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(II) A certain neutron star has five times the mass of our Sun packed into a sphere about 10 km in radius. Estimate the surface gravity on this monster.

Short Answer

Expert verified
The surface gravity of the neutron star is \( 6.64 \times 10^{12} \, \text{m/s}^2 \).

Step by step solution

01

Understanding the Given Problem

We need to estimate the surface gravity of a neutron star with a mass five times that of the Sun, packed into a sphere with a radius of 10 km. The gravitational force on the surface can be calculated using the formula for gravitational acceleration: \[ g = \frac{G \cdot M}{r^2} \] where \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)), \( M \) is the mass, and \( r \) is the radius.
02

Identify Known Quantities

The mass of our Sun \( M_{\odot} \) is approximately \( 1.989 \times 10^{30} \, \text{kg} \). Therefore, the mass of the neutron star \( M \) is 5 times that of the Sun, which is: \[ M = 5 \times M_{\odot} = 5 \times 1.989 \times 10^{30} \, \text{kg} = 9.945 \times 10^{30} \, \text{kg} \] The radius \( r \) of the neutron star is 10 km, or \( 10,000 \, \text{m} \).
03

Apply the Formula for Surface Gravity

Using the formula \( g = \frac{G \cdot M}{r^2} \), we substitute the known values:\[ g = \frac{6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \times 9.945 \times 10^{30} \, \text{kg}}{(10,000 \, \text{m})^2} \] Simplifying the expression gives:\[ g = \frac{6.674 \times 10^{-11} \times 9.945 \times 10^{30}}{10^{8}} \]\[ g \approx 6.64 \times 10^{12} \, \text{m/s}^2 \] Thus, the surface gravity of the neutron star is approximately \( 6.64 \times 10^{12} \, \text{m/s}^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Surface Gravity
Surface gravity refers to the gravitational acceleration experienced by an object situated on the surface of a celestial body. It is the force that holds everything together on a planet, a star, or in this case, a neutron star. For celestial objects, the surface gravity can vary significantly due to differences in mass and radius. For example, Earth's surface gravity is approximately 9.8 m/s², but that of a neutron star can be billions of times higher due to its extremely compact nature.The formula for calculating surface gravity, represented as \( g \), is:\[ g = \frac{G \cdot M}{r^2} \]- **\( G \)** is the gravitational constant.- **\( M \)** is the mass of the object.- **\( r \)** is the radius from the object’s center to its surface.This formula helps us understand how gravity behaves in different astronomical contexts, showing how even small changes in mass or radius can lead to huge differences in surface gravity.
Gravitational Acceleration
Gravitational acceleration is the rate at which an object accelerates due to the gravitational pull of another object. It is what we experience as weight. When you stand on a scale, it measures the gravitational acceleration pulling you downwards towards Earth. On a neutron star, gravitational acceleration has a much higher value because of its enormous mass and small radius. This means an object would weigh significantly more on a neutron star compared to Earth. The immense gravitational pull is a direct result of the neutron star compressing a mass much greater than the Sun's into a minuscule volume, creating an overwhelming acceleration force at its surface. In step-by-step calculations, gravitational acceleration is determined using the relevant formula, ensuring all values for mass and radius are precisely plugged in to ascertain the enormous gravitational pull experienced at the surface of these dense celestial entities.
Mass of Neutron Star
The mass of a neutron star is a crucial factor in determining its gravitational characteristics. Neutron stars are remnants of supernova explosions and contain mass approximately 1.4 to 3 times that of the Sun, condensed into a radius of just 10 to 20 kilometers. This massive compression results in extraordinarily high densities.In the exercise, the given neutron star has a mass five times that of the Sun. Since the Sun's mass is about \( 1.989 \times 10^{30} \text{ kg} \), the neutron star's mass would be functioned as:\[ M = 5 \times 1.989 \times 10^{30} = 9.945 \times 10^{30} \text{ kg} \] This immense mass contributes to the impossibly strong gravitational forces typical of neutron stars. Their mass magnitude plays a vital role in the equation determining surface gravity and impacts other physical phenomena associated with these dense stellar objects.
Gravitational Constant
The gravitational constant, denoted as \( G \), is a fundamental constant in physics that appears in Newton’s law of universal gravitation. It quantifies the strength of gravity in natural units and is a key factor in calculating gravitational forces between two masses.The value of \( G \) is approximately \( 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \). Despite being tiny, this constant is crucial in describing how masses interact within the universe.In any gravitational calculation, including that for a neutron star, \( G \) helps determine how strongly two bodies attract each other:\[ F = \frac{G \cdot M_1 \cdot M_2}{r^2} \] Here, it is pivotal in witnessing the gravitational acceleration at the neutron star's surface by multiplying it with the neutron star's mass and adequately dividing by the distance squared from the center to the surface. The gravitational constant thus stitches together all gravitational equations, ensuring that they accurately depict reality in astronomical contexts.

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Most popular questions from this chapter

(I) A jet plane traveling 1890 km/h (525 m/s)pulls out of a dive by moving in an arc of radius 5.20 km. What is the plane's acceleration in \(g\)'s?

(III) (\(a\)) Show that if a satellite orbits very near the surface of a planet with period \(T\), the density ( \(=\) mass per unit volume) of the planet is \(\rho = m/V = 3\pi /GT^2\). (\(b\)) Estimate the density of the Earth, given that a satellite near the surface orbits with a period of 85 min. Approximate the Earth as a uniform sphere.

A science-fiction tale describes an artificial "planet" in the form of a band completely encircling a sun (Fig. 5-50). The inhabitants live on the inside surface (where it is always noon). Imagine that this sun is exactly like our own, that the distance to the band is the same as the Earth-Sun distance (to make the climate livable), and that the ring rotates quickly enough to produce an apparent gravity of \(g\) as on Earth. What will be the period of revolution, this planet's year, in Earth days?

The \(rings\) \(of\) \(Saturn\) are composed of chunks of ice that orbit the planet. The inner radius of the rings is 73,000 km, and the outer radius is 170,000 km. Find the period of an orbiting chunk of ice at the inner radius and the period of a chunk at the outer radius. Compare your numbers with Saturn's own rotation period of 10 hours and 39 minutes. The mass of Saturn is 5.7 \(\times\) 10\(^{26}\) kg.

(II) If you doubled the mass and tripled the radius of a planet, by what factor would g at its surface change?

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