- #1
leok
- 4
- 0
picture a body falling toward a black hole.
at the center of the black hole you have an infinite amount of force pulling you toward the center: F=Gm1m2/x^2 while x=0. F=infinity
so at a certain distance away from the black hole you have a certain force pulling you in.
but as you approach the center your acceleration increases.
but as you accelerate your velocity increases
so you were once not moving at you initial position. after a few seconds you start moving due to the force pulling you in.
but what happens is you gain velocity so now your acceleration is increasing.
but since you are accelerating your velocity starts to jerk.
the jerk effects the acceleration and you get a cycle of one derivative influencing the next.
so the position vs time graph becomes infinitely differentiable.
if you look at it in terms of energy:
work=force x distance
so Kinetic energy would be the integral of f=Gm1m2/x^2
your velocity is the square root of 2 times the integral
if change in momentum=impulse
m x dv= f x t
we have m2 x our velocity = f x t
we have force for a given X but not a given T so we cannot integrate Gm1m2/x^2
so to be able to graph x vs t we need f vs t
please help if you have any insight
at the center of the black hole you have an infinite amount of force pulling you toward the center: F=Gm1m2/x^2 while x=0. F=infinity
so at a certain distance away from the black hole you have a certain force pulling you in.
but as you approach the center your acceleration increases.
but as you accelerate your velocity increases
so you were once not moving at you initial position. after a few seconds you start moving due to the force pulling you in.
but what happens is you gain velocity so now your acceleration is increasing.
but since you are accelerating your velocity starts to jerk.
the jerk effects the acceleration and you get a cycle of one derivative influencing the next.
so the position vs time graph becomes infinitely differentiable.
if you look at it in terms of energy:
work=force x distance
so Kinetic energy would be the integral of f=Gm1m2/x^2
your velocity is the square root of 2 times the integral
if change in momentum=impulse
m x dv= f x t
we have m2 x our velocity = f x t
we have force for a given X but not a given T so we cannot integrate Gm1m2/x^2
so to be able to graph x vs t we need f vs t
please help if you have any insight