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Problem 5 Easy Difficulty

Using $2.3 \times 10^{17} \mathrm{kg} / \mathrm{m}^{3}$ as the density of nuclear matter, find
the radius of a sphere of such matter that would have a mass equal to that of Earth. Earth has a mass equal to $5.98 \times 10^{24} \mathrm{kg}$ and average radius of $6.37 \times 10^{6} \mathrm{m}$

Answer

$r=182m$

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Video Transcript

in number five were asked to find the radius of a sphere of nuclear matter would have the same mass as the mass of the earth. So this is mass of the earth, and this is the density of the nuclear matter. Um, and that density is mass per volume. No So vented. The density of the nuclear matter covet density and to be the same as the mass of the earth. So this would be a massive birth. And this is the volume of that nuclear matter. And it's a sphere. So I know the volume of a sphere is 4/3 pi r cubed. So I'm just gonna rearrange my equation here. Well, let me substitute that in. I'm gonna get your cute boy. It's so I'll divide both sides by the density. So now I have r Cubed will be the mass of the earth divide by the density of the nuclear matter. Times 4/3 pi. That's it. I'm ready for my values. Mass of the earth. I was given 5.98 times, 10 to the 24th the density of given 2.3 times 10 to 17 times 4/3 times by and then I will just need to take the cube root of this. So que beautiful this and I get my radius is 100 and 82 meters, so that is very, very dense.

University of Virginia
Top Physics 103 Educators
Elyse G.

Cornell University

Andy C.

University of Michigan - Ann Arbor

Marshall S.

University of Washington

Farnaz M.

Other Schools