Question

The weight of a star is usually balanced by two forces: the gravitational force, acting inward, and the force created by nuclear reactions, 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. a) Consider a neutron star whose mass is twice the mass of the Sun and whose radius is \(10.0 \mathrm{~km} .\) If it rotates with a period of \(1.00 \mathrm{~s}\), what is the speed of a point on the Equator of this star? Compare this speed with the speed of a point on the Earth's Equator. b) What is the value of \(g\) at the surface of this star? c) Compare the weight of a 1.00 -kg mass on the Earth with its weight on the neutron star. d) If a satellite is to circle \(10.0 \mathrm{~km}\) above the surface of such a neutron star, how many revolutions per minute will it make? e) What is the radius of the geostationary orbit for this neutron star?

Short answer

a) To find the rotational speed at the neutron star's equator, calculate the following: \(v = 2\pi * 10,000 \: \mathrm{m/s}\) For the Earth's equator, calculate: \(v_\mathrm{earth} = \omega_\mathrm{earth} * 6,378,000 \: \mathrm{m/s}\) Compare the two rotational speeds. b) To find the value of g at the surface of the neutron star, calculate: \(g = \frac{G(2M_{\odot})}{(10,000)^2} \: \mathrm{m/s^2}\) c) Calculate the weight of a 1.00 kg mass on the neutron star and compare it to the weight of the same mass on Earth (g on Earth = \(9.81 \: \mathrm{m/s^2}\)). d) To find the satellite's revolutions per minute: 1. Calculate the orbital period (\(T_\mathrm{orbit}\)) using the formula: \(T_\mathrm{orbit} = 2\pi \sqrt{\frac{r_\mathrm{orbit}^3}{GM}}\) 2. Convert the orbital period to revolutions per minute. e) To find the radius of the geostationary orbit: 1. Use the formula for the orbital period and set \(T_\mathrm{orbit} = T\). 2. Solve for \(r_\mathrm{orbit}\) to find the geostationary orbit radius.

Step-by-step solution

Identify the rotation period and the radius of the neutron star.

The given rotation period is 1.00 s and the radius is 10.0 km.

Calculate the angular velocity of the neutron star.

Angular velocity (\(\omega\)) can be calculated by: \(\omega = \frac{2\pi}{T}\) Where T is the period (1.00 s). Plugging in the values, we get: \(\omega = \frac{2\pi}{1.00} \: \mathrm{rad/s}\)

Calculate the rotational speed at the equator.

Rotational velocity (v) can be calculated by: \(v = \omega r\) Where r is the radius of the neutron star (10.0 km). Converting the radius to meters: 10.0 km = \(10,000 \: \mathrm{m}\) Plugging in the values, we get: \(v = 2\pi * 10,000 \: \mathrm{m/s}\) Now, we'll find the speed of a point on the Earth's Equator.

Earth's radius and rotation period.

The Earth's radius is approximately \(6,378 \: \mathrm{km}\) and the rotation period is roughly 24 hours or 86400 s.

Calculate the angular velocity of the Earth.

Using the formula for angular velocity and the Earth's rotation period: \(\omega_\mathrm{earth} = \frac{2\pi}{86400} \: \mathrm{rad/s}\)

Calculate the rotational speed at the Earth's equator.

Using the formula for rotational velocity and the Earth's radius (converted to meters): \(v_\mathrm{earth} = \omega_\mathrm{earth} * 6,378,000 \: \mathrm{m/s}\) We can now compare the rotational speed of a point on the neutron star's equator with the speed of a point on the Earth's equator. b) Finding value of g at the surface

Newton's law of universal gravitation.

We can find the gravitational acceleration (g) using the formula: \(g = \frac{GM}{r^2}\) Where G is the gravitational constant, M is the mass of the neutron star, and r is its radius. The mass of the neutron star is given to be twice the mass of the Sun. Therefore, M = \(2M_{\odot}\), where \(M_{\odot}\) represents the mass of the Sun.

Calculate the value of g.

Plugging in the values into the formula, we get: \(g = \frac{G(2M_{\odot})}{(10,000)^2} \: \mathrm{m/s^2}\) c) Comparing the weight of a 1.00 kg mass on Earth and the neutron star Note that weight is related to gravitational force, which can be expressed as \(W = mg\). As we already found the value of g for the neutron star, we can calculate the weight of a 1.00 kg mass on the neutron star and compare it to the weight of the same mass on Earth. d) Orbiting satellite's revolutions per minute.

Use centripetal force formula.

