Sub-Orbital Earth Transits

Well, not having received any negative responses
to my original query in sci.space.policy, concerning
Virgin Galactic, here goes.

--------------------------------------------------------------
Moderators: This is highly speculative and unsupported by fact.
I am posting this because its mainly a technology question
about the feasibility of using sub-orbital trajectories to
do point-to-point travel around the Earth relying on the
planet's rotation to do the translation. I am interested in
knowing the what the potential problems are and if this could
possibly be made to work and if not, why not. I'm looking for
specific cites and references I can study. If this post is
too speculative for sci.space.tech, feel free to reject and
let me know the reasoning and a a better group to post to.
--------------------------------------------------------------

When I first posted my speculation that Sir Richard's ulterior
motive in founding Virgin Galactic as a space tourism adventure,
was really not about space tourism (that's mearly the cover venture
for bootstrapping the technology) but about hypersonic transport
between major cities along the pacific rim, it was pointed out
that flattening the parabolic trajectory to any significant degree
to enable transits between cities would cause the glider to come
in far too fast for a safe rentry without a significant TPS that
would add so much to the weight and cost as to make it largely
uneconomical. It also calls into question to some degree the
viability of the current 'shuttlecock' re-entry scheme devised
by Burt Rutan for SpaceShipOne and SpaceShipTwo.

Recently, in off-line discussions with a friend of mine, we came
up with a slightly different idea. Rather than flattening the
parabola, why not just extend it? So in a proposed SpaceShipThree
scenario it flies much, much, higher, but returns along pretty
much the same trajectory. The idea being that by keeping the
airspeed relatively low, by going nearly straight up and down,
but at an extremely high altitude you let the Earth's rotation
underneath you do the transit for you. That's a translation
of 15 degrees per hour at the equator. If you can stay up for
3 hours above the atmosphere you've translated east-to-west
by 45 degrees longitude at the equator in about a 2 hour flight.
One could imagine a fleet of SpaceShipThrees that make the
east-to-west transits in a circuit fashion ala the days of
the propeller driven China Clipper flights from the 1930's
and 1940's before jet travel.

The airspeed would remain low because you're accelerating down-
ward via gravity alone and the thought occurred to me that it
would be more comfortable for the passengers of such a 'business-
class' flight to avoid zero-g altogether, so the addition of
cabin 'ullage' rockets to maintain an 'upwards' .5g would not
only give the passengers an easier ride but also help keep
the airspeed for re-entry down. Everything else would remain
basically the same as it is for SpaceShipTwo today, where it
returns to being a glider at about 100kft.

I presume it would take a lot more fuel than SpaceShipTwo
carries today...

The real question behind all this is the practicality of
using an extremely high altitude, highly parabolic sub-orbital
trajectory to provide hypersonic transport.

Critiques and Comments?

Dave


======================================= MODERATOR'S COMMENT:
As Mr Spain has asked, let's focus on the technical aspects, not the business aspects. Thanks.

David Spain wrote:
If you can stay up for
3 hours above the atmosphere

And there's the rub. Doing that at less than orbital speed requires a
continuous expenditure of fuel. Lots of it.

Sylvia.

MODERATOR'S COMMENT:
As Mr Spain has asked, let's focus on the technical aspects, not the b
usiness aspects. Thanks.

The is however a combination business/tech aspects.

Present day airports system are not for the narrow transit windows
that sub-orbital transpost system tend to require.

VTVL can support large landing footprints (IE. they don't need to land
at thier original assign landing pad but can re-route).

HTHL and VTHL are often design for once around, even twice around hold-
offs for landing.

But none of them can loiter in a holding pattern for 1-2 hours like
our present plane systems allow when things go very wrong at a
destination airport.



To make sub-orbital flights workable we will need the flight
controllers at both ends approve that conditions in real-time with a
full knowledge about the near-term landing conditions. This I
believe will require extra communication systems to be put in place
that will cost money.

Example, before I approve a takeoff from my airport, I probably want
to see the radar weather map of where I am sending a craft to, and to
talk over anything odd with a local controller at the destination
port. And they probably would like to do likewise.

David Spain wrote:
If you can stay up for
3 hours above the atmosphere you've translated east-to-west
by 45 degrees longitude at the equator in about a 2 hour flight.

