Gravitational Potential Energy

A form of Potential Energy that depends on the configuration of masses in a System

Definition §

$$U_{grav}=-\frac{G{m}_1{m}_2}{r}$$

• $U_{grav}$ = Potential energy due to the Gravitational Force (in $J$)
• $G$ = Gravitational constant = $6.67430\times 10^{-11}$
• $m_1,m_2$ = Masses in system (in $kg$)
• $r$ = Distance between masses

Derivation §

Assume we have a system of two masses at rest, $m_1$ and $m_2$, separated by distance $r$ No other objects exist. We want to derive only potential energy, so we need to ensure this system has no moving masses, and no kinetic energy. The only internal force is the fundamental force of gravity, as the system has two masses.

We are looking for some configuration of this system that we will define as having zero potential energy. If we do that, any other configuration that affects $r$ will either have a lower or higher potential energy.

Physicists decided the best ‘zero-point’ for gravitational potential energy is when the gravitational force and the gravitational potential energy both equal each other. In our proposed system, $F_{grav}=0$ only when $\frac{1}{r^2}=0$, as masses must be grater than 0, and $G$ is a constant. The Limit of $\frac{1}{r^2}$ as $r\to \infty$ is 0, and so the force approaches zero if the two masses are an infinite distance apart.

In other words, just as we define $K=0$ when $v_{obj}=0$, we define $U_{grav}=0$ when $r=\infty$

So our ‘zero’ point is when the masses are infinitely far apart. Cool. In order to change the system from this ‘zero’ configuration to any other configuration, with a finite distance $r$, we need to do work on the system.

Now, we need to ensure the kinetic energy of the system, $K$ is zero during all configurations , else some of this work would result in a change in kinetic energy ($W=\Delta K$). We want all the work done to affect the potential energy, $U$, only. So we need to ensure $v$ is as little as possible.

To change our system from C1 to C2, let’s apply an external force equal in magnitude to the gravitational force, but in the opposite direction, on $m_2$. I.e. $F_{ext}=-F_{grav}$. Then $\sum F=ma=0\to a=0$. So $K=0$.

The formula for work done is:

Here, we want to calculate the work done by the external force to change the distance from $\infty$ to $r_f$.

The displacement, $\vec{x}$, can be expressed as a change in r, $d\vec{r}$

Because the force is non constant, we need to integrate the total work.

$$W=\int \vec{F}\cdot \text{d}\vec{x}=-\int |F_{ext}|\text{d}x\text{( as vectors are anti-parallel)}$$

$$W=-{\int }_{r_i}^{r_f}|F_{ext}|\text{d}r$$

$$W=-G{m}_1{m}_2{\int }_{r_i}^{r_f}\frac{1}{r^2}\text{d}r$$

$$W=-G{m}_1{m}_2{\left[\frac{-1}{r}\right]}_{r_i}^{r_f}=G{m}_1{m}_2(\frac{1}{r_f}-\frac{1}{r_i})$$

At the same time, we have a conservative, internal force of gravity acting on the mass - the gravitational force, $F_{grav}$. This gravitational force was acting parallel to the displacement. I could do the full calculations, but it’s fairly trivial that if the external force did positive work, then the gravity force did negative work on the system. I’ll prove it anyway, because I goddamn hate the statement “It’s trivial, or left as an exercise to the reader, etc.“’

$$W_c=\int \vec{F}\cdot d\vec{x}=\int |{\vec{F}}_{grav}|dx\text{(as vectors are parallel)}$$

$$W_c={\int }_{r_i}^{r_f}|F_{ext}|\text{d}r$$

$$W_c=G{m}_1{m}_2{\int }_{r_i}^{r_f}\frac{1}{r^2}\text{d}r$$

$$W_c=G{m}_1{m}_2{\left[\frac{-1}{r}\right]}_{r_i}^{r_f}=-G{m}_1{m}_2(\frac{1}{r_f}-\frac{1}{r_i})$$

Also, this shows that:

\sum W = W_{ext} + W_{c} = 0$$And by the [Work-Energy Theorem](Work-Energy%20Theorem.md):

\sum W =0 \to \Delta K =0

The change in potential energy, $U$, comes directly from here, as it is a result of the work done by a conservative force: ---

U = -Gm_{1}m_{2}(\dfrac{1}{r_{f}} - \dfrac{1}{\infty}) = -\dfrac{Gm_{1}m_{2}}{r}

--- Now, we can prove that $-\Delta U = W$

-\Delta U = -U_{f} + U_{i} = \dfrac{Gm_{1}m_{2}}{r_{f}} - \dfrac{Gm_{1}m_{2}}{r_{i}}

-\Delta U = W

In terms of $F_{c}$ - Conservative forces In plain English, $U$ of a system is the work exerted *by gravity* to allow the system to be in a state away from $U = 0$. $U$ is always negative because $r < \infty$ always. Consequently, the change in $U$ is simply the work done by this conservative force. $\Delta U$ then symbolises the system trying to 'go back' towards $U = 0$, which means a positive $\Delta U$ is gravity performing more work, while a negative $\Delta U$ means gravity performs less work. In terms of $F_{ext}$ - External force $U$ of a system is the amount of *external* work needed to be put in by an external force to move it back to $U = 0$. I.e. if we were to add external work equal to $U$ on a system to push the masses apart, we would theoretically have them an infinite distance apart. Change in $U$ means the external force is 'losing' the fight against gravity. If $\Delta U > 0$, the external force is performing less work. If $\Delta U < 0$, the external force is doing more work on the system.