Ch 06: Centripetal Forces & GravitationSee all chapters

Sections | |||
---|---|---|---|

Uniform Circular Motion | 27 mins | 0 completed | Learn |

Centripetal Forces | 53 mins | 0 completed | Learn |

Universal Law of Gravitation | 36 mins | 0 completed | Learn |

Gravitational Forces in 2D | 25 mins | 0 completed | Learn |

Acceleration Due to Gravity | 22 mins | 0 completed | Learn |

Satellite Motion | 45 mins | 0 completed | Learn |

Concept #1: Satellite Motion & Circular Orbit

**Transcript**

Hey guys. So in this video I'm going introduce satellite motion and circular orbit. Let's check it out. So a satellite is any object that orbits another. So if you have two objects are orbit each other for example the moon going on the earth and the earth going around the sun. These are usually obviously for astronomical objects then the satellite is the one that is orbiting the middle object. So for example the object in the middle we gonna call that big M and the object around it little m and that's because the or the object that's orbiting is the lighter one. So if the moon goes on the earth the moon is little m and the earth is the thing in the middle so it's big M if the earth is going around the sun so the Earth is a little m and the sun is Big M because again the sun is in the middle. OK. Now there are different types of satellite motion or different types of satellite orbits you might have an elliptical orbit or might have that circular orbit and the one we're going to focus on in this video is circular orbits. So it's important for you to know that the way you're going to get an elliptical orbit or circular orbit depends on the initial velocity of the satellite. So it says here for a satellite launch from the earth its orbit or really for any planet. Right. It doesn't have to be just the earth in its orbit which is the shape of its path. There's like a bunch of different shapes here will get there in the second depends on its initial speed. I also call this launch speed depend on how fast you throw something. It might take a different shape. OK. So let's check this out. I made this sort of velocity line here to show you how this works. So first of all imagine that there is a canon up here and it's pretty far up. Somehow they built a massive cannon. This is sort of a thought experiment and it's a classic thought experiment in physics to help explain this. But if you put a can here and you shoot Well first of all if you don't if you don't shoot it at all there's no velocity right so nothing happens. But if you shoot it it's going to move. Now if you shoot it with a little bit of velocity it's going to sort of just fall hit the earth with velocities with paths A or B and that's just projectile motion horizontal projectile motion we've done that. So you get projectile motion A and B. So that's nothing new. The part that gets weird is if you shoot this thing fast enough if you shoot this thing fast enough it's going to have a circular orbit around the earth and it basically never touches the ground. And that's because as it's falling towards the center of the earth the earth also sort of curves beneath it. So it just keeps looping forever. And that's in fact exactly how satellites work to get satellites around the earth. We sort of fly them out here and then we push them out with a certain velocity not obviously a very precise velocity. So that the satellite will then keep spinning, that's because the satellite has an initial velocity this way but the earth is pulling it that way. So it just keeps doing this right and circular motion uniform circular motion in fact. OK. So there's a magic speed at which if you were to throw something you would get a circular orbit and that's this. I'm going to call this Vcircular, okay at this exact number you would get circular. So let's say this number is 5000 ok just making that up 5000 meters per second. I mean is that if you do something with 4000 meters per second it's going to go really far but it's going to eventually land and hit the Earth. Right. And obviously this is assuming that there are no buildings on the way. It's not killing anyone right. So it's a it's a theoretical situation but it would it wouldn't make the full circle. If you throw something with five it would make an exact circle around the Earth. So if you throw it from one meter eventually go all on the earth and come back to you at one meter and just keep doing that it would be weird to see if you throw with more than 5000 let's say 6000 then it's going to have an elliptical orbit. So circular orbit is C over here in diagram, elliptical order is D ok elliptical order means it's sort of a distorted circle instead of a circle. Let's say that's a circle. It would look kind of like this. OK. So but we're not going to get into that now. And also there is such a thing as a speed that so fast that the object loses that the object escapes the earth it's too fast. The earth can't pull the object hard enough to keep it in a circle and the object escapes the earth. So let's say again I'm just making this up I'm going to scratch these up later. Let's say this is 10000. So if I throw something between five and 10 I get elliptical. So it's a big range if I do it faster than 10 it escapes the earth and never comes back. All right. Let me scratch these out because these are not the actual numbers for the earth and I don't want you guys to get confused there. So anyway that's the set up of what might happen depending how fast you throw something. And we're going to see this. And for some of you guys you're going to get into this as well. All right. So circular orbits are simpler. That's why we're going to focus on circular orbits. OK. Because they have a constant path. All right. Now planets have elliptical orbits. Right. But we're going to do is we're going to simplify by saying that they're circular enough or nearly circular. When you see the words nearly circle just means you're going to consider to be circular. OK so we're going to simplify planetary motion so that we can use more simple physics. And like I said for some of you we might get into elliptical orbits later and see how that works. So in circular orbits the orbital speed which is how fast you're going and the Heights which is h. This is height are related by this equation right here. Now let me quickly show you how to get to this equation because it's actually pretty easy. So let's pretend this is a satellite or you or whatever. So you would be moving this way at that point with a V 10 tangential velocity going around in circles so it's of the circle right which I can also call Vsat because you're basically behaving like a satellite now. And you would also have an acceleration towards the center Ac, by the way when you're over here you will have V this way and a this way. This is just uniform circular motion in such a way that you're forming obviously a circular path around this planet so let's say this is the earth. OK. So I want to write an equation that shows you what the velocity relationship with the velocity so you might remember velocity in uniform circular motion problems. We can find either using Ac equals V square over r or F equals in ma. So I'm going to do this, I'm gonna write sum of all forces in the central direction equals mAc. Now the only force here acting on you is the force of gravity, the force of gravity is pulling you in. So I'm going to put here force of gravity equals M and then we'll replace Ac with V square over r. So we can get this number now. The force of gravity is given by this equation here by universal law of gravitation. Big G, big M and little m over r square equals mV squared over r. I want to remind you big G is the universal gravitational constant right here, big M that would be this mass little m would be the satellite, r is the distance between the center of both. So r is actually this distance right here from the center of the earth to you little r is a distance not necessarily a radius and sometimes you going to be different things. OK. Notice at this distance little r over here. Is made up of two things. I'm sorry I'm going to call this little h over here, this is the height okay. Little r is a combination of your heights plus the radius of the earth. So hope you see that it's the radius of the Earth plus how far away from the earth you are. So I'm gonna write here the little r is the radius of the Earth plus your heights. And I want to remind you that little r is distance from the center.

