/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 A typical neutron star may have ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A typical neutron star may have a mass equal to that of the Sun but a radius of only \(10 \mathrm{~km}\). (a) What is the gravitational acceleration at the surface of such a star? (b) How fast would an object be moving if it fell from rest through a distance of \(1.0 \mathrm{~m}\) on such a star? (Assume the star does not rotate.)

Short Answer

Expert verified
Gravitational acceleration: \(1.327 \times 10^{12} \, \text{m/s}^2\). Speed after 1 m fall: \(1.63 \times 10^6 \, \text{m/s}\).

Step by step solution

01

Understand Given Information

We have a neutron star with mass equal to that of the Sun, which is approximately \(1.989 \times 10^{30}\) kg and a radius of \(10^4\) m (since \(10\) km is \(10,000\) meters). We need to find the gravitational acceleration at the surface of the star and the speed of an object falling 1 meter on the star's surface.
02

Calculate Gravitational Acceleration at the Surface

Using the formula for gravitational acceleration \(g = \frac{GM}{R^2}\), where \(G = 6.674 \times 10^{-11}\, \text{m}^3\, \text{kg}^{-1}\, \text{s}^{-2}\), \(M = 1.989 \times 10^{30}\, \text{kg}\), and \(R = 10,000\, \text{m}\), we find \[ g = \frac{6.674 \times 10^{-11} \cdot 1.989 \times 10^{30}}{(10,000)^2} \approx 1.327 \times 10^{12} \, \text{m/s}^2. \]
03

Use Kinematics to Find Final Velocity

When an object falls from rest through a distance of 1 m on the neutron star, use the kinematic equation \(v^2 = u^2 + 2as\), where \(u = 0\, \text{m/s}\), \(a = 1.327 \times 10^{12}\, \text{m/s}^2\), and \(s = 1\, \text{m}\). Calculate \(v\) as follows: \[ v^2 = 0 + 2 \times 1.327 \times 10^{12} \times 1 \] \[ v^2 = 2.654 \times 10^{12} \] \[ v \approx \sqrt{2.654 \times 10^{12}} \approx 1.63 \times 10^6 \, \text{m/s}. \]
04

Compile Answers

The gravitational acceleration at the surface of the neutron star is approximately \(1.327 \times 10^{12}\, \text{m/s}^2\). An object falling 1 meter on this star would reach a speed of approximately \(1.63 \times 10^6\, \text{m/s}.\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Gravitational Acceleration
Gravitational acceleration is the force that pulls objects towards the center of a massive body. It is dictated by the mass of the celestial object and the distance from its center to the point of interest, usually the surface. For any object, the gravitational acceleration \( g \) can be calculated using the formula:
  • \( g = \frac{GM}{R^2} \)
Here, \( G \) is the gravitational constant \( 6.674 \times 10^{-11}\, \text{m}^3\, \text{kg}^{-1}\, \text{s}^{-2} \), \( M \) is the mass of the celestial body, and \( R \) is its radius.
In the case of a neutron star with a mass equal to the Sun's and a radius of only 10 km, gravitational acceleration becomes extremely intense. Because\( R^2 \) appears in the denominator, the relatively small radius of the neutron star leads to a significantly larger gravitational acceleration compared to planets or stars with larger radii.
Kinematic Equations
Kinematic equations describe the motion of objects and are very useful for calculating unknown quantities in mechanics problems. One of these equations is:
  • \( v^2 = u^2 + 2as \)
where:
  • \( v \) is the final velocity,
  • \( u \) is the initial velocity,
  • \( a \) is the acceleration,
  • \( s \) is the displacement.
In our exercise, an object falls from rest (meaning \( u = 0 \)) over a short distance of 1 meter on the neutron star.
Using the kinematic equation, we can discover the final speed \( v \) just before it hits the surface. Given the extreme gravitational acceleration calculated earlier, objects on neutron stars can reach incredibly high speeds over very short distances.
Mass of the Sun
The mass of the Sun is a fundamental constant in astronomy, playing a crucial role in gravitational dynamics within the solar system. It is approximately \( 1.989 \times 10^{30} \) kilograms. This value is important because the gravitational pull exerted by a celestial body scales with its mass.
In calculating forces such as gravitational acceleration, using the Sun's mass provides a sense of the immense gravitational force present.
  • It's used as a baseline to compare other stellar objects, like neutron stars.
  • Helps in understanding the binding energy within the solar system's gravitational field.
For our neutron star, having a mass equivalent to the Sun but compressed into a much smaller radius results in an incredible gravitational force.
Radius of a Neutron Star
The radius of a neutron star is extraordinarily small compared to other stellar objects, typically around 10 km. Despite having a mass comparable to the Sun, a neutron star's volume is minute, leading to some of the densest objects in space.
  • This small radius causes the gravitational acceleration at the surface to be tremendous.
  • Neutron stars are formed from the remnants of supernova explosions and are predominantly composed of tightly packed neutrons.
Through these dense structures, the properties of matter under extreme conditions are often studied.
The compact size of neutron stars also has interesting implications on the motion of objects and light near them, due to the effects of their gravitational fields.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Moon effect. Some people believe that the Moon controls their activities. If the Moon moves from being directly on the opposite side of Earth from you to being directly overhead, by what percent does (a) the Moon's gravitational pull on you increase and (b) your weight (as measured on a scale) decrease? Assume that the Earth-Moon (center-to-center) distance is \(3.82 \times 10^{8} \mathrm{~m}\) and Earth's radius is \(6.37 \times 10^{6} \mathrm{~m}\).

(a) What will an object weigh on the Moon's surface if it weighs \(100 \mathrm{~N}\) on Earth's surface? (b) How many Earth radii must this same object be from the center of Earth if it is to weigh the same as it does on the Moon?

(a) What linear speed must an Earth satellite have to be in a circular orbit at an altitude of \(160 \mathrm{~km}\) above Earth's surface? (b) What is the period of revolution?

The Sun's center is at one focus of Earth's orbit. How far from this focus is the other focus, (a) in meters and (b) in terms of the solar radius, \(6.96 \times 10^{8} \mathrm{~m}\) ? The eccentricity is \(0.0167\), and the semimajor axis is \(1.50 \times 10^{11} \mathrm{~m}\).

Hunting a black hole. Observations of the light from a certain star indicate that it is part of a binary (two-star) system. This visible star has orbital speed \(v=270\) \(\mathrm{km} / \mathrm{s}\), orbital period \(T=1.70\) days, and approximate mass \(m_{1}=6 M_{s}\), where \(M_{s}\) is the Sun's mass, \(1.99 \times\) \(10^{50} \mathrm{~kg}\). Assume that the visible star and its companion star, which is dark and unseen, are both in circular orbits (Fig. \(13-47\) ). What integer multiple of \(M_{s}\) gives the approximate mass \(m_{2}\) of the dark star?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.