Question

Suppose a neutron star with a mass of about 1.5MSun and a radius of 10 kilometers suddenly appeared in your hometown. Part A How thick a layer would Earth form as it wraps around the neutron star’s surface? Assume that the layer formed by Earth has the same average density as the neutron star. (Hint: Consider the mass of Earth to be distributed in a spherical shell over the surface of the neutron star and then calculate the thickness of such a shell with the same mass as Earth. The volume of a spherical shell is approximately its surface area times its thickness: Vshell=4πr2×thickness. Because the shell will be thin, you can assume that its radius is the radius of the neutron star.) How thick a layer would Earth form as it wraps around the neutron star’s surface? Assume that the layer formed by Earth has the same average density as the neutron star. (Hint: Consider the mass of Earth to be distributed in a spherical shell over the surface of the neutron star and then calculate the thickness of such a shell with the same mass as Earth. The volume of a spherical shell is approximately its surface area times its thickness: . Because the shell will be thin, you can assume that its radius is the radius of the neutron star.) ≈6.4×103km ≈35cm ≈7mm ≈10km

Answer 1

Answer:

e = 6.67 10⁻³  m

Explanation:

For this exercise we use the definition density

      ρ = m / V

where ρ tell us to use the density of the neutron star, m is the mass of the Earth 5.98 10²⁴ km and V is the volume of the spherical layer

calculate the density of the neutron star

     ρ  = M / V

     

the volume of a sphere is

     V = 4/3 π r³

     

The mass of the star e

     M = 1.5 $M_{Sum}$ = 1.5 1,991 10³⁰

     M = 2.99 10³⁰ kg

the density is

     ρ  = 2.99 10³⁰ / [4/3 π (10 10³)³]

     ρ  = 7.13 10 17 kg / m³

we clear the volume of the layer

     V = m / ρ  

     V = 5.98 10²⁴ / 7.13 10¹⁷

      V = 8,387 10⁶ m³

now we can find the thickness of the layer with the formula that they give us

      V = 4π r² e

       e = V / 4π r²

       

calculate

      e = 8,387 10⁶ / [4π (10 10³)²]

     e = 6.67 10⁻³  m