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(4 pts total) The International Space Station is orbiting at an altitude of about 370 km above the Earth'$ surface; The mass of the earth is 5.976 x 102* kg an...

Question

(4 pts total) The International Space Station is orbiting at an altitude of about 370 km above the Earth'$ surface; The mass of the earth is 5.976 x 102* kg and the radius of the Earth is 378 * 106 m: (a) Assuming circular orbit, what is the speed of the International Space Station in its orbit?(b) Assuming circular orbit, what is the period of the International Space Station'$ orbit?

(4 pts total) The International Space Station is orbiting at an altitude of about 370 km above the Earth'$ surface; The mass of the earth is 5.976 x 102* kg and the radius of the Earth is 378 * 106 m: (a) Assuming circular orbit, what is the speed of the International Space Station in its orbit? (b) Assuming circular orbit, what is the period of the International Space Station'$ orbit?



Answers

An Earth satellite in a circular orbit is at an altitude of $3185.5 \mathrm{~km}$. The acceleration due to gravity at that distance is $4.36 \mathrm{~m} / \mathrm{s}^{2}$, and the mean radius of the Earth is $6371 \mathrm{~km} .(a)$ What is the radius of the orbit? (b) Find the speed of the satellite.

So we're told there's a satellite that's already ing 890 kilometers above the surface of the earth. Um, and so then, in part, A were asked to find out what the orbital speed of that silent might be. So there's a couple of ways we could approach this. We could do energy, but I think probably looking at this for a forced perspective will be a little bit quicker here. So because the satellite is moving in a circle, we know that subjected to a centripetal force and the only force that's acting on it is the force of gravity. So it must be gravity that's making that centripetal force. So we say force of gravity equals the centripetal force on the satellite, and the force of gravity will be gravitational, constant times, the mass of the Earth times, the mass of the satellite and then divided by R squared. And so are here is the total distance from the center of the earth. So it's the radius of the earth. Plus there's a tannery Danny climbers, and so that should equal the centripetal force, which always takes the form N V squared over R. So we can simplify this one of these ours goes away and so as well as the mass of the satellite. So the master Sally doesn't matter Here. Any satellite of any weight would have to have the same speed to be ordering at this height. And we can solve for the velocity then and we get that it should be the square roots of G mass of the earth Over are and so are here is raised the earth plus 8 90 kilometers. So times 10 to the third meters ends, we can evaluate that we should get that is 74 10 meters per second. So that's the speed of the satellite. And then in part, be were asked what the period will be in hours for the satellite. So for the period, we can say that the distance task to travel is one circumference, so to pie are so that's this circumference. Not a very good circle here, going around the whole thing. And so again are is raised the earth plus 8 90 s o the distance. It travels over it's average velocity. So that will just be two pi greatest. The earth plus 8 90 times 10 to the third meters and divided by that philosophy. We just found 7410 meters per second and we should find that comes out to be 6000 144 seconds and then we convert that two hours. We get that it's 1.71 hours.