Which isn't always the radius. Right. And this is height. Obviously R is the radius of the planets. So in physics uppercase letters. Are Used to mean a constant. They they they simply signify that this is a constant like big M the mass of the earth not changing, the radius of the earth not changing big G universal constant never changes. Right now r and h change you could move farther which would increase your h, therefore would increase your r OK. Cool. So let's finish this up real quick we're almost done. I can notice that this little m cancels with this little m and this r cancels with one of these. So I can just solve to ensure that V equals the square root of big G big M over r and that's what I'm going to put right here that's the first question we have for satellite motion that the speed of a satellite or that the velocity of satellite has magnitude of the square root of G m over little r, notice that it's little r not big or it's not the radius of the planet but it's the orbital radius. That's what that's called. This is called orbital radius which is the distance not from the surface but from the center of the planet, orbital radius or orbital distance is another way you can see that. OK. So this is also notice that this is the mass of the earth not the mass of the objects. In fact right here right here there's a little point that says Notice how satellites motion does not depend on its own mass and little m it only depends on the mass of the thing in the middle right the planet where the sun or the star that it's rotating around. So I get this equation and if I combine this equation right here with this equation here from uniform circle motion I can solve for t, t will equal if you move stuff around 2 pi r over t. So So t is 2pi r over V. Now if you do this and I get Vsats which is this guy here and I stick Vsats right here. Ok I'm going to end up with this really really ugly equation. OK. That's right that this is going to be 2 pi r to the three over two divided by the square root of Gm. Now some books some professors don't actually give you the equation and you only get as far as having to find a period using this. But to find a period of a satellite which means how long it takes to make a full rotation around whatever planet or or star it's rotating around. You can use either one of these equations obviously to use the the smaller one you have to already know V to use the big one you don't have to have the. So this is more straightforward but it's also more more messy. OK. So those are basically the three equations you would have and you need to know at least these two for some of you're going to have to know all three of them. OK. So an important point here is this for every height a the satellite may have there is an exact corresponding speed it must have to maintain that height. OK. If I want a satellite to be in a circular orbit at two kilometers above the earth there is a very specific velocity. It needs to begin, if it's slower it's not going to work it's not going to stay in that exact height. If it's high it's not going to work. So you might imagine that this is a very precise thing because you put satellites in space and you want them to just go away and get lost and you lose all your investment right. So that's pretty interesting. And I can show you that here R is a constant right. R is a constant so the only variables I have is h and little r OK. Let me just write it this way. V equals the square root of Gm over little r the square root of Gm. I can replace little R with R plus h and I can show you that everything here is a constant. The only thing that's not a constant is h. So the idea is that as h changes, Vchanges and there is a one to one mapping between the two meaning with if one changes the other one has an exact corresponding number. The same exact thing happens for Tsats. Notice how Tsat changes everything. Here's a constant except R and inside of R the only thing that's a constant is that a variable is h ok. So those three things are related. Ok orbital speed the period T and height h are inter-dependence, they are interdependent. I can also show you how they are going to affect each other in terms of you know if this grows what happens to that. So as height increases as the satellite gets farther and farther from the planet. If you want to have your satellite farther from the planet your velocity will decrease. And I can show you that here really simply. If my age increases over here then R obviously increases. And this is at denominator which means this will decrease your period if you look in the period equation. If your age increases your period increases. So it's the opposite. And this might makes and this should make sense because decreasing means you are slower. So your period will be greater which means you're slower right greater period takes longer. So you are slower. Right. So the closer you got to the earth the faster a satellite has to be. And that should make some sense because if you think about the satellite if it's not super fast it's going to fall into the earth. So if you get closer it can pull harder and it's closer. So you have to be moving faster to be able to continuously not fall into the earth and keep sort of not escaping but staying at that same constant rate anyway. Long story short we got these three equations here. Let's do a problem real quick. So here it says the International Space Station ISS is in a nearly circular orbit. Nearly circular means that obviously we're going to treat this a circular ordered at 370 kilometers above the earth. So you have to know your equations which also your variables. Which variable is 370 above the earth. It's your h that's your height above the surface of the earth. 370 kilometers or 370 thousand meters. Calculate its orbital speed. So a calculated orbital speed orbital speed is given by the square root of G m over r. Notice you don't have it yet. Right. So r is big R plus h and I have these numbers big R is the radius of the earth which I have up here. For you guys it's six point thirty seven times into the six so six point thirty seven times ten to the six plus h which is 370 thousand. And when you add this you get when you add this you get six point seventy four times ten to the six. OK. What really sucks about this stuff is that this topic is going to a lot of big numbers again.

