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I was looking in my Wolfram ChatGPT book and I happened upon the question, "How many calories in a cubic light year of ice cream?" The software used for the book could not answer so I tried with a newer version and got an answer which makes sense. Then I asked "Would a cubic light year of ice cream collapse into a black hole?" Depending upon when I ask and some random effects I don't know about, it gives different answers, none conclusive.

This is the type of question I always enjoy giving to students to get them to understand using units and algebra to get answers, useful or not. Meanwhile, I'm having trouble coming up with my own answers so I was wondering if I could get some help. Could I get people to look this over and see if I am thinking of this correctly?

First let's come up with an equation for ice cream. The density of ice cream depends upon many factors, mostly the amount of air whipped into it. To make the numbers come out more evenly, I'll give the density of water, 1000 $\frac{kg}{m^3}$. This is a bit more dense than ice cream but it's on the right order of magnitude so it will do. Google informs me that a light year is 9.46e+15 meters so the volume would be:

$V=({9.46(10^{15})})^3$ m^3

or 8.47e+47 m^3 and the mass would be:

$M={8.47(10^{47})}m^3\times{1000} kg/m^3$

M=8.47e+50 kg

The Schwarzschild radius, the radius which a given mass, if compressed below to that dimension will collapse into a black hole, is defined as:

$r_\text{s} = \frac{2 G M}{c^2}$

If we use the mass of ice cream with this formula, we get

$r_\text{s}=\frac{2G{8.47({10}^{50})}}{c^2}$ meters

where

  • c speed of light in vacuum 300e6 m/s
  • G gravitational constant 6.67e-11 Nm^2/kg^2

so:

$r_\text{s}=\frac{2(6.67(10^{-11}))(8.47(10^{50}))}{(300(10^6))^2}$

solving:

$r_\text{s}=1.25({10}^{24})$ meters

Meanwhile, the volume of a sphere is:

$V=\frac{4\pi{R}^3}{3}$

using our volume from before:

${8.47(10^{47})}m^3 = \frac{4\pi{R}^3}{3}$

Solving for R, we get:

$R_\text{ly}=(\frac{3(8.47(10^{47}))}{4\pi})^{\frac{1}{3}}$

$R_\text{ly}=5.87(10^{15})$ meters

If I am interpreting this correctly (and if my arithmetic is correct (two big ifs)), the answer to my question is:

is: $R_\text{ly} < r_{s}$

The two numbers were 5.87e+15 and 1.25e+24 so . . . .

YES, A cubic light year worth of ice cream WOULD collapse down into a black hole.

user1683793
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1 Answers1

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You're obscuring the obvious through a lot of math/calculations.

The mass of a cube is $\rho a^3$ where $a$ is the side length in meters.

The schwarzchild radius of a cube is therefore

$ r_s = \frac{2G\rho a^3}{c^2}$

If you want to set $r_s = a$ you get $ a = \sqrt{\frac{c^2}{2G\rho}}$

So you can easily calculate the side length necessary for any material given its density. The schwarzchild radius will scale with the cube of the side length.

A cubic light year of ice cream (or any material) is really heavy. It will collapse under its own weight into a black hole, just like a dead star would.

Señor O
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