Satellite Motion: Speed & Period - Video Tutorials & Practice Problems

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1

concept

Satellite Speed

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6m

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Hey, guys. So remember when we were talking about satellite motion? We said that there was a specific velocity a satellite needed. In order to go in a perfectly circular orbit, we're gonna need to know how to use that formula and calculate that velocity. So let's go ahead and cover that in this video. So we have a satellite that's out of some distance away from the earth. That center of mass distance is little are now at that distance. There is a gravitational force that constantly tries to pull it back towards the earth. So the question is, why doesn't just come crashing into the surface? Remember that this asked. The satellite is actually falling towards the earth, but it has a tangential velocity that basically keeps it going in a circle. So this thing is constantly falling around the earth, and the earth is trying to pull it backwards. But that tangential velocity keeps it going Eso that the earth is constantly curving beneath it, okay, And so this for a satellite in circular orbit that gravitational forces, actually what's causing it to go in a circle so that gravitational force keeps the satellite going in uniform circular motion and the relationship between that speed, which is the orbital speed and the distance, which is that little are is the V sat equation and that is the square root of capital G Capital M over. Little are so just want to remind you guys that that capital M is actually the mass of the big planet that's it's going around, not the mass of satellite. And that little art is not the radius. It's the orbital distance. It's that little our distance away. So sometimes you're gonna need to know where that equation comes from, so I can actually go ahead and work it out for you really quickly. So how do we get the velocity from uniforms? Circular motion? Well, remember that this if this is the forces acting on it, it's going uniforms, circular motion. And it has a centripetal acceleration, so we can actually start from F equals M A. To get to this, we know that f equals m A. But all these forces air centripetal, so the sum of all centripetal forces equals m A C. Now we know the only force that's acting on this is the force of gravity and that is m the A C becomes V squared over R So this is where this velocity actually comes from. Now we know what this force of gravity is. It's just g look big and little em over r squared as just Newton's law of gravity. And that's equal to M v squared over r. So to figure out what this V squared is, let's go ahead and simplify this equation. So I've got a little M that appears on both sides so I can cancel that out and I got a little are that also appears on both sides. And so what I'm left with is I'm left with capital G Capital M over our equals the square. So if I take the square roots, I'm just gonna get to that, uh, GM over r. I'm gonna get that lease at equation. So again, that is the relationship between the orbital speed and the orbital distance, which is your little are away from the surface. So I want you guys to remember is that for every value of our there is an exact speed. So there is an exact V sat in order to keep this thing going in circular motion. So, for instance, this satellite out here that's at some distance are has an exact V in order to keep it going in a perfect circle. Anything mawr or less than that, it's not gonna be perfectly circular. And also, if I wanted to change this orbit, if I wanted to go out farther or if I wanted to push this thing in closer, my V sat would have to change in order to keep this thing traveling in a perfect circle. So that's what that means. Alright, guys, let's go ahead and work out an example for this with the International Space station. So were asked to find the heights of the International Space station which travels around the earth. And we're told that the orbital speed is m per second in a nearly circular orbit. So any time you see this word nearly circular, you're just gonna assume that they're talking about a circular orbit so you can use all these equations to do that. Okay, so what are we told? We're actually trying to figure out what the height of this thing is, but remember that whenever we're trying to find little h or big are we're always gonna find always gonna find little our first and then we can relate it using the r equals big r plus h formula. So first we have to find Little are So how am I going to do that? Which equation? Um, I'm gonna use I'm trying to find what little our is, and I'm on. Lee told what the velocity of the satellite is so I can use the V sat equation in order to relate those two things because those are the only two variables that pop up in that equation. So let's start from the V sat equation. So I've got visa equals square root of GM over R. So now I want to actually get to what this are is so that I could basically get to what h is. Um eso I just have to go ahead and isolate that. I've got this are that's trapped in the denominator here s so I can lift the square roots by taking the square both sides. So I got visa squared, equals G m over our And now if I want to get our by itself, I basically want this thing to come up and I want the V sat to come down. So these things were just gonna trade places. So I've got that r equals gm over v sat squared. Now I just have to make sure I have all of this number, right? I have G, I have m and I have the sat squared and that capital M, because I'm going around the earth is just the mass of the Earth, which I have this table right here. So plugging all that stuff in I get 6.67 times 10 to the minus 11. I've got the mass of the earth 5.97 times 10 to the 24th. Whoops time, Senator, 24th and then divided by 7670. Just make sure that you square that in the denominator and you should get 6. times 10 to the sixth, and that's in meters. But just remember that we're not quite done yet because that was we've just only solved. We've only solved for little art. Now we have to go and plug it back into this equation to solve for H, which is the height. So we have little r equals big R plus h. So if we want to find a church, we have to sit. If the isolate h so h is equal to little ar minus big are so I got 6.77 times 10 to the sixth minus. What's this? What's this? Big are well this big are if we're talking about the earth is just the radius of the earth. So we're gonna have a track that 6.37 times 10 to the sixth and we should get four times 10 to the fifth, which is about 400 kilometers. You can actually go ahead and google this. If you google the height of the International space station orbit, you'll find that it is about 400 kilometers. On average, it goes up and down a little bit, but that's pretty much what it is on average, which is pretty cool. So you can use, um, you know F equals Emma and some simple equations to figure out how high the international space station actually orbits around the earth, which is pretty cool. Alright, guys, let me know if you have any questions with this stuff

