| Topic: |
Science > Physics |
| User: |
"Steven Gray" |
| Date: |
26 Jul 2003 09:59:34 AM |
| Object: |
Re: Orbital Angular Momentum and Energy |
"Old Man" <nomail@nomail.net> wrote in news:3f20b4e4_4@newsfeed:
Consider two equal energy orbits, one circular, and the other
elliptical. The orbits have the same energy but have different angular
momentum, the circular orbit having the highest angular momentum.
We wish to calculate the energy required to transfer the space
vehicle from one of these orbits to the other. The direction is
irrelevant. Why should energy have to be spent to perform a transfer
between two orbits of the same energy? Good question. Old Man
does not know the answer, but is convinced that the transfer cannot
be made without expending energy.
Because the only way we have of effecting orbital changes involves rockets.
Energy has to go into the expelled reaction mass. I suspect the answer is
going to depend on rocket parameters, such as exhaust velocity and thrust.
One way to make the transfer is as follows:
The circular orbit and the elliptical orbit intersect at two points.
Choose one of the intersection points. Since the energies and
potential energies (the radial vectors being identical) for the two
different orbits are equal at the intersection, the kinetic energies are
equal, meaning that the speeds are equal, but the velocity vectors
point in different directions. All that is required for the transfer
between orbits is a burn that changes the direction of the velocity
vector without changing the speed. An isosceles triangle supplies
the solution for delta-v.
And it could be accomplished without expenditure of energy, if there were
only a convenient pivot about which we could throw a rope <G>.
--
Steve Gray
sgray2@cfl.rr.com
.
|
|
| User: "Ken S. Tucker" |
|
| Title: Re: Orbital Angular Momentum and Energy |
26 Jul 2003 12:50:54 PM |
|
|
(Steven Gray) wrote in message news:<Xns93C46FE16BF47sgray2cflrrcom@65.32.1.7>...
points.
Neat question...
OLDMAN:
Consider a space vehicle in bound orbit about the Sun. In a given
plane that includes the Sun, allowed orbits are uniquely defined
(neglecting orbital orientation) by two parameters: the total energy
of the orbit, E = KE + PE = constant, where E < 0, and the orbital
angular momentum, L. The equation of motion can be written as
E = -GMm / | r | + (1 / 2) m [ d| r | / dt ]^2 + L^2 / 2m| r |^2
where, for any given orbit, E and L are constants. r is the radius
vector from the Sun, and m is the mass of the space vehicle. GM
is the gravitational mass parameter for the Sun.
The general shape of all bound orbits is elliptical. Amongst all
possible orbits of given energy, the circular orbit has the highest
angular momentum, and amongst all possible orbits of given
angular momentum, the circular orbit has the lowest energy.
Consider two equal energy orbits, one circular, and the other
elliptical. The orbits have the same energy but have different angular
momentum, the circular orbit having the highest angular momentum.
We wish to calculate the energy required to transfer the space
vehicle from one of these orbits to the other. The direction is
irrelevant. Why should energy have to be spent to perform a transfer
between two orbits of the same energy? Good question. Old Man
does not know the answer, but is convinced that the transfer cannot
be made without expending energy.
One way to make the transfer is as follows:
The circular orbit and the elliptical orbit intersect at two points.
Choose one of the intersection points. Since the energies and
potential energies (the radial vectors being identical) for the two
different orbits are equal at the intersection, the kinetic energies are
equal, meaning that the speeds are equal, but the velocity vectors
point in different directions. All that is required for the transfer
between orbits is a burn that changes the direction of the velocity
vector without changing the speed. An isosceles triangle supplies
the solution for delta-v.
The above maneuver between equal energy orbits (but with different
orbital angular momentum) requires the expenditure energy, and the
reverse transfer requires an identical and additional amount of energy
expenditure. It is not reversable.
Sorry OLDMAN, I disagree...
I may be cheating a bit, but I think it is reversible. To do this,
one must absorb the reaction mass used to deflect the orbit.
Imagine two Gemini spacecraft locked together with a spring
compressed against the other in a circular orbit. At some point
"N" (North) they release the energy stored in the spring, (BOING)
perpendicular to the orbital direction but still in the same plane.