All the gravitational topics you gonna have a lot of big numbers. Cool, so that's my R and I can now plug it in here so I can I'll find my V, V is the square root of G, G is 6.67 times 10 to the negative 11 with a bunch of units. And m the mass of the Earth five ninety seven times 10 to the 24th and you divide all of that by little r which is six point seventy four times ten to the six. Let's say you did this, you plug this into the calculator carefully and you get 7686 meters per second. By the way if you go on Google if you go on Google and you look up international space station you're going to see that the velocity is really really close to this number right. The actual velocity is really really close to this number. Obviously there are some small variations here and there. OK that's party let's do part B. I don't have a lot of space here but I don't need a lot of space. That should be enough for part B. I'm looking for that period orbital period. So what is Tsats, remember I can do this, I can use this equation up here or the bright green equation. The smaller one you should always try to use a smaller one obviously. And I know V. So that's what we're going to do. T equals 2 pi r over V. So it's 2 pi r is going to be this number down here. Six point seventy four times ten to the six and V is 7686. Once you do all of this you get 5510 seconds which I when I hadn't converted to hours on just to get a better idea you don't have to do this unless the problem tells you to. But a lot of problems will tell you to do this because they want you to kind of make sense of what's going on. One point fifty three hours. OK. So this means that the period of the International Space Station is 1.5 hours one and half hours which means that every hour and a half the space station actually makes a full circle around the Earth. So if you could see it you would be able to see it going by several times. Pretty cool. Then part C, D in here are conceptual part questions and it says what happens what would happen if you're slightly faster or slightly slower. Remember this velocity here this velocity here is the velocity to have circular motion to have circular orbit. And let me go back here and write this here. This is the velocity you need to have circular orbit you let me go up a little more. And I told you that there's a magic velocity at which this happens. And now we know that this magic velocity here. Is the square root of Gm. Over r. Ok that's this guy here. That's what's needed. OK. So that number means you're in perfect circle orbit but remember what I told you if are slightly faster than that. What happens is that you instead of having a perfectly circular orbit you're going to have an elliptical orbit. Now the farther you are from this number the more elliptical it's going to be the more stretched out of a circle it's going to be and the closer you are to that number the more of a circle you're going to be. So you don't really have to be perfectly at that number you just have to be close. So they have a very close to a perfect orbit a circular orbit. OK. What happens if it's much faster. Well if you look at the diagram on top of the page if your speed is much much faster than this then you're going to escape the earth. The Earth won't be able to pull you hard enough you just escape the earth. And what if it's much much slower then you're going to fall into the earth. And that sounds like bad news. In fact that's how they get rid of satellites when the satellite doesn't work anymore or whatever or they want to maintain anymore they might just leave it spinning around the earth. They could try to accelerate it and shoot it away from the earth. But very often these things sort of run out right and they start getting slower and slower because up there there's a little bit of friction. Very tiny amounts. And it's going to eventually start falling to the earth and these things take days and days sometimes of a longer than that spinning around the Earth until they splash somewhere hopefully in the water and not on top of a city or something. OK. So I want to make one last point here. And astronauts in the International Space Station are not weightless. Right. So there's this misperception it's a big conceptual point in physics. There's this misperception that this perception that if you're in space you're just floating around because you're weightless weightless means your weight equals zero which means weight is mg which would mean that the gravity at that point is zero and that's not true. OK. In fact the reason why they look the way they do is because they are constantly falling. So imagine you just fall the whole time. Right. Except they get used to it so they're not yelling anymore and they get used to that feeling. If you were an airplane and the airplane would have for free fall and you were to free fall in a controlled manner with it. You would. It would look like this. Right. You'd just be floating side of the airplane to be able to sort of air swim everywhere. In fact they have that now you can do that. This just cost a bunch of money. So they're not weightless they're constantly falling towards the earth but they're inside of the as they spin around as well. But they're inside of the of the satellite or whatever where the international space station. So they're falling together with this thing around the Earth. So looks like they're just floating there which they are. They're falling and floating but they are not weightless that effects when you're falling with something so it looks like you're flying also that looks like you're weightless is called apparent weightlessness apparent weightlessness but there is no such thing as weightlessness unless you were to go really far away from any objects really really far from any planets or stars or anything like that that maybe you would be able to get away from all these forces being having any impact on you. And then it says here. It says here in fact the acceleration there is 8.7. So the acceleration of the earth is 9.8. And if you go all the way up to international space station the gravitational acceleration is actually fairly close to to nine point it's eight point seven. They're experiencing something similar. It just feels different. Another reason why the difference differences that they're inside of this this space station so there's no air resistance if you will to jump. Let's say you're skydiving or whatever as you're falling there's all these air in your face. So it feels like you're falling right that's the normal sort of feeling of falling that we think of. But when you're out there's no air resistance you literally floating in the air but you are not weightless. You you're not weightless. There is gravitation there is gravity on you. Right. So this is also referred to zero gravity but that's again wrong. It's just apparent weightlessness or you know zero gravity with quote quotations around it. Because there's actually gravity. So just a quick conceptual point there to wrap it up let me know if you guys have any questions.

