/*! 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 34 A neutron star has a mass of \(2... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 34

A neutron star has a mass of \(2.0 \times 10^{30} \mathrm{kg}\) (about the mass of our sun) and a radius of \(5.0 \times 10^{3} \mathrm{m}\) (about the height of a good-sized mountain). Suppose an object falls from rest near the surface of such a star. How fast would this object be moving after it had fallen a distance of \(0.010 \mathrm{m} ?\) (Assume that the gravitational force is constant over the distance of the fall and that the star is not rotating.)

Short Answer

Expert verified
The object's final speed is approximately \(1.03 \times 10^6\, \mathrm{m/s}\).

Step by step solution

01

Identify the Values

First, we need to identify the given values in the problem. We have:- Mass of the neutron star, \( M = 2.0 \times 10^{30} \, \mathrm{kg} \).- Radius of the neutron star, \( R = 5.0 \times 10^{3} \, \mathrm{m} \).- Distance fallen, \( d = 0.010 \, \mathrm{m} \).We also know the gravitational constant, \( G = 6.674 \times 10^{-11} \, \mathrm{m^3 \, kg^{-1} \, s^{-2}} \).
02

Calculate Gravitational Acceleration

Next, calculate the gravitational acceleration \( g \) on the surface of the neutron star using the formula:\[g = \frac{GM}{R^2}\]Substitute the known values:\[g = \frac{6.674 \times 10^{-11} \, \mathrm{m^3 \, kg^{-1} \, s^{-2}} \times 2.0 \times 10^{30} \, \mathrm{kg}}{(5.0 \times 10^3 \, \mathrm{m})^2}\]After calculating, \( g \approx 5.34 \times 10^{12} \, \mathrm{m/s^2} \).
03

Use Energy Conservation to Find Final Velocity

To find the velocity of the object after falling a distance of \( 0.010 \, \mathrm{m} \), apply energy conservation. The potential energy lost is converted into kinetic energy:\[mgh = \frac{1}{2} mv^2\]Here, \( h = 0.010 \, \mathrm{m} \) and \( g \) is the gravitational acceleration calculated previously. The mass \( m \) of the object cancels out:\[gh = \frac{1}{2} v^2\]Solve for \( v \):\[v = \sqrt{2gh} = \sqrt{2 \times 5.34 \times 10^{12} \, \mathrm{m/s^2} \times 0.010 \, \mathrm{m}}\]Upon calculation, \( v \approx 1.03 \times 10^6 \, \mathrm{m/s} \).

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Key Concepts

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

Gravitational Acceleration
When an object is near a massive body, like a neutron star, it experiences a gravitational pull. This force is quantified as gravitational acceleration. Using Newton's law of universal gravitation, the gravitational acceleration \( g \) at the surface of a neutron star can be calculated with the formula:
\[ g = \frac{GM}{R^2} \]
Where:
  • \( G \) is the gravitational constant, about \( 6.674 \times 10^{-11} \, \mathrm{m^3 \, kg^{-1} \, s^{-2}} \)
  • \( M \) is the mass of the neutron star
  • \( R \) is the radius of the neutron star
By plugging in the given values, you calculate an extremely high gravitational acceleration \( g \approx 5.34 \times 10^{12} \mathrm{m/s^2} \), which reflects the immense gravity near the surface of a neutron star.
Energy Conservation
Energy conservation is a fundamental principle in physics. It states that the total energy in a closed system remains constant over time. In the context of the falling object, the potential energy it initially has is entirely converted into kinetic energy as it falls.
The equation representing this conversion is:
\[ mgh = \frac{1}{2} mv^2 \]
Here:
  • \( mgh \) represents the potential energy
  • \( \frac{1}{2} mv^2 \) is the kinetic energy
Due to the energy conservation principle, the mass \( m \) cancels out, simplifying the calculations for speed, no matter the object's mass.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. As the object falls toward the neutron star, it accelerates, converting potential energy into kinetic energy. The formula for kinetic energy is:
\[ KE = \frac{1}{2} mv^2 \]
Where:
  • \( KE \) is the kinetic energy
  • \( m \) is the mass of the object
  • \( v \) is the velocity of the object
This exercise demonstrates that as the object falls, its velocity increases rapidly, providing kinetic energy that is measurable, even if the distance fallen is small.
Potential Energy
Potential energy, in this scenario, is the energy held by the object because of its position in a gravitational field. It is given by the formula:
\[ PE = mgh \]
Where:
  • \( PE \) is the potential energy
  • \( m \) is the mass of the object
  • \( g \) is the gravitational acceleration
  • \( h \) is the height or distance fallen
In this exercise, as the object falls the short distance of \( 0.010 \, \mathrm{m} \), its potential energy diminishes and becomes kinetic energy, showing the power of gravitational forces in action.

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Most popular questions from this chapter

A student is skateboarding down a ramp that is 6.0 m long and inclined at \(18^{\circ}\) with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is \(2.6 \mathrm{m} / \mathrm{s} .\) Neglect friction and find the speed at the bottom of the ramp.

Scientists are experimenting with a kind of gun that may eventually be used to fi re payloads directly into orbit. In one test, this gun accelerates a 5.0-kg projectile from rest to a speed of \(4.0 \times 10^{3} \mathrm{m} / \mathrm{s} .\) The net force accelerating the projectile is \(4.9 \times 10^{5} \mathrm{N} .\) How much time is required for the projectile to come up to speed?

A girl is sledding down a slope that is inclined at \(30.0^{\circ}\) with respect to the horizontal. The wind is aiding the motion by providing a steady force of \(105 \mathrm{N}\) that is parallel to the motion of the sled. The combined mass of the girl and the sled is \(65.0 \mathrm{kg}\), and the coefficient of kinetic friction between the snow and the runners of the sled is \(0.150 .\) How much time is required for the sled to travel down a \(175-\mathrm{m}\) slope, starting from rest?

A person whose weight is \(5.20 \times 10^{2} \mathrm{N}\) is being pulled up vertically by a rope from the bottom of a cave that is \(35.1 \mathrm{m}\) deep. The maximum tension that the rope can withstand without breaking is 569 N. What is the shortest time, starting from rest, in which the person can be brought out of the cave?

Consult Multiple-Concept Example 10 for insight into solving this type of problem. A box is sliding up an incline that makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is \(0.180 .\) The initial speed of the box at the bottom of the incline is \(1.50 \mathrm{m} / \mathrm{s}\). How far does the box travel along the incline before coming to rest?
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