How long would it take a mass dropped from the ISS (~400km above

I will exclude friction since you specify nothing more about your mass.
With the second law of movement you get:
[eqn]\ddot r \left(t\right) \,=\, - \mathscr G \, \frac{M_\oplus}{r \left(t\right)^2}[/eqn]
It's a nonlinear differential equation. You can try to approximate it by taking the tangent of the term on the right and get a linear differential equation. At least that's what a non-autistic physicist would do.

It will never hit the ground if you ignore friction because it'd be in orbit.

>>8609565
Oh man, I found that question in one of my old high school physics textbooks...

>>8609565
>Newtonian notation

>>8609443
You cannot drop stuff from orbit. If you throw it down it will assume an elliptical orbit and come back half an orbit later. The best way to deorbit something is to throw it backward at the same velocity that you have. Whole story here:
http://www.science20.com/robert_inventor/why_a_ball_thrown_to_earth_from_orbit_boomerangs_can_astronauts_hit_earth_with_a_ball_arrow_or_bullet-158237

>>8610061
I know that, but this is from a high school physics textbook which basically ignores orbital dynamics

>>8610317
>basically ignores orbital dynamics

I hate when textbooks or professors do this shit. They skim through the subjects quickly and often pass over small details.

>>8609443
Why don't they just pose the question, "How long would it take an object to hit the ground fro 400km high?" Then you'd have to take in density and shape and air resistance and terminal velocity. That's where the good nit-picking comes into play.

>>8610495
What I assume your problem is, it's how to solve a free fall problem without a constant gravity. Well, asuming you want to use newtons universal Gravitation and you asume there the movement ocurrs radialy then you just have to consider conservation of energy. OKay, so because the movement ocurrs in one directio we put our frame of reference yadiyadiyaya and we get an expresión of the total energy of the form:
[math]E_{T}=\frac{mv^2}{2}-\frac{GMm}{x}=\frac{GMm}{x_{0}}[/math]
The last term just being the initial potentitial energy. So we just rearrange shit to get
[math]\sqrt{\frac{2GM}{x_{0}}+\frac{2GM}{x}}=\frac{dx(t)}{dt}\\
\therefore\\
t-t_{0}=\int_{t_{0}}^{t}\frac{1}{\sqrt{\frac{2GM}{x_{0}}+\frac{2GM}{x}}}dx [/math] Probably you can solve this integral, though I don't think you can solve for x to get a closed shit.

>>8610574
Can someone confirm?