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(a) Calculate the magnitude of the gravitational force exerted on a 425-$\mathrm{kg}$ satellite that is a distance of two earth radii from the center of the earth. (b) What is the magnitude of the gravitational force exerted on the earth by the satellite? (c) Determine the magnitude of the satellite’s acceleration. (d) What is the magnitude of the earth’s acceleration?

$1.04 \times 10^{3} \mathrm{N}$

$1.74 \times 10^{-22} \mathrm{m} / \mathrm{s}^{2}$

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Cornell University

University of Washington

Hope College

University of Sheffield

for the first item we have to use. The equation for the magnitude of the rev additional force, which is the following magnitude, is given by Newton's constant times the mass off the first object times the mass off the other object divided by the distance in between them squared. Then, using the data given by their problem, we have 6.6 to 7 times stand to minus 11 times the mass off the earth. Just 5.98 times 10 2 24 times the mass of the satellite, which is 425 kilograms. And this is divided by the distance between the satellite and this center off the earth on this distance is given by two times the radios off the earth. Then we have true times 6.38. I'm standing on the six squared and these results in the gravitational force of approximately 1000 and 40 new terms, which you can write as 1.4 times stand littered mutants. It means that the Earth is exerting a force on the spaceship attracting it. And at the same time, this spaceship is attracting Earth. If a force off the same magnitude but opposite direction. So on the second item, we have to calculate what is the magnitude off the rev additional force exerted on the earth by the satellite? It happens that the reputational force exerted on here by the satellite has the same magnitude as a reputational force exerted on the satellite. My dear, if there's a symmetry in the reputational force, so again we have 1.4 times 10 to the third new tools on the next item, we have to calculate the satellites acceleration. For that, we have to use Newton's second law. Let me choose Thies to be my reference frame and I call this access X. Then the net force exerted on the satellite is given by the satellites mast times its acceleration before the acceleration off The satellite is given by the net force exerted on it. Divided by its mass. The net force exerted on the satellite is the gravitational force that has been exerted to the left. Then we have minus the gravitational force divided by the mass off the satellite. And these is minus 1.4 times 10 to the third, divided by 425 which results in an acceleration off approximately minus two points 45 meters per second squared. Note that the minus sign indicates that the satellite is being accelerated to the left. Then for the last item, we have to calculate what is the acceleration off the earth again? We have to use Newton's second law, sir. The net force exerted on the earth is of course, to the mass off the earth times its declaration. Then the acceleration off the earth is given by the net force exerted on the earth divided bites mats, which again the net force exerted on the earth is the rev additional force. But this time it's exerted to the right. So it's positive we have 1.4 times standing, the third divided by the mass off the earth, which is 5.98 times stand toe into four. These results an acceleration of approximately 1.74 times 10 to minus 22 meters per second squared

Brazilian Center for Research in Physics