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Radial fall in a Newtonian gravitational field
My math and physics are rusty. I am trying to calculate the time an object takes to fall to a large body. Before you answer $1/2at^2$, the conditions where $1/2at^2$ apply is where $a$ is a constant. Constant acceleration is a fine approximation when the distance the object falls is $<<$ than the distance to the center of the large body. i.e. 100 meters on the surface of the Earth.
I'm looking at problems where distance fallen is 50% or more of the distance between objects.
Gravity varies with the inverse square of distance. Given two bodies, $m_s$ and $m_l$.
$$ F = m_sa = m_la $$ $$ F = Gm_sm_l/r^2 $$ $$ m_sa = Gm_sm_l/r^2 $$ $$ a = Gm_l/r^2 $$
Thus acceleration of $m_s$ varies with the inverse square of the distance to the center of $m_l$, and versa visa.
Given R is the radius of the $m_l$, defining the surface of $m_l$. Given $h$ is the distance $m_s$ is above $m_l$'s surface, subsitute $R+h$ for $r$.
$$ a = Gm_l/(R+h)^2 $$
This is as discussed at stackexchange 35878.
Given the initial velocity of $m_s$ is zero, what is the time, $t$, it takes $m_s$ to fall distance $h$?
An alternative question is give a $t$, how far with $m_s$ fall?
This is where I'm rusty. I don't believe one can't simply substitute $Gm_l/(R+h)^2$ for $a$ in $1/2at^2$, but I don't know what else to do.
The stackexchange articles, 15587 and 41741, discuss $a(x)$ where $a$ changes linearly with distance. Here $a$ changes with the inverse square of the distance.
I did a lot of web searching and did not find this topic discussed. The discussions are with $a$ being constant or changing in a linear manner.
Thanks for the help.
