Not considering the effects of atmospheric drag, is there any angle + velocity that will result in an unpowered projectile fired from roughly sea level on Earth attaining a stable orbit (around Earth)? If so, what would that be, and what would be the orbital altitude?
Intuition tells me that this would have to be an altitude of zero at a speed of pretty damned fast, but that sort of math isn’t my strong suit.
One simple approach would be to fire the projectile tangential to the earth, at a velocity that matches orbital velocity, yielding a reasonably stable orbit at a height above the earth’s surface equal to the height of the gun (it’s recommended you lower the gun after firing). If the gun is a few meters above sea level, this results in an orbital period just under 85 minutes, which corresponds to a speed a bit over 28000 kph.
This takes considerable advantage of the no atmospheric drag stipulation. A minor complication is that it may be hard to find an orbital path that low that doesn’t intersect some piece of terrain.
Also note that the earth’s rotational speed isn’t considered (but probably should be).
The question’s a bit more interesting if atmospheric drag is considered. I bet you could do it with a lunar encounter.
Any stable orbit which intersects the surface of the Earth once will intersect it again. So you have to have your launch point a bit above the surface (and then return to that height at least once every orbit).
With just the Earth and the Moon (and your projectile, of course), you could get your projectile into a chaotic orbit which stayed up for an arbitrarily long span of time, but eventually, the same sort of lunar interactions which got it into that situation are going to get it out of it, resulting in your projectile either crashing into the Moon or Earth, or (more likely) being ejected from the system. If you mean to use the atmosphere to stabilize the orbit, then you face the problem that any stable orbit that intersects the atmosphere once will intersect it again. You could, however, if you planned carefully and were very patient, attach a small rocket to your payload, launch into an unstable orbit involving the Moon and/or the atmosphere, and then use a very small amount of fuel to stabilize the orbit.
Great, that’s right at the heart of what I was asking.
Hmm, so on the moon if we were to find a perfectly flat spherical section of the planet (or we levelled the surface with bulldozers) for an orbital path, you could have something orbit the moon indefinitely at a few inches off the ground?
I guess I don’t see why not, just seems counterintuitive somehow.
Yes you could, but I doubt it would be indefinite - the influence of the earth would disturb the orbit, and with such a small margin of error it would likely hit the moon eventually. In a theoretical situation, in the absence of an atmosphere and assuming a uniformly shaped moon or a carved path as you say, you could keep it up indefinitely.
You really do like to provide answers that have absolutely no relevance to the questions, don’t you?
As per the OP, this is all assuming no air resistance (and a uniform, perfectly spherical Earth, and probably a few other things, too).
Assuming Xema’s math is correct, if you fire the projectile at 28,000 kph, it will settle in to a nice, circular orbit just above the surface of the Earth. If you fire it slower, the projectile will get closer to the surface until it eventually hits, like a bullet fired from a gun. If you fire the projectile faster, it will gain altitude (and lose speed) until it’s exactly opposite the point where you fired it. Then it will start descending and picking up speed until it returns to where it started. If you were out in space watching this happen, the path would form an ellipse. Here’s the slightly counter-intuitive part; even though that projectile was fired faster, it takes longer for it go around the Earth.
Not only will an object come back to the same height at which it was fired, it will come back to the same point. It’s just that the Earth is turning. Depending on where you are, by the time the object comes back to the launch site after 85 minutes, you may be a thousand miles to the east.
For an example of what Xema is talking about, try reading “The Holes Around Mars” by Jerome Bixby. (Warning: bad puns.)
You’ll need a completely smooth planet, or an orbit high enough to avoid microgravity perturbations from any mountains you pass.
Ben Bova wrote a humorous short story based on precisely this. The military base on the moon needed good computers to keep track of fired ordnance, so they could be certain of where to be so those old rounds didn’t hit them on one of their many returning orbits.
NO. No no no…
Simple orbital mechanics - anything “fired” into an orbit will return to where it was launched from on the next pass. If you launch it horizontal in a spot with no air and then move the gun out of the way, it will zoom past just grazing the spot where it “took off” from, next pass.
A horizontal launch means tangent to the earth’s surface, meaning this is the “low point” or perigee in the orbit. How high the apogee is depends on the velocity. If you fire at any other angle, it’s as if the shell is coming from a lower perigee deep in the earth.
By “fired” from a gun, assume you mean given an initial speed and direction. Unless that speed exceeds escape velocity, the object will swing around the central mass(earth), then return to where it came from.
The details - yes, air drag will eventually slow the satellite so it drops lower. Yes, in an Earth-moon system, the moon will perturb the orbit somewhat; the closer it gets to the moon, the bigger the effect. Even the sun, Jupiter and other planets will hav a tiny effect. The variations in density of the earth itself affect the regularity of the orbit somewhat. These are mostly secondary, minor effects.
Also, the earth rotates at about 24 hours per cycle. So the starting point will have moved if the shell comes back before or after 24 hours have passed, and the grazing point will be different.
This is why most spacecraft have a final “orbital insertion maneuver”. Once they reach orbital height and are close to orbital velocity, they give a roughly horizontal burst to change the direction so their orbit path does not take them back where they started. Of course, the rockets are boosting all the way up, and slowly tilting over so the trajectory is more horizontal than vertical once they are out of the main atmosphere.
However, even at low orbital heights there is some air drag; Skylab lasted what, about 10 years? Mir about 20, with occasional boosts. Surface area vs weight is relevant. Apparently too, solar activity will heat up the upper atmosphere, expanding it upward to increase density and resistance; which is why Skylab came down faster than anticipated, being relatively light for its cross section area.
The Moon’s centre of mass is actually 2 km closer to Earth than its geographical centre, so your orbit would have to be big enough to take that into account.
Quoth md2000:
To what are you objecting? What you say is correct, but it’s also more or less the same thing everyone else in the thread has been saying.
And I’m guessing that this “Bob Wolmer” fellow was a spammer who got disappeared? And that the spambot recognized this as an astronomy-themed discussion and gave a stock astronomy response, in an attempt to fit in?
Here’s the Wikipedia page Gravitation of the Moon. Just because it has a nice picture showing the variation of the Moon’s gravity at its surface.
I reported the post based on those criteria.
Unless, of course, you fire it at higher than escape velocity, and calculate your trajectory to lose just enough velocity to become stable in orbit as it leaves the atmosphere. Essentially, entering orbit after your initial trajectory precisely at the right speed, and angle to orbit thereafter.
Ignore friction, my ass! Use that bitch.
Tris
No good. As it leaves the atmosphere and enters orbit, it’s trajectory will come full circle to the same position and velocity, meaning that’s it’s leaving the atmosphere again, which means that it must have entered the atmosphere again at some point.
If md2000 is right, and the upper atmosphere expands with increased solar activity, you could try to time the launch to coincide with the maximum expansion. That way your object could leave the atmosphere, and by the time it completes an orbit the atmosphere will have contracted slightly.
Well, I was going to ask that question next, if everyone wasn’t already tired out from the first question 
So if we take atmospheric drag into account, does the OP then become a possible proposition for launch angles other than horizontal?