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ssm Synchronous communications satellites are placed in a circular orbit that is $3.59 \times 10^{7}$ m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?

0.223$m / s^{2}$

Physics 101 Mechanics

Chapter 4

Forces and Newton’s Laws of Motion

Newton's Laws of Motion

Applying Newton's Laws

Alyssa J.

November 10, 2021

Excellent demonstration of solving this problem

Rutgers, The State University of New Jersey

University of Washington

Simon Fraser University

University of Winnipeg

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the weight force that is acting you're the satellite is related to the gravitational force that is actually acting on the satellite by the following equation. From one side, the weight forced is equal to the mass, the satellite times, the acceleration of gravity where the satellite is located. On the other hand, the nature off the weight force or the force that produces the weight force is the gravitational interaction between the satellite and the earth, and the gravitational interaction produces a repetition of force. We just give him by Newton's constant times the mass off the earth times, the mass off the object that is under the influence off the gravitational field producer by the Earth. And finally we divide lose by the distance between the center of the earth on the object that is under the effect of gravity. Now, to calculate exhibition off gravity, we create these two equations so the weight is equals to the gravitational force. Then the mass times acceleration of gravity. Busy Costa G mass off your mask off the object divided by r squared. There is this simplification of the masses under get in equation for the graphic acceleration, G minds off the earth, divided by R squared now breaking the virus that were given by the problem we get the following gravitational acceleration G Eazy kohstuh 6.6 to 7 times 10 to minus 11 times the mass off the earth 5.98 times stand to 24 kilograms divided by r squared. Now R squared is the distance between the centre off the earth and the object there is orbiting the earth. So it is equals to the radios off the earth, plus the distance between the surface off the earth on the satellite. So this is divided by 6.38 times 10 to the sixth, plus 3.59 times stand to the servant and this is squared. It gives us a gravitational acceleration off approximately 0.223 meters per second squared.

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