The Physics Behind the Nonsense

Pinocchio’s Nose and Earth Orbit

Created March 3, 2014. Post Last Modified August 27, 2022

As I was watching videos on my YouTube account, this SciShow video was among the recommendations. Now, I love reading the comments as much as watching the video itself; there is usually a comment that simply makes me roll over laughing. Every so often though, something golden comes up, like the following question which has become one of the top comments at the time of publication.

If Pinocchio’s nose is 1 inch long, how many lies does Pinocchio need to tell to get into near earth orbit, remembering that at some point Pinocchio will fall over because his nose is too long?

The Short Answers

There are many possible answers, depending on how Pinocchio is positioned. If he was lying down with his nose pointing up, then one of the following would occur.

It would take around 6.5 million lies for his nose to get to low earth orbit, assuming the nose is massless.
If the mass of the nose changes with length, his face gets crushed after 240 thousand lies.

If Pinocchio was standing up with his nose level to the ground, either of the following may happen

About 57 million lies are required to get to low earth orbit. This assumes the nose is massless.
If the mass of the nose changes with length, he falls over somewhere around 85 lies. As he continues to lie, his nose will keep growing until it surrounds the earth, hugging the ground.

These are all the simple answers possible, and we’ll look at why on the next section.

The Long Answer: Part 1

First, we need to assume the diameter of Pinocchio’s nose, otherwise the calculations can have many different answers. So to make things easier, we will set the diameter to 0.50 inches (1.27 cm). Another assumption we will make is that Earth’s gravitational field strength is uniform for the distances covered. A short calculation will reveal that this assumption is a very good approximation, since the value of g at 160 km (low earth orbit) above the earth is 95% of that on the surface.

If Pinocchio lies down, his nose only needs to travel 160 km to get to low earth orbit.

Since an inch is defined to be 2.54 cm in length, we convert it and use the value to find the number of lies needed.

$\text{Nose length increment per lie} = 2.54 \text{ cm} = 2.54 \times {10^{-5}} \text{ km}$

$\text {Number of lies} = 160 \text{ km} \times \frac{1 \text{ lie}}{2.54 \times {10^{-5}}\text{ km}}$

$\text {Number of lies} = 6.53 \times 10^{6} \text{ lies} \approx 6,530,000 \text{ lies}$

For the other questions, let’s assume that Pinocchio is still a marionette made of wood. Most types of wood have a density of 700 kilograms per cubic metre, which is about half that of flesh. We can assume that Pinocchio is half the size of an 8-year-old boy. Since an average 8-year-old would have a mass of 30 kg, Pinocchio should have a mass of 8 kg, rounded off. Furthermore, Pinocchio’s nose 3.0 cm in length and 0.5 cm in diameter.

All solid materials, wood included, has a quantity called the ultimate compression stress (UCS). This quantity, which depends on the material, is a measurement of how much force a certain material can withstand, while being compressed, without breaking. Most woods have a UCS value of approximately 42 megapascals. As can be seen in the unit, the UCS is dimensionally identical to pressure.

Doing the calculations yields:

$\text{Number of lies} = \frac {\text{UCS}}{2.54 \text{ cm}\times \text{ density of wood} \times g}$

$\text{Number of lies} = \frac {42\times 10^6 \text{ kg m }^{-1} \text{ s}^{-2}}{2.54\times10^{-2}\text{ m}\times 700 \text { kg m}^{-3} \times 10 \text{ m s}^{-2}}$

$\text{Number of lies} \approx 236,221\text{ lies}$

This number is the number of lies needed in order for Pinocchio’s nose to crush his face from the sheer mass of the nose, assuming he is lying on the ground and facing upwards. Note that this is number much smaller than the value for a massless nose. This means in reality, Pinocchio’s face will be destroyed way before his nose reaches low Earth orbit. Calculating the overall length of his nose results to a length of

$\text{Nose Length} = \text {number of lies} \times \frac{2.54 \times 10^{-5}\text { km}}{1 \text { lie}}$

$\text{Nose Length} = 236,221 \text { lies} \times \frac{2.54 \times 10^{-5} \text { km}}{1 \text { lie}}=6 \text{ km}$

which is waaaaay lower than orbit.

The Long Answer: Part 2

The situation is slightly different when Pinocchio’s nose is horizontal to the ground. In this case, his nose has to travel 1,437 km, since it has to go through a thicker later of atmosphere, as shown in the image.

Calculating the number of lies is pretty straightforward though.

$\text {Number of lies} = 1,437 \text{ km} \times \frac{1 \text{ lie}}{2.54 \times 10^{-5} \text { km}} = \approx 56,565,912 \text{ lies}$

Finally, when Pinocchio topples depends on the length of his nose. He will fall over when the moment of the weight of his nose exceeds the moment of his body weight, as measured from his toes. The maths is tedious at this point, but suffice to say that Pinocchio will topple over when his nose reaches a length of 2.14 m. This length will make Pinocchio fall over at just over 84 lies. Hence, he will fall over when he tells his 85th lie.

One thing to note is that wood is a very flexible material. If Pinocchio
keeps lying in order to reach outer space, his nose simply bends on the
surface of the earth until he is stuck to the ground.

To sum up the answers, If Pinocchio tries to reach outer space by telling lies until his nose is long enough,

from a lying position, his face will be crushed by the sheer weight of his nose when he tells his 57 millionth lie.
from a standing or sitting position, he will fall over on the 85th lie, after which falls over and his nose simply bends with earth’s curvature.

In both cases, his nose does not reach Earth orbit.

Application

This is the reverse situation of a modern problem in engineering, the space elevator. The same issues arise between Pinocchio’s nose going to orbit and a structure that can be used to transport materials from the Earth’s surface to orbital positions. Click on the link for more information.

Final Thoughts

Keep the questions coming!