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A satellite of mass 5500 $\mathrm{kg}$ orbits the Earth and has a periodof 6200 $\mathrm{s} .$ Determine $(a)$ the radius of its circular orbit,(b) the magnitude of the Earth's gravitational force on thesatellite, and $(c)$ the altitude of the satellite.

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a) $7.3 \times 10^{6} m$b) $4.1 \times 10^{4} N$c) $9.2 \times 10^{5} m$

Physics 101 Mechanics

Chapter 6

Gravitation and Newton's Synthesis

Physics Basics

Newton's Laws of Motion

Applying Newton's Laws

Gravitation

Cornell University

Rutgers, The State University of New Jersey

Hope College

McMaster University

Lectures

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.

03:55

In physics, orbital motion is the motion of an object around another object, which is often a star or planet. Orbital motion is affected by the gravity of the central object, as well as by the resistance of deep space (which is negligible at the distances of most orbits in the Solar System).

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So here we can say that the force of gravity on the satellite is going to be equal to G times the mass of the Earth times the mass of the satellite divided by r squared. We know that the force net is going to be equal to the mass times c centripetal acceleration. So this will be M v squared over r. So we can say that V is equaling two pi r. This will be essentially based on any mass that is in uniform circular motion. So we can say that the force net substituting this inn would be four pi squared em are divided by t squared. At this point, we can say that four pi squared are divided by t squared with the equal to ah, the gravitational force over r squared. And so we can cancel out the mass of the satellite and say that the radius is going to be equal to the gravitational constant times The mass of the earth times the period of the satellite squared divided by four pi squared all to the 1/3 power on DSO for part A. When it's asking us what's the distance between the satellite and the center of the earth. We can say that for part of the distance between the satellite and the center of the earth is going to be 6.67 times 10 to the negative. 11th times 6.0 times 10 to the 24th kilograms The mass of the earth times 6200 seconds squared. That's the period of the satellite abided by for pie squared all to the 1/3 Power and R is giving us 7.304 times 10 No, to the sixth meters. So this would be our answer for part A. Now, for part B, they want us to find the force of gravity on the ah satellite. So this will be G times the mass of the Earth times the mass of the satellite divided by r squared. And this will be 6.67 times 10 to the negative 11th. The mass of the Earth is again 6.0 times 10 to the 24th kilograms times the mass of the satellite, 5500 kilograms and then this will be divided by R squared. So this will be seven 0.304 times 10 to the sixth meters Quantity squared, and we find that the force of gravity on the satellite is equal to 4.126 times 10 to the fourth Newtons. So this would be our final answer for Part B and then for part, see, we confined the multitude, the altitude above Earth's surface, and this would be really easy. This would simply be equal to the radius of the earth, so this would be the center from the Earth's surface to the center of the earth minus. Rather, it would be our the distance from the satellite to the center of the earth, minus the distance from the center of the earth to the surface of the earth. So essentially, this will be equal 27.304 minus 6.38 time 10 to the sixth, and this is giving us 9.24 times tend to the fifth meters. So that's how high above the earth's surface it is again. You simply need to sub subtract the distance from the satellite to the center of the earth. Ah, the minus the distance from the center of the earth to the surface of the earth s 9.224 times 10 to the fifth meters. That'll be your final answer for part C. That is the end of the solution. Thank you for watching.

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