For a satellite in circular orbit: \(F_\mathrm{centripetal} = F_\mathrm{gravitational}\) \(\frac{mv^2}{r_\mathrm{orbit}} = \frac{GM_\mathrm{satellite} m}{r_\mathrm{orbit}^2}\) m cancels out, and we can solve for \(v^2\): \(v^2 = \frac{GM}{r_\mathrm{orbit}}\)

Relate the orbital speed to the rotation period.

Use the formula: \(v = \frac{2\pi r_\mathrm{orbit}}{T_\mathrm{orbit}}\)

Solve for T (orbit).

Square the orbital speed formula and substitute for \(v^2\): \(\frac{4\pi^2 r_\mathrm{orbit}^2}{T_\mathrm{orbit}^2} = \frac{GM}{r_\mathrm{orbit}}\) Solve for \(T_\mathrm{orbit}\): \(T_\mathrm{orbit} = 2\pi \sqrt{\frac{r_\mathrm{orbit}^3}{GM}}\)

Find the revolutions per minute.

The orbit is given to be 10.0 km above the neutron star's surface. Thus, \(r_\mathrm{orbit} = 10,000 + 10,000 = 20,000 \: \mathrm{m}\). Plugging in the values for G, M, and \(r_\mathrm{orbit}\), calculate the period and convert it to revolutions per minute. e) Radius of the geostationary orbit.

Define the geostationary condition.

In a geostationary orbit, the rotational period of the satellite matches the rotational period of the neutron star.

Calculate the geostationary orbit radius.

Using the formula for the orbital period (from part d), set \(T_\mathrm{orbit} = T\) (period of neutron star) and solve for \(r_\mathrm{orbit}\). Once all the calculations have been made, we will get the results for each part of the exercise.

Key concepts

Gravitational Collapse

Gravitational collapse is a phenomenon that occurs when a star's internal pressure is no longer able to resist its own gravity, leading to a contraction that can result in the formation of compact objects like white dwarfs, neutron stars, or black holes, depending on the mass of the progenitor star.

For a star like the Sun, nuclear fusion at its core produces an outward pressure that balances the force of gravity pulling everything inward. Over time, as the nuclear fuel is exhausted, this pressure diminishes. If the star is massive enough, it can't maintain the balance and gravity wins, compressing the core dramatically. In the process, electrons and protons may be squeezed together to form neutrons, resulting in a neutron star.

A teaspoon of neutron star material would weigh about 100 million metric tons here on Earth due to its incredibly dense and compact nature. The gravitational collapse is therefore fundamental not only to the life cycle of stars but also to the creation of some of the most extreme states of matter in the universe.

Nuclear Reactions in Stars

Stars generate energy and counteract gravitational collapse through nuclear reactions at their cores, primarily through the fusion of hydrogen into helium. These nuclear reactions release incredible amounts of energy, which creates the pressure needed to support a star against its own gravity.

Fusion processes in stars are a delicate balance between energy production and energy requirements for maintaining structural integrity against gravitational pull. As stars consume their nuclear fuel, the fusion reactions eventually slow down and, for massive stars, cease to be effective at countering gravitational forces, leading to a gravitational collapse. The balance between gravity and nuclear reactions determines a star's life span and ultimate fate, influencing the types of remnants it leaves behind, such as neutron stars.

Angular Velocity

Angular velocity is a measure of how fast an object rotates or spins around an axis. In the context of a neutron star, angular velocity is crucial because these stars can spin extremely rapidly as a result of their compact size and conservation of angular momentum.

When a gigantic star collapses into a neutron star, its core's spin rate increases tremendously, akin to how a figure skater's spin rate increases as they pull their arms in. This is due to the conservation of angular momentum, expressed mathematically as \( L = I\omega \), where \( L \) is the angular momentum, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity. Since the moment of inertia decreases during collapse, the angular velocity increases, resulting in neutron stars that can have spin periods of just milliseconds.

Gravitational Acceleration

Gravitational acceleration, often denoted as \( g \), is the acceleration due to gravity experienced by objects in the gravitational field of a massive body like a star or planet. It's given by the formula \( g = \frac{GM}{r^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the object creating the gravitational field, and \( r \) is the distance from the center of the object to the point where \( g \) is being measured.

In the case of a neutron star, \( g \) is extremely high due to the star's mass being comparable to the Sun's packed into a sphere with a radius of only about 10 kilometers. This high gravitational acceleration has profound effects on the star's surroundings and anything that comes close to it, such as intense tidal forces or rapid orbital motions for satellites.