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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.)

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

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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.

Brazilian Center for Research in Physics