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'[OT]:Mir'
2001\03\22@235607 by Jinx

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I've heard the question asked several times but never an answer

Why is so much trouble being taken to crash Mir when it seems
possible to send it out of orbit away from Earth ? Is it just a
question of fuel ? There've been many cargo trips to Mir, just
wondered why they didn't do it that way

Thought I'd better ask now as there's only an hour before the old
rustbucket's due over NZ and well, you never know. Although the
planned braking bursts are said to be going OK and on schedule
so no one's kacked their dacks yet about an unexpected visit

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2001\03\23@002851 by David Duffy

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Jinx wrote:
>I've heard the question asked several times but never an answer
>
>Why is so much trouble being taken to crash Mir when it seems
>possible to send it out of orbit away from Earth ? Is it just a
>question of fuel ? There've been many cargo trips to Mir, just
>wondered why they didn't do it that way

Maybe it's better not to have it floating around to crash into later?
There's already way too much space junk out there.

>Thought I'd better ask now as there's only an hour before the old
>rustbucket's due over NZ and well, you never know. Although the
>planned braking bursts are said to be going OK and on schedule
>so no one's kacked their dacks yet about an unexpected visit

Look out !!  Duck !!  Whoosh !!  Splash !!
Regards...

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2001\03\23@003533 by Mike Morris

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Velocity required for Earth orbit is about 17,500 MPH.  Escape velocity is
around 25k MPH. So yes, it takes a lot of horsepower... ie, *lots* of gas
(ok, LOX + Hydrogen) and *big* engines (in good condition :)  to burn it.
Mir has (or had) neither.

Speaking of Earth orbit, anyone know if a PIC's been in space yet?

- Mike

At 04:57 PM 3/23/2001 +1200, Jinx wrote:
{Quote hidden}

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2001\03\23@010432 by Jinx

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> Velocity required for Earth orbit is about 17,500 MPH.  Escape velocity is
> around 25k MPH. So yes, it takes a lot of horsepower... ie, *lots* of gas
> (ok, LOX + Hydrogen) and *big* engines (in good condition :)  to burn it.
> Mir has (or had) neither.

Thanks

> Speaking of Earth orbit, anyone know if a PIC's been in space yet?

The Muppets used to do it all the time. Oh, wait. That was pigs
>
> - Mike

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2001\03\23@014056 by Jinx

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> Look out !!  Duck !!  Whoosh !!  Splash !!
> Regards...

You rotter. But just once in my life I'd like to say "Cool, two suns"

Two moons you can get any time you drive past a bus full of rugby
supporters

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2001\03\23@045947 by Alan B. Pearce

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>Two moons you can get any time you drive past a bus full of rugby
>supporters

ROTFL

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2001\03\23@080550 by t F. Touchton

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part 1 1827 bytes content-type:text/plain; charset=us-asciiI had one fly on a space shuttle mission.  Was used in a circuit that mixed
bacterial cultures under zero gravity (or close to it!).  Basically, the PIC
just monitored a couple of switches and controlled 2 motors, but it worked!!!!!!


|--------+-------------------------->
|        |          Jinx            |
|        |          <joecolquitt@CLE|
|        |          AR.NET.NZ>      |
|        |                          |
|        |          03/23/01 01:04  |
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 |       cc:     (bcc: Scott Touchton/US/UNIPHASE)                            |
 |       Subject:     Re: [OT]:Mir                                            |
 >----------------------------------------------------------------------------|





> Velocity required for Earth orbit is about 17,500 MPH.  Escape velocity is
> around 25k MPH. So yes, it takes a lot of horsepower... ie, *lots* of gas
> (ok, LOX + Hydrogen) and *big* engines (in good condition :)  to burn it.
> Mir has (or had) neither.

Thanks

> Speaking of Earth orbit, anyone know if a PIC's been in space yet?

The Muppets used to do it all the time. Oh, wait. That was pigs
>
> - Mike

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part 2 2521 bytes content-type:application/octet-stream; (decode)

part 3 154 bytes
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2001\03\23@080953 by Jinx

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From: Scott F. Touchton

> I had one fly on a space shuttle mission.  Was used in a circuit
> that mixed bacterial cultures under zero gravity (or close to it!).
> Basically, the PIC just monitored a couple of switches and
> controlled 2 motors, but it worked!!!!!!

Nice one. So "Touchton, we have a problem" was never heard ?

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2001\03\23@081426 by t F. Touchton

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part 1 1732 bytes content-type:text/plain; charset=us-asciiNo... but the original power source was Zinc-Air batteries...
That got pretty far along before someone realized this was taking place in the
cargo bay, and the presence of air was not guaranteed.
Alkalines to the rescue!


|--------+-------------------------->
|        |          Jinx            |
|        |          <joecolquitt@CLE|
|        |          AR.NET.NZ>      |
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|        |          03/23/01 08:11  |
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 |       cc:     (bcc: Scott Touchton/US/UNIPHASE)                            |
 |       Subject:     Re: [OT]:Mir                                            |
 >----------------------------------------------------------------------------|





From: Scott F. Touchton

> I had one fly on a space shuttle mission.  Was used in a circuit
> that mixed bacterial cultures under zero gravity (or close to it!).
> Basically, the PIC just monitored a couple of switches and
> controlled 2 motors, but it worked!!!!!!

Nice one. So "Touchton, we have a problem" was never heard ?

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part 2 2414 bytes content-type:application/octet-stream; (decode)

part 3 154 bytes
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2001\03\23@083727 by rchock, Steve

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Mike:

I live in Phoenix, Arizona and the local university (ASU)
was working on some satellites that had PICs in them. Microchip
even did an article on them. Check out
http://www.microchip.com/10/school/student/satproj/index.htm

Steve

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2001\03\23@121540 by Glenn Mitchell

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I bet that some of the AMSAT amateur radio satellites had PICs onboard.
http://www.amsat.org

Glenn Mitchell.

{Original Message removed}

2001\03\23@162510 by M. Adam Davis
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The orbit altitude of an object is determined by its rotational speed
above the earth.  If you want a higher orbit, start spinning around the
earth faster, lower, slow yourself down.  A space station like MIR was
at a very low orbit to begin with, and its size would make it so it
could orbit much further from the earth than any satellite and still
orbit earth (ie, not break away).

The reason they dropped it was because they had just barely enough fuel
since this last cargo ship to brake it enough (while still under their
control) to drop out of orbit.  It would have taken significantly more
fuel to speed it up enough to get it out of earth orbit, not to mention
that if they didn't make it it might start doing an elliptical orbit,
and becoming more hazardous than it was when it just occupied its own
orbit.

They didn't have enough money to send up another freighter, and if they
waited too much longer the fuel would have been used up just keeping it
in its current orbit (so it doesn't hit anything else, drift higher or
lower, and has a predictable path - critical in space)

-Adam

(actually, I'm just sending this to make sure my cable modem's smtp
server is working...  Yay for broadband!  And birthdays!  And loving
wives!  Yay!)

Jinx wrote:
{Quote hidden}

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2001\03\23@164153 by Russell McMahon

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I estimate that it would have taken about 100 supply ship loads of fuel to
get it out to near Lunar orbit.


     Russell McMahon
_____________________________


{Original Message removed}

2001\03\23@164204 by Russell McMahon

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The former Russian Space Station is now a MIR drop in the ocean.

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2001\03\23@174022 by Jinx

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From: Russell McMahon

> The former Russian Space Station is now a MIR drop in
> the ocean.

Russell, pop your head out of the window and listen for prolonged
groaning from over New Lynn way

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2001\03\23@174644 by David VanHorn

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At 10:40 AM 3/24/01 +1200, Jinx wrote:
>From: Russell McMahon
>
> > The former Russian Space Station is now a MIR drop in
> > the ocean.

