Question

If an object is dropped to the earth's surface,prove that path is semi cubical parabola ?

Answer 1

Answer:

For distances small enough that your measurements can safely assume that the earth is flat and gravity is constant throughout, classic mechanical motion equations show that time and distance (or height) take the form of a second-degree equation. Which is the equation of a parabola with various linear transformations. Not a semicubical parabola.

If you use a set of coordinates so that you can see the earth’s rotation, and therefore you have projectile motion, where the horizontal part is at a fixed velocity. Its path then describes a parabola, not a semicubical parabola. Even if the object perfectly bounces on the ground, it is still a set of intersecting parabolas, not a semicubical parabola.

All this, again, under the assumption that the distances involved are small enough that the earth is assumed flat, and gravity is constant throughout.

I’m not that familiar with the actual equation if distances are far enough that the earth has to be modeled as spherical, gravity is not constant, and earth rotates around its axis. I do know that, for this case, trajectories take the form of an ellipse or arc thereof, as we’re dealing with actual orbits. Also that one of Kepler’s law state that:

The total orbit times for planet 1 and planet 2 have a ratio in which the square of the orbital period is proportional to the cube of their (semi)major axis.

Even though this law introduces both a cube and a square relationship, needed for a semicubic parabola to develop, I still don’t see how to pull that up to turn such a curve out of dropping an object onto the earth.

hope it will help uu