As a result each will assume an elliptical orbit, with Gemini U
going Up and Gemini D going, hmmmm...Down.
180 degrees from N the orbits will cross at point S (South).
At this point a rendevous is accomplished that recompresses
the spring and recovers BOING energy and re-orbits the craft
back into a circular orbit, thus reversing the orbital transfer
with no energy expenditure (aside from spring inefficiencies).
I cheated by adding BOING energy in the first maneuver
that separated Gemini U and D, but I think I can uncheat.
When Gemini U and D are at maximum separation which
is at the E (East) or W (West) point of the orbit, a tether
joining the vehicles may be used to recompress the springs,
by removing energy BOING from the Gemini's orbits and
restoring it to the vehicles. The vehicles will now have
energy equal to the original circular orbit, but instead each
will occupy an elliptic orbit of equal energy.
I've implied two Gemini's of equal mass for simplicity,
but the whole point of the post was to show that no energy
is lost if one can reabsorb the reaction mass used for the
orbital transition.
Old Man has not shown, but
beleives that the maneuver discribed herein is the minimum required
transfer energy.
In a similar manner, in QM, for the hydrogen atom, there is no direct
way to force a transition between degenerate states (equal energy)
that possess different orbital angular momentum. one must first
excite the atom into a higher energy state, and then allow it to decay
into the state with differing orbital angular momentum. Maybe.
That is, unless there is a way of absorbing orbital angular momentum
without absorbing energy, there is no way. Old Man is probably going
to get flamed for making silly rules, but it sounds so good!
Can a system of bound particles absorb orbital angular momentum
without absorbing energy? Old Man says no. Anybody wanna bet?
After thought, with hang-over:
Well no. L must be a good quantum number. Otherwise, because
of parity, L = 0 and L = 1 can't be mixed, but L = 0, 2, 4 , ... can be
mixed, and L = 1, 3, 5, ... can be mixed. Oh Well. This is getting
confusing. Goodbye! [Old Man]
Prior to considering further arguments, I'll wait and see if I cheated
to much!
STEVEN GRAY:
And it could be accomplished without expenditure of energy, if there were
only a convenient pivot about which we could throw a rope <G>.
It was Steven's comments that suggested to me some kind of tether.
Regards
Ken S. Tucker
.
|
|
|
| User: "Steven Gray" |
|
| Title: Re: Orbital Angular Momentum and Energy |
26 Jul 2003 03:02:02 PM |
|
|
(Ken S. Tucker) wrote in
news:2202379a.0307260950.669af0d3@posting.google.com:
The above maneuver between equal energy orbits (but with different
orbital angular momentum) requires the expenditure energy, and the
reverse transfer requires an identical and additional amount of energy
expenditure. It is not reversable.
Sorry OLDMAN, I disagree...
I may be cheating a bit, but I think it is reversible. To do this,
one must absorb the reaction mass used to deflect the orbit.
Imagine two Gemini spacecraft locked together with a spring
compressed against the other in a circular orbit. At some point
"N" (North) they release the energy stored in the spring, (BOING)
perpendicular to the orbital direction but still in the same plane.
As a result each will assume an elliptical orbit, with Gemini U
going Up and Gemini D going, hmmmm...Down.
180 degrees from N the orbits will cross at point S (South).
At this point a rendevous is accomplished that recompresses
the spring and recovers BOING energy and re-orbits the craft
back into a circular orbit, thus reversing the orbital transfer
with no energy expenditure (aside from spring inefficiencies).
I don't know that the orbits will have the same period, nor do I think
they'll intersect 180 degrees away. They will, of course, intersect at the
point of separation, but the two craft may not be there at the same time.
I don't, however, have any problem with your basic idea that energy can be
recovered.
--
Steve Gray
sgray2@cfl.rr.com
.
|
|
|
| User: "Old Man" |
|
| Title: Re: Orbital Angular Momentum and Energy |
26 Jul 2003 05:45:00 PM |
|
|
Steven Gray <sgray2@cfl.rr.com> wrote in message
news:Xns93C4A3281AF4sgray2cflrrcom@65.32.1.7...
dynamics@vianet.on.ca (Ken S. Tucker) wrote in
news:2202379a.0307260950.669af0d3@posting.google.com:
The above maneuver between equal energy orbits (but with different
orbital angular momentum) requires the expenditure energy, and the
reverse transfer requires an identical and additional amount of energy
expenditure. It is not reversable.
