Often in discussions about blackholes, the question arises of observer experience when falling into a blackhole. Due to the extreme tidal forces experienced along such a path, the observer would be stretched out - a process colloquially referred to as spaghettification. This week I would like to look at a happier use for spaghetti near blackholes. If an observer in a stable circular orbit around the supermassive blackhole at the center of our galaxy were to use spaghetti as a tether, how much spaghetti would be required? Restated, the question of interest this week is:

How much spaghetti would be required to form the smallest stable circle around the supermassive blackhole at the center of the Milky Way?

The term for such an orbit (of a massive test particle) is commonly called the innermost stable circular orbit, or ISCO for short. For today’s post, we’ll pretend the “S” stands for spaghetti.

ISCO Radius

To keep things simple, we take the case of a non-spinning blackhole, which is described by the Schwarzschild solution to the Einstein equations. The radius of the spaghetti orbit in this case would be $3$ times the blackhole radius $R_S$, which is also called the Schwarzschild radius. ¹

$$R_{\mathrm{ISCO}} = 6 \frac{G M}{c^2} = 3 R_{S}$$

In the above, $G$ represents Newton’s gravitational constant, $c$ represents the speed of light, and $M$ represents the mass of the blackhole around which our spaghetti tether has been formed.

Spaghetti Distance

Next, we’ll need a relationship between the length of spaghetti $\ell$ and the mass of that spaghetti $m$. If we assume a fixed cross-sectional area $\sigma$ then we can use the familiar relationship between the density $\rho$ and volume $V$ of matter:

$$m = \rho V = \rho \sigma \ell \implies \ell = \frac{m}{\rho \sigma}$$

We can estimate the cross-sectional area of the spaghetti strands $\sigma$ by using the radius of a spaghetti stick $r_s$ as $\sigma = \pi r_s^2$. Similarly, we can estimate the density of spaghetti in terms of a familiar spaghetti box!

$$\rho = \frac{m_{\mathrm{box}}}{V_{\mathrm{box}}} = \frac{m_{\mathrm{box}}}{\ell_{\mathrm{box}} h_{\mathrm{box}} w_{\mathrm{box}}}$$

Where $\ell_{\mathrm{box}}, h_{\mathrm{box}}, w_{\mathrm{box}}$ are the length, height, and width of a spaghetti box, respectively.

Solution

Putting the above expressions together, we arrive at the innermost spaghetti circular orbit mass:

$$\begin{aligned}m &= \ell \rho \sigma = \left(2 \pi \frac{6 GM}{c^2}\right) \left( \frac{m_{\mathrm{box}}}{\ell_{\mathrm{box}} h_{\mathrm{box}} w_{\mathrm{box}}} \right) \left(\pi r_s^2\right)\\\\ &= \frac{12 \pi^2 r_s^2 G M m_{\mathrm{box}}}{c^2 \ell_{\mathrm{box}} h_{\mathrm{box}} w_{\mathrm{box}}}\\ \end{aligned}$$

Plugging in some numbers:

• Mass of a (galactic center) blackhole: $M \approx 4e6 M_{\odot} = 8 \times 10^{36}$ kg
• Radius of spaghetti strand: $r_s \approx 2 \times 10^{-3}$ m
• Newton’s Gravitational constant $G \approx 7 \times 10^{-11} \mathrm{\ m^{3} kg^{-1} s^{-2}}$
• Mass of spaghetti box: $m_{\mathrm{box}} \approx 5 \times 10^{-1}$ kg
• Dimensions of spaghetti box: $(\ell_{\mathrm{box}}, h_{\mathrm{box}}, w_{\mathrm{box}}) \approx (25, 3, 7) \times 10^{-2}$ m
• Speed of light: $c \approx 3 \times 10^8\ \mathrm{ms^{-1}}$

The estimate is hidden below. Press the button below to see solution!

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What a large amount of pasta! To put this number in context, This amount of spaghetti could feed every human on Earth approximately $3$ times.