The problem is the "If you can stay up for 3 hours above the atmosphere"
part: Any orbit which will keep you up for 3 hours will
(a) go pretty high (== substantial radiation exposure from the inner
Van Allen belts), and
(b) reenter at a rather large velocity
== need fancy/expensive/heavy heat-shield
== maybe significant g-loads on reentry

*If* we were to approximate the Earth's gravitational field as uniform
and the Earth's surface as flat, then we could estimate the trajectory
as follows: Staying up for 3 hours implies going up for 1.5 hours,
then down for 1.5 hours. If you free-fall at 1g acceleration for
1.5 hours, you travel a distance of 1/2 * g * t^2 = about 150,000 km,
and reach a speed of g*t = 54 km/second. These numbers are clearly
nonsense (they're about 1/2 way to the moon and twice the Earth's
escape velocity respectively). So, we need to do a proper Keplerian
(elliptical/parabolic) orbit calculation, which is more work than I
want to do for this posting. But until I see numbers to the contrary,
I strongly suspect that the result will *not* be be a low-altitude
(& hence low-radiation) trajectory, nor will it end with a low-speed
re-entry.


the thought occurred to me that it
would be more comfortable for the passengers of such a 'business-
class' flight to avoid zero-g altogether, so the addition of
cabin 'ullage' rockets to maintain an 'upwards' .5g would not
only give the passengers an easier ride but also help keep
the airspeed for re-entry down.

How much rocket fuel does it take to maintain 0.5g acceleration for
3 hours? Let's work it out: specific impulse in seconds is numerically
equal to the e-folding time of the mass of a 100%-fuel-fraction rocket
maintaining a 1 g acceleration, so (still assuming 100% fuel fraction)
1/2 g at an impulse of 400 seconds implies a mass e-folding time of
800 seconds -- a mass ratio (ratio of mass at start of "ullage" burn
to mass at end of "ullage" burn) of
exp(3 hours * 3600 seconds/hour / 800 seconds) = exp(13.5) = 700,000.
So for each kilgram of structure + payload, you'd need 700 tonnes
of fuel.

--
-- "Jonathan Thornburg [remove -animal to reply]"
Dept of Astronomy, Indiana University, Bloomington, Indiana, USA
"Space travel is utter bilge" -- common misquote of UK Astronomer Royal
Richard Woolley's remarks of 1956
"All this writing about space travel is utter bilge. To go to the
moon would cost as much as a major war." -- what he actually said

David Spain wrote:

The idea being that by keeping the
airspeed relatively low, by going nearly straight up and down,
but at an extremely high altitude you let the Earth's rotation
underneath you do the transit for you.

David, what you are in effect suggesting is using rockets to hover.
The propellant consumption would be prohibitively expensive. The TPS
problems of a ballistic trajectory would be far more tractable.

Jim Davis

On Dec 9, 7:56 am, David Spain
wrote:

The airspeed would remain low because you're accelerating down-
ward via gravity alone and the thought occurred to me that it
would be more comfortable for the passengers of such a 'business-
class' flight to avoid zero-g altogether, so the addition of
cabin 'ullage' rockets to maintain an 'upwards' .5g would not
only give the passengers an easier ride but also help keep
the airspeed for re-entry down. Everything else would remain
basically the same as it is for SpaceShipTwo today, where it
returns to being a glider at about 100kft.


Wrong and not viable. Complete disregard for physics. This is a
sounding rocket trajectory. The entry speed is going to be the same
as the speed of the vehicle at motor burnout. To reach a high
altitude (a 20 minute lob goes more than 1300km alt and velocity of
more than 18k kmph) going to require higher velocity. The SS2 can
not handle this



Also, hypersonic is Mach 5 and greater.

Me writes:

Also, hypersonic is Mach 5 and greater.

Hypersonic transport was a poor choice of wording, but
I presume accurate relative to the ground speed not the
airspeed.

Dave

David Spain wrote:

Recently, in off-line discussions with a friend of mine, we came
up with a slightly different idea. Rather than flattening the
parabola, why not just extend it? So in a proposed SpaceShipThree
scenario it flies much, much, higher, but returns along pretty
much the same trajectory. The idea being that by keeping the
airspeed relatively low, by going nearly straight up and down,
but at an extremely high altitude you let the Earth's rotation
underneath you do the transit for you. That's a translation
of 15 degrees per hour at the equator. If you can stay up for
3 hours above the atmosphere you've translated east-to-west
by 45 degrees longitude at the equator in about a 2 hour flight.
One could imagine a fleet of SpaceShipThrees that make the
east-to-west transits in a circuit fashion ala the days of
the propeller driven China Clipper flights from the 1930's
and 1940's before jet travel.

