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)

In article ,
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?

Not that bad, because there's a different strategy you've missed. You can
do *any* LEO plane change with about 6.6km/s. The first 3.3km/s turns the
orbit into a very elliptical one with an extremely high apogee. Coast up
to apogee. Orbital velocity there is negligible, so a negligible burn can
accomplish *any* plane change there. Coast back down. The other 3.3km/s
converts the elliptical orbit back into a circular one.

Add a little if you're impatient and prefer a somewhat less extreme apogee
(you save almost nothing on the two big burns, and the actual plane change
costs a bit more). Subtract some if you can use a lunar flyby to do the
plane change. Subtract a whole bunch if you use aerobraking to do most of
the final apogee lowering (but be warned that if you do it gradually, it
takes a long time and you'll make many passes through the Van Allen belts).
--
"Think outside the box -- the box isn't our friend." | Henry Spencer
-- George Herbert |

"Monte Davis" wrote in message
...
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?
(Assume a spherical Earth, homogeneous frictionless elephant, etc)

Yes, your intuition is spot on.

But you could save a little by either making the plane change elsewhere.
Recall, escape speed from a circular orbit is root 2 times the orbital
speed, so after a 2.8 km/s burn in your flight direction, you find yourself
with a very high apogeee after a while, *then* when suitably far from the
earth, perform the plane change for near zero delta-v cost, and then make a
de-orbit burn of another 2.8 km/s, leaving you in a nice polar orbit. It'll
take a bit longer than the brute-force powerered turn that you describe, and
would be far less theatrical, but would save around 1.5 km/s.
This option, for LEO to GEO is termed a super-syncrhonous plane change for
obvious reasons, and is not uncommon.

The other fun way is to deorbit and make the plane change in atmosphere. Use
aerodynamic lift to rotate your velocity vector, raise your apogee with a
burn and then circularize as normal. Much higher 'wow' factor but this is a
non-spherical cow solution as you'll be riding in a (sadly) non-existing
vehicle.

-James Garry

On Tue, 22 Mar 2005, in sci.space.science,
Monte Davis said:

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's not quite that bad, it's the vector difference between the two.

Delta-v = 2 * v sin(alpha/2)

(assuming v1=v2=v)


delta-i delta-v/v
------- ---------
30° 52%
60° 100%
90° 141%
120° 173%
150° 193%
180° 200%

But look how horrifying even a "small" 30° change in inclination is.
That's why they make such a big deal over what the inclination of space
stations should be, and why you can't just get from one to the other
even though they're both "in orbit".

--
Del Cotter
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