It did make a big splash :)

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2001\03\23@231258 by mmucker

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> The orbit altitude of an object is determined by its rotational speed
> above the earth.  If you want a higher orbit, start spinning around the
> earth faster, lower, slow yourself down.

Actually, I think this is backwards.  I'm no expert, but after reading Gene
Krantz' autobiography, I reacll that NASA had lots of problems with stuff
like this until they realized that at a lower orbit, a body must revolve
around the earch much faster to keep it in orbit.  Any experts out there who
can clarify?

> A space station like MIR was
> at a very low orbit to begin with, and its size would make it so it
> could orbit much further from the earth than any satellite and still
> orbit earth (ie, not break away).

Does the size of a body have any effect on its orbit?  I know that force is
a constant times em-one times em-two over r-squared, but the relationship of
masses (earth and Mir) is probably close enough to the relationship of Earth
to, say, a golf ball, that there would be no difference.

Just curious.

-Matt

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2001\03\26@040056 by Jeszs

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Object of mass m rotating around a gravitation field of strength 'mu'
so-called Earth:
Centigugal force = m * V * V / R                (V is velocity and R the
radius)
Weight = mu * m / ( R * R )

In order to have equilibrium: weight = centigugal force
That means:
   V = sqrt( mu / R )

Conclusion:
higher orbit -> slower vehicle
lower orbit -> quicker vehicle
mass -> second order effect. Only if atmospheric drag is important (let us
say R < 450 km), the ratio between mass and front surface should be
considered.

ah, the question! when you are at certain altitude and want to go lower you
need to slow down your vehicle (you go to an elliptical orbit with apogee at
the initial altitude and perigee at the new target altitude, lower) and
then, when at perigee, to speed up again to reach new circular orbit of that
radius.

--------------------
Jeszs Gonzalo
Lesn (SPAIN)
--------------------
{Original Message removed}

2001\03\26@160616 by Barry Gershenfeld

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>From: "Matthew Mucker" <mmuckerspamspam_OUTAIRMAIL.NET>
>> Actually, I think this is backwards.  I'm no expert, but after
>> reading Gene Krantz' autobiography, I reacll that NASA had
>> lots of problems with stuff like this until they realized
>> that at a lower orbit, a body must revolve around the earch
>> much faster to keep it in orbit.  Any experts out there who
>> can clarify?

>Object of mass m rotating around a gravitation field of strength 'mu'
>so-called Earth:
>Centigugal force = m * V * V / R                (V is velocity and R the
>radius)
>Weight = mu * m / ( R * R )
(snip)
>Jeszs Gonzalo

In english:  At a higher orbit you are traveling faster.  As in,
if you were standing still and watched it go by.  (So-called,
linear velocity).  But at the same time, it takes longer to
complete an orbit, so it "goes around" slower (so-called,
angular velocity).

This is a favorite story of mine, because the best and brightest
missed the point the first time.  Not that I knew anything, I
read about it in the paper back then, and I hope I remember
it right.

On one of the early Gemini flights they did a rendezvous experiment.
As I recall, a lot of the Gemini program was aimed towards working
out all the particulars about rendezvous-and-docking for the
subsequent Apollo (moon shot) program.  This flight predated
the later docking with other capsules, adapters, Russian
ships, etc.  What they did was to put lights on the booster
stage which remained in orbit (or near-orbit) along with the
flight capsule.  Early in the mission they practised the
rendezvous maneuver with this booster.  It didn't work.
Basically the complaint was "the more I tried to catch
up with it, the farther away it got".  So they went home
and did some head scratching, and presumably looked at
those formulas again, and said, "Oh...now we get it".

Essentially you slow down, drop to a lower orbit, which
lets you swing ahead of the thing you want to catch up with.

Orbital mechanics is interesting.

Barry

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2001\03\26@162543 by Dale Botkin

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On Mon, 26 Mar 2001, Barry Gershenfeld wrote:

> flight capsule.  Early in the mission they practised the
> rendezvous maneuver with this booster.  It didn't work.
> Basically the complaint was "the more I tried to catch
> up with it, the farther away it got".  So they went home
> and did some head scratching, and presumably looked at
> those formulas again, and said, "Oh...now we get it".
>
> Essentially you slow down, drop to a lower orbit, which
> lets you swing ahead of the thing you want to catch up with.

I get it...  don't go faster, take a shorter route - right?

> Orbital mechanics is interesting.

Race cars too.  Sometimes you have to slow down to go faster.

Dale
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2001\03\26@164209 by O'Reilly John E NORC

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If I remember right from college (university for those of you outside the
US), there was something about the area of the arc the orbiting object
sweeps out in a given amount of time is constant and based on the masses of
the two bodies.  Basically, if you take the radius of the orbit r, multiply
by the angular velocity a / t , it is constant.  r * a / t is constant.  So,
as you increase r, the angular velocity decreases, and vice versa.  I'm sure
there are hundreds of other equations that can be derived from this, but I
haven't touched this stuff in 7 years.

On a similar note, the reason that NASA launches from Florida is that
launching to the east coast takes advantage of the rotation of the Earth.
It is also closer to the equator, which spins faster.  If they were to
launch from the west coast, they would have to launch over my apartment.
Or, if they launched the space shuttle to the west, they would have to lose
about 50,000 pounds of payload to get into orbit.

John

{Original Message removed}

2001\03\26@170100 by John Pfaff

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If I'm right (and who knows, I just may be), that applies to a single body
in an eliptical orbit.  When the orbiting body is closer (ie the earth and
the sun), it moves faster, when it's farther away it slows down, but in any
given period of time (t) it will sweep and equal arc area.  Whether or not
you can apply it to two distinct bodies, I don't know.

----- Original Message -----
From: "O'Reilly John E NORC" <RemoveMEOReillyJETakeThisOuTspamCORONA.NAVY.MIL>
To: <spamBeGonePICLISTspamBeGonespamMITVMA.MIT.EDU>
Sent: Monday, March 26, 2001 4:37 PM
Subject: Re: [OT]:Mir


> If I remember right from college (university for those of you outside the
> US), there was something about the area of the arc the orbiting object
> sweeps out in a given amount of time is constant and based on the masses
of
> the two bodies.  Basically, if you take the radius of the orbit r,
multiply
> by the angular velocity a / t , it is constant.  r * a / t is constant.
So,
> as you increase r, the angular velocity decreases, and vice versa.  I'm
sure
> there are hundreds of other equations that can be derived from this, but I
> haven't touched this stuff in 7 years.
>
> On a similar note, the reason that NASA launches from Florida is that
> launching to the east coast takes advantage of the rotation of the Earth.
> It is also closer to the equator, which spins faster.  If they were to
> launch from the west coast, they would have to launch over my apartment.
> Or, if they launched the space shuttle to the west, they would have to
lose
> about 50,000 pounds of payload to get into orbit.
>
> John
>
> {Original Message removed}

2001\03\26@193930 by Brandon Fosdick

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Barry Gershenfeld wrote:
{Quote hidden}

Actually, higher circular orbits have a slower tangential velocity than
lower circular oribits.
Vcirc = sqrt(GM/r) where G is the gravitational constant, M is the mass
of the orbited body and r is the radius of the orbit.

Incidentally the mu referred to above is mu=GM, the gravtiational
paramenter for the body in question. In general the constant isn't
specifically tied to Earth, it can be computed for any mass.

AIAA (http://www.aiaa.org) has a very good book written by Chobotov on this
subject. If I had it with me I could give you the complete reference. I
think it was simply called "Orbital Mechanics".

{Quote hidden}

Hill's equations weren't widely known at the time, and less widely
understood. Hence the initial problems.