Okay, so this problem is asking us to find the velocity. A satellite that is traveling in a circular orbit a certain distance above the earth soon as we see gravity of this is acceleration and acceleration in a circular orbit is modeled by the equation a equals B squared over r squared. So this is a good So I'm sorry, just be squared over R. So this will be a good starting point for us. And we will combine this with the gravity form of Newton's second law. So Newton's second law is f equals I m A. We're going to use the gravitational force, which is K g Big M little am over r squared. The G is the gravitational constant. Big M and Ms a case will be the mass of the Earth. But it took the stands for the larger of the two masses, and small m will be the satellite or the smaller to mass is in question. So now we can put all three of these equations together, so this expression here will go in for this, and this expression here will go in for this. And so we get m B squared over r equals the g big m them over r squared. So now we are looking for the velocity. So we're gonna have to solve this equation for this velocity right here. So to do that, we will first multiply both sides by our this will cancel. This are out. And one of these leaving us with m b squared equals big G big M little over our Sorry about my little ab. And now we can divide by m on both sides to get rid of the M So those will cancel each other out and we're left with the squared equals big G big M over our now Just one last step to do is take the square root. And so this will go to be equals the square root, uh, gravitational, constant times the mass of the earth divided by the radius to the satellite. And now that is all the information that we need to be able to solve our A. So with this G is equal to 6.673 speaking times, 10 to the negative 11 Big M is going to equal 5.972 times 10 to 24 and are equals the radius of the earth, plus our distance above the earth, which for this case is 700 kilometers, which is equal to 7.8 times under the five meters. Add that to the radius of the earth and we get 7.1. I times tend to the six years. So this is in meters. This isn't you May meter squared or kilograms square. Mm. Sorry. Mascot stuff. Kilograms squared and this is in kilograms. So you just plug these three numbers into here and we should get e equals 7465. And what we are assuming at this moment that the units are going to be meters per second because that's what we know velocity should be. But now we need to go and check to make sure that that's where it's equal. So we will go through and we will set this equation up here with our circle but a blue. So this is our velocity equation. And now we will go through. And we will do this with just the units, though, so we're not gonna worry about the numbers now. So and the gravitational constant. So first of all, it's all gonna be in a square root so B equals the square root of mutants. Meters squared over kilograms squared times, kilograms all that over meters. So this kilogram will cancel with one of those, and this meter will cancel with one of those some. Now we're left with Newton's times meters over, killer grabs and it's square rooted still. Well, that doesn't look anything like meters per second. So we're have to go a little deeper and break down the Newton's into their minutes. Newton is a kilogram times a meter square, no kilograms times meter over seconds squared. And then we still have times a meter over a kilogram and it's all square. It's still so this meter and this meter will combine after these to go and we have the square root of meters squared over seconds squared. You take those square roots, you get meters over seconds. It works out. We're good. So this is our answer for velocity Now moving on to part B. We need to figure out what the period of the orbit is for that satellite. Now that would be found using the equation for a circular period which is given by T. The period equals two pi r and we used to pie for a full circle of rotation. That wasn't a circle. It wouldn't be that simple over velocity. So now we have a velocity equation already. So let's just put that into this equation since we know this represents the velocity of our satellite. So what we're going to have is this is gonna equal to hi are times the square root. And since cities on the bottom, we can just flip this over instead of adding more ratios into our problem. So this will just be are over big G Big M. Now the only thing we need to do is combine these ours together. So when you combine those ours together, you get to because this is our to the 1/2. And if so, this would be our t to the 2/2, which equals R to the one to over two puts one over to you get 3/2 all over the square, root of gravitational, constant times, the mass of the earth. This is our equation for the period that simple. So we got to dio now you just pop your numbers in there and figure it out again. And after shutting through all the numbers, you should get something like 6000 and nine Teen. Now what we're looking for our units to be for a period is one over seconds. Which means it has one revolution every year. Revolutions per second. Yeah, yeah, yes, Yes. One over seconds is what we're looking for. So now we'll go through and do the dimensional analysis again for this one. So on top, we have the radius, which wasn't meters. And it is to the three halves. So t equals to Pai has no units, so we don't need to worry about that when we're doing our units. So this is meters to the three hands divided by, and we're going to use the full in form of G. So big square. Yeah. So we're going to have so first is the newton. So you've got kilograms times, meters over seconds, squared times meters, and then the rest of our units here is meter squared over kilograms squared. So all of that together is our units for G. It's why we like to use a little bit simpler. One like this and then we're going to be multiplying by mass, which is just Kipper grips. So Kilogram cancels one of the kilograms kilograms cancel kilograms. And, um, I got one too many MP's here. Oh, yeah, that end was unnecessary. It's in this one here. So now we're left with I m to the three halves over the square root of m, cubed seconds squared square root of em. Cute is the same as M to the three halves. So this will cancel with this and you are left with one over the square root of one over seconds squared, which means you will just invert this. This is the square root of seconds squared and you take the square root of that which equals seconds. So, yes, we were good. It's not one over seconds. It is just seconds that we are looking for. There we go. Now. If we want Teoh have something that makes a little more sense to us since we don't typically count 6000 seconds for things, divide that by 60 get 100 0.3 minutes, which is also equal to 1.67 hours, meaning that that satellite travels around the earth once every 1.67 hours. And if you divide that bites one or that means it goes around about 14 times a day, So thanks for watching.

