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A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force exerted on the spacecraft by the earth balance that exerted by the moon? This point lies on a line between the centers of the earth and the moon. The distance between the earth and the moon is $3.85 \times 10^{8} \mathrm{m},$ and the mass of the earth is 81.4 times as great as that of the moon.

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Physics 101 Mechanics

Chapter 4

Forces and Newton’s Laws of Motion

Newton's Laws of Motion

Applying Newton's Laws

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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.

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in these grass wonder to forces acting on the spacecraft. One force is exerted by the room on the spacecraft and another force is exerted by the earth on the spacecraft. The big question then is when does this happen? When is the force exerted by the year from the spacecraft is equals to the force exerted by the room on that spacecraft. So when those it happened to do that to some things equation, we have to remember how to tell plate the magnitude off the reputational force and remember that the equation is the following. The magnitude of the gravitational force is equals to the Newton. Constant times the mass off the massive body times the mass off the other body divided by the distance between these two bodies squared. So the force that is exerted by the earth on the spacecraft is given by the Newton constant times the mass off the earth times the mass off the spacecraft divided by the distance between the earth on the spacecraft squared. So are E s squared. On the other side of this equation, we have the gravitational force exerted by the moon on the spacecraft. In this case, we have the Newton constant times the mass off the moon times the mass off the spacecraft divided by the distance between Dimon. Unless spacecraft squared as the Newton Constant is the same, we have one simplification and at the same time the mass off the spacecraft is the same. So we have another simplification. Then we get the following relation. The mass off the earth divided by the distance between the earth on the spacecraft squared Is it close to the mass off the moon, divided by the distance between the room on the spacecraft squared. Then we can send the mass of the moon to the other side, dividing and send their distance between the earth and the spacecraft through the other side, multiplying together following the mass off the earth divided by the mass off the moon Is it cost too? The distance between the earth on the spacecraft squared divided by the distance between the moon on the spacecraft squared But note that the mass of the earth is equals to the mass off the moon Times 81.4 Therefore the mass off the earth divided by the mass off the moon. Is it close to a 21.4 true, 8 to 1.4 is a close to their the stents between the spacecraft and the earth squared, invited by the distance between the moon under spacecraft squared. Now to continue serving his equation. Let me organize the board now, not the following. We can write the distance between the spacecraft and the boom as follows. So this is the full distance between the earth from the moon, and this is the distance between a boom on the spacecraft. So if we pick up the full distance and subtract from the full distance, the distance between the spacecraft and hear what is left is the distance between this spacecraft and the moon. So we can write the distance between the spacecraft and the moon as the distance between the Earth and the moon. Minus that, these tests between this spacecraft on earth too 81.4 is equals to the distance between the spacecraft and the earth squared, divided by the four distance minus the distance between the spacecraft and your squared. Now we have to solve this equation for the distance between this spacecraft and the earth. How can we do that? We begin by standing this term to the other side, multiplying so 81.4 times Are you a M minus? R E s squared is equals two R s squared. Now we take the square it off both sides off this equation to get the following the square root off into 1.4 times are the M minus R E s is the close r e s the reform The square root off 8 to 1.4 times our a m minus x squared off 1.4 times are as easy close to R E s Now send this term to the other side together following square it off to 1.4 times r e m is equals to R E s times one plus square root off to 1.4. Now send this term to the other side Dividing to get spirit off 81.4 times R e m divided by one plus square root off 81.4 Easy of course to R. E s. This is exactly what we want to complete. Now let me organized import to finish the question. Finally, we substitute the value off R E M to get the following r E s is it goes to the square it off 81.4 times 3.85 times 10 38 divided by one plus the square root off 8 to 1.4 And this gives us our ES as approximately three 0.47 times 10 to the eight meters. So when the spaceship is that these these tests from the earth the magnitude of the gravitational force exerted on the spaceship by the moon is bigger than the magnitude off the reputational force exerted by the earth on the spaceship.

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