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2001\03\26@193938 by Brandon Fosdick

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O'Reilly John E NORC wrote:
>
> If I remember right from college (university for those of you outside the
> US), there was something about the area of the arc the orbiting object
> sweeps out in a given amount of time is constant and based on the masses of
> the two bodies.  Basically, if you take the radius of the orbit r, multiply
> by the angular velocity a / t , it is constant.  r * a / t is constant.  So,
> as you increase r, the angular velocity decreases, and vice versa.  I'm sure
> there are hundreds of other equations that can be derived from this, but I
> haven't touched this stuff in 7 years.

Kepler's Sencond Law: Equal Area, Equal Time
Simply, if you measure the area swept out by an orbit in a given amount
of time, and then look at the area swept out by a different part of the
same orbit in the same amount of time, the two areas will be equal. It's
independant of the mass of either of the bodies.

> On a similar note, the reason that NASA launches from Florida is that
> launching to the east coast takes advantage of the rotation of the Earth.
> It is also closer to the equator, which spins faster.  If they were to
> launch from the west coast, they would have to launch over my apartment.
> Or, if they launched the space shuttle to the west, they would have to lose
> about 50,000 pounds of payload to get into orbit.

Launching to the East from closer to the equator takes advantage of
Earth's spin. Its not dependant on the coast (or longitude), that just
helps with safety concerns. Shuttle Launch Complex 6 (SLC6) at
Vandenberg AFB, in California, is the alternate launch site for the
shuttle, although its never been used for a shuttle launch. It was
intended for high inclination orbits back when the shuttle was to be
more closely associated with national security.

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2001\03\27@005338 by Dan Michaels

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Brandon Fosdick wrote:

>> On a similar note, the reason that NASA launches from Florida is that
>> launching to the east coast takes advantage of the rotation of the Earth.
>> It is also closer to the equator, which spins faster.  If they were to
>> launch from the west coast, they would have to launch over my apartment.
>> Or, if they launched the space shuttle to the west, they would have to lose
>> about 50,000 pounds of payload to get into orbit.
>
>Launching to the East from closer to the equator takes advantage of
>Earth's spin. Its not dependant on the coast (or longitude), that just
>helps with safety concerns. Shuttle Launch Complex 6 (SLC6) at
>Vandenberg AFB, in California, is the alternate launch site for the
>shuttle, although its never been used for a shuttle launch. It was
>intended for high inclination orbits back when the shuttle was to be
>more closely associated with national security.
>

They have launched satellites 100s, if not 1000s, of times from VAFB.
It's used for polar orbits, while FLA is used for equitorial orbits.
With polar orbit, you fly over every bit of the earth's surface at
least once per day, IIRC - with equitorial orbit, not so often.

In Florida you launch to the east, and in Calif you launch to the
south. In both cases, the rockets head off over the ocean, instead
of over people's heads. And as mentioned, when launching east from
FLA, the earth's spin helps a lot. Geosynchronous satellites, no
doubt, all go out of FLA.

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2001\03\27@010948 by Dan Michaels

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Barry Gershenfeld wrote:

>>> Actually, I think this is backwards.  I'm no expert, but after
>>> reading Gene Krantz' autobiography, I reacll that NASA had
>>> lots of problems with stuff like this until they realized
>>> that at a lower orbit, a body must revolve around the earch
>>> much faster to keep it in orbit.  Any experts out there who
>>> can clarify?
>
.......
>In english:  At a higher orbit you are traveling faster.  As in,
>if you were standing still and watched it go by.  (So-called,
>linear velocity).  But at the same time, it takes longer to
>complete an orbit, so it "goes around" slower (so-called,
>angular velocity).
>

I don't know any of the equations [I failed orbital mechanics
in college], but it seems to me that geosynchronous satellites
are out at 22,300 miles and circle the earth once per day, while
satellites 200 miles up circle about once every 90 minutes.
Therefore, their velocities are gonna be approx:

from C = 2*pi*R

geo orbit: C = 2*pi*22,300 = 140,000 miles
          140,000 miles/24 hours = 5800 MPH

close in orbit: C = 2*pi*[4000+200] = 26,000 miles
            26,000 / 1.5 hours = 17,000 MPH

Hmmm, 17,000 MPH sounds about like the number I had always
heard used in relation to gaining orbit.

How about the moon? 240,000 miles out, takes 30 days to orbit.

moon's orbit: C = 2*pi*240,000 = 1,500,000 miles
     [1,500,000 miles/30 days] / 24 hours/day = 2,000 MPH

Looks like you go a lot faster for closer in orbits. No doubt,
because you have to keep your centrifugal force high, else
you come right on down. [hey, Alice, you're the orbital
mechanic - ??????]

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2001\03\27@012007 by Sean H. Breheny

face picon face
I think the confusion is this: if you want to go to a higher orbit, you
need more energy. Speeding up provides this energy, which is then
transferred to potential energy as you slow down and go higher. Another way
to look at it is that as you speed up, your outward tendency (centrifugal
force) increases beyond what is necessary to maintain the current orbit, so
you climb outward, gravity slows you down, and you reach equilibrium again
at a higher, slower orbit.

Sean

At 01:09 AM 3/27/01 -0500, you wrote:
{Quote hidden}

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2001\03\27@032127 by Alan B. Pearce

face picon face
>> Essentially you slow down, drop to a lower orbit, which
>> lets you swing ahead of the thing you want to catch up with.

>I get it...  don't go faster, take a shorter route - right?

Ah, so it is like work "the faster I go the behinder I get"...

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2001\03\27@122440 by Alice Campbell

flavicon
face
{Quote hidden}

Gravity pulls stuff in, and the closer you are to a gravity
well, the faster you fall.  so the faster you fall, the
faster you have to run either tangentially or away to keep
from being captured. Yeah, you have to brake to get to a
lower orbit, which would seem like slowing down, but youre
deeper in the gravitational well now and you have to run
faster to keep from being sucked in.  In this sense, bigger
orbits are 'cheaper' because it takes less force to fiddle
with them.  Closer in, there is always a big energy penalty
fiddling with orbits.  And of course, it all depends on the
mass of your planet, an orbit 200 miles above the moon is
slower than an orbit 200 miles above the earth.

alice

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2001\03\27@124925 by Dipperstein, Michael

face picon face
> From: Alice Campbell [acampbellSTOPspamspamspam_OUTscsengineers.com]

...

{Quote hidden}

Every time this discussion comes up, I manage to keep things straight by
remembering that an object in a circular orbit is always falling towards the
object it's orbiting.  The only reason the object doesn't hit the thing it's
orbiting is because its tangential velocity is such that the object is moving
along the curvature of the orbit as it falls.

If the tangential velocity is increased, the orbiting object moves further from
the object it was orbiting; it's moving forward faster than it's falling.
Decreasing the tangential velocity, the orbiting object moves closer to the
object it was orbiting; it's falling faster than it's moving forward.

To borrow liberally from "The Hitchhiker's Guide to the Galaxy", orbiting is
falling and missing the ground.

-Mike

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2001\03\27@132809 by jamesnewton

face picon face
So when you brake to get lower, does your fall toward the earth somehow
increase your tangential velocity? Or do you have to do a second burn at the
lower altitude to speed back up on the tangent? Is this the correction of an
elliptical orbit back into a circular orbit that was discussed earlier?

Or is it that the direction of the "braking" is towards the earth? i.e.
thrusters pointed away from the planet. I always understood that braking for
a lower orbit was done by pointing your thrusters in the direction of travel
and boosting "backwards." But I've never before heard about a second boost
at the bottom and never thought about what it is that increases your
tangential velocity when you get there...

...I always thought that lower orbits only APPEAR to be traveling faster
because they have less distance to cover along the circumference of a
smaller circle...

