Gravitational Potential Energy

Suppose that we have an object $A$ with mass $M$ and an object $B$ with mass $m$. We would like to know the gravitational potential energy of object $B$. For simplicity, we will assume that the universe consists of only these two objects (this is actually an acceptable assumption in many cases).

We want to calculate the work that we do in taking object $B$ a distance $R$ away from object $A$.

Define $\vec{F_{gr}}$ as the gravitational force acting upon object $B$. Clearly, the force that we exert to overcome gravity is equal to $-\vec{F_{gr}}$.

We use the formula for potential energy: $$U=\int_{\infty}^{P}\vec{F}\cdot\vec{dr}$$ Now, we can substitute. We can let $F=-\vec{F_{gr}}$ and replace the point $P$ with the distance $R$. We will call $U_{gr}$ our desired gravitational potential energy.

Since gravity is conservative (I will not prove that here), we may choose any path we wish. The natural choice is a straight line. Thus, $\vec{F}\parallel\vec{dr}$ at all times, so we may simplify the dot product.

Our new integral is the following: $$U=\int_{\infty}^{R}|\vec{F_{gr}}||\vec{dr}|$$ Now, we can reverse the integration and use Newton's Law of Universal Gravitation: $$U=\int_{R}^{\infty}-|\vec{F_{gr}}||\vec{dr}|$$ $$U=\int_{R}^{\infty}-\frac{mMG}{r^2}dr$$ $$U=[\frac{mMG}{r}]^{\infty}_{R}$$ $$U=-\frac{mMG}{R}$$ So there you have it! That is the formula for gravitational potential energy!

Notice that this value is always negative. This is no mistake, since gravity is an always attractive force. Thus, the object $B$ is being pulled toward $A$, and you are actually always doing negative work on object $B$.

Now let us say that you are on the surface of a planet with radius $R$. Now, you are on a smaller scale everywhere close to the surface, and you really do not want to deal with the universal gravitational constant $G$. You are working with small-scale problems, and the gravitational strength will be roughly the same throughout. Thus, the gravitational acceleration $g$ of the planet will suffice. Now, what is a good approximation of the potential energy?

Calculating the approximate gravitational acceleration: $$g=\frac{MG}{R^2}$$ Now, let us calculate the change in potential energy from the distance $r_1$ to the distance $r_2$: $$\Delta U=\frac{mMG}{r_1}-\frac{mMG}{r_2}$$ $$\Delta U=mMG\frac{r_2-r_1}{r_1r_2}$$ But wait! $r_2-r_1$ is the change in height above the planet, $\Delta h$! Moreover, both $r_1$ and $r_2$ are so close to the radius of the planet $R$ that the quantity $r_1r_2$ is roughly equal to $R^2$! Thus, we make the following approximation: $$\Delta U=mMG\frac{r_2-r_1}{r_1r_2}$$ $$\Delta U\approx mMG\frac{\Delta h}{R^2}$$ $$\Delta U\approx mg\Delta h$$ This is an amazing result! The change in gravitational potential energy is approximately equal to the product of the mass of the object, the gravitational acceleration, and the change in height. This is what you will use in most Earth-based mechanics problems, but be sure to use the exact formula shown earlier in problems with a large scale.

This exercise will present a GREAT use of energy, and demonstrates the power of energy.

A ball of mass $m$. is on top of a frictionless ramp of height $H$, and the angle of inclination is $\theta$.

The ball rolls off the ramp onto a flat surface. What's the ball's speed when it hits the surface?

You can probably see an approach using horse diagrams, computing acceleration and velocity, and finally figuring out the answer. But that's ugly!!! Here's a much nicer approach.

The ball is elevated, but is not moving. Therefore, it only has gravitational potential energy.

Good work! Now let's do the other half: when the ball is off the ramp.

Though the ball is now moving, it is no longer elevated. Therefore, it only has kinetic energy, and no gravitational potential energy.

Now, let's finish. Since the ramp is frictionless, there was no work done on the ball during the process. Therefore the starting and ending energies are the same!

Setting the 2 equations equal, we get that $\frac{1}{2}mv^2 = mgH \implies \frac{1}{2}\not{m}v^2 = \not{m}gH \implies v^2 = 2gH \implies v = \sqrt{2gH}$.

Not only does the mass not matter, but the angle $\theta$ does not matter either. How cool is that! And there were almost no computations involved in the process. :)