>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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>
> 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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>
> >From: "Matthew Mucker" <TakeThisOuTmmuckerEraseMEspam_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.

>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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> 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 - ??????]

> 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.

> 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.

>
> 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...

>
> 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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>
>

>
>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.

> From: Dan Michaels [EraseMEoricom@spam@uswest.net]

> 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
> >

>
> 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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>
>

> >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.

> >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.
> ===================
>

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

>
>> >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.
>> ===================
>>
>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.
>

> >...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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>
> 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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>
> 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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> zero.
>
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>
>