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{Original Message removed}

2001\03\27@133033 by Dan Michaels

flavicon
face
Dipperstein, Michael wrote:
......
>If the tangential velocity is increased, the orbiting object moves further from
>the object it was orbiting; it's moving forward faster than it's falling.
>Decreasing the tangential velocity, the orbiting object moves closer to the
>object it was orbiting; it's falling faster than it's moving forward.
>

Well that explains the following - he, he :):

>
>geo orbit: C = 2*pi*22,300 = 140,000 miles
>           140,000 miles/24 hours = 5800 MPH
>
>close in orbit: C = 2*pi*[4000+200] = 26,000 miles
>             26,000 / 1.5 hours = 17,000 MPH
>

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2001\03\27@133421 by Russell McMahon

picon face
 From: "Dan Michaels" <@spam@oricom@spam@spamspam_OUTUSWEST.NET>
> Barry Gershenfeld wrote:
> >In english:  At a higher orbit you are traveling faster.

Slower - see below for Orbital Mechanics None-Oh-One which explain the basic
relationship twixt orbital height and velocity.


{Quote hidden}

Yes.

>>No doubt, because you have to keep your centrifugal force high, else
> you come right on down.



ORBITAL MECHANICS NONE-OH-ONE :-)
________________________________

The equations relating orbital height and velocity are quite simple but not
quite intuitive.
If you are even a little interested try to follow this as it really only
takes TWO equations and will add to your general appreciation of the world
at large.

A satellite in orbit is subject to centripetal (aka centrifugal) force.

   Centripetal (centrifugal) force is

       Fc = mV^2/R            ... 1

   The derivation of this is not TOO hard but outside the scope of this
description.

In orbit this is EXACTLY balanced by weight = mg

where g is the local gravity.

Now weight (force exerted by attraction between masses) =

   F = G x m x M / R^2        ... 2

   ie weight (force of attraction) is proportional to the masses involved
and inversely
   proportional to the square of the distance between them.

   (which is what Newton deduced)

m = mass of "satellite" (or you)
M = mass of other body (here the earth)
G = a constant (universal gravitational constant)
R = radius = distance from centre of earth.

For practical purposes GM = K = a constant.

So

2 again

.     weight = force of gravity = mg = mK/R^2


Combining 1 & 2 gives

1 & 2 -->     mV^2/R =  mK/R^2        ...3

or
       V^2 = K/R
or
       V = sqrt(K/R)
or
       R = K / V^2

ie orbital velocity is inversely proportional to the distance from the
centre of the earth.
or
   orbital radius is inversely proportional to the inverse of the square of
the velocity

The lower you go the FASTER you go.
The net energy goes up due to potential energy increasing.
(You CAN'T just apply PE = mgh as you can on the surface of the earth as g
changes with height for large changes.
(We cheat by ignoring the change in g for small heights of up to a few km or
miles.)
(For falling from orbit you have to integrate the result of increasing g
with decreasing altitude which is outside the scope of this discussion
(fortunately :-) )).

This leads to some interesting space craft action.
Two craft orbit side by side.
Craft 1 fires RETRO rockets.
It ACCELERATES in velocity and dives down and ahead of craft 2 !!!
(It also much more greatly drops the altitude of the opposite side of its
orbit but that's for another day.)



regards



     Russell McMahon
_____________________________


{Original Message removed}

2001\03\27@140320 by Sean Breheny

face picon face
Hi Russell,

Net energy goes UP as you descend?! Are you sure? I don't have time right
now to work out the math, but think of the following: if you are at a
high orbit, you can get to a lower orbit by just breaking. Breaking
doesn't have to use up any energy (for example, if you were in a large
atmosphere which extended to orbital altitudes, you could break by just
extending a bunch of fan blades and
actually GENERATE energy during the breaking). So, while going from a
higher orbit to a lower one you can actually extract energy, so it seems
to me as if you must be dropping in net energy (PE + KE).

Yes, you can't use PE=mgh, but higher H still means higher PE, it just
goes nonlinearly. You also have to take KE into account (which
complicates things a bit) but I bet if you actually work it out you get a
higher total E at a higher orbit.

Sean


On Wed, 28 Mar 2001, Russell McMahon wrote:

> The lower you go the FASTER you go.
> The net energy goes up due to potential energy increasing.
> (You CAN'T just apply PE = mgh as you can on the surface of the earth as g
> changes with height for large changes.
> (We cheat by ignoring the change in g for small heights of up to a few km or
> miles.)

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2001\03\27@140333 by Alice Campbell

flavicon
face
> So when you brake to get lower, does your fall toward the earth somehow
> increase your tangential velocity? Or do you have to do a second burn at the
> lower altitude to speed back up on the tangent? Is this the correction of an
> elliptical orbit back into a circular orbit that was discussed earlier?
No.  You lose sideways velocity when braking.  You have to
speed up again to catch up when you get to the lower orbit,
that or re-enter the planets atmosphere.

The most efficient way to transfer between 2 circular orbits
is via an elliptical orbit that has its major axis the same
as the outer orbit and its minor axis the same as the desired
inner orbit.  Then a quick control burn to speed up at the
lower orbit, and you're in.  The name of this transfer, by
the way, is the Hohman Transfer Orbit, no doubt after
somebody of the same name.
{Quote hidden}

No, because of the higher gravitational potential they hafta be faster, its not linear because the field is 1/R^2 not 1/R.

The actual shape of a gravitational well is a hyperboloid of
one sheet.  At the L.A County Natural History museum, they
have a display that you put a coin in a slot, and it spirals
down a curiously shaped conic and ends up inside.  That is a
model of the shape of a gravity well, and the speedup of the
coin as it nears the center is analogous to this speedup as
you move in toward a planet.  In order to keep the coin from
spiraling in, you would need to add enough tangential energy
to overcome the friction .  Otherwise, it gets sucked in.
>
> ---
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> .....jamesnewtonspam_OUTspampiclist.com 1-619-652-0593
> PIC/PICList FAQ: http://www.piclist.com or .org
>
> {Original Message removed}

2001\03\27@142156 by Alice Campbell

flavicon
face
> Hi Russell,
>
> Net energy goes UP as you descend?! Are you sure? I don't have time right
> now to work out the math, but think of the following: if you are at a
> high orbit, you can get to a lower orbit by just breaking. Breaking
> doesn't have to use up any energy (for example, if you were in a large
> atmosphere which extended to orbital altitudes, you could break by just
> extending a bunch of fan blades and
> actually GENERATE energy during the breaking). So, while going from a
> higher orbit to a lower one you can actually extract energy, so it seems
> to me as if you must be dropping in net energy (PE + KE).
Well, you can brake to GET to a lower orbit, but you wont BE
in the lower orbit without speeding up again.  Otherwise you
re-enter.



{Quote hidden}

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2001\03\27@142207 by Bob Ammerman

picon face
> The most efficient way to transfer between 2 circular orbits
> is via an elliptical orbit that has its major axis the same
> as the outer orbit and its minor axis the same as the desired
> inner orbit.  Then a quick control burn to speed up at the
> lower orbit, and you're in.  The name of this transfer, by
> the way, is the Hohman Transfer Orbit, no doubt after
> somebody of the same name.

Ah yes, good old Mr. Orbit. :-)

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

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2001\03\27@142810 by David VanHorn

flavicon
face
At 02:03 PM 3/27/01 -0500, Sean Breheny wrote:
>Hi Russell,
>
>Net energy goes UP as you descend?!

And potential (from altitude) goes down.
It has to go somewhere, and you're the only thing convenient to dump it in.


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2001\03\27@145241 by Dan Michaels

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face
James Newton wrote:
>So when you brake to get lower, does your fall toward the earth somehow
>increase your tangential velocity? Or do you have to do a second burn at the
>lower altitude to speed back up on the tangent? Is this the correction of an
>elliptical orbit back into a circular orbit that was discussed earlier?
>
>Or is it that the direction of the "braking" is towards the earth? i.e.
>thrusters pointed away from the planet. I always understood that braking for
>a lower orbit was done by pointing your thrusters in the direction of travel
>and boosting "backwards." But I've never before heard about a second boost
>at the bottom and never thought about what it is that increases your
>tangential velocity when you get there...


