How come projectiles have parabolic path?

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Discussion Overview

The discussion revolves around the nature of projectile motion and the reasons behind the parabolic trajectory observed in projectiles. Participants explore the implications of gravitational forces, the assumptions made in classical mechanics, and the relationship between gravitational direction and trajectory shapes.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants note that the parabolic trajectory of projectiles assumes a constant gravitational force, which is a good approximation for everyday objects.
  • Others argue that the gravitational force actually varies slightly with altitude, suggesting that the true trajectory is a segment of an ellipse.
  • One participant points out that in a geosynchronous orbit, the gravitational force does not change, yet circular motion is observed, proposing that the direction of gravity's variation is crucial for elliptical and circular solutions.
  • Another participant suggests that measuring the trajectory of a ballistic object would reveal it as a segment of an ellipse with its focus at the center of the Earth.
  • Some participants mention that the gravitational gradient is spherical, which could imply a more complex path than a simple parabola.
  • There is a reference to Newton's law of gravitation leading to conic sections as solutions to the two-body problem, indicating a broader context for understanding projectile motion.
  • Participants discuss the approximation of the Earth's surface as flat for local analysis, which simplifies the understanding of gravitational forces acting perpendicular to this plane.

Areas of Agreement / Disagreement

Participants express various viewpoints on the nature of projectile motion, with no consensus reached on the implications of gravitational variation or the exact nature of the trajectory. Multiple competing views remain regarding the relationship between gravitational force and trajectory shape.

Contextual Notes

Some limitations in the discussion include assumptions about gravitational constancy, the effects of altitude on gravitational force, and the simplifications made when modeling the Earth's surface as flat.

xAxis
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I wonder no one posted this question before. I understand we don't take the Earth curviture into equation of motion, but still projectiles do have parabolic trajectory.
Doesn't parabolic trajectory imply unbounded orbit?
 
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The parabolic trajectory assumes that the gravitational force on the projectile is constant. For everyday objects, like a ball thrown into the air, this is an exceedingly good approximation. However, if you really want to be exact about it, you would have to consider that the force on the projectile actually decreases (imperceptibly slightly) as it moves upward and increases again as it comes down. If you include this, you'll find that the trajectory is really a very short segment of an ellipse.
 
Ah, but for a geosynchronus orbit, g does not change at all, yet we get a circular solution to the motion of a geosynchronus satellite!

I would argue that it is the fact the variation in the DIRECTION of g, and not it's magnitude that results in elliptical and circular solutions. In other words, we don't get circular/elliptical solutions because we assume the direction of g remains constant.

If we assume the direction of g can vary (like we do for circular motion) than we can arrive at the said elliptical/circular solutions.

Claude.
 
Claude Bile said:
Ah, but for a geosynchronus orbit, g does not change at all, yet we get a circular solution to the motion of a geosynchronus satellite!

I would argue that it is the fact the variation in the DIRECTION of g, and not it's magnitude that results in elliptical and circular solutions. In other words, we don't get circular/elliptical solutions because we assume the direction of g remains constant.

If we assume the direction of g can vary (like we do for circular motion) than we can arrive at the said elliptical/circular solutions.

Claude.

In fact, to get elliptical solutions you need both. [tex]\vec{g}[/tex] must always point to one particular place and it must vary as [tex]\frac{1}{r^2}[/tex], with r the distance to that point.
 
You can even ignore all the points made above and still solve the paradox.

If you were very, very careful to measure the (apparfently parabolic) trajectory of a ballistic object, you would discover that it is actually a segment of an ellipse with its focus 4000 miles away - at the center of the Earth.

If the Earth were magically replaced with a black hole of 1 Earth mass, the object would follow an elliptical orbit (well, if you ignore the gradient).
 
Last edited:
Vf= Vi + at

Acceleration is a "squared" quantity.

The graph of a square is parabolic
 
Agnostic said:
Vf= Vi + at

Acceleration is a "squared" quantity.

The graph of a square is parabolic
Of course, the gravitational gradient around the Earth is spherical, and thus would require a graph with polar coordinates. Thus making the path an ellipse...:biggrin:
 
the Newton's law of gravitation could result in a parabolic path (when potential energy=kinetic energy and when some other conditions are meet). In fact, the general solution is all conic sections as a result of the solution of the two body problem.
 
Usually, near the surface of the earth, you approximate the local surface of the Earth with it's tangent, or actually on infinite plane, and then assume that the gravitational force is prependicular to this plane. And as it was pointed out, the Newtons gravitational force is spherically symmetric and the force acts to the radial direction. And this is needed for the Kepler's laws.
 
  • #10
WHen i first looked at this thread i thought "hey that's something easy," but then got confused when reading other answers :P

Los Bobos said:
Usually, near the surface of the earth, you approximate the local surface of the Earth with it's tangent, or actually on infinite plane, and then assume that the gravitational force is prependicular to this plane.

That makes sense. Earth is big:smile:
 
  • #11
You may want to look at the date of the last post. This may no longer be an issue.

Zz.
 

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