Right, but you noted a difference between a path that's a consequence of constant-G (parabolic) and a path that's a consequence of decreasing-G (elliptical).RandallB said:But I didn’t note a difference between a constant G-field and a decreasing G-field.
It's got nothing to do with air resistance, in fact I'm assuming there are no forces besides gravity acting on the object. In a constant G-field the acceleration in the vertical direction should be constant (because the gravitational force on a mass m will always be -mg), so you have:RandallB said:I agree that is the question being asked. And conceptually GR and Newton both do not predict parabolic trajectories here. Conceptually the declining horizontal speed due to air resistance does. I don’t see where a G-field that doesn’t increase during the fall helps.
The only conceptual approximation I see that could create a parabolic without the air resistance would be to assume a flat surface to earth with an infinite deep fall.
I’m not sure what a constant G-field would do – assuming adequate height and speed, maybe something like a elliptic orbit with a
large precession or maybe a spiral.
[tex]a(t) = -g[/tex]
Integrate that with respect to t to find vertical velocity as a function of time, and you get:
[tex]v(t) = -gt + v_0[/tex]
Integrate again with respect to t to find vertical position as a function of time:
[tex]y(t) = (-g/2) t^2 + v_0 t + y_0[/tex]
([tex]v_0[/tex] and [tex]y_0[/tex] represent the velocity and height at t=0)
If you graph position vs. time, this gives you a parabola. And if the object had an initial horizontal velocity as well as a vertical velocity, then the horizontal velocity would be constant since the gravitational force only acts in the vertical direction, so the object would describe a parabolic path through space as well.