Now you do need an orbital mechanic.

However, as an EE, I suspect that a "small" retro burn will slow you
down a bit, and then gravity will pull you downward, thereby increasing
your speed [from vector physics], and you will go into lower, faster
orbit. Too large a retro burn, and you keep coming down like Mir.

OTOH, you can probably also do a forward burn while pointing slightly
downwards and go into lower faster orbit, but too much of this, and
the orbit is probably gonna be highly ellipsoidal in shape.

If you are in 200 mile, 17,000 MPH parking orbit and want to go out
to 23,000 mile, 5800 MPH geosynch orbit, you "first" speed up to 25,000
MPH or so [ie, near escape velocity] to pull away from earth's gravity,
and then make a "retro" burn at 22,000 miles to slow down to 5800 MPH
and let gravity catch back up. Again, probably takes some fine adjustments
to make the orbit more circular rather than highly ellipsoidal.

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2001\03\27@150208 by Dan Michaels

flavicon
face
Alice wrote:
>> So when you brake to get lower, does your fall toward the earth somehow
>> increase your tangential velocity? Or do you have to do a second burn at the
>> lower altitude to speed back up on the tangent? Is this the correction of an
>> elliptical orbit back into a circular orbit that was discussed earlier?
>No.  You lose sideways velocity when braking.  You have to
>speed up again to catch up when you get to the lower orbit,
>that or re-enter the planets atmosphere.
>

If you do only a "small" retro burn, will not you just go into slightly
lower and faster orbit, rather than simply plunging straight for New
Zealand? The increased force of gravity pulling downwards combined
with forward momentum would actually speed you up as you come down
- no?
=============

>The most efficient way to transfer between 2 circular orbits
>is via an elliptical orbit that has its major axis the same
>as the outer orbit and its minor axis the same as the desired
>inner orbit.  Then a quick control burn to speed up at the
>lower orbit, and you're in.  The name of this transfer, by
>the way, is the Hohman Transfer Orbit, no doubt after
>somebody of the same name.

This is no doubt how they get from low-earth to geosynchronous
orbit.
===================

{Quote hidden}

Ha - I just knew Alice was really an orbital mechanic :).

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2001\03\27@150625 by Sean Breheny

face picon face
Hi Alice,

I see, that's where I was incorrect. So, was Russell correct, the total
energy for a lower orbit is higher than for a higher altitude one? If so,
don't you need more fuel to get to a higher orbit? Where does the extra
energy go, just into the kinetic energy of the ejected gas?

Thanks,

Sean


On Tue, 27 Mar 2001, Alice Campbell wrote:
> Well, you can brake to GET to a lower orbit, but you wont BE
> in the lower orbit without speeding up again.  Otherwise you
> re-enter.

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2001\03\27@150831 by Sean Breheny

face picon face
Hi Dave,

But we were talking about total E (KE + PE), not just KE.

Sean


On Tue, 27 Mar 2001, David VanHorn wrote:

> And potential (from altitude) goes down.
> It has to go somewhere, and you're the only thing convenient to dump it in.
>
>
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2001\03\27@151657 by Barry Gershenfeld

picon face
>But I've never before heard about a second boost
>at the bottom and never thought about what it is that increases your
>tangential velocity when you get there...

In my simple world, it's like this.  If you're not firing any
jets your orbit will not change.  If you are at some altitude
and you do a quick burn you have changed your velocity in
some way.  The moment you cut off your engine you are in that
constant orbit, whatever it may be.  If you fired toward your
direction of travel then you will be going down closer to the
earth.  That effectively means that you've just lowered the
altitude you'll be at on the other side.   But you will return
to where you are right now; you have to! (unchanging orbit).  So
what you actually did was move the opposite point.  From here
you can see that you have to travel around to that opposite
point and do another burn, in order to lower the altitude you're
at right now.

Corollary:  I think that if you jettison your garbage and
don't change your orbit it will come right back to you one
orbit later, at the point you tossed it.

Barry

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2001\03\27@151701 by Dipperstein, Michael

face picon face
{Quote hidden}

There's a difference between falling and maintaining an orbit.  Decreasing
tangential velocity will cause the orbiting object to break orbit and begin to
fall towards the object it was orbiting.  As it gets closer, it falls faster.
To reestablish a lower orbit, the object has to increase it's tangential
velocity above what it was at the higher orbit.

-Mike

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2001\03\27@152241 by Barry Gershenfeld

picon face
>moon's orbit: C = 2*pi*240,000 = 1,500,000 miles
>      [1,500,000 miles/30 days] / 24 hours/day = 2,000 MPH
>
>Looks like you go a lot faster for closer in orbits. No doubt,
>because you have to keep your centrifugal force high, else
>you come right on down. [hey, Alice, you're the orbital
>mechanic - ??????]

Well, OK, but if I want an infinitely large orbit that
means I have to come to a stop??  The other magic number I
recall is 25,000 MPH (escape velocity), which I had assumed
was an infinite orbit.

Barry

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2001\03\27@160545 by jamesnewton

face picon face
Ahhh... makes sense.. too much reading Science Fiction on my part and not
enough reading Science.

Somebody could do a really nice web page explaining all this stuff...

...if only I had time. I'll host it if needed. Takers?

"The Science of Orbital Science Fiction"

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2001\03\27@164418 by Alice Campbell

flavicon
face
> Hi Alice,
>
> I see, that's where I was incorrect. So, was Russell correct, the total
> energy for a lower orbit is higher than for a higher altitude one? If so,
> don't you need more fuel to get to a higher orbit? Where does the extra
> energy go, just into the kinetic energy of the ejected gas?
>
Youre not gonna like this.  Some of the energy goes into
slowing down the rotation of the earth, so the total angular
momentum of the earth=sattelite system is conserved.


{Quote hidden}

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2001\03\27@164646 by O'Reilly John E NORC

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face
Ok, here's a good monkey wrench to throw into the discussion.  I remember a
while back NASA was doing some experiments with generating electricity from
the magnetosphere (or one of those spheres anyway) by putting some
thingamajig on the end of a copper wire and sending it to some higher orbit.
Something like a mile higher.  I don't remember, but it was a ways.  I know
one of the things they had to figure out was how to keep the thingamajig
above the shuttle.

For your consideration.

John

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2001\03\27@165030 by Dale Botkin

flavicon
face
On Tue, 27 Mar 2001, Barry Gershenfeld wrote:

{Quote hidden}

But from what I've been reading, if you jettison it by booting it out the
hatch toward the rear, it will be travelling slower than you...  which
means it drops to a lower orbit, speeds up, and passes you.  Unless you
chuck it out the torpedo tubes, in which case it's going faster than you
so it settles into a higher, slower orbit...  Oh, my, this DOES get
confusing, doesn't it?

Dale
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discoveries, is not "Eureka!" (I found it!) but "That's funny ..."
               -- Isaac Asimov

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2001\03\27@165045 by Dipperstein, Michael

face picon face
> From: Barry Gershenfeld [barryEraseMEspam@spam@zmicro.com]
>
> >moon's orbit: C = 2*pi*240,000 = 1,500,000 miles
> >      [1,500,000 miles/30 days] / 24 hours/day = 2,000 MPH
> >
> >Looks like you go a lot faster for closer in orbits. No doubt,
> >because you have to keep your centrifugal force high, else
> >you come right on down. [hey, Alice, you're the orbital
> >mechanic - ??????]
>
> Well, OK, but if I want an infinitely large orbit that
> means I have to come to a stop??  The other magic number I
> recall is 25,000 MPH (escape velocity), which I had assumed
> was an infinite orbit.

I never understood escape velocity.  What is the direction of it's vector, and
how is it determined?  Is the 25,000 MPH figure an initial velocity, away from
the Earth, with a deceleration due to Earth's gravity, and not other
acceleration?

