 There is relatively little empty space between atoms in solids and liquids, so that the average density of an atom is about the same as matter on a macroscopic scale—approximately 10^3 kg/m3. The nucleus of an atom has a radius about 10^-5 times that of the entire atom, and contains nearly all the mass of the atom.
One possible remnant of a supernova, called a neutron star, can have the density of a nucleus, while being the size of a small city. What would be the radius, in kilometers, of a neutron star with a mass 10 times that of the Sun? The radius of the Sun is 7 × 10^8 m and its mass is 1.99 × 10^30 kg. 


To solve this problem, we need to calculate the radius of a neutron star with a mass 10 times that of the Sun, given that the density of a neutron star is comparable to the density of an atomic nucleus.

### Step 1: Determine the mass and density of the neutron star
- The mass of the Sun, \( M_{\odot} \), is:
  \[
  M_{\odot} = 1.99 \times 10^{30} \, \text{kg}
  \]
- The mass of the neutron star, \( M_{\text{NS}} \), is 10 times the mass of the Sun:
  \[
  M_{\text{NS}} = 10 \times M_{\odot} = 10 \times 1.99 \times 10^{30} \, \text{kg} = 1.99 \times 10^{31} \, \text{kg}
  \]

The density of the neutron star is approximately the same as the density of an atomic nucleus, which is around:
\[
\rho_{\text{NS}} = 10^{17} \, \text{kg/m}^3
\]

### Step 2: Use the formula for the volume of a sphere to find the radius

The volume \( V \) of a sphere is related to its radius \( R \) by the formula:
\[
V = \frac{4}{3} \pi R^3
\]
The mass \( M \) of the neutron star is related to its density \( \rho \) and volume \( V \) by:
\[
M = \rho \times V
\]
Substitute the expression for the volume of a sphere:
\[
M = \rho \times \frac{4}{3} \pi R^3
\]
Solve for \( R \):
\[
R^3 = \frac{3M}{4\pi\rho}
\]
\[
R = \left( \frac{3M}{4\pi\rho} \right)^{1/3}
\]

### Step 3: Plug in the values for mass and density

Substitute \( M = 1.99 \times 10^{31} \, \text{kg} \) and \( \rho = 10^{17} \, \text{kg/m}^3 \):
\[
R = \left( \frac{3 \times 1.99 \times 10^{31}}{4 \pi \times 10^{17}} \right)^{1/3}
\]

Let’s calculate this next.

The radius of the neutron star would be approximately \( 36.2 \, \text{km} \).

Would you like any additional details or further explanations on this?

Here are five related questions you might find interesting:
1. How does the density of a neutron star compare to that of other stellar remnants like white dwarfs?
2. What physical processes occur during the formation of a neutron star?
3. What role do neutron stars play in the study of gravitational waves?
4. How do scientists measure the mass and radius of neutron stars?
5. Can a neutron star form from a supernova in any type of star?

**Tip**: The immense density of neutron stars is such that a single cubic centimeter of neutron-star material would weigh about the same as a large mountain on Earth!

There is relatively little empty space between atoms in solids and liquids, so that the average density of an atom is about the same as matter on a macroscopic scale—approximately 10^3 kg/m3. The nucleus of an atom has a radius about 10^-5 times that of the entire atom, and contains nearly all the mass of the atom. One possible remnant of a supernova, called a neutron star, can have the density of a nucleus, while being the size of a small city. What would be the radius, in kilometers, of a neutron star with a mass 10 times that of the Sun? The radius of the Sun is 7 × 10^8 m and its mass is 1.99 × 10^30 kg.

Learn how to calculate the radius of a neutron star with a mass 10 times that of the Sun. Using concepts from astrophysics, including the volume of a sphere and density, this problem is ideal for high school students in Grades 11-12. Key formulas include mass and volume relations and the radius calculation formula. The estimated radius is approximately 36.2 kilometers.

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