its speed will just asymptote to zero by the time its position
from the earth asymptotes to infinity.
Franz Heymann
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| User: "Robert J. Kolker" |
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| Title: Re: Escape Velocity |
12 Dec 2003 08:10:38 AM |
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Franz Heymann wrote:
No. If it starts with a speed equal to the numerical value of the escape
velocity, its speed will just asymptote to zero by the time its position
from the earth asymptotes to infinity.
Isn't -asymptote- a noun. The verb form should be -to be asymptotic to-
Bob Kolker
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| User: "Richard Henry" |
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| Title: Re: Escape Velocity |
12 Dec 2003 10:41:51 AM |
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"Robert J. Kolker" <bobkolker@attbi.com> wrote in message
news:brci90$1uhm3$2@ID-76471.news.uni-berlin.de...
Franz Heymann wrote:
No. If it starts with a speed equal to the numerical value of the
escape
velocity, its speed will just asymptote to zero by the time its position
from the earth asymptotes to infinity.
Isn't -asymptote- a noun. The verb form should be -to be asymptotic to-
Forming a verb from what was previously considered tobe a noun is a modern
literary/journalistic process know as "verbing". "Verbing" is itself an
example of verbing.
Which should lead us to Douglas Hofstader's (un?)answered question -
What question is its own answer?
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| User: "Franz Heymann" |
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| Title: Re: Escape Velocity |
12 Dec 2003 02:22:51 PM |
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"Richard Henry" <rphenry@home.com> wrote in message
news:jdmCb.1676$7D3.638@fed1read02...
"Robert J. Kolker" <bobkolker@attbi.com> wrote in message
news:brci90$1uhm3$2@ID-76471.news.uni-berlin.de...
Franz Heymann wrote:
No. If it starts with a speed equal to the numerical value of the
escape
velocity, its speed will just asymptote to zero by the time its
position
from the earth asymptotes to infinity.
Isn't -asymptote- a noun. The verb form should be -to be asymptotic to-
Forming a verb from what was previously considered tobe a noun is a modern
literary/journalistic process know as "verbing". "Verbing" is itself an
example of verbing.
If it is not in the OED it is not an English word.
There is no verb "to verb" in the OED.
Which should lead us to Douglas Hofstader's (un?)answered question -
What question is its own answer?
Franz Heymann
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| User: "Richard Henry" |
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| Title: Re: Escape Velocity |
12 Dec 2003 04:36:44 PM |
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"Franz Heymann" <notfranz.heymann@btopenworld.com> wrote in message
news:brd82r$be5$1@hercules.btinternet.com...
"Richard Henry" <rphenry@home.com> wrote in message
news:jdmCb.1676$7D3.638@fed1read02...
"Robert J. Kolker" <bobkolker@attbi.com> wrote in message
news:brci90$1uhm3$2@ID-76471.news.uni-berlin.de...
Franz Heymann wrote:
No. If it starts with a speed equal to the numerical value of the
escape
velocity, its speed will just asymptote to zero by the time its
position
from the earth asymptotes to infinity.
Isn't -asymptote- a noun. The verb form should be -to be asymptotic
to-
Forming a verb from what was previously considered tobe a noun is a
modern
literary/journalistic process know as "verbing". "Verbing" is itself an
example of verbing.
If it is not in the OED it is not an English word.
There is no verb "to verb" in the OED.
When the OED opts to include a new word, is the new word then "worded"?
.
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| User: "Franz Heymann" |
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| Title: Re: Escape Velocity |
13 Dec 2003 08:48:52 AM |
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"Richard Henry" <rphenry@home.com> wrote in message
news:1qrCb.1928$7D3.510@fed1read02...
"Franz Heymann" <notfranz.heymann@btopenworld.com> wrote in message
news:brd82r$be5$1@hercules.btinternet.com...
"Richard Henry" <rphenry@home.com> wrote in message
news:jdmCb.1676$7D3.638@fed1read02...
"Robert J. Kolker" <bobkolker@attbi.com> wrote in message
news:brci90$1uhm3$2@ID-76471.news.uni-berlin.de...
Franz Heymann wrote:
No. If it starts with a speed equal to the numerical value of the
escape
velocity, its speed will just asymptote to zero by the time its
position
from the earth asymptotes to infinity.
Isn't -asymptote- a noun. The verb form should be -to be asymptotic
to-
Forming a verb from what was previously considered tobe a noun is a
modern
literary/journalistic process know as "verbing". "Verbing" is itself
an
example of verbing.
If it is not in the OED it is not an English word.
There is no verb "to verb" in the OED.
When the OED opts to include a new word, is the new word then "worded"?
No. Writed. {:-))
Franz Heymann
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| User: "Franz Heymann" |
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| Title: Re: Escape Velocity |
12 Dec 2003 10:38:05 AM |
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"Robert J. Kolker" <bobkolker@attbi.com> wrote in message
news:brci90$1uhm3$2@ID-76471.news.uni-berlin.de...
Franz Heymann wrote:
No. If it starts with a speed equal to the numerical value of the
escape
velocity, its speed will just asymptote to zero by the time its position
from the earth asymptotes to infinity.
Isn't -asymptote- a noun. The verb form should be -to be asymptotic to-
You are quite right. My apologies for misusing the English language.
Franz Heymann
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| User: "Robert J. Kolker" |
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| Title: Re: Escape Velocity |
12 Dec 2003 10:46:38 AM |
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Franz Heymann wrote:
You are quite right. My apologies for misusing the English language.
Do you see what happens when you hang around near Christian Fundies?
Bob Kolker
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