Concept #2: Circular Orbit Problems (Practice Intro)

**Transcript**

Hey guys so I want to go over a few more things in circular orbits of satellites I want to start off with giving your practice probably right before I do that I want to cover something real quick so protip number one over here if you're ever solving for the speed necessary to put an object in orbit that's basically just asking for your Vsats circular which is this equation right here OK So the idea is that that's the speed of satellite way from the Earth would need to go like this but it's the same speed that if I were to throw something horizontally it would theoretically go around the entire earth and come back to the same point if there's no air resistance if it didn't hit a building if it didnÕt it anyone right so that's what their questions are asking speed to put an object in orbit and circle or it would be simply looking for this the SAT over here OK So anyway this first question deals exactly with that says how fast would have to throw an object horizontally from the ground for it to become a low orbit satellite around the earth lots of words here.

So you throw something you're going to throw that way in fact direction that you throw it doesn't really matter but if you were to throw let's say a horizontal like you would go like this what speed do you need so that it would go all the way around and come back at the same height again ignoring all these other things so I want you to try this and also find how long it would take to go around the earth what's the orbital period I pretty straightforward got the equations I also gave you here all the gravitational constants that you might need so let's give this a shot.

Practice: (a) How fast would you have to throw an object, horizontally from the ground, for it to become a low-orbit satellite around the Earth?

(b) What orbital period (in hours) would it have?

