Math Problem Statement

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.

Solution

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, MM_{\odot}, is: M=1.99×1030kgM_{\odot} = 1.99 \times 10^{30} \, \text{kg}
  • The mass of the neutron star, MNSM_{\text{NS}}, is 10 times the mass of the Sun: MNS=10×M=10×1.99×1030kg=1.99×1031kgM_{\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: ρNS=1017kg/m3\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 VV of a sphere is related to its radius RR by the formula: V=43πR3V = \frac{4}{3} \pi R^3 The mass MM of the neutron star is related to its density ρ\rho and volume VV by: M=ρ×VM = \rho \times V Substitute the expression for the volume of a sphere: M=ρ×43πR3M = \rho \times \frac{4}{3} \pi R^3 Solve for RR: R3=3M4πρR^3 = \frac{3M}{4\pi\rho} R=(3M4πρ)1/3R = \left( \frac{3M}{4\pi\rho} \right)^{1/3}

Step 3: Plug in the values for mass and density

Substitute M=1.99×1031kgM = 1.99 \times 10^{31} \, \text{kg} and ρ=1017kg/m3\rho = 10^{17} \, \text{kg/m}^3: R=(3×1.99×10314π×1017)1/3R = \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 36.2km36.2 \, \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!

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Math Problem Analysis

Mathematical Concepts

Astrophysics
Volume of a Sphere
Density
Scientific Notation

Formulas

M = ρ × V
V = (4/3) π R^3
R = (3M / 4πρ)^(1/3)

Theorems

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Suitable Grade Level

Grades 11-12