Can the displacement in a gravitational field by given by a 3rd degree polynomial?

Question

I mean a function of the form

x(t) = at^3 + bt^2 + ct + d.

Why or why not?

Answer 1

if you assume newtonian's theory, then the solution for displacement comes out to be a conic section, either a ellipse, or a hyperbola or a parabola. I think we can't express a conic as a polynomial equation. Please correct me if Iam wrong.

Answer 2

Time is almost an omnipotent unit of inverse-square law force fields, where force carries time along.

a, b, c, d are not absolute constants, but vary with related parameters. Say an areoplane, when moves with velocity (v) - it faces a resistance proportion to some power of V. But when it moves with 2V velocity, air resitance varies with cube of velocity. These situations do arise, when we take real picture into consideration, and not the ideal pictures - like let there be vacuum, let the string of pendulum be of negligible mass, neglect friction, etc in physics.

Here, a, b (acceleration), c (velocity), d (distance) have difference dimensions or units say respectively, as shown below (units of each term on each side of = must be same)

x(t) = at^3 + bt^2 + ct + d diferentiate wrt "t"

velocity = 3at^2 + 2bt + c (variable velocity)

acceleration = 6at + 2b (ist term-variable accelation)

This is helpful when work done is to be calcuated under gravitational field, Here you would be given Force F(t) then you will calculate dx by differentiating displacement equation along with (dt) and then total work done from t=0 to t=t can be calculated under real scenario,

You find so many such steps in engineering

Answer 3

It can if the acceleration values. Gravitational acceleration values for example in a free fall from a very high altitude, comparable to the Earth's radius.