Orbital Motion - Physics - Gravity Course (Class 5) | Edouard RENY | Skillshare

Orbital Motion - Physics - Gravity Course (Class 5)

Edouard RENY, Music Producer & Tutor in Physics

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9 Lessons (42m)
    • 1. Orbital Motion - Introduction

      2:21
    • 2. Orbital Motion - Lesson (Video 5.1)

      8:02
    • 3. The Speed of the Earth - Exercise (Video 5.2)

      4:56
    • 4. The Mass of Jupiter - Exercise (Video 5.3)

      3:51
    • 5. The Moons of Saturn - Exercise (Video 5.4)

      8:47
    • 6. The 3rd Law of Kepler - Lesson (Video 5.5)

      6:43
    • 7. Geostationary Orbits - Exercise (Video 5.6)

      3:04
    • 8. A Visit to the Kuiper Belt - Exercise (Video 5.7)

      2:45
    • 9. The 3rd Law of Kepler used in Animations - Application Example (Video 5.8)

      1:11

About This Class

Have you ever wondered why the Moon, although attracted by the Earth, never falls on it? Instead, our Moon describes a circular path around it. How come? This class will make you understand why.

By taking this class, you will gain an understanding of Orbital Motion. Orbital Motion combines the laws of gravity with the principles of circular motion (We have reviewed respectively these topics in Classes 3 and 4 of this course). This concept is detailed in the first video (5.1) where you will learn what is an orbit and realise how the velocity of an orbiting object can define its orbit.

The three videos that follow are training exercises showing fun stuff you can do with orbital motion equations, like determining the speed of the Earth, calculating the mass of Jupiter or exploring the intricacies of the Saturnian system. This latter exercise will allow you to discover a celestial Law: The 3rd Law of Kepler that links the period of revolution of a body to the distance from the massive object it is orbiting around.

To exemplify this, we will fly far beyond Neptune, deep into the Kuiper belt, where dozens of mysterious draft planets live their frozen lives. We will also play closer to Earth, like calculating the height at which a TV satellite must be launched!

Finally, for those interested in animation, Video 5.8 shows how to use the 3rd law of Kepler to design the motion of a bunch planets orbiting a star.

*** Content of the class ***

Video 5.0: Introduction

Video 5.1: Lesson – Orbital Motion

Video 5.2: Training Exercise – The Speed of the Earth

Video 5.3: Training Exercise – The Mass of Jupiter

Video 5.4: Training Exercise – The Moons of Saturn

Video 5.5: Lesson – The 3rd Law of Kepler

Video 5.6: Training Exercise – Geostationary Orbits

Video 5.7: Training Exercise – A Visit to the Kuiper Belt

Video 5.8: Application Example – the 3rd Law of Kepler at the service of animation

All training exercises are provided as pdf files and formatted so that you can print the exercise and work on them away from the computer. Answers are provided also in a pdf document.

 

*** This class is part of a larger course named “Gravity, The Basics”. ***
 
“Gravity, The Basics” explores the elementary notions of Newtonian gravity.

Class 1: “Linear Motion” (because being comfortable with this notion will allow you to make the most of the full course). This class presents a step-by-step technique to solve all motion problems when the acceleration is constant. This class can be taken by itself.

Class 2: “Newton’s Universal Law of Gravitation”, which you can be seen as a doorway to the deeper dive we carry out in the next classes.

Class 3: “Gravitational Fields”, the core of this course: This class teaches first what is a field, and then dives into a deep description of gravitational fields, gravitational field strength and gravitational field lines. It is packed with many exercises aimed at making the student comfortable with these notions.

Class 4: “Circular Motion”, prepares you for the section on orbital motion. You are presented with a detailed explanation of fundamental quantities that occur in circular motion (angles, angular velocity, centripetal acceleration and forces). This class can be taken by itself.

Class 5: “Orbital Motion”, for you to master the motion of bodies in circular orbits around a massive object: This class blends notions taught in class 3 (gravitational fields) with notions presented in class 4 (circular motion), in order to get a good grasp of what is an orbit, and how to manipulate easily orbital motion concepts.

Class 6: “Wrapping-up and Gravity Quiz”. Once you have viewed all 5 classes, it’s time for the exam! After a 5-minute summary of all notions presented in “Gravity, The Basics”, 12 exam-like questions are presented and corrected in detail.

