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At a distance $H$ above the surface of a planet, the true weight of a remote probe is one percent less than its true weight on the surface. The radius of the planet is $R$ . Find the ratio $H / R$ .
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Physics 101 Mechanics
Forces and Newton’s Laws of Motion
Newton's Laws of Motion
Applying Newton's Laws
University of Sheffield
University of Winnipeg
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.
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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So this question all we have to do is use the equation for the magnitude off the gravitational force that a planet is exerting on some object. So that equation is the following. The gravitational force is given by Newton's constant times the mass off the planet times the mass off the object divided by the distance between the centre off the planet and the object squared. Now, using these equation, we will substitute this week's here. And so the question. So the weight off the object at a distance age from the surface off the planet is equals to 99% the weight off that same object at the surface off the planet. 0.99 because it's equivalent to 99% which says that the weight off that object at a distance age is 1% less than the weight off that same object on the surface off that planet. So the distance age from the surface. The weight is given by G Mass off the planet mass off the object divided by the radios off the planet, plus the distance between the object on the surface off the planet. On the other hand, we have 0.99 times The Newton constant times the mass of the planet times the mass off the object divided by the radios off the planet and this distances r squared. So we simply find a common terms on both sides off this equation. And we're left with one divided by r plus age squared is it goes to zero going to 99. Do I buy r squared? Then we send this term to the other side, this term to the other side to get r squared. Is it close to zero points? 99 times are plus age squared. Now take the square it off both sides to get our is he goes to 0.99 under a square root times are plus age Now we can send they can write it as square root off 0.99 times are plus square root off zero going to 99 times age Then we can proceed in Send this term to the other side To get the following miners square root off 0.99 times are busy goes to the square it off 0.99 sometimes age them We can write it us Our times one miners square root off 0.99 is equals to the square root off 0.99. Thanks age Then we can send these are to the other side dividing to get one minus square it off 0.99 is equal to the square it off 0.99 times age divided by r Now we send this term to the other side dividing to get one minus square it up 0.9 to 9 divided by the square it off 0.99 Is the course to the ratio age divided by r Then these results in the following 0.5 is equals to age divided by r and this is the final answer for this problem.
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