Falling to Earth from Orbit Suppose we ascend to a height $h$ away from the surface of the earth and drop an object. High school physics tells us that it will accelerate at a speed of $9.8 \frac ms$ and will hit the ground in $\sqrt{\frac h{4.9}}$ seconds. However, if we ascend high enough, the gravitational force from the earth is less, and so the acceleration is less, and it will take longer to hit the ground. What is the true equation for how far the object has fallen and how long it will take to reach the ground?
Here's what I have so far: for generality, let the earth or other planet have a radius of $r$ and surface acceleration due to gravity of $G$.
We can derive from the variables above that the force acting on the object at height $h$ is $mG\left(\frac r h\right)^2$. Thus it will accelerate at $G\left(\frac r h\right)^2 = Gr^2 \frac 1 {h^2}\frac m{s^2}$. I want to integrate this with respect to time, and yet my variable is in meters. I need some related rate between the two, but that related rate would be how many seconds it takes to fall some number of meters, which is exactly what I am looking for to begin with. How can I proceed?
 A: Since you're a math major, the way to think about this is that Newton's Second Law is an ODE.  If $h$ is the distance between the object and the center of the Earth, the force on the object is (as you have shown)
$$
F = - m g \frac{r^2}{h^2}
$$
where $r$ is the radius of the Earth.  (Note the minus sign, since the force is attractive.  Also, $g$ is more standard for the gravitational acceleration;  $G$ is the gravitational constant, which is different.) And since $F = m a$, we have
$$
m \frac{d^2h}{dt^2} = - m g \frac{r^2}{h^2} \quad \Rightarrow \frac{d^2h}{dt^2} = - \frac{g r^2}{h^2}
$$
Now you can break out your skills from your ODEs class to solve this ODE for $h(t)$, subject to the initial conditions of $h(0) = h_0$ (some initial position) and $h'(0) = 0$ (the vertical velocity of the particle is zero.)
I will warn you that (IIRC) there is not a closed-form expression for $h(t)$ in terms of elementary functions that solves this equation. But there is an (ugly) expression for $t(h)$—and since you want to find the amount of time for a given $h_0$, that's probably more useful to you anyhow.