 

*** Level of this class ***

This class is suited for end high school and entry level University students taking Physics. Any person interested in Physics and in need of a refresher about orbital motion and gravity will also enjoy this class.

Transcripts

1. Orbital Motion - Introduction: orbital motion combines the laws of gravity with the principles off circle emotion. We have reviewed these topics in classes three and four of this course to study orbital motion. We will be traveling to planets and stars and former exploration. We will realize that a law emerges. The Third Law of Kepler, the third law of key player links the period of revolution of a planet to the distance from its star. We will dive into orbital motion of yes on formal lessons and many interesting exercises for you to train off. When writing these exercises, I tried to make them as interesting as possible. For example, we will fly far beyond Neptune deep into the Cuba belt where dozens of mysterious dwarf planets lived there. Frozen knives. We will also visit the moon's off the giant planets of our solar system. For example, Lancelot, a moon of Saturn that hides under its quest of ice and ocean and maybe even primitive extraterrestrial life. This class is part of a global course named Gravity. The basics it is actually the last class was for more lessons. After this, the quiz class will give you a chance to test your newly acquired skills on were closed. This course the level of this class is end of high school, so it is a great tool for high school students taking physics and preparing exams. It is also suited for any person that wishes to revive their knowledge and gravity and orbital motion. Now tie to new see details on Enjoy the Life. 2. Orbital Motion - Lesson (Video 5.1): imagine a man on the top of a very high mountain on earth the foreseeable, always entirely with a moderate speed. The ball feels the force of gravity plea down towards the surface, or more precisely, to more the center of the earth. And at the same time, the ball covers a certain always gentle distance. In a second experiment, the same man throws a ball with a larger speed. If the earth was flat, the ball would spend as much time in the air as in the first case, because the acceleration downwards is independent from the office rental speed. But the earth is not flat, the curvature of the earth's resources, the ball staying longer in the air, thus covering a larger distance. The man throws 1/3 bowl with an even larger speed. The effect of the curvature of the earth is even more pronounced. Now the ball spends enough time in the air to travel a large fraction off the perimeter off the planet. I suppose now that the man throws the ball, it a very large speed. The ball experiences continuously the force of gravity. Purple Nikola trades motion, putting it to walls of ground However, it's always until speed is so large that it never can reach the ground due to the curvature of the earth. The board never has enough time to lose altitude needed to reach the surface. The tragic tree of the ball is now describing a perfect circle around the earth, called the Air Force. The force of gravity. The acceleration of the ball is a wave perpendicular to its motion. The ball is now in circular motion around the earth. It is a circular orbit. Now think about the national In In orbiting station, both the astronaut and a space station line free fall with the same acceleration due to gravity. This is why the astronaut appears to be floating around the station even though the force of gravity is acting with him. Imagine that you have a planet off Mass begin around which there was a satellite in orbit. I satellite here off mass Little. The distance between the satellite on the center off the planet off mass began is off, as we have seen in order for a satellite remain in orbit, it needs to have a certain velocity. So it isn't circle emotion say it has also essentially little force, which will be equal to. And these questions are what is this force? But this forces a gravitational force between the two masses. So we can just write the universal law of gravitation on equated to the mathematical form off a centripetal force and B squared of law. Because the force of gravitation is creating a circle emotion, it is a centripetal force. Now look at this. We have a pretty interesting relationship here on you. See that we can do some simplifications. We can hear the EMS and kill one of the girls. Giving you b squared equals G m of all. And if I take out the square square with this, I get the magnitude of the velocity needed. It was a satellite to remain in a circular orbit at a distance r from the center of the object it is in orbit around. This is a very interesting relationship. Can't, for example, allow us to calculate the speed of the moon? Yeah, Let's do it. Actually, I'm just gonna put the data and you do it. I will write down here the formula for the velocity required for an object to be in orbit around a planet n at a distance off from its center, and now we present the earth on the moon in orbit around the earth of one. The distance between the two centers is 384,000 kilometers, and you might need the math of