2

Problem

Problem

Suppose that you used some geometry and kinematics to estimate that the Earth goes around the Sun with an orbital speed of approximately 30,000 m/s (60,000 mph), and that the Sun is approximately 150 million kilometers away from the Earth. Use this information to estimate the mass of the Sun.

A

1.42×10^{15}kg

B

2.02×10^{29}kg

C

2.02×10^{30}kg

D

1.42×10^{17}kg

3

example

Find speed of second satellite, given speed of first

Video duration:

4m

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Hey, guys, let's check out this problem here. We're told to satellites are orbiting this planet and were given a whole bunch of numbers. So let's go ahead and just start drawing a diagram. So my diagram here is gonna involve some planet and I have two satellites in circular orbits, so let me go. Oops. Let me go ahead and drop my circular orbits. I've got one right there and we got another one that's gonna be right there. Let's say cool. So now I've got a mass. I've got a satellite that's over here. I'm told the mass of that satellite is equal to 68 and I'm told that the orbital radius, the orbital radius is six times 10 to the eighth. Remember that is that little our distance. And that is the distance between the centers of mass. That's not big R. And that's not a church. This is our But because the first one, because I have two satellites, I'm gonna call this our one. And then I'm told that the velocity, the orbital velocity, which is that tangential velocity of you one is equal to 3000, right? Cool. So I have another satellite that is some other distance away. So I'm gonna call this m two. And I know the mass is 84 then I've got the orbital radius of that thing. So then I've got here. The orbital radius is equal. Thio are too. And I know what that number is, right? That's nine times 10 to the eighth. Now I'm asked to figure out what is the orbital velocity off the second satellite. So really V two is my target variable. So let's start there. So I've got V two as my target variable. So how do I get that? Remember, from our equations, we're gonna use the V sad equation. We're looking for orbital velocity. So I've got V two is just gonna be the square roots of g times, Big M. And because I'm looking for V two, I'm gonna be plugging in our too into this. So let's take a look. This is my target variable right here. Let's take a look. If I have everything else in the problem, G is just a constant. Then I've got the mass of the planets, but I don't have that variable, so I've got the radius of the orbital radius, but I don't have Big M. So how do we go about solving for that? Because I'm gonna need that. I've never told it in the equation. So let's go over here and try to solve for big M. Let me go ahead, scroll down. So I've got Big M equals something. How do we get big? We can use either Forces Look at our table of equations right here. We can either use forces or we can use accelerations or we can use satellite motion. But I'm not told anything about forces in the problem. I don't have any forces between two objects and I don't have any any accelerations here. I know that it's circular motion and I know that I'm working with satellites, So let's use the other V sat equation. So remember, we have two satellites here, so let's look at the other satellite. The first one V one equals square roots of G times. Big M, divided by little are one. Now the are one of the other satellite. So if I take a look at this, I have the V one. I have G is just a constant and I have our one, so I can actually go ahead and figure out what this Big M is. So I'm gonna go ahead and rearrange. I've got this thing in the square root so I have to get rid of that square root by squaring both sides. I've got V one squared equals g times Big M over our one. Now I'm just gonna move everything over and sulfur are on sulfur m So I've got the one squared are one divided by G equals m. So I actually have everything there that orbital velocity for that first one is 3000. Then I've got a square that the orbital radius of the first satellite, we're told is six times 10 to the eighth and then we're just divided by the gravitational constant. And if you do that, you should get 8.1 times 10 to the 25th, and that's in kilograms. So now that we have that mass right here, we can actually plug it back into this equation for and sulfur V two. So let's go ahead and do that. So I've got V two equals square roots. And then I've got 6.67 times 10 to the minus 11. Then I've got the massive this thing 8.1 times 10 to the 25th. Now I've got two divided by the Oriole radius of the second satellite, which I'm told is 9.8. Sorry, nine times 10 to the eighth, and that's in meters. So you go ahead and do that. You get the velocity is equal to 2450 m per second. So we've got that number right here. Notice that this number is actually lowered. By the way, this is the final answer. Notice that this number is actually lower than the orbital velocity of the of the first satellite. And that's because it's a farther distance away. So we can see from the V sad equation that if our orbital distance is farther than your velocity is going to be lower. So that actually just makes some sense. Let me know if you guys have any questions with this

