Powered re-entry

Allen Meece wrote:
That acceleration is only 8 G-s, quite survivable.
So, this thread has concluded that reentry doesn't have to cook a spacecraft
if the space craft retains enough fuel to stop itself in mid-orbit and drop
straight down. And release a drogue chute to slow it a bit and then release a

The "stop itself in mid-orbit" is orders of magnitude harder than
the "drop straight down" bit.
You need the same amount of propellant to slow down as you do to speed
up.
You have burnt all the propellant you used to speed up, so you'r kind
of stuck.

parachute or better yet, a paraglider to fly the rest of the way down.
Alternatively, the space craft can de-orbit by spreading huge wings to
catch the sparse air molecules in orbit and *slowly* descend, taking a few
*days* to scrub off speed before entering the thicker, lower air.

No, you can't.

Once you start slowing down, you need to provide lift to stop yourself
falling into the atmosphere.

In order to get lift, you inherently get drag.
At hypersonic speeds, the lift you get is only 2-3 times drag, so you
inherently are going to slow down at around 2-3 times slower than
you would otherwise.

The original post said if you could kill your orbital speed of 18,000
mph you'd drop straight down like a rock and burn up.

This is bogus.

First, minimum orbital height is the surface of the Earth, and
according to;

V = SQRT(G*M/r)

Where G = gravitational constant - 6.67e-11 N m2 / kg2
M = mass of Earth - 5.98e24 kg
R = Earth radius - 6.378e6 m
r = R + a
a = altitude
V = velocity - 18,000 mph -- 8.05e3 m/sec

The fastest orbital speed occurs at a = 0 so, computing that speed we
have

V = 7908 m/sec -- 17,683 mph

Which is just under 18,000 mph.

Clearly, if we slow from orbital speed to zero speed at the surface of
the Earth, we don't fall at all to reach the surface.

So, we're talking low orbits. At an altitude of 200 miles (a=322,000
m) - we have an orbital velocity of;

V = SQRT(GM/r)
= SQRT(3.99e14 / (6.378e6 + 0.322e6))
= 7,715.7 m/sec
= 17,252 mph

If we kill this orbital velocity at an altitude of 200 miles that
means the vehicle would fall like a rock a vertical distance of 200
miles, starting at zero velocity.

Ignoring air and gravity changes with altitude for a minute we can
calculate at worst what the terminal velocity might be;

V = g0 * time
D = 1/2 * g0 * time ^2

So,

time = V/g0
and
D = 1/2 * g0 * V^2 / g0^2
= V^2 / (2*g0)

rearranging;

V = SQRT(D * 2 * g0)

where
V = terminal velocity (m/sec)
D = distance (m)
g0 = gravitational acceleration = 9.82 m/sec/sec

So;

V = SQRT( 322,000 * 2 * 9.82)
= 2,514 m/sec
= 5,623 mph

This is about the speed of the SR-71 blackbird.

There are many models that relate heating to velocity

http://www.ese.ic.ac.uk/userfiles/PD...skell_2002.pdf

Tmax = A * V^(.75)

Where A is a whole host of functions, density, the way the density
changes with altitude, angle of penetration of the atmosphere, density
of the object, object shape, and so forth.

But, by taking a ratio of speeds we can get rid of this A factor and
get a ratio of temperatures.

(5,263 / 17,252)^(.75) = 0.4314

So, by stopping the vehicle's orbital velocity at 200 miles altitude
we reduce Tmax to 43% of its full value.


But, this begs the question. If you can cancel the orbital speed, why
not cancel the vertical speed too? And the point is if you can do the
first, you can do the second equally well.

You'd need a density profile of the atmosphere and a gravitational
field description to obtain a minimum energy/minimum heating velocity
profile for a rocket capable of slowing and descending without
recourse to atmospheric braking. This would be the reverse of an
optimal take off trajectory - a Goddard trajectory. It would also use
Calculus of Variations in its solution.

If I had more time this holiday weekend I'd do that, but I do not
alas.

In any case, if you can stop a vehicle in orbit with a rocket - you
can equally well land a vehicle with minimal heating using that same
rocket. Then all you need are aluminum or plastic vehicles.

"James M. Devine" wrote in
:

Ian Stirling wrote:
The "stop itself in mid-orbit" is orders of magnitude harder than
the "drop straight down" bit.
You need the same amount of propellant to slow down as you do to speed
up.
You have burnt all the propellant you used to speed up, so you'r kind
of stuck.
Actually, The vast majority of the energy spent is to get into orbit
(increase altitude). Only a small relative proportion of it goes into
lateral motion.

Incorrect. Most of the energy goes into kinetic energy (velocity), not
potential energy (altitude). For a 200 km orbit, for example, 94% of the
energy change is kinetic.


--
JRF

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