It would seem like an object already in orbit would have an escape velocity
based on it's orbit.

-Mike

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2001\03\27@171042 by Rick Mann

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on 3/27/01 1:48 PM, Dipperstein, Michael at spamBeGonemdippersKILLspamspam@spam@HARRIS.COM wrote:

> I never understood escape velocity.  What is the direction of it's vector, and
> how is it determined?  Is the 25,000 MPH figure an initial velocity, away from
> the Earth, with a deceleration due to Earth's gravity, and not other
> acceleration?

That is correct; it is the initial speed required, directly away from the
center of mass, to escape the gravity well in absence of other acceleration.

> It would seem like an object already in orbit would have an escape velocity
> based on it's orbit.


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2001\03\27@171053 by Rick Mann

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Sent that too soon...


on 3/27/01 1:48 PM, Dipperstein, Michael at spamBeGonemdippers@spam@spamHARRIS.COM wrote:

> I never understood escape velocity.  What is the direction of it's vector, and
> how is it determined?  Is the 25,000 MPH figure an initial velocity, away from
> the Earth, with a deceleration due to Earth's gravity, and not other
> acceleration?

That is correct; it is the initial speed required, directly away from the
center of mass, to escape the gravity well in absence of other acceleration.

> It would seem like an object already in orbit would have an escape velocity
> based on it's orbit.

An object in a higher orbit, therefore, requires a smaller initial velocity
to escape Earth's gravity. That's why it's more efficient to launch
deep-space missions from a space station (rather than trying to accelerate
all of the required fuel to the surface escape velocity).

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2001\03\27@174212 by Alice Campbell

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Hi dan,

> Alice wrote:
> >> So when you brake to get lower, does your fall toward the earth somehow
> >> increase your tangential velocity? Or do you have to do a second burn at the
> >> lower altitude to speed back up on the tangent? Is this the correction of an
> >> elliptical orbit back into a circular orbit that was discussed earlier?
> >No.  You lose sideways velocity when braking.  You have to
> >speed up again to catch up when you get to the lower orbit,
> >that or re-enter the planets atmosphere.
> >
>
> If you do only a "small" retro burn, will not you just go into slightly
> lower and faster orbit, rather than simply plunging straight for New
> Zealand? The increased force of gravity pulling downwards combined
> with forward momentum would actually speed you up as you come down
> - no?
> =============
>
well, it of course depends on what direction you point the
nozzle.  pointing it forwards slows you down, pointind it
back speeds you up, and circularizing an orbit involves a
tangent of some kind.


{Quote hidden}

No, i think its more fuel-efficient to do the 'up' part and
then the 'park_it' part in as few steps as practical.  The
transfer part is just for changing orbits.


{Quote hidden}

space cadet?

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2001\03\27@180134 by Tony Nixon

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I wonder if the Russians had this exact same discussion in the tea room
just before they launched Sputnik.

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2001\03\27@204557 by Dan Michaels

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Alice wrote:

>>
>> If you do only a "small" retro burn, will not you just go into slightly
>> lower and faster orbit, rather than simply plunging straight for New
>> Zealand? The increased force of gravity pulling downwards combined
>> with forward momentum would actually speed you up as you come down
>> - no?
>> =============
>>
>well, it of course depends on what direction you point the
>nozzle.  pointing it forwards slows you down, pointind it
>back speeds you up, and circularizing an orbit involves a
>tangent of some kind.
>

That's why I said "retro" burn, and also suggested elsewhere that
a forward but down-pointing burn should also take you into a lower
orbit, but may end up more elliptical. A small retro burn will
probably not change the shape of the orbit all that much.
===============

{Quote hidden}

Maybe can do today, but I seem to recall earlier they went first to
parking, then to geosynchronous. Considering you have "one" liftoff
point - in FLA - but you might want to park in geosyn over "any"
point of the earth, seems it might be easier to get there by leaving
a parking orbit at precisely the correct time to most efficiently
get to your final destination.

Also, since you are not launching from right on the equator, but you
eventually end up precisely over the equator, you might want to use
the parking orbit for re-alignment. Do a 2nd burn once in parking
orbit to get you going precisely in orbit over the equator, and
then the boost burn at precisely the right time to take you out
to 22,300 miles to precisely the final destination over the earth.

Seems all of this would be much more difficult going straight from
the gnd from a point not on the equator up to 22,300 in a single
shot. Here you have to figure out both how to get equitorial and
also over to the correct final destination above the earth.
==============


>>
>> Ha - I just knew Alice was really an orbital mechanic :).
>>
>space cadet?
>

Naw, Alice, exactly "not" what I meant -[can I tell them about
your Palm IR games now? :)].

best regards,
- dan
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2001\03\27@222845 by Russell McMahon

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> Net energy goes UP as you descend?! Are you sure?

No!!!
Looking back, that's what I said but it's not what I meant :-(
I think I nay have destroyed something while editing.

What I was attempting to say was that the normal formula for net energy

       E = mgh + 0.5mV^2



doesn't apply directly as g decreases with h^2 suggesting that energy
DECREASES with increasing altitude.
In fact the changing g with altitude causes the simple equation to be
inapplicabvle and you need to integrate the energy change across the change
in altitude .
> On Wed, 28 Mar 2001, Russell McMahon wrote:
>
> > The lower you go the FASTER you go.
> > The net energy goes up due to potential energy increasing.
> > (You CAN'T just apply PE = mgh as you can on the surface of the earth as
g
> > changes with height for large changes.
> > (We cheat by ignoring the change in g for small heights of up to a few
km or
> > miles.)
>
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2001\03\27@222856 by Russell McMahon

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> Corollary:  I think that if you jettison your garbage and
> don't change your orbit it will come right back to you one
> orbit later, at the point you tossed it.

Yes it will, but not how many people think.
If two orbits intersect at  a point and there are no subsequent energy
changes (losses or inputs) then the orbits will intersect at the same point
on all subsequent orbits.
The only way to escape jettisoned garbage in a lossless system is to do a
burn  AFTER abandoning the garbage.
In practice, anywhere near the earth, garbage is liable to have a MUCH
higher drag per mass ratio and therefore its subsequent orbits will not
intersect,



Russell McMahon

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2001\03\27@222908 by Russell McMahon

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> >Looks like you go a lot faster for closer in orbits. No doubt,
> >because you have to keep your centrifugal force high, else
> >you come right on down. [hey, Alice, you're the orbital
> >mechanic - ??????]
>
> Well, OK, but if I want an infinitely large orbit that
> means I have to come to a stop??  The other magic number I
> recall is 25,000 MPH (escape velocity), which I had assumed
> was an infinite orbit.


Yes.
As I noted, for a given mass orbiting a given planet in a circular orbit,
V^2 x R is constant.
At infinite R, V^2 tends to zero.

The energy required to change R by  a certain amount drops with increasing
distance. At a very large distance it takes very little effort (as you would
expect) to move further away."Escape velocity" is the velocity for a
projectile AS IT LEAVES THE PLANETS SURFACE (centre?) which would JUST allow
the projectile to coast out to infinity arriving at zero velocity (after an
infinite time). For earth escape velocity is 7 miles per second =~ 25000 mph
as noted



There have been a few other erroneous things said about orbit in a few other
posts (at least one by me :-) ).
Here's a comment on a few of them:


- In the absence of air, if a body is in orbit and you decelerate it
"slightly" it will STILL be in orbit. It will remain so as long as the new
orbit does not intersect the planets surface (and is still in orbit until
the moment of impact)( and even then if you dig it an appropriate lossless
tunnel to pass through).

- If in a circular orbit and you fire an essentially instantaneous
retrograde burn (rockets fire forward along the orbital path), you will
   - decrease the net energy
   - immediately drop in altitude
   - immediately go FASTER
   - now be at the Apogee (highest point) of an elliptical orbit
   - lower the Perigee (lowest point) even more, which will now be exactly
on the other side of the planet.