4

Problem

Problem

You throw a baseball horizontally while on the surface of a small, spherical asteroid of mass 7×10^{16} kg and diameter of 22km. What is the minimum speed so that it just barely goes around the asteroid without hitting anything?

A

8.9 m/s

B

14.6 m/s

C

115.3 m/s

D

20.6 m/s

5

concept

Satellite Period

Video duration:

6m

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Hey guys, We saw that the orbital velocity of satellite was how fast the satellite travel in its orbit. There's another variable you need to know to solve problems. And that's the orbital period, which the time that it takes to complete one orbit. So we saw that the satellite has a tangential velocity as it's going around in its orbit, but it takes some amount of time to actually complete one full orbit. That's called T, and we actually relate that velocity to the period just by using circular motion and basic mathematics. How do we get t? How do we get time from velocity? We know from cinematics that V equals distance over time. What's the distance that this thing completes in one orbit? Well, it's traveling in a circle of radius R. So that's just the circumference of that circle two pi r. And if you do that one full circle, then this T just becomes a capital T. And this equation is really useful for us in satellite motion. So I'm actually gonna write it here. Two pi r over tea. So now we can actually solve for this thi this capital T right here most of the time. You'll see it in his T squared form. This is four pi squared R cubed, divided by GM. You could actually get to it pretty quickly by using this equation. So let's go ahead and do it real quickly. V sat equals two pi r over tea. So what happens is if you want tea, you could get t equals two pi r over v sat. So we have this V sat, but we have already an equation that will tell us we have another equation for V sat. We could actually stick that in here, and we're gonna get two pi r divided by square roots of GM over R So you get this nasty formula with square roots and fractions. So what will happen is if you square it, it will become a lot cleaner. Two becomes the four pi becomes pi squared, are becomes r squared. And what happens to the square root is that when you square it, the whole thing just goes away. Um, or the square root goes away. So you get GM over a little are so we've got this situation. We have a fraction of a fraction and what happens is that this denominator of the bottom fraction will actually go up and merge with that top Little are so what'll happen is four pi squared r cubed over gm So you get that equation. This equation actually has a name. It is called Kepler's Third Law. Kepler was a guy who is studying the motions of planets in our solar system and you notice that the relationship between the orbital period of all the planets and the distances from the sun it's pretty cool. So we've got these three equations and these are all of our V sat equations are sorry, none of visa our satellite motion equations. So we've got the speed That's V SATs. We have the period t and the distance, and they're all related by these three equations, and they're all sort of dependent on each other. So what happens is the distance increases. Alright, So happens is our increases we can use. How does V change? So we have these two V sad equations, and if the little are is changing, how does this equation change? It's a little bit tricky to tell just by using this one, because you have all three variables present in these and we don't know how t changes yet. So instead, let's use the other V sat equation. So as you are increases in the denominator Thievy sat has to decrease. All right. And again, let's not use this, uh, this equation 40 because we have a relationship between just t and R. So