Don’t ask me why this was on my mind, I have to deal with my thoughts and now so do you
Anyways, have you ever experienced a Poseidon’s Kiss? In simple terms, it’s when you’re using the toilet, and the stool dropping into the bowl causes a jet of water to rush up and splash right where everything originated. Well, I’ve wondered for some time exactly how high this jet of water could go. Unfortunately, that’s a very difficult science problem involving surface tension, hydrodynamics, etc.
But I am still an engineer, which means I can just make assumptions until I am able to solve a problem and then just draw a box around it and call it done. So, I will instead solve the much easier problem of How much water could the average stool send into orbit, assuming no frictional losses? Great question! Let me tell you.
So first, how fast are we talking? To achieve Low Earth Orbit (LEO), our liquid payload needs to reach roughly 7,800 m/s. For perspective, that’s about 23 times the speed of sound. If your toilet water actually hit this speed, the resulting shockwave would likely deconstruct the porcelain, and your bathroom, into a fine ceramic mist long before the water left the bowl. To maximize the stool-powered water launcher, assume perfect energy transfer and (critically) neglect atmospheric drag. Additionally, assume the following values:
The Energy Source
- Mass ($m_{s}$): $0.4\text{ kg}$ (a healthy, robust contribution).
- Drop Height ($h$): $0.3\text{ meters}$ (from the “launch pad” to the water line).
- Gravity ($g$): $9.8\text{ m/s}^2$.
With this, the total potential energy ($U$) is calculated as:
\[\begin{split} U &= m_{s} \cdot g \cdot h \\ U &= 0.4 \cdot 9.8 \cdot 0.3 \\ U &\approx 1.18\text{ Joules} \end{split}\]Orbital Velocity
The kinetic energy ($K$) required to achieve orbit depends on that staggering velocity of $7,800\text{ m/s}$. To find the mass ($m_{w}$) of water that could hit orbital speeds using our $1.18\text{ J}$ of energy, we use the kinetic energy formula:
\[K = \frac{1}{2} m_{w} v^2\]Rearranging for $m_{w}$:
\[\begin{split} m_{w} &= \frac{2K}{v^2} \\ m_{w} &= \frac{2 \cdot 1.18}{7,800^2} \\ m_{w} &\approx 3.88 \times 10^{-8}\text{ kg} \\ m_{w} &\approx 0.0388\text{ mg} \end{split}\]That is roughly 0.038 milligrams of water. For context, a single standard raindrop is about $50\text{ mg}$. This is approximately half the weight of a single grain of salt. So, your morning routine could theoretically launch a miniscule mist into orbit, or at least the amount of dignity remaining after the splash.
Takeaways
Next time you experience this aquatic phenomenon, don’t be annoyed. Be proud! You aren’t just a victim of fluid displacement; you are a localized energy generator capable of accelerating matter to speeds that would make NASA sweat!
Pro-tip: To decrease the “Delta-V” of the return splash, consider deploying a “landing pad” of toilet paper to dampen the kinetic transfer. Credit: Poop Splash Elimination - Smarter Every Day 22