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Ch 13: Gravitation

Chapter 13, Problem 13

In its orbit each day, the International Space Station makes 15.65 revolutions around the earth. Assuming a circular orbit, how high is this satellite above the surface of the earth?

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welcome back everybody. We have an observer who is standing on the surface of the earth. Now, they are looking up at the planet mars and they see a little satellite that is rotating around MArs now. We are told that this satellite makes 21.96 revolutions every single week around MArs. And we are asked to find what the height of the orbit is above MArs surface. Well, we can use this relation here, right that the radius of orbit or the distance from the satellite to we're sorry, the distance of the satellite from the center of MArs is equal to the radius of mars plus this desired height. Now, just some facts about MArs, we have that the radius of MArs is equal to 3. Times 10 to the six m. And we also know that the mass of Mars is 6.4, 2 times 10 To the 23rd kg. So we know what the radius is. We're trying to find the height. That means we need to find this radius of orbit. Well, what we're gonna do in order to find that is we are going to use kepler's laws that states that T squared is equal to four, I squared times r radius of orbit cubed all over Newton's gravitational constant times the mass of mars rearranging this equation here, we get that our radius is equal to the cube root of the gravitational constant times the mass of MArs times our uh period T squared. All over four pi squared, but we don't. So we have all of this except we don't know what T. Is. Well we can use this information right here to find what T. Is. So let's go ahead and do that. Well t or or the amount of time it takes for one full revolution is given by some time period which we're dealing with weeks over one revolution. Right? But we're looking at 21. revolutions. Okay so we need to transfer this to some number in terms of seconds per revolution. Right? Well in one week we Have seven days and in one day we have 24 hours and in one hour we have 3600 seconds. If you multiply all this out we get better period is 27,540 seconds per revolution. Great. So now that we have our tea properly, let's go ahead and plug in our tea. And all of our other values into our equation for R. We have that are is the cube root of gravitational constant of 6.67 times 10 to the negative times 6.42 times 10 to the 23rd times are period of 27,540. All divided by four pi squared which when you plug this into your calculator you get that are are is 9.37 times 10 to the six m. Great. Now that we have our our we can go back to our original equation up here. I'm actually gonna subtract r. M. From both sides, and we get better. Height is equal to the radius of orbit minus the radius of mars. This is equal to 9.37 times 10 to the six minus 3.39 times 10 to the six, which gives us 5.98 times 10 to the six m or 5000, 980 kilometers corresponding to answer choice. C thank you all so much for watching. Hope this video helped. We will see you all in the next one.
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