it's gonna be easier to relate. Those to buy this equation now are is in the numerator here. So if our goes up than T has to increase as well. So that means that as r increases, your velocity decreases. But your period increases. This should make some sense because as you're going farther away, the force of gravity gets weaker on you, which means you don't have to go is fast to stay in a circle. On the other hand, if you were to go really close to the earth, the earth would be pulling on you really, really hard. Would you have to go really fast not to crash into the earth because you're traveling in a larger circle as R increases. The time that it takes also should be increasing, and that's basically it. So let's go ahead and take a look, at example, and use all of these equations together. So we've got the orbital period and speed of the international space station. So first thing is, we're gonna be calculating the orbital period. So let's take a look at our equations, so t equals What? So you've got these equations right here? So I've got V equals two pi r over t. The problem is, I don't know what the velocity of this thing is. Actually, that's part B. So I can't use this equation just yet, So I don't have V. But I can't use this equation because this just relates t and R. So I've got t squared equals I've got four pi r Sorry. Four pi squared are cubes over g times m Now I don't have our but I can relate are using big R plus h right? I know what the radius of the earth is and I also have the height. So I know what actually little are is so I could go ahead and use this equation to solve for t So I got t squared equals. Then I've got four pi squared. So what is this? Our distance? We'll remember this is This is just the radius of the earth. 6.37 times, 10 to the sixth. Oops. Plus this, 400 kilometers above the earth's surface. So that means that H is equal to 400 and then we've got 400 kilometers, so it's actually 400,000 400,000 m. So we've got four times, 10 to the fifth meters, and you're gonna have to cube that, actually. Just be careful. You gotta cube that. And then you've got a divided by 6.67 times 10 to minus 11 and then multiplied by the mass of the earth. 5.97 times, 10 to the 24. If you do that, you're gonna get some crazy number, which is 3.8 times 10 to the seventh. But we're not quite done yet, because remember, you have t squared here, so you have to square both sides square root both sides. Just make sure that's the last step. You're not forgetting that. So if you square root both sides, you're gonna get 5550 seconds, which is actually about 1.5 hours. This is pretty cool. You can actually google this and you'll find the International Space station does orbit about 1.5 hours. Takes 1.5 hours to travel around the entire earth, which is pretty awesome. It's going insanely fast. So that was part a eso now in part B were actually figuring out what the velocity is. So in part B, what is V sat right? Cool. Let's take a look at our equations here. I have this equation and I have this equation. But in this case right here, I actually have what are is. And now I've just figured out what the velocity is, so I can actually use either one of these equations. I'm just gonna use the simpler one. So if I have both equations, then Visa equals two pi. I know what the R is that our distance here is just 6.37 times 10 to the sixth, plus four times 10 to the fifth and then divided by the period, which is 5550 seconds. And if you do that, you should actually get, um, 7660 m per second, which you can also Google and you'll find that it's just about this number, so let me know if you guys have any questions with this.

6

Problem

Problem

A satellite orbits at an orbital period of 2 hours around the Moon. What is the satellite's orbital altitude?

