Path of a thrown object?

[mvronsky]

It seems as if many textbooks say that the path of a thrown object is a parabola, and fail to mention it is in fact an elipse with one focus at the center of the earth. I have not yet seen a single textbook in which anything is mentioned of the fact that a parabola only very closely estimates the actual trajectory of the thrown object. I have also only been able to find a few websites online discussing this problem.

Fun story: I asked my math teacher this question, whether the path of a thrown object is a parabola or a highly eccentric ellipse. Obviously, she said a parabola. After I tried to prove my point by outlining Newton's cannonball she promptly denied Newton's cannonball and said that an orbit that launches on one side of a large body such as earth, and then goes 3/4 around the earth before hitting the ground, is not possible.

 

[Quick_Nick]

If the specifics are firing directly sideways, and going 3/4 around, I agree that should not be possible. (if you're assuming an ellipse)
I don't know what you expect. Textbooks (K-12, right?) don't immediately describe the world in terms of quantum mechanics either. They give reasonably simple and accurate models, and maybe build from there if it makes sense to.

 

[mvronsky]

I didn't imply zero vertical velocity.

---------- Post added at 06:10 AM ---------- Previous post was at 06:09 AM ----------

Additionally, I don't expect them to go into detail or even list formulas, I just expect them to acknowledge that this is a very precise approximation.

 

[MattBaker]

Not really the best in physics here but:

and said that an orbit that launches on one side of a large body such as earth, and then goes 3/4 around the earth before hitting the ground, is not possible.

She's right. If we assume a flat surface of the Earth/no atmosphere etc. and your orbit hits the ground at 3/4 that would be the lowest point of your orbit up to then (not necessarily your periapsis since that could be later on). But that's not possible, the periapsis should be after 1/2 of the orbit. Or if you accelerated enough and made the cannon your periapsis at your starting position. Unless you're launching from space, make course corrections or dug your cannon a few kilometers into the ground the scenario you describe is impossible and your teacher was right.

I can already see I got :ninja:'d

 

[mvronsky]

But I never implied no vertical velocity. It would be possible if you didn't launch perfectly horizontally, right?

---------- Post added at 06:22 AM ---------- Previous post was at 06:15 AM ----------

Nevermind these previous comments.
I now understand my flaw. But still, she dodn't acknowledge the whole ellipse thing

 

[asbjos]

But I never implied no vertical velocity. It would be possible if you didn't launch perfectly horizontally, right?

Take a powerful spaceship in Orbiter to the Moon and try to launch in one burn on such a trajectory that you will crash into the surface after half an orbit.
You will soon see that your periapsis will be a point on the Moon never more than 180 degrees from your launch location.

 

[Quick_Nick]

I now understand my flaw. But still, she dodn't acknowledge the whole ellipse thing

You're in the United States. Teachers have some of the lowest salaries and frequently don't require much more than a basic education themselves. I've seen much worse; consider your teacher one of the better ones if that's the biggest error they make. (Not that it should be this way...)

 

[Zatnikitelman]

I don't know, after seeing this topic, I immediately went into Orbiter, lifted the DG off from Brighton and got my self into an "orbit" with a re-intercept of the moon's surface roughly 3/4 of the way around from where I was. I may be missing something in the question here, it's late, but I can play with this some more tomorrow. Since I only hovered 0.01km off the surface and burned horizontally till I had the orbit, the size of the vessel may mean a practical intercept of something less than 3/4 of the way around, but my center of mass would not by the numbers hit the surface until 270 degrees around my orbit from where I was.

 

[Hielor]

Newton's cannonball posits that the cannonball is fired horizontally. If you're mentioning Newton's cannonball and then claiming that it's not firing horizontally, you're not talking about Newton's cannonball.

The "thrown object follows a parabola" simplification exists with the "gravity is down" assumption and is only valid in small enough situations where that assumption is more or less true. If your path is large enough that the curvature of the Earth becomes an issue, and you can no longer assume that gravity is acting in a constant direction, then paths will indeed no longer be parabolas.

I highly doubt that you can throw a ball fast enough for the curvature of the Earth to come into play.

Edit: Since you're such a mathematical genius, I challenge you to determine at what distance across the surface of the Earth the differences between a parabolic path (in the "gravity is down" assumption) and an elliptical path (in the "gravity is inward since you're on the surface of a sphere") become mathematically significant. I'd be willing to bet that the "no atmospheric drag" assumption introduces significantly more error than the "parabolic path" assumption.

 

[dgatsoulis]

Take a powerful spaceship in Orbiter to the Moon and try to launch in one burn on such a trajectory that you will crash into the surface after half an orbit.
You will soon see that your periapsis will be a point on the Moon never more than 180 degrees from your launch location.

That's not the correct way to test this. Remember that in Newton's canon all the dV is instantaneous, the canon fires horizontally and the starting altitude is not 0.

Mvronsky already pointed out that he never implied a horizontal launch, so even with a 0 starting altitude the trajectory works out and you can arrive back at the surface with a traveled angular distance > 180°.

