# How come projectiles have parabolic path?

1. Nov 15, 2006

### xAxis

I wonder noone 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?

2. Nov 15, 2006

### Parlyne

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.

3. Nov 16, 2006

### Claude Bile

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.

4. Nov 16, 2006

### Parlyne

In fact, to get elliptical solutions you need both. $$\vec{g}$$ must always point to one particular place and it must vary as $$\frac{1}{r^2}$$, with r the distance to that point.

5. Nov 16, 2006

### DaveC426913

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: Nov 16, 2006
6. Nov 16, 2006

### Agnostic

Vf= Vi + at

Acceleration is a "squared" quantity.

The graph of a square is parabolic

7. Nov 16, 2006

### DaveC426913

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...

8. Nov 16, 2006

### tim_lou

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.

9. Nov 17, 2006

### Los Bobos

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. Jul 27, 2007

### laurelelizabeth

WHen i first looked at this thread i thought "hey that's something easy," but then got confused when reading other answers :P

That makes sense. Earth is big

11. Jul 27, 2007

### ZapperZ

Staff Emeritus
You may want to look at the date of the last post. This may no longer be an issue.

Zz.