When you arrive at Perigee. if you do nothing you will continue in a mirror
image of the path your have just followed from Apogee and return to Apogee
and repeat ad infinitum.

If you fire a Posigrade (rockets fire behind you along the path of your
orbit) burn at Perigee you can "circularise" your orbit (Alice's Hohman
transfer orbit.)
This is the normal way to get to eg Mars when not using planetary flybys.
(Not to be confused with air points Fly Buys :-)).

To transfer a satellite from LEO (low earth orbit) to GSO (geosynchronous
orbit) a tangential posigrade burn is fired. The orbit radius increases and
velocity falls (!). As the burn occurs over  a period of time and may take
place in several stages the simple explanations do not apply directly.

Note that when the Shuttle releases the external tank BOTH are in identical
highly elliptical orbits with an Apogee of 100's of km and a Perigee of
around 80 to 100 km. The Shuttle subsequently makes  a SMALL burn at Apogee
to bring its next Perigee up. The external tank doesn't and next time it
dips to Perigee it re-enters due to losses from air drag.

Note that ALL closed orbits are elliptical - a circle is a special case of
an ellipse. If you are in a hyperbolic or a parabolic "orbit" then y'aint
coming back now y'hear.

For those who think the idea of slowing down as you fire posigrade sounds
incredible, try this simple experiment.
This is NOT orbital mechanics but is of similar unintuitive value:

Predict what will happen when you do the following, then try it.

- Ride a bicycle
- Place palms flat on each handlebar without actually grasping bars
- As you ride, press GENTLY on eg RH handlebar with palm of your hand.
- Predict what will happen.
- Try it
- Explain how it works!
- Wonder how you never noticed this effect
- Marvel


STAGE 2 - only vaguely related - only for the foolhardy and young.

DO THIS ON SOFT GRASS !!!!
DO NOT DO THIS ON A MOTORCYCLE
DO NOT DO THIS AT SPEED.
DO NOT DO THIS !!!! :-)

- Ride Bicycle
- Cross hands on handlebars so you hold left grip with right hand and right
grip with left hand.
- Try not to fall off.
- :-)




Enjoy

               Russell McMahon

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2001\03\28@110711 by Alan B. Pearce

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>...I always thought that lower orbits only APPEAR to be traveling faster
>because they have less distance to cover along the circumference of a
>smaller circle...

Try remembering (as a child) having a weight on the end of a piece of string,
and whirling it round your head. As the string gets longer, the radial (angular)
speed gets slower, but I am not sure if the linear speed stays the same, and I
do not have time to work it out right now.

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2001\03\28@110848 by James Newton

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Your description nicely links falling toward the planet and velocity changes
in an elliptical orbit with the orbit changing questions we have been
discussing.

Is there a simulation designed to teach one how to pilot your "spacecraft"
from orbit to orbit and to take off, slingshot to other planets and back?

P.S. Having been both young and foolhardy once (still?) I have personally
done your STAGE 2 at speed on a freeway on a motorcycle. It's not hard once
you learn how bicycle example works, which I was taught at a motorcycle
safety course. Obviously, only parts of that class sunk in. <WANE GRIN> And
I have the road rash to prove it (not from that incident or many others,
instead, I ran over a curb and damn near killed myself). And my wife (girl
friend at the time) has scars to prove it too. And she still married me.
Love is blind.

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{Original Message removed}

2001\03\28@155239 by James R. Cunningham

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The speed varies.  In high orbit, you go slower.  For example, the moon at an
average distance of about 239,000 miles orbits the earth in about 29 days, so
travels on the average about 0.6 miles/second.  An object in low earth orbit does
about 5 miles/second or thereabouts.  By comparison, earth escape velocity is about
7 miles/second and solar orbital velocity at 1 A.U. (about 93,000,000 miles) is
about 18.5 miles/second (not that that has anything to do with objects orbiting the
earth.  All of these numbers are from memory, so don't quote me too loudly.  As an
aside, though the moon is in orbit about the earth, and therefore sometimes closer
to the sun (new moon) and sometimes further (full moon) its path around the sun is
always convex toward the sun.

Jim

Alan B. Pearce wrote:

{Quote hidden}

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2001\03\28@160912 by David VanHorn

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At 03:11 PM 3/28/01 -0800, James R. Cunningham wrote:
>The speed varies.  In high orbit, you go slower.


I think this is the core of the confusion..

On the surface of the earth, you're traveling eastward at about 515 MI/Hr
at the equator. At 22500 (Geosync) you are traveling in the same direction
at something like 2944 MPH. (or 3463 if that 22500 is above the earth's
surface.

In both cases, your relative velocity to a point on the earth's surface, is
zero.

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2001\03\28@172938 by Sean H. Breheny

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Hi Dave,

Isn't it 1000 mph on the earth's surface at the equator? The circumference
is about 24,000 miles, and it takes 24 hours to go around once, or am I
missing something?

Sean


At 04:06 PM 3/28/01 -0500, you wrote:
On the surface of the earth, you're traveling eastward at about 515 MI/Hr
>at the equator. At 22500 (Geosync) you are traveling in the same direction

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2001\03\28@173834 by David VanHorn

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At 05:31 PM 3/28/01 -0500, Sean H. Breheny wrote:
>Hi Dave,
>
>Isn't it 1000 mph on the earth's surface at the equator? The circumference
>is about 24,000 miles, and it takes 24 hours to go around once, or am I
>missing something?

I may have flunked trig..
Circ is PI * R right?? Or is it PI * D?
Anyhoo, the proportions should be right, and the numbers within an order of
magnitude (physicist handwaving occurs)


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2001\03\28@175351 by James Paul

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All,

Excuse me for butting in, but I think the original question was
something about gravity and orbital speed (Velocity).  Someone
was saying something about the moon and it's relative velocity
through space.  And knowing that the moon is about 240000 miles
from earth and it revolves around the earth, therefore 240000 miles
is the radius.  And the radius times 2pi is the circumference of
the basic circular orbit of the moon around the earth.  Now, knowing
this (circumference) and knowing it takes 30 days approximately for
the moon to orbit the earth once, you can calculate the velocity.
Which is (((2*pi*240000)/30)/24) = moons velocity around the earth.
This figures out to be just less than 2100 mph.  Note that these are
just approximations.  Nothing is exact in this calculation.
Am I correct here? Correct me if I'm wrong.
(Like I really have to worry that no one will)

                                           Regards,

                                             Jim










On Wed, 28 March 2001, David VanHorn wrote:

{Quote hidden}

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2001\03\28@180222 by James Paul

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All,

Oh yeah, I forgot.  Isn't the circumference of a circle a part of
geometry and not trig?  Just an observation.

                                               Regards,

                                                 Jim


On Wed, 28 March 2001, David VanHorn wrote:

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2001\03\28@181442 by David VanHorn

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At 03:01 PM 3/28/01 -0800, James Paul wrote:
>  All,
>
>  Oh yeah, I forgot.  Isn't the circumference of a circle a part of
>  geometry and not trig?  Just an observation.

I really should have stayed awake more in that class :)
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2001\03\28@183401 by James R. Cunningham

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David VanHorn wrote:

> At 03:11 PM 3/28/01 -0800, James R. Cunningham wrote:
> >The speed varies.  In high orbit, you go slower.
>
> I think this is the core of the confusion..

Er uh, what's the core of the confusion?  I'm afraid I've missed your point.

> On the surface of the earth, you're traveling eastward at about 515 MI/Hr
> at the equator.

Diameter is about 7980 miles, more or less.  The earth rotates in about 23
hours 56 minutes (not 24 hours).  So tangential velocity on the surface at the
equator is about 1048 mph, more or less (I didn't use the exact diameter or the
exact rotational period).  But this has nothing whatever to do with orbital
velocities.

