Q. 83

Expert-verifiedFound in: Page 334

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

During most of its lifetime, a star maintains an equilibrium size in which the inward force of gravity on each atom is balanced by an outward pressure force due to the heat of the nuclear reactions in the core. But after all the hydrogen “fuel” is consumed by nuclear fusion, the pressure force drops and the star undergoes a gravitational collapse

until it becomes a neutron star. In a neutron star, the electrons and protons of the atoms are squeezed together by gravity until they fuse into neutrons. Neutron stars spin very rapidly and emit intense pulses of radio and light waves, one pulse per rotation. These “pulsing stars” were discovered in the 1960s and are called pulsars.a. A star with the mass M = 2.0 x 10^{30} kg and size R =7.0 x 10^{8} m of our sun rotates once every 30 days. After undergoing gravitational collapse, the star forms a pulsar that is observed by astronomers to emit radio pulses every 0.10 s. By treating the neutron star as a solid sphere, deduce its radius.b. What is the speed of a point on the equator of the neutron star? Your answers will be somewhat too large because a star cannot be accurately modeled as a solid sphere. Even so, you will be able to show that a star, whose mass is 10^{6 }larger than the earth’s, can be compressed by gravitational forces to a size smaller than a typical state in the United States!

a) Radius of the neutron star is r= 137.0 km

b) Speed of a point on the equator of the neutron star is v=8.6 x 10^{6} m/sec

M = 2 x 10^{30} kg

R =7.0 x 10^{8} m

One rotation is in 30 days

First find the moment of inertia. Consider this sphere so

Substitute the values we get

Find angular velocity

Angular momentum is constant form the momentum conservation law

We can get final angular velocity

Now find momentum

From law of momentum conservation

Substitute values

Moment of inertia is given by

Substitute the value and solve for r

M = 2 x 10^{30} kg

R =7.0 x 10^{8} m

One rotation is in 30 days

linear velocity of a point on the equator of the neutron star can be calculated as

94% of StudySmarter users get better grades.

Sign up for free