Inclination change: worst case

Dumb question: If one were in circular LEO and crazy enough to want a
90-degree change of inclination, what's the cost in delta-v? Is it, as
intuition is telling me, equivalent to throwing away all your ~9 kps
and re-acquiring it all?

I'm aware of the corrections for energy spent gaining altitude and
fighting drag on the initial launch, and whatever freebie you may have
gotten from the earth's rotation in a W-E launch. But is the
broad-brush picture "almost as hard as getting to orbit in the first
place"..? And for a still crazier 180-degree change, "twice as
hard"..?

(Assume a spherical Earth, homogeneous frictionless elephant, etc)

Monte Davis wrote:
Dumb question: If one were in circular LEO and crazy enough to want a
90-degree change of inclination, what's the cost in delta-v? Is it,
as
intuition is telling me, equivalent to throwing away all your ~9 kps
and re-acquiring it all?

At first glance, a brute force 90-degree plane change would require
1.414x (square root 2 times) as much delta-V as it took to get to
orbit, so 7.8x1.4=11km/s. You have to kill all that horizontal velocity
and replace it with orbital velocity perpendicular to your first.

It would, in fact, be less costly in delta-V to do something like
accelerate to escape velocity (~3.3km/s above orbital velocity), drift
up to about the L1 point or swing around the moon, and drop back to
Earth in a new, desired orbit. I think.

I await correction.

Mike Miller, Materials Engineer

On Tue, 22 Mar 2005 02:01:34 +0000, Monte Davis wrote:

Dumb question: If one were in circular LEO and crazy enough to want a
90-degree change of inclination, what's the cost in delta-v? Is it, as
intuition is telling me, equivalent to throwing away all your ~9 kps
and re-acquiring it all?

Yes, but you should use vector math. So it's not as bad as you think.

I'm aware of the corrections for energy spent gaining altitude and
fighting drag on the initial launch, and whatever freebie you may have
gotten from the earth's rotation in a W-E launch. But is the
broad-brush picture "almost as hard as getting to orbit in the first
place"..? And for a still crazier 180-degree change, "twice as
hard"..?

(Assume a spherical Earth, homogeneous frictionless elephant, etc)

Even with vector math, it's twice the delta v. At some point you can climb out
of the gravity well, make the inclination change at apogee (velocity is
low) then come back down again cheaper.

--
Craig Fink
Courtesy E-Mail Welcome @

"Monte Davis" wrote in message
news
Dumb question: If one were in circular LEO and crazy enough to want a
90-degree change of inclination, what's the cost in delta-v? Is it, as
intuition is telling me, equivalent to throwing away all your ~9 kps
and re-acquiring it all?

It is, but you can play some games here. What you can do is enter a very
highly elliptic orbit, so that you can do your plane change when furthest
away from the earth where your velocity is very small. Once this is done,
you do another burn to re-enter your circular LEO orbit. If you do the math
you'll find that this is far cheaper than doing the 90 degree plane change
in LEO.

Jeff
--
Remove icky phrase from email address to get a valid address.

" wrote in
ups.com:


Monte Davis wrote:
Dumb question: If one were in circular LEO and crazy enough to want a
90-degree change of inclination, what's the cost in delta-v? Is it,
as
intuition is telling me, equivalent to throwing away all your ~9 kps
and re-acquiring it all?

At first glance, a brute force 90-degree plane change would require
1.414x (square root 2 times) as much delta-V as it took to get to
orbit, so 7.8x1.4=11km/s. You have to kill all that horizontal velocity
and replace it with orbital velocity perpendicular to your first.

It would, in fact, be less costly in delta-V to do something like
accelerate to escape velocity (~3.3km/s above orbital velocity), drift
up to about the L1 point or swing around the moon, and drop back to
Earth in a new, desired orbit. I think.

I await correction.

None needed - as others have posted, you're correct.

