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Great question! You're essentially asking: if all matter in the universe were compressed such that every particle is only a Planck length apart from its neighbors, what would be the size of the resulting sphere?
Let’s break this down and calculate it.
We assume each particle occupies a cube of side ℓₚ.
So, volume per particle = ℓp3≈(1.616×10−35)3=4.22×10−105ℓₚ^3 \approx (1.616 \times 10^{-35})^3 = 4.22 \times 10^{-105} m³.
Total volume:
V=N⋅ℓp3=6×1079⋅4.22×10−105≈2.53×10−25 m3V = N \cdot ℓₚ^3 = 6 \times 10^{79} \cdot 4.22 \times 10^{-105} \approx 2.53 \times 10^{-25} \text{ m}^3
Volume of a sphere is:
V=43πr3⇒r=(3V4π)1/3V = \frac{4}{3} \pi r^3 \Rightarrow r = \left( \frac{3V}{4\pi} \right)^{1/3}
Plug in the volume:
r=(3⋅2.53×10−254π)1/3≈(6.04×10−26)1/3≈3.9×10−9 mr = \left( \frac{3 \cdot 2.53 \times 10^{-25}}{4\pi} \right)^{1/3} \approx \left( 6.04 \times 10^{-26} \right)^{1/3} \approx 3.9 \times 10^{-9} \text{ m}
If all matter in the observable universe were compressed such that particles are one Planck length apart, the resulting sphere would have a radius of about 3.9 nanometers, or a diameter of ~7.8 nanometers.
That’s about the size of a small virus — mind-bogglingly small considering it contains the entire universe's mass!
Great question! You're essentially asking: if all matter in the universe were compressed such that every particle is only a Planck length apart from its neighbors, what would be the size of the resulting sphere?
Let’s break this down and calculate it.
Step 1: Basic Constants
- Planck length (ℓₚ) ≈ 1.616×10−351.616 \times 10^{-35} meters
- Estimated mass of the observable universe ≈ 1×10531 \times 10^{53} kg
- Assume all matter is made of nucleons (protons/neutrons)
- Mass of one nucleon ≈ 1.67×10−271.67 \times 10^{-27} kg
- So, total number of nucleons:
N=1×10531.67×10−27≈6×1079N = \frac{1 \times 10^{53}}{1.67 \times 10^{-27}} \approx 6 \times 10^{79}
Step 2: Volume if each particle is 1 Planck length apart
We assume each particle occupies a cube of side ℓₚ.
So, volume per particle = ℓp3≈(1.616×10−35)3=4.22×10−105ℓₚ^3 \approx (1.616 \times 10^{-35})^3 = 4.22 \times 10^{-105} m³.
Total volume:
V=N⋅ℓp3=6×1079⋅4.22×10−105≈2.53×10−25 m3V = N \cdot ℓₚ^3 = 6 \times 10^{79} \cdot 4.22 \times 10^{-105} \approx 2.53 \times 10^{-25} \text{ m}^3
Step 3: Radius of the equivalent sphere
Volume of a sphere is:
V=43πr3⇒r=(3V4π)1/3V = \frac{4}{3} \pi r^3 \Rightarrow r = \left( \frac{3V}{4\pi} \right)^{1/3}
Plug in the volume:
r=(3⋅2.53×10−254π)1/3≈(6.04×10−26)1/3≈3.9×10−9 mr = \left( \frac{3 \cdot 2.53 \times 10^{-25}}{4\pi} \right)^{1/3} \approx \left( 6.04 \times 10^{-26} \right)^{1/3} \approx 3.9 \times 10^{-9} \text{ m}
Final Answer:
If all matter in the observable universe were compressed such that particles are one Planck length apart, the resulting sphere would have a radius of about 3.9 nanometers, or a diameter of ~7.8 nanometers.
That’s about the size of a small virus — mind-bogglingly small considering it contains the entire universe's mass!
omfg is the sound of that marker on the paper painful to listen to for anyone else?
like nails on a chalk board