he weight of a star is usually balanced by two forces: the gravitational force, acting inward, and the force created by nuclear reaction, acting outward. Over a long period of time, the force due to nuclear reactions gets weaker, causing the gravitational collapse of the star and crushing atoms out of existence. Under such extreme conditions, protons and electrons are squeezed to form neutrons, giving birth to a neutron star. Neutron stars are massively heavy-a teaspoon of the substance of a neutron star would weigh 100 million metric tons on the Earth. Consider a neutron star whose mass is twice the mass of the Sun and whose radius is 12.1 km. (The mass of the Sun is 1.99 ⋅ 1030 kg1.99 · 1030 kg ) If it rotates with a period of 1.73 s, what is the speed of a point on the Equator of this star? (You may enter your calculation using scientific notation.) 4.395E4 Numeric ResponseEdit Unavailable. 4.395E4 correct.m/s
Question
he weight of a star is usually balanced by two forces: the gravitational force, acting inward, and the force created by nuclear reaction, acting outward. Over a long period of time, the force due to nuclear reactions gets weaker, causing the gravitational collapse of the star and crushing atoms out of existence. Under such extreme conditions, protons and electrons are squeezed to form neutrons, giving birth to a neutron star. Neutron stars are massively heavy-a teaspoon of the substance of a neutron star would weigh 100 million metric tons on the Earth. Consider a neutron star whose mass is twice the mass of the Sun and whose radius is 12.1 km. (The mass of the Sun is 1.99 ⋅ 1030 kg1.99 · 1030 kg ) If it rotates with a period of 1.73 s, what is the speed of a point on the Equator of this star? (You may enter your calculation using scientific notation.) 4.395E4 Numeric ResponseEdit Unavailable. 4.395E4 correct.m/s
Solution
To calculate the speed of a point on the Equator of the neutron star, we need to use the formula for the speed of an object moving in a circular path, which is v = 2πr/T, where r is the radius of the circle and T is the period of rotation.
Given:
- Radius (r) = 12.1 km = 12.1 x 10^3 m (since 1 km = 10^3 m)
- Period (T) = 1.73 s
Substituting these values into the formula, we get:
v = 2π(12.1 x 10^3 m) / 1.73 s
Solving this equation will give us the speed of a point on the Equator of the neutron star.
Similar Questions
he weight of a star is usually balanced by two forces: the gravitational force, acting inward, and the force created by nuclear reaction, acting outward. Over a long period of time, the force due to nuclear reactions gets weaker, causing the gravitational collapse of the star and crushing atoms out of existence. Under such extreme conditions, protons and electrons are squeezed to form neutrons, giving birth to a neutron star. Neutron stars are massively heavy-a teaspoon of the substance of a neutron star would weigh 100 million metric tons on the Earth. What is the value of g at the surface of this star? (You may enter your calculation using scientific notation.)
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