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A satellite moves in a circular orbit at an altitude of 2,750 mabove a planet's surface. If the planet has a mass of 5.33 x1018 kg, and a radius of 3.80 x104 ...

Question

A satellite moves in a circular orbit at an altitude of 2,750 mabove a planet's surface. If the planet has a mass of 5.33 x1018 kg, and a radius of 3.80 x104 m, what isthe satellite’s orbital period?

A satellite moves in a circular orbit at an altitude of 2,750 m above a planet's surface. If the planet has a mass of 5.33 x 1018 kg, and a radius of 3.80 x 104 m, what is the satellite’s orbital period?



Answers

An earth satellite moves in a circular orbit at a speed of $5500 \mathrm{m} / \mathrm{s} .$ What is its orbital period?

For this problem, we can first start with the fact that the velocity is going to be equal their circumference of the supposedly circular orbit divided by the orbits period. And next, we can use Kepler's law, which states that t squared is equal to four pi squared over G m times are cute. Next what we can do its first take this equation here and solve for R And so we will get is that r is equal to V T over two pi. And now that we have that, we can plug it into the second equation here, and we should get that t squared is equal to four pi squared, divided by G n multiplied by the cubed times t cute, divided by two pi Cute, which is going to be pie cubed times and eight. So now we can start canceling some things first, this left hand side becomes a one and we can cancel off, uh, the three here and just leave the tea. And so this pie squared cancels and we're left with one factor of pie, and this eight becomes a two, and so we can bring this GM over to the other side here and we'll get GM equal to V Cubed T over two pi. And so now we can solve for T. And we should get that t is equal to two pi g m divided by the Cube. And so now we can just solve because we know V and M is going to be the mass of the earth. And so what we should get for a final answer is about 15,110 seconds.

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.


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