I am (just for fun) trying to calculate the velocity with which the moon would impact the earth if it stopped orbiting and simply headed towards earth.
I know the force is $\frac{G m_1 m_2}{r^2}$.
How do I calculate a final velocity resulting from a varying acceleration which in turn is dependant on a varying force which in turn is dependant on a varying distance.
This question is easilly answered by considering the gravitational potential of earth, and invoking conservation of energy.
The potential is $V(r) = -G\frac{m_1 m_2}{r}$ and the kinetic energy of the moon will be given by
$$ K = V(r_{moon}) - V(r_{impact}) $$ where $r_{moon}$ is the current distance from the center of the earth to the center of the moon, and $r_{impact}$ is the distance between the centres when the moon impacts the earth.
Instead of forces, you work with conservation of potential and kinetic energy.
In this case, the kinetic energy that gives this force is $-\frac{Gm_1 m_2}{r}$. The difference in potential energy is transformed into kinetic energy.
With forces, you'd have to go the long way around (integrate the acceleration over time to get the velocity and figure out the time needed to get to the earth).
To begin with, $v dv= a ds$ is the differential equation that relates velocity with variable acceleration. $a=(G*\mathrm{massOfEarth})/s^2$
Substitute it in the first equation and integrate for $v$, 0 to $v$ and $s$, distance between the moon and earth to 0
Hope this is right
