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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 m? (Assume that the gravitational force is constant over the distance of the fall and that the star is not rotating.)

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$3.3 \times 10^{5} \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 4

Forces and Newton’s Laws of Motion

Newton's Laws of Motion

Applying Newton's Laws

Cornell University

Rutgers, The State University of New Jersey

University of Washington

Lectures

03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

03:43

In physics, dynamics is the branch of physics concerned with the study of forces and their effect on matter, commonly in the context of motion. In everyday usage, "dynamics" usually refers to a set of laws that describe the motion of bodies under the action of a system of forces. The motion of a body is described by its position and its velocity as the time value varies. The science of dynamics can be subdivided into, Dynamics of a rigid body, which deals with the motion of a rigid body in the frame of reference where it is considered to be a rigid body. Dynamics of a continuum, which deals with the motion of a continuous system, in the frame of reference where the system is considered to be a continuum.

02:17

A neutron star has a mass …

02:06

01:44

A typical neutron star may…

02:08

05:32

A star $2.5$ times the mas…

05:09

Under some circumstances, …

02:43

\bullet Under some circums…

02:40

05:44

II Under some circumstance…

03:36

we begin this question by calculating the acceleration of gravity on the surface. Off these nutrients start so on the surface, off that neutral star, the acceleration of gravity is given by the Newton constant times The mass off the star divided by the radius squared off that star if you forgot the new town Constant is a close to 6.6 to 7 times stand to minors. 11 new terms meters squared, kilograms squared. Then on the surface off that store, acceleration of gravity is equals to 6.67 times. Stand to minus 11 times two time. Stand to the authority divided by the radios squared so five times stand to the third squared. Okay, what do we do if that well, that is approximately eight. These give us surface repetition acceleration off approximately 5.3 time Stan to the 12 meters per second squared. Note that it's way bigger than the gravity of the surface off the earth. This is because neutron stars are super down stars, so they have huge gravitational acceleration. Now we proceed to calculate what is the acceleration off this object that is falling on the surface off that star so there is some object falling. Therefore there is a wait forcefully. Now that weight force is given by the mass off the object times the acceleration of gravity on the surface off that star. So the weight force is It goes to the mass of the object. Times 5.3 times stand on 12. On the other hand, according to Newton's second Law, the Net force acting on that object is equals to the object's mass times the object acceleration. So the net force is composed only by the weight force. Therefore, the mass off the object times 5.3 times stand to the 12 is equals to the mass off the open tractor times acceleration. Therefore, acceleration off these objects is the close to 5.3 times Stand 12. Now we proceed to complete What is the velocity off this object after falling 0.1 meters toward the surface of the planet? First, let me organize the board. Okay. Now, to complete the final velocity, we will use Torricelli's equation, which says the following the final velocity squared is he goes to the initial Velocity Square plus true times and exploration times, displacement So in this situation, the initial velocities because 20 on the displacement is actually displacement in the height, so final velocity squared is equals to two times. That's elevation of 5.3 times 10 to 12 times 0.1 Now, taking the square it we get a final velocity that is equal to the square. Root off truth times 5.3 times. Stand to the 12 times stand to the miners to, and these give us a velocity off approximately tree 0.3 times stand to the 50 meters per second.

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