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A 2150 kg satellite used in a cellular telephone network is in a circular orbit at a height of 780 $\mathrm{km}$ above the surface of the earth. What is the gravitational force on the satellite? What fraction is this force of the satellite's weight at the surface of the earth?

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$2.59 \times 10^{8} \mathrm{m}$

Physics 101 Mechanics

Chapter 6

Circular Motion and Gravitatio

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Lectures

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In physics, kinematics is the description of the motion of objects and systems in the frame of reference defined by the observer. An observer has to be specified, otherwise the term is meaningless.

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In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

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A 2150 kg satellite used i…

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A satellite used in a cell…

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A communications satellite…

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A satellite has a mass of …

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A $320 \mathrm{kg}$ satell…

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If the satellite in Proble…

grant ish in force is given us f g f subscript TV. It is equal to G Oh, that's captain T times M e times am divided play r squared Where are is the distance of the satellite from the center of the art. So a distance from thie center off the earth two. You satellite? Yeah, light and ah m is thie mass of the earth. So that's myself, all right. And him is thie myself off the satellite, right? And these are gravitational constant. So to get the distance from Xander, be our earth, we need to know the radius off. There are their distance from the center to its surface, which we didn't notice. Our E and our is six find 38 times sent the power six meter. So that means the total are or smaller should be this distance ar e plus the height of the satellite from the surface of the earth. It is that's a age. So if we try to get the smaller will have will get 6.38 times kind of us six meter class and which is 7.8 times Temple bar, five meter. So we're converting everything in meters just to be consistent. And then that should be 7.16 times. Tend to the par six meter right and m e, or Mass of the Earth is given as 5.97 times 10 to the bar training for Kensi. So from there on also, that's Adlai. Its mass is given us 21 five serial kg, so we know all the information. We know the gravitational constant so we can solve for the force that's acting on satellite. So force will be six. Find 673 times 10 to the power minus 11 Newton meter squared like a D squared. Then we have 5.97 times 10 to the power. 24 Katie times 2150 K G. Divided by 7.16 times 10 to the par six meter squared. So this's our square. So we're squaring out the distance from here. We get the force as 1.67 time's tender bond for noon. Now the weight of the satellite at an art surface should be the wait W, which is mass times gravity. Now we know masses to 150 Katie and the gravity is 9.8 meters per second squared. And from here we get the weight as 2.11 times 10 to regard for Munich. So if you want to find the factional, the fraction that should be FT over the weight. So that 1.67 times 10 to the par four divided by 2.11 times and bar for and that is serial 0.791 And if we write that in terms of percentage So the gravitation force at a certain distance eso a certain height which was, um, 7.8 and 10 or five or 7 80 kilometer is ah, 79.1% off W or the weight when the satellite is on the surface of the earth. Thank you.

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