Minimum fuel launch termination point

http://www.braeunig.us/space/orbmech.htm

This article on orbital mechanics seems excellent!!
I had a question on one statement, whcih is just befoe the "Click here
for example problem #3.8"

It says, "Please note that in practice spacecraft launches are usually
terminated at either perigee or apogee, i.e. =90. This condition
results in the minimum use of fuel"

I'm thinking a spacecraft in orbit has the same energy (kinetic and
potential total) at each point in the orbit.. And that there is only
a set amount of energy available to provide to the spacecraft to put
it at that orbit. Thus to achieve a designated orbit, the amount of
fuel used would be the same regardless of where the thrust terminated.
So why would it matter at what point in the orbit the launch is
terminated? For simplicity, it appears the author is not considering
air resistance on initial launch.

Anyone with any thoughts? Stan

wrote in message
...
: http://www.braeunig.us/space/orbmech.htm
:
: This article on orbital mechanics seems excellent!!
: I had a question on one statement, whcih is just befoe the "Click here
: for example problem #3.8"
:
: It says, "Please note that in practice spacecraft launches are usually
: terminated at either perigee or apogee, i.e. =90. This condition
: results in the minimum use of fuel"



That's kinda funny, a self-fullfilling prophecy. If the burn
ends when you are not at 90 degrees to the Earth's surface
then you are not at perigee or apogee. If you thrust at apogee
or perigee then you'll have a new perigee or apogee.

: I'm thinking a spacecraft in orbit has the same energy (kinetic and
: potential total) at each point in the orbit.. And that there is only
: a set amount of energy available to provide to the spacecraft to put
: it at that orbit. Thus to achieve a designated orbit, the amount of
: fuel used would be the same regardless of where the thrust terminated.
: So why would it matter at what point in the orbit the launch is
: terminated? For simplicity, it appears the author is not considering
: air resistance on initial launch.
:
: Anyone with any thoughts? Stan

1) Energy is relative because E = 1/2mv^2 and v is relative.
For example a shuttle docking at a space station has a low
relative energy to the space station. Relative to the Earth it has
a high energy and will heat up on re-entering atmosphere, losing
its energy to radiation (heat). If it did not it would impact with
high energy.

2) Key to your question is the term "designated orbit".
You can shoot straight up vertically (minimum fuel) and then
fall straight back down, reaching zenith just as the space station
passes overhead, but the space station will be going too fast
horizontally to dock with it.
In other words you'll have high energy relative to the space station.
You need the same horizontal velocity as the space station
for the designated orbit. Remember that orbits are elliptical,
so even though you can go higher than the space station
and cross over to go to a lower perigee than the station has,
you have to match velocity as well. That burns more fuel
than needed.

On Fri, 03 Aug 2007 14:18:07 GMT, "Androcles"
wrote:

1) Energy is relative because E = 1/2mv^2 and v is relative.
For example a shuttle docking at a space station has a low
relative energy to the space station. Relative to the Earth it has
a high energy and will heat up on re-entering atmosphere, losing
its energy to radiation (heat). If it did not it would impact with
high energy.

2) Key to your question is the term "designated orbit".
You can shoot straight up vertically (minimum fuel) and then
fall straight back down, reaching zenith just as the space station
passes overhead, but the space station will be going too fast
horizontally to dock with it.
In other words you'll have high energy relative to the space station.
You need the same horizontal velocity as the space station
for the designated orbit. Remember that orbits are elliptical,
so even though you can go higher than the space station
and cross over to go to a lower perigee than the station has,
you have to match velocity as well. That burns more fuel
than needed.

I really don't follow your ideas. So why does one need to finish the
burn at apogee or perigee in order to burn minimum fuel?

wrote in message
...
: On Fri, 03 Aug 2007 14:18:07 GMT, "Androcles"
: wrote:
:
: 1) Energy is relative because E = 1/2mv^2 and v is relative.
: For example a shuttle docking at a space station has a low
: relative energy to the space station. Relative to the Earth it has
: a high energy and will heat up on re-entering atmosphere, losing
: its energy to radiation (heat). If it did not it would impact with
: high energy.
:
: 2) Key to your question is the term "designated orbit".
: You can shoot straight up vertically (minimum fuel) and then
: fall straight back down, reaching zenith just as the space station
: passes overhead, but the space station will be going too fast
: horizontally to dock with it.
: In other words you'll have high energy relative to the space station.
: You need the same horizontal velocity as the space station
: for the designated orbit. Remember that orbits are elliptical,
: so even though you can go higher than the space station
: and cross over to go to a lower perigee than the station has,
: you have to match velocity as well. That burns more fuel
: than needed.
:
: I really don't follow your ideas. So why does one need to finish the
: burn at apogee or perigee in order to burn minimum fuel?

For the same reason you can stop pedalling your bicycle when you've
reached the top of a hill (apogee). The only way to go is down, but if
you want to pedal downhill and go faster at the bottom (perigee)
then by all means do so.
You can also stop pedalling at the bottom of the hill if you have
built up enough speed to coast to the top. A traditional roller
coaster lifts the vehicle to the top of the first hill and then lets it
go, no further energy is needed for the rest of the ride.
http://static.howstuffworks.com/gif/roller-coaster8.jpg
Other more modern types use a horizontal catapult.
http://en.wikipedia.org/wiki/Flight_...Kings_Dominion)

http://tinyurl.com/2zj7j5

So.. apogee, height but no speed. perigee, speed but no height.

Dear pstanle...:

On Aug 3, 6:26 am, wrote:
http://www.braeunig.us/space/orbmech.htm

This article on orbital mechanics seems excellent!!
I had a question on one statement, whcih is just befoe
the "Click here for example problem #3.8"

It says, "Please note that in practice spacecraft
launches are usually terminated at either perigee or
apogee, i.e. =90. This condition results in the
minimum use of fuel"

I'm thinking a spacecraft in orbit has the same energy
(kinetic and potential total) at each point in the orbit..
And that there is only a set amount of energy available
to provide to the spacecraft to put it at that orbit. Thus
to achieve a designated orbit, the amount of fuel used
would be the same regardless of where the thrust
terminated.

Thrust 90 deg from the vehicle path does not contribute *directly* to
path energy. It may alter the path only by "precessing" it.

So why would it matter at what point in the orbit the
launch is terminated? For simplicity, it appears the
author is not considering air resistance on initial launch.

Anyone with any thoughts? Stan

I would simply assume that thrust is always oriented along the path
(for maximum change in energy), and apogee and perigee is where this
thrust has no radial (towards or away from the orbited body)
component. So it does not resist / assist gravity, but is confined to
altering path energy only.

My guess, anyway...

David A. Smith