> At 22500 (Geosync) you are traveling in the same direction
> at something like 2944 MPH. (or 3463 if that 22500 is above the earth's
> surface.

From memory, geosync orbit is about 22,300 miles above the CENTER of the earth
(don't hold me to the 22,300, I didn't look it up either, but it's about 18,300
above the surface). So 2*22300*pi/23.933=5854 mph, roughly.  I think you may be
using the radius to compute circumference when you intended to use diameter.

> In both cases, your relative velocity to a point on the earth's surface, is
> zero.

True, except that the object in geosynchronus orbit will describe an analemna
on the sky as it moves north and south during its orbit. And the surface
velocity has nothing to do with orbital mechanics.

Cheers,

Jim

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2001\03\28@195838 by David VanHorn

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>
>True, except that the object in geosynchronus orbit will describe an analemna
>on the sky as it moves north and south during its orbit. And the surface
>velocity has nothing to do with orbital mechanics.

I think people are thinking of orbital speed as somehow relative to a point
on the earth..

It would be interesting to run the numbers, and see what the forward
velocity is in the case of GEO and LEO, and see if lower is faster in an
absolute sense, or just with respect to a point on the ground.


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2001\03\28@201527 by Barry Gershenfeld

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>- If in a circular orbit and you fire an essentially instantaneous
>retrograde burn (rockets fire forward along the orbital path), you will
...
>    - now be at the Apogee (highest point) of an elliptical orbit
>    - lower the Perigee (lowest point) even more, which will now be exactly
>on the other side of the planet.
...
>If you fire a Posigrade (rockets fire behind you along the path of your
>orbit) burn at Perigee you can "circularise" your orbit (Alice's Hohman
>transfer orbit.)


Problem.
My retrograde firing lowered the other side of the orbit, creating
a perigee.  Now I want to lower the apogee to make the new smaller
orbit circular.  So I go around to perigee and fire again.
Posigrade?
Drops the Apogee?  Not in my simple mind. (Here we go again...)

Barry

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2001\03\28@211321 by James R. Cunningham

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Lower is faster in an absolute sense.  But one of the interesting tidbits of
orbital mechanics is that you 'speed up' in order to 'slow down'.  If you are in a
low level circular orbit, and wish to move to a high level circular orbit then you
first do a burn to increase your speed to the perigee speed of an elliptical orbit
whose apogee is at the altitude of the desired new circular orbit.  Because of the
speed lost while climbing out of the gravity well, when you arrive at apogee, you
will be moving too slow to maintain altitude, so must burn again to increase your
speed to that required to maintain a circular orbit at the higher altitude.  At
which time your two forward burns will have allowed you to achieve a slower orbit
at a higher altitude.  I've always thought that to be fascinating.  It works in
reverse on the way back down.

Jim

David VanHorn wrote:

> It would be interesting to run the numbers, and see what the forward
> velocity is in the case of GEO and LEO, and see if lower is faster in an
> absolute sense, or just with respect to a point on the ground.

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2001\03\28@212124 by James R. Cunningham

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Barry Gershenfeld wrote:

> >- If in a circular orbit and you fire an essentially instantaneous
> >retrograde burn (rockets fire forward along the orbital path), you will
>  ...
> >    - now be at the Apogee (highest point) of an elliptical orbit
> >    - lower the Perigee (lowest point) even more, which will now be exactly
> >on the other side of the planet.
>  ...
> >If you fire a Posigrade (rockets fire behind you along the path of your
> >orbit) burn at Perigee you can "circularise" your orbit (Alice's Hohman
> >transfer orbit.)

As an aside, a Hohman transfer orbit minimises delta v and fuel burn, but not
travel time.

> Problem.
> My retrograde firing lowered the other side of the orbit, creating
> a perigee.  Now I want to lower the apogee to make the new smaller
> orbit circular.

At this point, you are at perigee and traveling too fast for a circular orbit
because of the speed gained while descending in the gravity well.  So while at
perigee, you fire retrograde to slow down to the speed for a circular orbit at
the perigee altitude.

>  So I go around to perigee and fire again.
> Posigrade?
> Drops the Apogee?  Not in my simple mind. (Here we go again...)

Try firing retrograde on the other side of the planet while at perigee for the
transfer orbit..

Jim

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2001\03\28@225039 by Russell McMahon

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> At 03:11 PM 3/28/01 -0800, James R. Cunningham wrote:
> >The speed varies.  In high orbit, you go slower.
>
>
> I think this is the core of the confusion..


No - just a part of the greater confusion :-)
The motion of an obsrever on the earth's surface does need to be taken
account of BUT the increase in velocity as the orbital radius decreases is
real and absolute.

Geosynchronous orbit is the radius at which the APPARENT motion of a
satellite relative to a surface observer is zero.
Below geosynchronous orbit a West to East travelling satellite in an orbit
at the equator will APPEAR to travel Eastwards. Above geosynchronous orbit
(and there are few of these) it will APPEAR to travel Westwards. This is
entirely due to matching the velocity of the planets surface. If the earth
had a different period of revolution (length of day) the satellite moions
would be essentially unaffected BUT geosynchronous orbit would be at a
different altitude than it is now ! :-)

Does that add to the confusion? :-)



RM

>
> On the surface of the earth, you're traveling eastward at about 515 MI/Hr
> at the equator. At 22500 (Geosync) you are traveling in the same direction
> at something like 2944 MPH. (or 3463 if that 22500 is above the earth's
> surface.
>
> In both cases, your relative velocity to a point on the earth's surface,
is
{Quote hidden}

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2001\03\29@005319 by Kari Lehikko

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James,

> Is there a simulation designed to teach one how to pilot your "spacecraft"
> from orbit to orbit and to take off, slingshot to other planets and back?

Have you ever played David Braben's (the coder of Elite) 'FRONTIER'
-series? AFAIK it is the most realistic space flight game. You can
manually try to do slingshot orbits and other things. It also simultes
the boredom during spaceflight :)

There are also some space shuttle simulators that you could check out.
Anyhow, if I feel like piloting a spacecraft, I'll fire up my old 486
and do some serius slingshoting from planet to planet. As I am at work
now, I am unable to check out the game-house that published
'FRONTIER'-series, Sorry.

Hmm... I imagine that a real world simulation would be like:

1. Enter new values to your flight computer. (Planet and the altitude of
the wanted orbit)
2. Push the 'calculate slingshot orbit' -key.
3. Wait a bit.
4. Push the 'change orbit according to calculated values' -key.
5. Wait until you have reached the orbit you wanted.

At least I assume that NASA would not give the flight calculations of a
zillion dollar spacecraft in the hands of a human.

I wouldn't.  :)

Regards,

- Kari -

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2001\03\29@070519 by Russell McMahon

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> > Is there a simulation designed to teach one how to pilot your
"spacecraft"
> > from orbit to orbit and to take off, slingshot to other planets and
back?


Microsoft Space Simulator is fairly real.
No longer available new AFAIK.
Too real to sell well as a game I suspect.

RM

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2001\03\29@082543 by Jeszs

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Satellite ToolKit at http://www.stk.com is good and has a free version really
interesting.
Do not miss the 'Satellite Primer' document, an excellent review of the
orbital mechanics as discussed in the list these days.

I am still interested in the question: any PIC in space currently?

--------------------
Jeszs Gonzalo
Lesn (SPAIN)
--------------------

{Original Message removed}

2001\03\29@130823 by Dan Michaels

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At 08:31 AM 3/29/01 +0300, you wrote:
>James,
>
>> Is there a simulation designed to teach one how to pilot your "spacecraft"
>> from orbit to orbit and to take off, slingshot to other planets and back?
>

Don't know about that, but NASA does have a Solar System Simulator, for
virtual voyagers:

http://space.jpl.nasa.gov/


- dan michaels
http://www.oricomtech.com
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