The airspeed would remain low because you're accelerating down-
ward via gravity alone and the thought occurred to me that it
would be more comfortable for the passengers of such a 'business-
class' flight to avoid zero-g altogether, so the addition of
cabin 'ullage' rockets to maintain an 'upwards' .5g would not
only give the passengers an easier ride but also help keep
the airspeed for re-entry down. Everything else would remain
basically the same as it is for SpaceShipTwo today, where it
returns to being a glider at about 100kft.

As others have mentioned, using rocket motors to hover is very expensive.
But even if you had fairy dust for your rocket engines, enabling them to give
you your half g acceleration for a few hours, you still don't gain anything
by using them the way you propose, fo two reasons.

1- When you take-off, you have a horizontal speed given to you by Earth's
rotation. If you want to let Earth's rotation bring you your destination under
you, you first have to cancel that horizontal motion you have. There is
no advantage in doing so in the direction against Earth's rotation. Instead
of accelerating westwardly to cancel your eastwardly motion and let Earth
rotate under you motionless you could just as easily accelerate eastwardly
and go to a an eastern destination. In fact, because of orbital mechanics
it is more efficient to go eastwardly.

2- You assume that because you didn't accelerate to high speeds you
won't need much for heat shield. Well, if you do the math for the flight
plan you seem to be proposing your in for a big surprise. If you want
to stay out of the atmosphere for a few hours with only your initial
vertical speed and a 0.5 g acceleration, you will see that you need
to go very high. So high that when you will fall back to the atmosphere
you will be coming in at very high speed. I didn't bother computing at
what speed you would be coming in. Let's say it is at 5.6 km/s. That
speed is just a guesstimate, but it is the right order of magnitude
and it fits well with Earth's atmosphere's scale height, because Earth's
atmosphere doubles in density every time you go down 5.6 km.
When you hit the atmosphere you need to stop your rocket motor
or else good luck with dealing with both the atmospheric entry heat
and engine heat. So at about a height of 100 km you will have a
deceleration of 1 g and a speed of 5.6 km/s. One second later, you
went down 5.6 km, the atmosphere is twice as dense and you now
have a 2 g deceleration. Another second, the atmosphere is again
twice as dense and you have a 4 g deceleration. A third second
and you have an 8 g deceleration. You are now going at about
5.5 km/s, so it now takes about 1.02 seconds to double the pressure
and the deceleration. So after 4.02 seconds of re-entry you now
have a deceleration of 16 g, this will cause you to faint, thankfully
because you will be cremated by the incredible heat within a few
seconds.

A vertical re-entry is *much* more difficult then a very shallow angle
re-entry. So much so that unless you come in at a much slower
speed, it will kill you.


Alain Fournier

"Jonathan Thornburg [remove -animal to reply]" writes:

How much rocket fuel does it take to maintain 0.5g acceleration for
3 hours?

When does the cabin start to experience zero 0? Is it as soon as the ascent
rocket cuts out or at apogee and the return leg until you hit the atmosphere?

Ignoring cabin zero 0 for the moment, and say you're only trying to reduce
descent velocity (again relative to your airpseed), wouldn't you only need to
maintain .5g during the descent leg, starting at apogee, 'airspeed' = 0, which
does not exceed 1.5 hours?

By your prior estimate it would seem more like:

exp(1.5 hours * 3600 seconds/hour / 800 seconds) = exp(6.75) = 854.

That would yield, by your own estimate, 854 kilograms of fuel per kilogram of
structure + payload, correct?

Because we're dealing with an exponential It could be even less assuming the
...5g burn is not required for the full 1.5 hours on the return leg, esp. after
you again reach the atmosphere and can return to being a glider. I'm not
saying its practical, but it seems we could do better than your original 700
metric tonnes estimate.

Dave

Jim Davis writes:

David, what you are in effect suggesting is using rockets to hover.
The propellant consumption would be prohibitively expensive. The TPS
problems of a ballistic trajectory would be far more tractable.


Hi Jim,

Not hover, migtigate. .5g will not keep you from re-entry, only slow
your retry airspeed. But the fuel fraction to weight ratio might be
way too high, I'm not an expert here, that's why I am asking the
question.

For a configuration like the SpaceShipOne/Two, it's not only just
a question of TPS but also of max airspeed that the 'shuttlecock'
recovery system can accomodate. I suspect if this gets so high,
to the point of needing a TPS, you'll be going so fast you
literally won't be able to 'lower the booms'!

Dave