the earth also, which is 5.98 by 10 to 24 kilograms. So find the velocity of the moon, and why not? Also the angular velocity of them pulls a video. So what did you find? Let's calculate this for V. Let's just plug in the numbers GM R 6.67 by 10 to the minus 11 but applied by the massive via kilograms divided by the distance between the centres off the moon and the earth in meters. So few 184,000. I tend to the three and this gives me V equals 1000 and 19 meters per second that they can round up to 1.2 kilometers the second. And what about the angular velocity of the move? The moon isn't circle emotion. Seven. News V equals omega are and find Omega, which is V of R. So 1000 and 19 divided by people 84 by 10 to the eighth, which is in meters. And I find 2.65 by 10 to the minus six. Reagan's second. And why stop now? I could find other interesting things. Like, for example, the lengths off the lunar cycle. Yeah, the period of revolution off the moon. I have only got so I need space here. Let me the hell. I know that omega is two pi over tea, so t would be to pie of what? Omega. So to pie. Divided by 2.65 by 10 to the minus six. And let me grab my calculator because this was not planned. So I got to buy pie divided by 2.65 10 to minus six, giving me in seconds 2.37 by 10 to 6 seconds, which doesn't mean much to me. So I'm going to put it days by dividing this number by 24 then by 3600 and I get 27 point full the bears, if you're still here following this course, this means that you gained quite some knowledge in gravity and circular motion. So I invite you to tighten your seat belt and join me in the next episode? Yes. We will be visiting the country systems of Jupiter and Saturn and thanks to our knowledge, will be able to calculate some off their characteristics. 3. The Speed of the Earth - Exercise (Video 5.2): and exercise appears on the screen, pulls a video and work on the question. When you are ready to resume the video on view the collection, as we all know the Earth revolves of understand. We are given the math off the sun, the math off the earth as well as the distance between the Earth and the sun. We are asked to calculate the linear and angular speeds off the earth of understand. This is a direct application off the formula we have seen in the previous video. This one, the is speed off. The object in orbit G is a universal gravitational constant and is a mass around which the object is norm It and the is a distance between two between here, the sun and the earth. So let's plug in the numbers the mass around which the body is in orbit. Well, here's the earth around the sun. So here is the mass of the sun, and I divide by the distance between the tooth in meters. So here is 150 million kilometers 1,000,000,000 meters on the result, I find when a plug this in the calculator is 29,747 meters per second. Have you seen if you configures three Therefore it will be 29.7 kilometers per second. The speed of the earth around the sun is around 30 kilometers per second. We are also asked to determine the angular speed of the earth around the sun. The earth is in circular motion so we can use circle emotional questions like this one. Which leg? The linear speed, the radius of the circle emotion on the Anglo speak so we can rearrange and like in numbers V will be the speed we just calculated linear speed. On off will be the distance between the sun here 150 bilion meters on we find 1.98 by 10 to the minus seven virgins per second question be were asked calculate the length of an earth year in days. This is a cycle motion. A circular motion is a psychic motion so we can use Saglik equations that we have seen in previous videos like for example, this one. There's a direct relationship between the angular speed and the period of the cycle. So here the time it takes for the earth to go around the sun. That is a year. So we can you arrange and plug in numbers. T is two pi divided by 1.98 by 10 to the minus seven legends for second and I find it answer in seconds, which is 3.168 by 10 to the seven seconds this is the duration of a year in seconds. I wanted 10 days, so I just need to divide this number by the number of seconds in a day that would give me so 3.168 10 to 7 seconds, divided by 24 hours in today, 60 minutes and now and 60 seconds in a minute on the also I find is 367 days. It's not 365 days because, well, the data provided his only on three significant figures to on We rounded a little bit during the calculations. 