# Potential Energy per Kilogram, on Earth and on the Moon

###### tags: `Jean-Baptiste`

# Gravitational potential at the surface of the Moon


The problem inputs are

$$
\begin{equation}
\tag{1}
\begin{cases}
\frac{M_{Earth}}{M_{Moon}} = 81 \\
\frac{r_{Earth}}{r_{Moon}} = 3.7 \\
E_{P\,Earth} = -63\ \mathrm{MJ\ kg^{-1}}
\end{cases}
\end{equation}
$$

We also know

$$
\begin{equation}
\tag{2}
E_P = - \frac{GMm}{r}
\end{equation}
$$

but we don't know $m$ (the mass of the object whose gravitational potential we wish to compare).

However, applying $(2)$ to both the Earth and the Moon situations, and keeping the same $m$, we derive

$$
\begin{equation}
\tag{3}
\begin{cases}
E_{P\,Earth} = - \frac{GM_{Earth}m}{r_{Earth}} \\
E_{P\,Moon} = - \frac{GM_{Moon}m}{r_{Moon}}
\end{cases}
\end{equation}
$$

The ratio of $E_{P\,Moon}$ to $E_{P\,Earth}$ works out to

$$
\begin{align*}
\frac{E_{P\,Moon}}{E_{P\,Earth}} &= \frac{\frac{GM_{Moon}m}{r_{Moon}}}{\frac{GM_{Earth}m}{r_{Earth}}}\\
&= \frac{GM_{Moon}m}{r_{Moon}} \times \frac{r_{Earth}}{GM_{Earth}m}\\
\end{align*}
$$

Simplifying by both $G$ and $m$, we have[^1]

$$
\begin{align*}
\frac{E_{P\,Moon}}{E_{P\,Earth}} &= \frac{M_{Moon} \times r_{Earth}}{r_{Moon} \times M_{Earth}} \\
&= \frac{M_{Moon}}{M_{Earth}} \times \frac{r_{Earth}}{r_{Moon}} \\
&= \left(\frac{M_{Earth}}{M_{Moon}}\right)^{-1} \times \left(\frac{r_{Earth}}{r_{Moon}}\right) \\
\end{align*}
$$

and therefore

$$
\begin{equation}
\boxed{
E_{P\,Moon} = E_{P\,Earth} \times \left(\frac{M_{Earth}}{M_{Moon}}\right)^{-1} \times \left(\frac{r_{Earth}}{r_{Moon}}\right)
}
\end{equation}
$$

From the problem data, this works out to

$$
E_{P\,Moon} = (-63 \times \frac{1}{81} \times 3.7) \ \mathrm{MJ\ kg^{-1}}
$$

$$
\boxed{
E_{P\,Moon} = -2.877 \ \mathrm{MJ\ kg^{-1}}
}
$$

## Epilogue


<style>
img[alt="Saturn V"], img[alt="LEM"] {
max-width: 40%;
}
</style>

From data available on [Wikipedia](https://en.wikipedia.org/wiki/Saturn_V) (for Saturn V and the CSM) and the [NASA website](https://nssdc.gsfc.nasa.gov/nmc/spacecraft/display.action?id=1969-059C) (for the LM):

| Stage | Wet (full) mass | Dry (empty) mass | Propellant mass
| -------- | -------- | -------- | -------- |
| [S-IC](https://en.wikipedia.org/wiki/Saturn_V#S-IC_first_stage) | 2,214,000 kg | 137,000 kg | 2,077,000 kg |
| [S-II](https://en.wikipedia.org/wiki/Saturn_V#S-II_second_stage) | 480,000 kg | 36,000 kg | 444,000 kg |
| [S-IVB](https://en.wikipedia.org/wiki/Saturn_V#S-IVB_third_stage) | 119,000 kg | 10,000 kg | 109,000 kg |
| **Total (Saturn V)** | | | **2,630,000 kg** |
| [Apollo LM ascent stage](https://nssdc.gsfc.nasa.gov/nmc/spacecraft/display.action?id=1969-059C) | | 2,445 kg | 2,376 kg
| [Apollo CSM](https://en.wikipedia.org/wiki/Apollo_command_and_service_module) | 28,800 kg | 11,900 kg | 16,900 kg |
| **Total (Apollo return to Earth)** | | | **19,276 kg** |

... which works out[^2] to a *practical* $E_{🚀\, Earth}$ more than 130 times larger than ${E_{🚀\, Moon}}$. This is called, [the tyranny of the rocket equation](https://www.nasa.gov/mission_pages/station/expeditions/expedition30/tryanny.html) (not to mention atmospheric drag).

[^1]: [Strikethrough doesn't appear to be possible in MathJax](https://stackoverflow.com/questions/62089276/strikethrough-in-mathjax), sorry.
[^2]: Assuming both sets of propellants have similar performance, as measured by their energy-to-mass ratio. We neglected the weight of the descent stage of the LM (which stayed on the Moon); but we suspect that the difference would be negligible if we didn't.