Are all trajectories parabolic?

I've heard this before, and it makes sense. Is it true?

This being: Due to the effects of gravity and inertia, an object in movement not being acted on by any other force will form a parabolic curve with its trajectory.

• This is only true for the case of a constant gravity field and without friction, in all other cases you are looking at more complicated solutions like ellipses and hyperbolas for the Kepler problem and ballistic curves for constant gravity but with different assumptions about friction. – CuriousOne Jun 8 '15 at 0:11

Are all trajectories parabolic?

No, most trajectories are ellipses. If an object has less than escape velocity, it is an ellipse. If the object just has escape velocity, it has a parabolic trajectory. If it is greater than escape velocity, it is hyperbolic.

For Earth, the escape velocity is just over 11.1 km/s (6.89 miles/s). This is faster than a high speed rifle bullet.

Apollo 10 set the record for the highest speed attained by a manned vehicle at 39,897 km/h (11.08 km/s or 24,791 mph) during the return from the Moon on May 26, 1969. This is just shy of the speed for escape velocity.

• I suspect the OP is asking about the usual "near the surface of the Earth" constant $\vec{g}$ approximation... – dmckee Jun 8 '15 at 1:16
• @dmckee My statement is still correct. It's just that with air resistance, the object usually needs the force applied over time to it to get it up to escape velocity. Oh, did you mean your comment to go with Bruce's answer. – LDC3 Jun 8 '15 at 1:25
• Certainly it is correct, but without a little bit more explanation it is likely to confuse the OP. You really need to tell him that the nearly parabolic arch of a fountain jet (a case where air resistance is minimized) is actually the apogee end of a very pointy ellipse... – dmckee Jun 8 '15 at 1:28

the parabolic curve is an approximation only valid when you can consider g as a constant, but g is a function of the height: $g=GM_{earth}/(R_{earth}+h)^2$. Regardless of the angle you trow the object, if the speed is below the escape velocity of Earth (most likely, unless you use a rocket), and in absence of friction, your trajectory will be exactly an ellipse, not a parabola.

• Absent air resistance it'd be giant rail guns all the way to escape velocity ;-) – Steve Jessop Jun 8 '15 at 12:23

The assumptions of a parabolic trajectory is:

1. (gravity) force is a /constant/, /downwards/ force,
2. /no other forces/ acting on the object,
3. the object under observation does not affect the (gravity) field,
4. classical mechanics

In practice, this approximation is almost true with gravity when:

1. there is only one significant mass that imparts any significant gravity,
2. the distance traveled by the object under observation is fairly short,
3. the mass of the object is miniscule relative to the mass of the object imparting the dominant gravity,
4. speed is sufficiently small,
5. friction is negligible.

In reality, the gravity field is not just a constant downwards force. Instead, gravity is a field that is dependant on the masses of objects around it. In the large space, there are multiple "centers of gravities". The magnitude of gravity changes as the object travels further from the center of gravity, rather than being constant; the direction of the gravity field changes over large distances, not just downwards.

In reality, there's frictions, and multiple centers of gravities. Also any objects that have mass affects the gravity field, imparting its own gravity by an equal amount of force in opposite direction.

In reality, the force of gravity does not traverse over space instantaneously; instead it travels at the speed of light.

In reality, "center of gravity" is a lie; objects have volume and the shape of the object affects the shape of the gravity field.