Sorry OLDMAN, I disagree...
I may be cheating a bit, but I think it is reversible. To do this,
one must absorb the reaction mass used to deflect the orbit.
Imagine two Gemini spacecraft locked together with a spring
compressed against the other in a circular orbit. At some point
"N" (North) they release the energy stored in the spring, (BOING)
perpendicular to the orbital direction but still in the same plane.
As a result each will assume an elliptical orbit, with Gemini U
going Up and Gemini D going, hmmmm...Down.
180 degrees from N the orbits will cross at point S (South).
At this point a rendevous is accomplished that recompresses
the spring and recovers BOING energy and re-orbits the craft
back into a circular orbit, thus reversing the orbital transfer
with no energy expenditure (aside from spring inefficiencies).
I don't know that the orbits will have the same period, nor do I think
they'll intersect 180 degrees away. They will, of course, intersect at
the
point of separation, but the two craft may not be there at the same time.
I don't, however, have any problem with your basic idea that energy can be
recovered.
Agreed. [Old Man]
--
Steve Gray
sgray2@cfl.rr.com
.
|
|
|
|
|
| User: "Old Man" |
|
| Title: Re: Orbital Angular Momentum and Energy |
26 Jul 2003 05:39:11 PM |
|
|
Ken S. Tucker <dynamics@vianet.on.ca> wrote in message
news:2202379a.0307260950.669af0d3@posting.google.com...
sgray2@cfl.rr.com (Steven Gray) wrote in message
news:<Xns93C46FE16BF47sgray2cflrrcom@65.32.1.7>...
points.
Neat question...
OLDMAN:
Consider a space vehicle in bound orbit about the Sun. In a given
plane that includes the Sun, allowed orbits are uniquely defined
(neglecting orbital orientation) by two parameters: the total energy
of the orbit, E = KE + PE = constant, where E < 0, and the orbital
angular momentum, L. The equation of motion can be written as
E = -GMm / | r | + (1 / 2) m [ d| r | / dt ]^2 + L^2 / 2m| r |^2
where, for any given orbit, E and L are constants. r is the radius
vector from the Sun, and m is the mass of the space vehicle. GM
is the gravitational mass parameter for the Sun.
The general shape of all bound orbits is elliptical. Amongst all
possible orbits of given energy, the circular orbit has the highest
angular momentum, and amongst all possible orbits of given
angular momentum, the circular orbit has the lowest energy.
Consider two equal energy orbits, one circular, and the other
elliptical. The orbits have the same energy but have different angular
momentum, the circular orbit having the highest angular momentum.
We wish to calculate the energy required to transfer the space
vehicle from one of these orbits to the other. The direction is
irrelevant. Why should energy have to be spent to perform a transfer
between two orbits of the same energy? Good question. Old Man
does not know the answer, but is convinced that the transfer cannot
be made without expending energy.
One way to make the transfer is as follows:
The circular orbit and the elliptical orbit intersect at two points.
Choose one of the intersection points. Since the energies and
potential energies (the radial vectors being identical) for the two
different orbits are equal at the intersection, the kinetic energies are
equal, meaning that the speeds are equal, but the velocity vectors
point in different directions. All that is required for the transfer
between orbits is a burn that changes the direction of the velocity
vector without changing the speed. An isosceles triangle supplies
the solution for delta-v.
The above maneuver between equal energy orbits (but with different
orbital angular momentum) requires the expenditure energy, and the
reverse transfer requires an identical and additional amount of energy
expenditure. It is not reversable.
Sorry OLDMAN, I disagree...
I may be cheating a bit, but I think it is reversible. To do this,
one must absorb the reaction mass used to deflect the orbit.
Imagine two Gemini spacecraft locked together with a spring
compressed against the other in a circular orbit. At some point
"N" (North) they release the energy stored in the spring, (BOING)
perpendicular to the orbital direction but still in the same plane.
As a result each will assume an elliptical orbit, with Gemini U
going Up and Gemini D going, hmmmm...Down.