To generalize a bit, an orbital plane change *without* an altitude change
is a purely geometric problem. In the general case, the velocity vectors of
the old and new orbits form an isoceles triangle, so the delta-V is
2*sin(theta/2) times the orbital velocity. There are a couple of special
cases, as you and others have noted: a 60-degree plane change forms an
equilateral triangle, so delta-V equals orbital velocity. A 90-degree plane
change forms a right triangle, so delta-V equals sqrt(2) times orbital
velocity. And a 180-degree plane change simply reverses the direction of
velocity, so delta-V equals two times orbital velocity.

In all the above cases, it is seen that delta-V is a function of orbital
velocity, so it is advantageous to perform plane changes at the point in
the orbit where velocity is lowest (i.e. apogee). As others have posted, in
the limiting case this means achieving escape velocity, performing the
plane change at infinity for zero cost, then recircularizing at perigee.
Escape velocity is sqrt(2) times circular orbit velocity, so in this
limiting case, the delta-V is 2*(sqrt(2) - 1) times the orbital velocity.
Of course, it takes infinite time to get to infinity, so this is not
terribly practical. So, as you say, it's more practical to transfer into a
very elliptical orbit, such as L1 or lunar orbit. The delta-V cost in this
case is not much higher than going to infinity, but the transfer time is
much more reasonable, on the order of a week.

--
JRF

Reply-to address spam-proofed - to reply by E-mail,
check "Organization" (I am not assimilated) and
think one step ahead of IBM.

Craig Fink wrote:

At some point you can climb out
of the gravity well, make the inclination change at apogee (velocity is
low) then come back down again cheaper.

Thanks, Craig (and all). Imagine of heel of hand hitting forehead
doh!

Here I am wrestling with the dynamics just 300 km up, and you guys are
all getting high :-)

JRS: In article . com,
dated Tue, 22 Mar 2005 04:24:23, seen in news:sci.space.science,
posted :

Monte Davis wrote:
Dumb question: If one were in circular LEO and crazy enough to want a
90-degree change of inclination, what's the cost in delta-v? Is it,
as
intuition is telling me, equivalent to throwing away all your ~9 kps
and re-acquiring it all?

At first glance, a brute force 90-degree plane change would require
1.414x (square root 2 times) as much delta-V as it took to get to
orbit, so 7.8x1.4=11km/s. You have to kill all that horizontal velocity
and replace it with orbital velocity perpendicular to your first.

But separately they need Root2 times the delta-V of doing them together,
and that way one can be less hasty for a given loss of altitude.

It would, in fact, be less costly in delta-V to do something like
accelerate to escape velocity (~3.3km/s above orbital velocity), drift
up to about the L1 point or swing around the moon, and drop back to
Earth in a new, desired orbit. I think.

Let orbital velocity V have magnitude 1.0 units. Then the right-angle
plane change, done be a rapid burn at 45 degrees to original-forwards,
needs delta-V of Root2 units, as you say.

Burns to/from escape velocity needs dV of Root2-1, and plane change is
free at infinity. That dV, applied directly, gives a plane change of
2(arcsin(Root2-1) which is about 48.94 degrees,

Lesser changes than that are better done directly, greater ones better
via infinity. With corrections for practicalities, using lunar gravity,
etc.

OTOH ISTR being corrected on this or a similar matter before; possibly
it is better for angles near that to burn to a higher but finite apogee,
change plane there, and return.

--
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Web URL:http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms & links;
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Dr John Stockton schrieb:

OTOH ISTR being corrected on this or a similar matter before; possibly
it is better for angles near that to burn to a higher but finite apogee,
change plane there, and return.


That should be right. Just some rough calculation:
If you wanted to reverse your LEO orbit, you'd need 16km/s.
To go from LEO into a highly elliptic orbit, touching the moon's, you'd
need some 3km/s. At lunar distance, your velocity would be lower than
1km/s (the moon's orbital speed). To reverse this, you'd need 2km/s.
Drop back to perigee and brake away the 3km/s from the beginning. Thus,
to reverse your orbit with this technique, requires 8km/s.