Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. Suppose the mass of a certain spherical neutron star is twice the mass of the Sun and its radius is 13.0 km. Determine the greatest possible angular speed the neutron star can have so that the matter at its surface on the equator is just held in orbit by the gravitational force. (The mass of the Sun is 1.99 1030 kg.)

Sagot :

Answer:

The required angular speed the neutron star is 10992.32 rad/s

Explanation:

Given the data in the question;

mass of the sun M[tex]_S[/tex] = 1.99 × 10³⁰ kg

Mass of the neutron star

M[tex]_N[/tex] = 2( M[tex]_S[/tex] )

M[tex]_N[/tex] = 2( 1.99 × 10³⁰ kg )

M[tex]_N[/tex] = ( 3.98 × 10³⁰ kg )

Radius of neutron star R[tex]_N[/tex] = 13.0 km = 13 × 10³ m

Now, let mass of a small object on the neutron star be m

angular speed be ω[tex]_N[/tex].

During rotational motion, the gravitational force on the object supplies the necessary centripetal force.

GmM[tex]_N[/tex] = / R[tex]_N[/tex]² = mR[tex]_N[/tex]ω[tex]_N[/tex]²

ω[tex]_N[/tex]² = GM[tex]_N[/tex] = / R[tex]_N[/tex]³

ω[tex]_N[/tex] = √(GM[tex]_N[/tex] = / R[tex]_N[/tex]³)

we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²

we substitute

ω[tex]_N[/tex] = √( (  6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)

ω[tex]_N[/tex] = √( 2.65466 × 10²⁰ / 2.197 × 10¹²

ω[tex]_N[/tex] = √ 120831133.3636777

ω[tex]_N[/tex] = 10992.32 rad/s

Therefore, The required angular speed the neutron star is 10992.32 rad/s