Example #1: Circular Orbit Problems

**Transcript**

All right so here I have another tip when you solve these problems which is very important if you're ever solving for r or h you're going to first find little r. and then use this equation which is the equation relates the two to be able to find big R or h OK So let me show you this real quick notice how these two big equations here all of these equations actually. Have little r in them and everything is in terms of little r's if you wanted to find so that's why you want to find little r first if you want to find r and H. one way you could go about it which is the wrong way some telling you basically what not to do would be to replace little r for R the equation for example v equals the square root of G.M. over r you could rewrite it this way. Big R+h and try to find the h but guess what that's going to be much harder than if you just find r first and then solve for whichever other variable you're looking for using this tiny question so I like to think of this as leave little r alone OK don't split it up find that first and then find the other two its gonna be much easier especially if you look at this equation here right little r two the three over two so this would become just that one part would become r plus H. to the power of three over two and good luck getting a number out of there right so find r first and then do it so lets do this question real quick it says how high above the earth's surface must an object moving at five thousand meters per second be in order to have circular orbit we're playing we're working just with circular orbits here so that's a constant a given.

The satellite has a velocity of five thousand v equals five thousand circular orbit tells me that this equation Works but I want to know how high above the earth that is H. If you look at my variables over here in the top right in orange there's no h that are readily available the only h shows up here so I'm going to find r first and then get H. because that's better that way that's all just covered so I'm going to find r first and then use it to find H K based on information I have if I have V. and I want r you should use the V question right here OK V. equals to square of G.M. over r. And if you want to solve for R. you can square both sides so this becomes the squared equals G.M. over are. And then you move some stuff around so you get r equals GM over the square OK G.M. over the square if I plug this in over here or you put all the numbers OK you put all the numbers you have so you put your G. here six point six seven times ten to the negative eleven you put your M. here. Five. Five point ninety seven times ten to the twenty fourth and you put V. here this v will be five thousand and you multiply all this carefully calculated please try this to make sure you know how to do you get in the right numbers that you're plugging this in correctly you should get to one point five nine times ten to the seventh OK one point five nine times ten to the seven.

But I donÕt want r I want h so what do i do well r is big R plus H. so h if you move things around is little r minus big R So let's do that here h is little r one point five nine times ten to the seven minus big R which is five I'm sorry six point three seven it's a constant times ten to the six. And if you do this you get nine point five three times ten to the six is how high you have to be or how far away from the surface of the earth OR same thing you would have to be. That said hopefully you got this I just want to make one more point little r is bigger then R right even though it's a small variable versus a big veriablr little letter versus uppercase letter it's actually a greater number because it is R plus h and less obvious you can say that h is negligible at which point they become the same thing all right so that said there is a practice problem here I want you guys to give this a shot so let's do tha

Practice: A satellite in circular orbit takes 30 hours to go around the Earth. Calculate its height above the Earth.

Example #2: Circular Orbit Problems

**Transcript**

I think I saw one last point I want to make is when you have a satellite when a problem tells that a satellite stays in place relative to the earth right stays in place so you hear it your house let's say you could see satellites flying around and you would notice while the guys always directly above me right so that's what it means to stay in place is always relative to you now the earth spends it means that the satellite spins with you so that it's always on top of you right satellite spins with you this is called giosynchronous orbit so it's orbits giosynchronous you see gross orbit kind of why we're there and the basic idea here that you need to know is that that means that the period of the satellite is the same as the period of the earth OK or if it's a different planet it would be the period of that plant so the puter the satellite is the same as the period of the planets to rightly about you do this in telecommunications satellites satellite cell phone satellites and all these kinds of things because if you want people in the city down here to be able to talk to each other and you have to be satellites appear and if the satellite spinning faster then all the sudden you can't get the satellite anymore right so if you want to always work for a certain area have to have the satellite apart now for example if you have a spy satellite that you want to sort of fly constantly fly over other places and take pictures or something and you might want that satellite to be there faster than the Earth or slower than the Earth probably faster so you can get a bunch of laps in one day in you get a bunch of pictures OK but so often satellites telecommunications satellites in general T.V. or whatever you want them to be directly above the area and then all you have to do is have a bunch of them sort of equally spaced or whatever so you can hit up a bunch of places.