For this problem on the topic of gravitation, we want to know what orbital to be must be given to a satellite in order for it to be in a circular or about 780 kilometers above the surface off the earth. We also want to know the period off this orbit. Now we will apply in second law to the motion of the satellite and obtain an equation that relates the orbital's BV to the orbital radius. R. Now we can see the ladies off the orbit is our which is H R E. Now H is the altitude off the satellite or the height above the surface of the Earth and R E is the radius off the earth. So our is able to age plus R E. Now we know that it's a height off 760 kilometers or 7.8 times 10 to the 3 m last year, radius of the earth which is six point 38 times 10 to the sixth meters. So the distance off the satellite to the center of the earth, at which point the mass is located. So I assume the masses all concentrated at the center of the earth in this distance are is 7.16 times 10 to the sixth meters. Now we know that if we take the some off all the forces, we'll call our vertical direction. Why in the diagram, some of all the forces in the Y direction must equal to m times the acceleration off the satellite in the Y direction. Now he forces in the Y direction due to gravity. So f g is equal to the mass of the satellite M times. It's radio acceleration. Now we know this force can be written as G and the master satellite times the mass of Earth e off the distance our square we ask where are the distance from the satellite and sent off earth and for the centrifugal motion This is equal to M v squared over R. Since the orbit is circular. So if we rearrange the equation, we can see the masses canceled and we're left with the orbital speed off satellite. The to be the square, root off the gravitational constant g times the mass of the earth off the distance Our all of these unknown so we can find this orbital speed. So this is the square root off G, which is 6.673 times 10 to the minus 11 in S I units, which is Newton meter squared for K G. Squid multiplied by the mass of the Earth, which we know is five 0.97 times 10 to the 24 k g. All divided bye. The distance from the satellite to the center of the earth which is 7.16 times 10 to the power 6 m. So calculating to get the orbital speed off the satellite to be 7.46 times 10 to the power three meters a second. So that's the speed that the satellite needs to be given in order to stay. I didn't orbital height off 700 80 kilometers about the surface of the earth. Now, in part B, we want to calculate the period of this orbit so we know the period off the circular orbit is equal to be distance off one orbit, which is two pi r over SBV and so this is two pi times 7.16 time stand to the six meters divided by this BV, which is 7.46 times 10 to the 3 m per second and so calculating we get the period off one orbit of the satellite to be 6000 and the D seconds, which is approximately one 0.68 hours.

For this problem. On the topic of gravitation, we went to no the orbital speed and the period off the orbit off a satellite in order to be in circular orbits that 780 kilometers above the surface of the earth. Now, firstly, we know we need to apply and you can second law to the motion of the satellite and obtain an equation that relates the orbital speed V to the orbital radius. Now, if you look at our first diagram, we can see the radios off. The orbit is actually are and our is equal to the altitude of satellite above the surface of the Earth, plus the radius of the earth itself. That's where the mass off the earth is concentrated. So this is 780 kilometer kilometers, which is 7.8 times 10 to the 3 m plus the ladies of the Earth R E, which is 6.38 times 10 to the 4 m. And this gives us the radius off the orbit from the center of the Earth to be 7.16 times 10 to the four meters. Now we've also drawn a free body diagram for the satellite has shown the satellite is marked by its mass. M. It's radio explorations acting toward the center of the earth as well as the force you to Gravity e. Or rather, if g so that's the gravitational force acting on the satellite due to the earth. So now, for part A, we need to calculate the orbital speed. Uh huh. Now we will look at meeting second Law, which says that the forces acting on the satellite along our wide direction must equal to, um, times a Y. Now they force acting on the satellite is due to gravity. So that's F G. And this must equal to mm times the radial acceleration towards the center of the earth. A r a d Are they forced you to gravitation from Newton's Universal Law is G and M E M is the mass of the satellite, and M is the mass of the earth over r squared. And this must equal to m the squared off our where the the orbital speed on satellite, which is essentially what we want to calculate now, rearranging this equation, we consult a V so we is equal to he square root off the gravitational constant G times the mass of the earth times the separation between the satellite in the sent off the are Since all these values are known, we can calculate this orbital speed. This is the square root off G, which is 6.673 times 10 to the minus 11. And that's mutant. Meet a squid. What k G squared? That s I units. I was the mass of the cut which is 5.97 times 10 to the 24 k. G. This is all divided by the distance between the center of the Earth and the satellite, which we calculated to be seven 0.16 times 10 to the power 6 m. So calculating with the orbital speed of the satellite to be seven 0.46 times 10 to the power three meters per second. So that's our answer part A. We found the orbital speed off our satellite now for part B, we asked asked to calculate the period off the orbiting hours. So the period of the overall satellite yeah, is given by T and we know T is equal to two pi r where r is the latest of the orbit divided by the speed of the Orbit V and these values again that all known so we can calculate this period. That's two pi times are, which is seven 0.16 times 10 to the power 6 m all over the which we calculated to be 7.46 times 10 to the power of three meters per second. And so we get the period of the satellite's orbit to be 6000 and 30 seconds, which translates to one 0.68 hours.


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