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 lay

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

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Nguyệt Ánh 4 years 2021-07-21T08:01:16+00:00 1 Answers 102 views 0

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  1. Farah
    1
    2021-07-21T08:02:56+00:00
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    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

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