So if I tell you that a satellite has geosynchronous motion or geosynchronous orbit. It's just a really fancy way of me telling that the period of the satellite is the same as the period of the planet and you should know the period of the earth the purity of the earth around itself is twenty four days I mean twenty four hours Whoops So the Earth takes twenty four hours spin around self technical little bit less than that but in physics problems you can just use twenty four hours this motion is called rotation and the motion of the earth around the sun has a period of three hundred sixty five days so this is one day and that's the definition of a day right and this is one year that's the vision of a year this is called translation rotation in translation OK So that's the idea in I want to quickly do one example here and it says how high above the earth was my age my height most you place a satellite so that it is constantly flying to Brett Lee over the same spot this is called This is where juice increase orbit is right flying directly over the same spawning like wow that's juice you can it's orbit now let's you know remember the word it doesn't matter as long as you remember the most important part which is that the period of the satellite is the same as the poor of the earth so in this case we're talking about a satellite picture of the earth around itself OK satellite goes around the same time that it takes for the earth to go around self so we're talking about the period of the earth around the earth which is twenty four hours I can use hours so I have to convert and this gives me eighty six thousand and four hundred seconds so I know the period now I want to know how high the satellite has to go OK the rest of this guys it's really very similar to this problem we worked out over here but let's do it real quick you're basically going to use. Because to find age you have to first get little or or. Is big or plus page but I'm first going to get lore and then I'm going to get my age so this means that age will be little or minus bigger so the first thing I have to find little age if I want middle age the distance and I have T.V. I mean he said equation which is this one.

And remember you can rewrite this in this green version over here so two square equals four pyrite squared or a cube over G.M. and now I'm looking for our SO Our if I move everything around them is kind of quickly because we did earlier I end up with this stuff up here OK So T. squared G M over for power squared and it's the cube root of this I'm going to show at this time a little bit different but it's there cube root of this whole thing if you want to plug the sim it's the same thing as before except now your time is eighty six you plug this in and you should get a four point twenty two times ten to the seven and this allows you to find your age H. equals or which is four to two ten seven minus big or which is the radius of the earth six thirty seven times ten to the six and if you do this the age that you get is three point fifty eight times ten to the seven meters OK that's the final answer and once again if you were if you were to go on Google and look for geosynchronous orbit around the earth just Google geosynchronous orbit around the earth you're going to see that the height of geosynchronous orbit is this one OK this is the height needed for geosynchronous orbit it's going to have a number of very versatile are to this so it's pretty cool you can actually use you know physics one basic basics physics one to figure out where you have to put satellites all spins around the same plane and that's pretty cool I think so anyway I mean if you guys have any questions and I hope you found it helpful.

Two satellites, one in geosynchronous orbit (T = 24 hrs) and one with a period of 12 hrs, are orbiting Earth. How many times larger than the radius of Earth is the distance between the orbits of the two satellites. MEarth = 5.98 × 1024 kg, G = 6.67 × 10–11 N·m2, g = 9.81 m/s2, REarth = 6.38 × 106 m
A. 0.51
B. 2.5
C. 6.6
D. 5.7
E. none of the above.

What is the distance between the Sun and the Earth? Note the mass of the sun is 1.99 x 1030 kg.

The international space station is in orbit about 400 km above the surface of the Earth. For astronauts to appear to be weightless on the ISS, what must the speed of the space station be? Note that the radius of the Earth is 6371 km and the mass is 5.97 x 1024 kg.

A satellite is in a geosynchronous orbit around the Earth, meaning that it always appears above one spot on the Earth as the Earth rotates. What would the altitude of this satellite's orbit need to be? Note that the mass of the Earth is 5.97 x 1024 kg and the radius of the Earth is 6,371 km.

How fast would a pitcher have to throw a 145 g baseball on the moon to put it in orbit just above the surface of the moon? Note that the mass of the moon is 7.34 x 1022 kg and the radius is 1.74 x 103 km.

Planet X has radius 4.00 x 106 m and mass 5.00 x 1024 kg. A satellite is in a circular orbit around planet X. If the speed of the satellite is 5400 m/s, how much time does it take the satellite to complete one orbit?

Your spaceship is in a circular orbit around planet X. The radius of the orbit is 9.00 x 106 m and the speed of the spaceship in its orbit is 2800 m/s. The radius of the planet is 4.00 x 106 m. What is the mass of planet X?

You are a member of a group of scientists who travel to planet Bubba, a planet that orbits a bright star in our southern sky with a mass of 5.62 x 1023 kg. The radius of the planet is Rb = 5.0 x 106 m. A satellite is in a circular orbit around planet Bubba. In its orbit the satellite is a distance of 9.00 x 106 m above the surface of the planet. What is the speed of the satellite?

Enter your friends' email addresses to invite them:

We invited your friends!

Join **thousands** of students and gain free access to **55 hours** of Physics videos that follow the topics **your textbook** covers.