<mmeron@cars3.uchicago.edu> wrote in message
news:RFObc.4$H77.4896@news.uchicago.edu...

In article <c4oc1j$eio$1@titan.btinternet.com>, "Franz Heymann"

<notfranz.heymann@btopenworld.com> writes:

"LV Poker Player" <lvpokerplayer@aol.com> wrote in message
news:20040403150036.04303.00000581@mb-m29.aol.com...

From: "Franz Heymann"

I reckon the question is harder to answer than you

perhaps

thought, because of the range of possibilities which have

to

be studied. For instance, you might fire a rocket in any
direction whatsoever to transfer just enough momentum to

the

craft to put it on an orbit which intersects the earth.

I am just trying to find an answer to what I consider a

relatively simple

question, which I will ask again.

We have two identical masses in circular Earth orbits, but

at different

heights. One is in low orbit, the other is in a much

higher orbit.

We make the most fuel efficient course change in order to

intersect the surface

of the Earth and hit it. We ignore air resistance for

this purpose.

Ok, how much delta v is needed? Do we just calculate

orbital velocity, then

fire enough delta v to cancel it?

Certainly not.

Actually, for the case which started this thread, which is dropping
from orbit with a major radius *much larger* larger than the radius

of

the orbited body, canceling the orbital velocity at apo... whatever

is

*precisely* the right thing to do, to first order. Any corrections

to

this are higher order in (radius_of _primary/major_radius_of orbit).

Yes. You and Erik Max Francis are right. I was trying to be too
clever. I had in mind not simply cancelling the azimuthal component
of the velocity, but firing a rocket in such a way as to maintain the
KE, and just turning the direction of the velocity to point radially
inwards.

But that, of course, involves a "delta v" whose magnitude is
sqrt 2 times larger. Silly idea.

mmeron@cars3.uchicago.edu wrote in message news:<RFObc.4$H77.4896@news.uchicago.edu>...

In article <c4oc1j$eio$1@titan.btinternet.com>, "Franz Heymann" <notfranz.heymann@btopenworld.com> writes:

"LV Poker Player" <lvpokerplayer@aol.com> wrote in message
news:20040403150036.04303.00000581@mb-m29.aol.com...

From: "Franz Heymann"

I reckon the question is harder to answer than you

perhaps

thought, because of the range of possibilities which have

to

be studied. For instance, you might fire a rocket in any
direction whatsoever to transfer just enough momentum to

the

craft to put it on an orbit which intersects the earth.

I am just trying to find an answer to what I consider a

relatively simple

question, which I will ask again.

We have two identical masses in circular Earth orbits, but

at different

heights. One is in low orbit, the other is in a much

higher orbit.

We make the most fuel efficient course change in order to

intersect the surface

of the Earth and hit it. We ignore air resistance for

this purpose.

Ok, how much delta v is needed? Do we just calculate

orbital velocity, then

fire enough delta v to cancel it?

Certainly not.

Actually, for the case which started this thread, which is dropping
from orbit with a major radius *much larger* larger than the radius of
the orbited body, canceling the orbital velocity at apo... whatever is
*precisely* the right thing to do, to first order. Any corrections to
this are higher order in (radius_of _primary/major_radius_of orbit).

COMMENT:

Yes, but because of the 1/2 power dependence, it takes going a LONG
way out before it's not worth doing. Farther than I thought. For
example, on reexamining the problem, you can save enough to be worth
doing on the Earth/Sun system; on the order of 10%.

Take a look at that equation again:

V(deorbit) = sqrt(1/N) * [ 1- sqrt (2/1+N)]

The first term is the entire orbital V, and so the savings factor you
get from having to aim only at an extended body to deorbit, rather
than a point, is that factor sqrt sqrt(2/(1+N)). For the sun, N is on
the order of 200 to get something down to where it will surely burn up
on first pass, so you're looking at the square root of 1/100 = 10%
savings in fuel. Even 10% savings is nothing to sneeze at with V for
earth orbit so high.

On the other hand, you see that aiming stuff for 2 solar radii where
it will burn up over a bunch of passes, doesn't save you as much as
you might think—just another 5% or so.

Other matters:

As was pointed out, the equation above has a maximum, which can be
found by differentiating it and setting equal to zero. This gives you
to solve:

[(n+1)/2 ]^3 = [N+1/2]^2, which is a nasty cubic, but empirically the
root we want is indeed about at N = 5.9 radii. This is about 20,000
miles above the ground, almost to geosync (the period is 20 hours). It
takes 19% of low orbit V to circularize at that orbit, and another 19%
to get down from it. That's 38% of LEO velocity, when escape V from
low orbit is only another 41.4% ! So it you add in the Hohman deltaV
necessary to get up to Clarke orbit, you find it's quite a lot harder
to get up to there, circularize, and then come back down, than it is
to escape from Earth entirely. So we won't be sending astronauts to
fix comsats real soon. It's harder to do repair a comsat with
astronauts, than to do an Apollo 13 style lunar free return. And just
about as hard as an Apollo 8 style lunar orbit mission.

So it you add in the Hohman deltaV necessary to get up to Clarke
orbit, you find it's quite a lot harder to get up to there,
circularize, and then come back down, than it is to escape from
Earth entirely. So we won't be sending astronauts to fix comsats
real soon.

Thank you. It looks like I was not right, but the simplistic "just
kill the
orbital velocity and drop straight down" was not right either. I'm
willing to
call this one a draw, especially if it stops the ad hominems.

It _is_ right, you're just misunderstanding what's involved

I admit to being surprised that a really high orbit, you know one way
out
beyond the moon, where you might be taking several months for a
circuit, takes
so little energy to kill. Yes, I know that a ridiculously high orbit
would be
subject to perterbations from other planets, but we can just look at
the
theory.

It's because you're assuming all the energy is in kinetic energy,
whereas most of it is in potential energy.