The best way to test this in Orbiter, is to get the Delta-Cannon and attach a DG on it. Do this on any moon where the orbital velocity at 0 alt is less than 1000 m/s because that's a limitation of the Delta Canon's output velocity.

On Oberon for example, at a starting lattitude of 1.0°, heading 90° and launch Pitch of 5°, you arrive back at the surface almost exactly 270° if you use 545 m/s as the canon's output velocity.

@mvronsky
You say that you never implied a horizontal launch, but did you explicitly state it as a prerequisite for the initial launch condition?
Newton's canon specifically states a horizontal launch.

If not, then your teacher is correct. At a 0 starting altitude and with a perfectly horizontal launch where all the dv is applied in 0 seconds, it is impossible.

If however, you specifically stated that you need not launch horizontally, then your teacher is wrong. You can arrive back at the surface at any angular distance from the launch site, depending only on the initial launch pitch and applied dv.

Here is a pic of the Oberon launch:

The spacecraft arrives back at the surface almost exactly 270° degrees away from the launch site.

Here is the scenario for you to try out. You will need Orbiter 2006P1 (Haven't tested with 2010P1, but I don't think the Delta-Canon works) and [ame="http://www.orbithangar.com/searchid.php?ID=621"]Kultch's Delta Canon, release 2.[/ame]

BEGIN_DESC
Delta Cannon is ready to shoot by DeltaGlider. Oberon
END_DESC

BEGIN_ENVIRONMENT
  System Sol
  Date MJD 53480.3133665278
END_ENVIRONMENT

BEGIN_FOCUS
  Ship GL-01
END_FOCUS

BEGIN_CAMERA
  TARGET GL-01
  MODE Extern
  POS 6.80 120.60 1.16
  TRACKMODE TargetRelative
  FOV 70.00
END_CAMERA

BEGIN_HUD
  TYPE Surface
END_HUD

BEGIN_MFD Left
  TYPE Surface
  SPDMODE 1
END_MFD

BEGIN_MFD Right
  TYPE Orbit
  PROJ Ship
  FRAME Equator
  ALT
  REF Oberon
END_MFD


BEGIN_SHIPS
GL-01:DeltaGlider
  STATUS Landed Oberon
  POS 45.00  1.0
  HEADING 90.00
  FUEL 1.000
  GEAR 0 0.0
  NOSECONE 0 0.0000
END
DeltaCannon:DeltaCannon
  STATUS Landed Oberon
  POS 45.00  1.0
  HEADING 90.00
  BUGEL 0 0.0000
  HEADDIR 0.00
  PITCHDIR 0.05
  OUTPUTSPEED 545
  CANNONBALL_NAME GL-01
END
END_SHIPS

All you need to do is to tap the numpad+ and the spacecraft will be in the trajectory you see in the pic above.

 

[asbjos]

That is actually incorrect.

I hate it when I'm wrong! Sorry.


On the other hand, "Mvronsky" said that it was his maths teacher. I think it's wrong to accuse the teacher of being wrong when it's not her area of knowledge.
Accuse the maths teacher if she doesn't know basic maths, accuse the physics teacher if she/he doesn't know basic physics.

 

[dgatsoulis]

I hate it when I'm wrong! Sorry.


On the other hand, "Mvronsky" said that it was his maths teacher. I think it's wrong to accuse the teacher of being wrong when it's not her area of knowledge.
Accuse the maths teacher if she doesn't know basic maths, accuse the physics teacher if she/he doesn't know basic physics.

Sorry for the misunderstanding, I re-worded that sentence. I was referring to the testing method.

 

[Thorsten]

The "thrown object follows a parabola" simplification exists with the "gravity is down" assumption and is only valid in small enough situations where that assumption is more or less true. If your path is large enough that the curvature of the Earth becomes an issue, and you can no longer assume that gravity is acting in a constant direction, then paths will indeed no longer be parabolas.

Since we're nit-picking here, it also requires a 'gravity is constant as function of altitude' assumption, which also isn't strictly true in nature.

Really strictly speaking, general relativity has effects like frame drag, so on some level it actually matters that Earth rotates and drags spacetime locally with it, so the trajectory along the rotation will differ from the trajectory against the rotation. And Earth isn't really a sphere either, so the gravity field has higher moments. So no trajectory will really come out as an ellipse either.

That's what physics is all about - select the approximation of the problem that will do well enough, given the accuracy you need. And for throwing objects, a parabola does usually well enough. I guess artillery gunners include drag and wind into the game if they actually want to hit something.

I'd be willing to bet that the "no atmospheric drag" assumption introduces significantly more error than the "parabolic path" assumption.

If you shoot a projectile out of the dense lower atmosphere, drag is going to mess a lot with your path for any realistic ballistic coefficient. Same on re-entry of the projectile.

But if anyone wants to take the bet for a given muzzle velocity and ballistic coefficient and guess what the deviation would be, my atmospheric entry simulation code can do cannons as well

 

[Urwumpe]

External ballistics... its really hard to believe that mechanical computers managed to compute them accurately, if you know that they also involve coriolis force and Magnus effect.