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

Question

14-04-2020
Physics

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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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moya1316 moya1316 · Answer 1 · 2020-04-19T14:12:13+02:00

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 [tex]M_{Sum}[/tex] = 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

tallinn tallinn · Answer 2 · 2022-03-16T11:31:19+01:00

When a layer would Earth form because it wraps round the neutron star surface e is = 6.67 10⁻³ m

Calculate the density of the Neutron star

For this exercise, we use the definition of density

ρ = m / V

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

ρ = M / V

the volume of a sphere is

V = 4/3 π r³

The mass of the star e

M = 1.5 = 1.5 1,991 10³⁰

M = 2.99 10³⁰ kg

Then the density is

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

Now, ρ is = 7.13 10 17 kg / m³

Then we clear the degree of the layer

The formula is V = m / ρ

V = 5.98 10²⁴ / 7.13 10¹⁷

Therefore, V = 8,387 10⁶ m³

Now we will find the thickness of the layer with the assistance of the formula that they furnish us:

then V = 4π r² e

e is = V / 4π r²

Now we calculate

Then e = 8,387 10⁶ / [4π (10 10³)²]

Therefore, e = 6.67 10⁻³ m

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