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3 years ago

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Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S

uppose 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.)

1 answer:

Vlada [557]3 years ago

5 0

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 $_S$ = 1.99 × 10³⁰ kg

Mass of the neutron star

M $_N$ = 2( M $_S$ )

M $_N$ = 2( 1.99 × 10³⁰ kg )

M $_N$ = ( 3.98 × 10³⁰ kg )

Radius of neutron star R $_N$ = 13.0 km = 13 × 10³ m

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

angular speed be ω $_N$ .

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

GmM $_N$ = / R $_N$ ² = mR $_N$ ω $_N$ ²

ω $_N$ ² = GM $_N$ = / R $_N$ ³

ω $_N$ = √(GM $_N$ = / R $_N$ ³)

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

we substitute

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

ω $_N$ = √( 2.65466 × 10²⁰ / 2.197 × 10¹²

ω $_N$ = √ 120831133.3636777

ω $_N$ = 10992.32 rad/s

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

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Answer:

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Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

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