A

1.9×10^{6}m

B

1.2×10^{5}m

C

1.7×10^{6}m

D

3.2×10^{5}m

7

example

Find mass of planet, given moon speed and period

Video duration:

4m

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Hey, guys, let's see if we can work it out. This problem together. So we have a moon in orbit around a planet, right? So let's go ahead and draw diagrams. So we have a planet like this and we have a moon that is orbiting around in a circular orbit. Just pretend that that's perfectly circular, right? And this moon has a has a speed around its planet, and we're told that velocity is equal to 7500. And we're also told that it takes 28 hours to go all the way around. So all the way one rotation, you know that T is equal to 28 that is ours. So using that information, how can you find out what the mass of this planet is? So we're finding the mass of the planet. That's the thing in the center. So we're really looking for capital M so capital and big is our target variable. So how do we go about doing that? Well, let's see. We've got a whole bunch of equations evolving forces and gravitation, gravitational acceleration. But we don't have any information about forces and gravitational acceleration, but we do have the motion of satellites. We're gonna use all these equations over here, so let's see, I've got orbital velocity and I also have orbital period. So really, I can use, like, any one of these equations to start off. I'm gonna use the V sad equation. So we've got visa equals, Let's see square root of g times, big M big M Big M over our and let's see, um, I know what the velocity of the satellite is and I'm looking for big end, which is my target variable gravitational constant since the number. So if I If I can figure out what this little our distances, I could find that. But I actually don't have any information about the orbital distance or the height or anything like that. So let's see, Maybe this isn't the approach. Let's see if I could start off with my tea sat equation with the T squared one. So let's start with the tea sat squared equation. Let's see if we have better luck there. So you've got four pi squared r cubed over g g times. Big M again. That Big M is our my target variable here. So I know what the orbital period is But again there's that are that I don't have that orbital distance. So in both of these approaches here, both these equations that I've seen, I ended up with the same unknown variable. I have too many unknowns. So there's gotta be something else I could do to solve for this little our distance. Let's see, what's the one equation I haven't used yet? So I'm gonna go over here and solve for a little r The one I haven't used it is this one. The V sat equals two pi r ver t So I've got visa equals two pi r divided by teeth. Now, in this situation for this equation, I have to knowns and Onley one unknown. So now I can use that to Seoul for my little are So let's see, I've got moving everything over to the left side. I'm gonna get visa times t divided by two pi is gonna give me Little are so that means that that little are here If you just go ahead and plug, everything in is going to be let's see 7500 times the period. Now that T is expressed in hours so first, I need to multiply by 3600 to get into seconds. If you do that, you should get 1.8 times 10 to the fifth. That's in seconds. So I'm just gonna go ahead and plug it in 1.8 times 10 to the fifth. Now, I've just got a divide by two pi. So if I divide by two pi, I'm gonna get the orbital distance that little are That's 1.2 times 10 to the eighth. So now what I could do is now I can take this. Little are that I found I can plug it back in tow either this t squared equation because now I only have this unknown. Or I can plug it back into the V sat equation so that I still only have one known. So really, the choice is up to you. I'm gonna just keep going with the visa approach. So I've got now my my little our distance. So I've got ve sat equals square roots. Whoops. Square root of GM over. Little are So now I'm just gonna go ahead and salt for that big M because I already have everything else, right? So I've got ve sat and I'm gonna square both sides because it's because I want to get rid of the square root. So they've got V sat squared, equals G Mm over our And then I just moved the are over everything over to the left side. So I get V sat squared are divided by G equals Big M. If you plug all that stuff in, what you're gonna get is 7500 squared times the radius 1.22 Sorry. 1.2 times 10 to the eighth. And then we've got to do divided by the gravitational constant. So it times 10 to the 11th. And if you do that, you should get the mass of the planet, which is 1.1 times 10 to the 26. And that's in kilograms. That's it for this one. Let me know if you guys have any questions with that

8

Problem

Problem

A distant planet orbits a star 3 times the mass of our Sun. This planet of mass 8 × 10^{26} kg feels a gravitational force of 2 × 10^{26} N. What is this planet's orbital speed and how long does it take to orbit once?

A

1 × 10^{5}m/s; 2.5 × 10^{6}s

B

1.29 × 10^{5}m/s; 2.5 × 10^{6}s

C

1 × 10^{5}m/s; 1.94 × 10^{6}s

D

1.29 × 10^{5}m/s; 1.94 × 10^{6}s

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