Tricky maneuver! Yes, with springs, the motion is periodic and
reversible for each vehicle when they are considered as separate
systems, which is permissible, but the orientation of the spring
axis determines the extent of changes in angular momentum and
energy of the separate orbits. The total angular momentum of
the system is not changed by spring action between the two
vehicles.
Providing a relative delta-v in the radial direction between the two
vehicles does not change the orbital angular momentum of either
vehicle, but it does change the energy and kinetic energy of the
orbit for each vehicle. In the case of a radial impulse from a circular
orbit, the kinetic energy of both vehicles is increased by an equal
amount.
So, in order to provide changes in orbital angular momentum, we'll
assume that, from an initial circular orbit, the impulse is applied at
right angles to the radius vector and parallel to the initial velocity
vector. In this case, the kinetic enery and angular momentum of the
leading vehicle are incresed, while the kinetic enery and angular
momentum of the trailing vehicle are decreased. Furthermore,
energy has been expended.
180 degrees from N the orbits will cross at point S (South).
At this point a rendevous is accomplished that recompresses
the spring and recovers BOING energy and re-orbits the craft
back into a circular orbit, thus reversing the orbital transfer
with no energy expenditure (aside from spring inefficiencies).
The two orbits will certainly intersect at 360 degrees. However, if
the semi-major axis of the two orbits are not equal, their periods
will be out of synchronization, and they will not re-arrive at the
intersection at the same time. If the impulse is perpendicular, as
initially assumed, then the two orbits will be shifted in orientation, but
with equal energy, orbital angular momentum, and period. I think.
Regardless, the interconnecting spring can be permanently attached
so that the vehicles are periodically pulled back together. In this case,
after one spring cycle, every thing will look as it did initially, and no
energy will have been expended. Sounds right. Nothing ventured,
nothing gained. Unless I'm missing somthing, the orbital angular
momentum and energy are unchanged. The process is periodic and
reversable.
So, the question is: in the case of an impulse that is applied parallel
to the initial velocity vector, is the angular momentum of each vehicle
changed? The answer is yes. Was energy expended? The answer
is yes. For perpendicular impulse, is the angular momentum of each
vehicle changed? The answer is no. Was energy expended? The
answer is yes. Finally, are changes in orbital angular momentum for
each vehicle reversible by spring action? the answer is yes.
[Old Man]
I cheated by adding BOING energy in the first maneuver
that separated Gemini U and D, but I think I can uncheat.
When Gemini U and D are at maximum separation which
is at the E (East) or W (West) point of the orbit, a tether
joining the vehicles may be used to recompress the springs,
by removing energy BOING from the Gemini's orbits and
restoring it to the vehicles. The vehicles will now have
energy equal to the original circular orbit, but instead each
will occupy an elliptic orbit of equal energy.
I've implied two Gemini's of equal mass for simplicity,
but the whole point of the post was to show that no energy
is lost if one can reabsorb the reaction mass used for the
orbital transition.
Old Man has not shown, but
beleives that the maneuver discribed herein is the minimum required
transfer energy.
In a similar manner, in QM, for the hydrogen atom, there is no direct
way to force a transition between degenerate states (equal energy)
that possess different orbital angular momentum. one must first
excite the atom into a higher energy state, and then allow it to decay
into the state with differing orbital angular momentum. Maybe.
That is, unless there is a way of absorbing orbital angular momentum
without absorbing energy, there is no way. Old Man is probably going
to get flamed for making silly rules, but it sounds so good!
Can a system of bound particles absorb orbital angular momentum
without absorbing energy? Old Man says no. Anybody wanna bet?
After thought, with hang-over:
Well no. L must be a good quantum number. Otherwise, because
of parity, L = 0 and L = 1 can't be mixed, but L = 0, 2, 4 , ... can be
mixed, and L = 1, 3, 5, ... can be mixed. Oh Well. This is getting
confusing. Goodbye! [Old Man]
Prior to considering further arguments, I'll wait and see if I cheated
to much!
STEVEN GRAY:
And it could be accomplished without expenditure of energy, if there
were
only a convenient pivot about which we could throw a rope <G>.
It was Steven's comments that suggested to me some kind of tether.
Regards
Ken S. Tucker
.
|
|
|
|
|
|
Related Articles |
|
|
