<p>Volume is mass divided by density.</p>

<p>Earth's mass = 5.972x10²⁴ kg.</p>

<p>Neutron star's density = 10¹⁷ kg/m³.</p>

<p>The sought volume of a neutron star's piece with Earth's mass is then Earth's mass divided by neutron star's density, which makes 59720000 m³. This is a cube with the side length of 390.87 m or a sphere with radius of 242.48 m.</p>

CC BY-SA 4.0
2018-08-20T11:58:04.043Z
The densities I'm finding are more like 3 x 10^17 kg/m^3.
Mike G
CC BY-SA 4.0
2018-08-20T12:04:50.363Z
With that density of a neutron star we'd get a cube with the side length of roughly 271 m or a sphere with radius of 168 m.
SergiusPro
CC BY-SA 4.0
2018-08-20T15:05:49.660Z
Understood, but if the density isn't known with 1 significant digit, then the volume isn't known with 4 significant digits.
Mike G
CC BY-SA 4.0
2018-08-20T15:56:07.800Z
This assumes that the density of the piece would remain constant, which I doubt. In general, more massive neutron stars are smaller, and less massive ones are larger. Of course, that doesn't tell us what would happen to relatively small chunks of a neutron star, but it does tell us that we shouldn't assume uniform density.
HDE 226868
CC BY-SA 4.0
2018-08-20T19:10:38.247Z
Earth-mass chunks of neutron star material don't exist and cannot exist.
ProfRob
CC BY-SA 4.0
2018-08-20T11:34:18.590Z
2018-08-20T11:45:34.367Z
2018-08-20T11:45:34.367Z
SergiusPro
<p>The question has no answer. The <strong>minimum</strong> mass for stable neutron matter is about 0.1 solar masses (<a href="https://physics.stackexchange.com/questions/143166/what-is-the-theoretical-lower-mass-limit-for-a-gravitationally-stable-neutron-st/143174#143174">see my answer here</a>) and so is much more massive than the Earth.</p>

<p>As neutron stars become less massive they also become less dense, so your question is unanswerable at that level too, since we cannot put a figure on the density of something that cannot even exist in principle.</p>

<p>If your question is merely what radius is a spherical object that has a similar  density to a typical neutron star, can you not do
$$\frac{4\pi r^3}{3} \rho = M_{\rm Earth}$$
to find $r$?</p>

CC BY-SA 4.0
2018-08-20T19:06:10.900Z
2018-08-20T19:06:10.900Z
ProfRob
<p>Imagine a neutron star could be broke into pieces.</p>

<p>How big would be the piece matching Earth's mass?</p>

CC BY-SA 4.0
2018-08-21T17:00:16.010Z
Why are people answering this question and not upvoting it? If a question is sufficiently good enough to warrant an answer, then it certainly warrants an upvote.
zephyr
<earth><neutron-star>
How big should be a piece of neutron star to reach Earth's mass?
J. Doe
# How big should be a piece of neutron star to reach Earth's mass?

(Asked by: username_0 on 2018-08-20)

Imagine a neutron star could be broke into pieces.

How big would be the piece matching Earth's mass?

**Tags:** earth, neutron-star

> **Comments:**
> **username_1**: Why are people answering this question and not upvoting it? If a question is sufficiently good enough to warrant an answer, then it certainly warrants an upvote.  

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## 2 answers

### Answer by username_5 on 2018-08-20. Score: 6 | Accepted answer

The question has no answer. The **minimum** mass for stable neutron matter is about 0.1 solar masses ([see my answer here](https://physics.stackexchange.com/questions/143166/what-is-the-theoretical-lower-mass-limit-for-a-gravitationally-stable-neutron-st/143174#143174)) and so is much more massive than the Earth.

As neutron stars become less massive they also become less dense, so your question is unanswerable at that level too, since we cannot put a figure on the density of something that cannot even exist in principle.

If your question is merely what radius is a spherical object that has a similar density to a typical neutron star, can you not do
$$\frac{4\pi r^3}{3} \rho = M\_{\rm Earth}$$
to find $r$?


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### Answer by username_2 on 2018-08-20. Score: 3

Volume is mass divided by density.

Earth's mass = 5.972x10²⁴ kg.

Neutron star's density = 10¹⁷ kg/m³.

The sought volume of a neutron star's piece with Earth's mass is then Earth's mass divided by neutron star's density, which makes 59720000 m³. This is a cube with the side length of 390.87 m or a sphere with radius of 242.48 m.

> **Comments:**
> **username_3**: The densities I'm finding are more like 3 x 10^17 kg/m^3.  
> **username_2**: With that density of a neutron star we'd get a cube with the side length of roughly 271 m or a sphere with radius of 168 m.  
> **username_3**: Understood, but if the density isn't known with 1 significant digit, then the volume isn't known with 4 significant digits.  
> **username_4**: This assumes that the density of the piece would remain constant, which I doubt. In general, more massive neutron stars are smaller, and less massive ones are larger. Of course, that doesn't tell us what would happen to relatively small chunks of a neutron star, but it does tell us that we shouldn't assume uniform density.  
> **username_5**: Earth-mass chunks of neutron star material don't exist and cannot exist.  

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