Re: Ballistic entry into circular orbit?
- From: tadchem <tadchem@xxxxxxxxxxx>
- Date: Mon, 18 Feb 2008 07:07:48 -0800 (PST)
On Feb 17, 12:45 am, nos...@xxxxxxxxxx wrote:
Hello all.
Would someone please help me figure out the correct way to analyze a
problem? Namely, is it possible to literally shoot something into
circular orbit?
Assuming a vacuum, basic orbital mechanics says the orbit intersects
the gun's location.
Posit an objectl in a highly elliptical orbit about a point-like Earth
mass, perigee at the gun 4,000 miles out. To circularize the orbit at
4,300 miles, we have to a) reduce the apogee from God knows what to
4,300 miles and b) increase the perigee from 4,000 miles to 4,300
miles. Can we do that?
One way to increase an orbit's perigee is to add energy, but you have
to do so when the satellite is at apogee. Obviously, you can't do
that by introducing atmospheric drag into the model because you'd be
subtracting energy, not adding it. Worse, the atmosphere is at the
wrong end of the orbit. On the other hand, I don't know if that's the
only way to increase an orbit's perigee.
By the same token, you can remove energy when the satellite is at
perigee to reduce its apogee. That you can certainly do with
atmospheric drag. Unfortunately, while that'll pull the apogee down,
you're still stuck with a perigee at the gun. Again, I don't know if
that's the only way to futz with the apogee.
For example, what happens if you add or subtract energy at a point in
the orbit somewhere between apogee and perigee? Can you thereby take
some apogee altitude and trade it for some perigee altitude?
One thing I know for sure: any object 300 miles up will be in circular
orbit as long as is has 25,000 fps tangential velocity, zero radial
velocity, and is under no accelerations other than gravity.
The sixty-four million dollar question is: given a real launch angle
and velocity, can you employ real atmospheric drag to make a ballistic
projectile launched from the surface achieve that state?
My first thought was to divide the muzzle velocity into radial and
tangential components - or vertical and horizontal, near enough.
Use drag and sqrt(2gh) to figure the vertical component of the muzzle
velocity needed to get the thing to 300 miles.
Use drag and surface velocity at your latitude to figure the
horizontal component needed to produce a 25,000 fps surplus after
passing though the atmosphere.
Muzzle velocity is then the root of the sum of the squares of the
component velocities.
Aim point altitude is the arctangent of the ratio of the component
velocities, vertical to horizontal.
Given strong enough materials, couldn't you thus put a projectile into
just about any dynamic state you wanted to?
BUT
Orbital mechanics says the perigee will intersect the gun.
I can't see how to reconcile those two ideas. One of 'em is obviously
wrong, or at least incomplete.
Any help with this would be appreciated; thank you.
--
Dave Typinski
The trajectory for orbital insertion is a segment of an elliptical
orbit. The apogee of this ellipse corresponds to the radius of the
circular orbit into which you want the projectile inserted - i.e. the
ellipse in internally tangent to the circle.
If allowed to continue under its own inertia, the projectile will
continue on the original elliptical trajectory, moving closer to the
earth. It simply will not have enough speed to maintain the circular
orbit.
You will need to add speed to the projectile *at the time it becomes
tangent to the circular orbit*, and in sufficient amount to maintain
the new orbit.
The angular momentum of the circular orbit is higher than that of the
elliptical orbit.
A good introductory text on celestial mechanics will give you the
relative velocity V of two masses M and m in a gravitationally bound
(G) orbit with a semi-major axis S and when they are at a distance r:
V^2 = G*(M + m) * (2/r - 1/S),
a consequence of the conservation of energy (kinetic + potential) of
the system.
For an elliptical orbit, the semi-major axis S is half the sum of the
periapsis R(p) (perigee for earth-bound calculations) and the apoapsis
R(a) (apogee for Earthers)
S = (R(p) + R(a))/2
For a circular orbit, R(p) = R(a) = S. For an elliptical orbit R(p) <
R(a).
When r = R(a) then
V^2 = G*(M + m) * [1 / R(a)] for the circular orbit, and
V^2 = G*(M + m) * [2/R(a) - 2/(R(p) + R(a))] for the elliptical orbit
The difference in velocities must be made in during the "orbital
insertion burn" at the apogee of the elliptical orbit or the
projectile will continue back towards earth on the elliptical path.
"This isn't exactly rocket science - well, actually, it *is* rocket
science." - Craig Kluever
Tom Davidson
Richmond, VA
.
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