I was just wondering how the following equation is derived:$$-G\frac{Mm}{r_1}+G\frac{Mm}{r_2}=\frac{1}{2}mv_1^2-\frac{1}{2}mv^2_2~?$$
I already understand why the work $W$ done in moving an object with mass $m$ by a force exerted by a mass $M$ from point $r_1$ to $r_2$ is given by
$$\Delta U = -G\frac{Mm}{r_1}+G\frac{Mm}{r_2}$$
yet the couple of proofs I've seen for $W=-\Delta KE$ assume that the acceleration is constant. How could one prove the equation in its more general form?
If you're familiar with calculus & vectors, then it is relatively straightforward to generalize the work-energy theorem to include forces that vary in direction and magnitude. Here's how:
Start with Newton's Second Law: $$ m \frac{d\vec{v}}{dt} = \vec{F} $$ Take the dot product of both sides with $\vec{v}$: $$ m \vec{v} \cdot \frac{d\vec{v}}{dt} = \vec{F} \cdot \vec{v} $$ Recognize that according to the chain rule, $\vec{v} \cdot (d\vec{v}/dt) = \frac{1}{2} d( \vec{v} \cdot \vec{v})/dt$: $$ \frac{m}{2} \frac{d}{dt} \left( v^2 \right) = \vec{F} \cdot \vec{v} $$ Bring $m/2$ inside the derivative: $$ \frac{d}{dt} \left(\frac{1}{2} m v^2 \right) = \vec{F} \cdot \vec{v} $$ Integrate both sides of the equation from some initial time $t_i$ to some final time $t_f$: $$ \left(\frac{1}{2} m v^2 \right)_{t = t_f} - \left(\frac{1}{2} m v^2 \right)_{t = t_i} = \int_{t_i}^{t_f} (\vec{F} \cdot \vec{v}) dt $$ Recognize that the integral on the right is the line integral of the force along the path taken, and that the left-hand side is the change in the kinetic energy: $$ \Delta (KE) = \int_{\vec{r}_i}^{\vec{r}_f} \vec{F} \cdot d\vec{r} $$ where $\vec{r}_i$ and $\vec{r}_f$ are the object's initial and final position. This last integral is the general definition for the work done on an object: $$ W \equiv \int_{\vec{r}_i}^{\vec{r}_f} \vec{F} \cdot d\vec{r} $$ You can hopefully see that this works out to be the more-familiar $W = \vec{F} \cdot \Delta \vec{r}$ for the case where $\vec{F}$ is constant.
The right hand side of your equation is the change in Kinetic Energy of the mass.
This can be found by doing a Work done = Force x Distance formula, with a variable force, by integrating Newtons Force law, like this
$$\int^{r_2}_{r_1}\frac{GM}{r^2}dr $$
