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AttemotiQuestlon 31 of 31intoan M( IS perigee nearest point the: Earth: ils slellite elliptical orbit around the Earth: When the satellite height above Lhe ground i...

Question

AttemotiQuestlon 31 of 31intoan M( IS perigee nearest point the: Earth: ils slellite elliptical orbit around the Earth: When the satellite height above Lhe ground is hp 2019,0 Km , and mcwing with ~Prd ol % 350 km/s. The grvitational constant cqjuake 6.67 m'+kg and the mass o Eaurth equals 5.972 10?4 Kg: When the satellite reaches its apgee. at its farthest [Oint from the Earth; what is its height ha above the ground? For this problen; choose gravitational potential energy Ol the satellile

Attemoti Questlon 31 of 31 intoan M( IS perigee nearest point the: Earth: ils slellite elliptical orbit around the Earth: When the satellite height above Lhe ground is hp 2019,0 Km , and mcwing with ~Prd ol % 350 km/s. The grvitational constant cqjuake 6.67 m'+kg and the mass o Eaurth equals 5.972 10?4 Kg: When the satellite reaches its apgee. at its farthest [Oint from the Earth; what is its height ha above the ground? For this problen; choose gravitational potential energy Ol the satellile inlinite distance [romn Furth: 83148 annaret



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A satellite of mass $m,$ originally on the surface of the Earth, is placed into Earth orbit at an altitude $h .$ (a) Assuming a circular orbit, how long does the satellite take to complete one orbit? (b) What is the satellite's speed? (c) What is the minimum energy input necessary to place this satellite in orbit? Ignore air resistance but include the effect of the planet's daily rotation. Represent the mass and radius of the Earth as $M_{E}$ and $R_{E},$ respectively.

Here. Although the multitudes were given, We have to find the orbital radio so we can say that our sub a would be equal to 6370 kilometers above the surface, plus the radius of the earth, 6370 kilometers. And this is equaling 272,000 another 12,740 kilometers. We can then say our sub B, this was going to be equal to 19,110 kilometers plus 6370 kilometers. This would give us 25,000 480 kilometers. Now, at this 0.4 party, we confined the ratios of potential energies. You sub be divided by use of A. This is giving us negative g times am times and divided by our sub B divided by negative G Time's and time's M spotted by our sub A and essentially all of these cancel out and this is equaling the radius of a divided by the radius of be that we had just calculated. We don't even need to change it into meters because of the units are also gonna count are also going to Ah, cancel out. And essentially, we have 1207 Sorry, 12,740 divided by 25,480 this is essentially one over two. So this would be the ratio of the potential energies. Four part B. We know that we're going to be using Equation 13 28 13 38 rather, And we can then say that the ratios of kinetic energies would be equal to the gravitational, constant times the mass of the essentially this Matthew Satellite times the mass of the Earth. There are two by two times the orbital radio of satellite be divided by got gravitational, constant times the mass of the satellite mass of the earth divided by two times the orbital radio of satellite A. And so again, this is gonna be equal to the radius. The orbital rate radius of satellite aid. About about orbital radius of satellite B, this is equal in again one over two. This would be our answer for part B from essentially from equation 13 40 for part C. We know that here it's gonna be clear that the satellite with the largest valley of our has the smallest value of absolute absolute rather the smallest value of the magnitude of the total energy. And since the values of energy are negative in this case, we can then say that the smallest value of E corresponds to the largest energy E, where we can say smallest value of the absolute value of B, uh, corresponds, essentially largest energy. Yeah, So we can say here, satellite be has being largest energy. And again, this is because essentially both of the both of the values, the values of the of the energy are negative. So the smallest value of the absolute value of the energy would equal the largest energy value. Um, the largest true energy value. So here satellite be had the largest energy and then for part D, we confined the difference. The change in the energy would be equal to the energy subete minus the energy. So a and this is gonna be equal to negative G times sometimes am to fight it by two multiplied by one over our sabian minus won over our sub A and again, uh, remember to convert these two meters. So now when we calculated the change in energy. These have to be converted two meters and not kilometers. So we find that the change in energy is that gonna be equal to 1.1 times 10 to the eighth. Jules, this would be our final answer for a party. That is the end of the solution. Thank you for watching.

So we're told that we have a 400 kg satellite that is 1500 kilometers above the surface of the earth. The celebration of gravity at that point is 643 m per second squared. And we know that the orbital speed is 25.6 times tender, the third kilometers per hour. Um, so they ask us, um, the linear momentum of the satellite, which is kind of uh, they gave us a whole lot of information, but we only need this and this, this stuff doesn't come into play there. We needed the angular momentum, then we would need and some some things this value and the radius of the earth. But just because they want the linear momentum, that's just M. Times V. And they gave us the linear velocity. So all we need to do is be careful and convert this two m per second. So we have 400 kg times 25 6 times 10 to the six m per hour. And then we have one hour equals 3600 seconds. And we just multiply this all through. And we get to a point, let's see here, did I? This should be attended to something here. And what is that? Let's see here. Let's calculate this out. 400 times 2.5. Listen here for uh huh. 400 times 25 6 times 10 to the sixth. I all divided by 3600. And that's what we had to four. And so that we have 123456 and then six. And in the sixth, power kilogram meters per second squared that right? Well, let's see here. That's probably looks about right here. Uhh so what we have is that's just our linear momentum. I'm not sure. You know, there's nothing really more to this problem than that. It's a little bit it's a kind of a simple problem. and that's um you know, they gave us more information than we actually need it. So again, it's not, you know, there's nothing wrong with having more information than you need really.

We have gravitational for G mass off arts times massive the object divided by Are you just age, age, height of of the art squared equals and the square over R and R is our e t s h. So from here we find that the sport equals g m e over r e thus h So that's basically the answer for part B. Now he is to buy r over V. It was to buy our ive thus Age divided by I ve ese squared off G i m e I'm squared off our e thus age So from here we find t ik was to buy our ive thus age race to the part of 3/2 delighted by spread of g m e.

Everyone. So what we have here is a satellite that's moving in an elliptical orbit within east interest city of E equal to 0.25 or as to determine is the speed when it is at its maximum distance and minimum distance from the earth we are known were we know that the minimum distances to mega meters and the concepts that will be we will be applying to solve for these values are the laws of orbit, orbital motion and the definition of these interesting. So let's go ahead and get started. De centris ITI is equal to C H squared all over g times m e The mass of the earth where c is equal to one over are not times one minus g Emmy or are not d not squared and age is equal toe are not be nuts. So plugging in these quantities gives us that e is equal toe are not V not squared all over g and e minus. What we can do is isolate v not and this gives us v not is equal to g times e e plus one all over or not in this case are not is our minimum distance so are not will be equal with that too. Mega meters The two times sense of the sixth meters, plus the radius of the earth as well. So plus r e. And what this does is it gives us an are not that's equal to 8.4 times 10 to the 6 m. So now what we can do is software velocity at B and we'll say that VB is equal to remember g time Emmy. He plus one all over or not. This gives us a VB that's equal to 7.71 kilometers per second. Next, what we can do is we can. What we need to do is solve for our distance to the maximum point. And what we can do is we can say that are a is equal to are not all over two times g over m e times are not peanut minus one And what this does is this gives us are a equal to 14 times 10 to the 6 m. Now we can use proportions to solve for our velocity at a we can just say that the velocity at a is going to be the ratio between the distance to a excuse me, the distance to pee or our shortest point divided by the distance to a the effigy and that times of quantity. What time's VB? And this gives us a velocity equal to 4.63 kilometers per second? And that's also our final answer.


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