A neutron star has a density equal to that of the nuclear matter. Assuming the staar to be spherical, find the radius of a neutron star whose mass is `4.0xx10^30`kg (twice the mass of the sun ).

Explanation

To find the radius of a neutron star with a given mass and density, we can follow these steps:

### Step 1: Identify the given values
- Mass of the neutron star, \( M = 4.0 \times 10^{30} \) kg
- Density of nuclear matter, \( \rho = 2.4 \times 10^{17} \) kg/m³

### Step 2: Use the formula for density
The density \( \rho \) is defined as mass \( M \) divided by volume \( V \):
\[ \rho = \frac{M}{V} \]
From this, we can rearrange the formula to find the volume:
\[ V = \frac{M}{\rho} \]

### Step 3: Calculate the volume of the neutron star
Substituting the values of mass and density into the volume formula:
\[ V = \frac{4.0 \times 10^{30} \text{ kg}}{2.4 \times 10^{17} \text{ kg/m}^3} \]
Calculating this gives:
\[ V = \frac{4.0}{2.4} \times 10^{30 - 17} \text{ m}^3 = \frac{4.0}{2.4} \times 10^{13} \text{ m}^3 \]
\[ V \approx 1.6667 \times 10^{13} \text{ m}^3 \]

### Step 4: Relate volume to the radius of a sphere
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Setting this equal to the volume we calculated:
\[ \frac{4}{3} \pi r^3 = 1.6667 \times 10^{13} \text{ m}^3 \]

### Step 5: Solve for the radius \( r \)
Rearranging the equation to solve for \( r^3 \):
\[ r^3 = \frac{1.6667 \times 10^{13} \text{ m}^3 \cdot 3}{4 \pi} \]
Calculating this:
\[ r^3 = \frac{5.0001 \times 10^{13}}{12.5664} \approx 3.9789 \times 10^{12} \text{ m}^3 \]
Now, taking the cube root to find \( r \):
\[ r \approx (3.9789 \times 10^{12})^{1/3} \approx 15.7 \times 10^{3} \text{ m} \]
\[ r \approx 15.7 \text{ km} \]

### Final Answer
The radius of the neutron star is approximately \( 15.7 \) km.

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