Will this be orbital or sub-orbital?

ryan

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I did something like this with an add-on I recently got. It is basically a straight up and straight down flight, but it takes more than 24 hours for the parabola to be flown (meaning that the earth made a complete rotation on it's own axis below it).

Like this - just a quick thing in paint. :lol:

whatif.jpg


Would this be considered oribtal or suborbital? Technically it flew "around the earth", but not in the manner we think. More of a situation of "The place comes to you" not "You go to the place".
 
It looks like a parabola, so I would say its a sub-orbial flight. However, that may not be the only criteria for a sub-orbital flight.

N.
 
Technically anything is an orbit, but in reality an orbit is normally defined as something that is stable whereas a suborbital trajectory is inherantly unstable.

Seeing as this impacts the earth I would volunteer that it is indeed unstable ;)
 
I would say sub-orbital because if you wanted to get in orbit, the path you were on had to be "trapped" in space for a period of time to claim that it is a regular orbit.
 
Something like that.
 
What I want to know is what the re-entry is like. Is it steep or does it make a smooth re-entry?
 
What I want to know is what the re-entry is like. Is it steep or does it make a smooth re-entry?

From the look at it, it will be the opposite of smooth. It will enter the dense parts of the atmosphere very fast. I could calculate peak heating and maximum deceleration for a ballistic reentry inside a Excel file, but my first guess is, that it is already beyond 20G and 3500K.
 
Nevermind, the re-entry can be smooth as long as the angle isn't too steep or too shallow.

Nice orbit idea and question. I feel like taking the DG up for a little spin.
 
Well, even more important, it will be a huge ellipse. I just made a quick test - apogee is at about 85000 km, when you have the periapsis radius close to zero. For getting this trajectory, you need 10 km/s at 100 km.

Of course, you will have the same 10 km/s again when you reenter. Needless to say, I just destroyed another Deltaglider, when I took the final 100 km of travel in only 10 seconds.

Just put it like that: If you want to land in one piece, you need to slow down from 10000 m/s in a distance of 100 km.

100 km = 0.5 * a * t²
v = a * t -> t = v/a
100 km = 0.5 * a * (10000/a)² = 0.5 * 10000²/a
10,000² / 200,000 = a


a = 500 m/s² ( or roughly 50 G)

I am not sure it can be survived...You need to survive this acceleration for 20 seconds.

EDIT: I just made a test using a periapsis altitude of about 0 km and a orbit period of 85000 seconds. Again a one way loop around Earth, but this time, reentry can be survived. The critical problem is just being able to control the reentry - I just skipped out because i did not bank fast enough down to negative lift. Started hot reentry over the Californian peninsula, skipped out shortly before New Orleans. Had a good view of the cape when I flew directly above it.

This approach also takes less fuel - I had 45% left this time because of the more shallow ascent, instead of 33%.
 
The speed would just be to massive for re-entry.
 
But guys does all spaceflight have to be orbital or sub-orbital.
I don't quite understand the question. Do you mean is all spaceflight divided into two categories, orbital and sub-orbital? The answer to that question is no. You can all kinds of trajectories, including chaotic ones - see Interplanetary Transport Network.

For the simpler case of two bodies, you can have four main types of trajectories: circular (Ecc=0), elliptal (0<Ecc<1), parabolic (Ecc=1) and hyperbolic (Ecc>1). In the real world, circular and parabolic are very difficult, if not impossible to acheive. In the circular case, other perterbations soon make the orbit elliptical (eg non-spherical gravity, mascons, etc). In the parabolic case, where your velocity exactly equals escape velocity, other influences soon change your trajectory (eg, gravity of other bodies). Of the remaining two, only an elliptical trajectory could be considered orbital, however it could also be sub-orbital if it intersects the other body, per your example above.
 
I think that if you go straight up, you do carry with you the velocity of the Earth's rotation.(The reason most spacecraft launch east) If you assume that this velocity is about 465.11m/s I think that you can get into an orbit of about 146M x 200km if you go DIRECTLY up from the equator with 0 inclination.
NOTE: Keep in mind that before and after apogee you will appear to be moving west because you are moving slower than the Earth's rotation. You will stay in one spot above the Earth at apogee because you are moving at the same rate as the Earth rotates.
 
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