4. The Mass of Jupiter - Exercise (Video 5.3): you is the first moon off Jupiter that's been discovered. I think it was by Galileo in 16 10 something like this. It is also the closest off the big moons of Jupiter, So it's just 422,000 kilometers away from the center of Jupiter. You actually can see you in the sky. You take a pair of good be Nicholas for them to Mars Jupiter, and you will see little dots of on Jupiter. Neither the moon's you back at least two and ghani net. If you observe quite regularly these moves, you should even be able to get a good idea off the period of revolution of them. So for your it's 1.77 days, and if you know the magnification of your binoculars, you should be able to get a night idea of the distance between you on Jupiter, not a three scene. If you could figures. Of course, that kind of an idea. Now what's amazing is that with this data the distance between you and Jupiter and the period of revolution of fuel, you are able to calculate the mass of Jupiter. Now it's very difficult to determine the mass of Jupiter cannot just take Jupiter in the bathroom, put it on the balance and check out. That's right. No, you need to find alternative fruits. It might be easier to determine the mass of Jupiter by looking at the influence off the gravitational field it creates on the objects which are rounded like its moods to determine the mass of Jupiter. We will be using the linear speed of you as intermediary. We know that. All right, we've seen this equation in the previous video. We're here. This is a linear speed of your M is the mass of Jupiter and Isa distance between between you. That's feed is distance of the time. Your covers a full perimeter in one period. That is, in 1.77 days. So we can write two pi r divided by t. We don't see we know Argies a constant. We can find that you just rearrange the equation Web. So for that I was square both sides and then we arranged to find him. GTE's quit and Clegg in numbers for buys. Glad but guy by off Cuba's size gonna be 422 1000 kilometers. It's 1,000,000 meters chipped. If I did by the universal gravitation constants on the periods quit line days in second period. So it's 1 77 multiplied by the number of seconds in the day. And it's squid and I find for Jupiter 1.9 by 10 to the 27 kilograms I checked on Google. This is correct. 5. The Moons of Saturn - Exercise (Video 5.4): Ancelotti's is an amazing that moon of Saturn. Jay's is of salty water emerges from its surface that makes us believe that under the crust off ice there is a liquid ocean. And actually many believe that it is one of the most promising places in the solar system where we could find extra terrace for life. Probably very primitive bacteria stuff, but pretty interesting. Nonetheless, I think they even plan to send a probe of their that would fly within the JSA and collect some samples on allies them on the spot for organic molecules and maybe more. Anyway, back to our exercise. We know that produces in circle emotional wild Saturday. We have requested to draw its acceleration and velocity vectors so it is in circular motion , say it will have an acceleration towards the center of the circle on. The velocity will be perpendicular to the acceleration that is tangent to the path question being prove that for any moon in orbit around the planet off mass n, the product of the square with angular speed by the Cuba of its orbital radius is constant . This means that we want to prove that only ask read our Cube did a constant Let's start with the universal law of gravitation, the way they would idea to start with this. Oh, but yeah, that's a force announcing, but answer that is in circular motion. So this is also a centripetal force that doesn't have me too much, because I don't feel omega out that there's another mathematical form for centripetal force that uses the angular speed. Yep, you got it. I've got only got Go ask. So let me read by this and other same time killed the massive on sale at I'm done. If I put the Ark square on this side, I get when we get Square Cube equals G. M. G is a universal, very additional constant and and the mask for Saturday. In that case, we don't change. So this is also a constant for this situation. Problem solved. The next question is about tighter. Titan also is quite an amazing moon. It's actually, I think, one of the biggest in the solar system and is the only one that has an atmosphere and quite an atmosphere. It's actually more dense than the atmosphere is not made off carbon dioxide and water line , like our close here. It's made of methane and ethane, and some of it is liquid on the surface. So basically, you have a brain system on Titan. It actually rains, not water. But if a man and methane and you got lakes and rivers, Actually, they sent a probe. You know, sometime ago, Cassidy, who again and again has landed and took pictures. And when you look at the pictures, it's quite amazing. There's coasts. There are rivers. Uh, I don't know. It looks like the coast of France a bit. It's They read that Google title. If you want to know about it, it's quite amazing Moon. Anyway, in our case, we are given the distance of Titan to Saturn as well. At the time it takes for one revolution a while Saturn, 15.9 days. And we are asked to determine the naso Saturn using the form that we developed so well. The massive section would be this so we can just revenge it for the mass of Saturn and equals only get squared on Cuba tree. We can't cabinet on the best crab because we know the period and we know that only guys two pi over tea. So let's just like to pi over teen here yet, um equals two pi of a T squared our cube of a G on. We can plug in the numbers to buy of 15.9 days, so we need to put it in seconds and forget 15.9 with Dr 86,400 above seconds in the day we square this we moved to fly by the distance that titan Saturn Cube and we divide by the universal gravitational constant. Just six points 67 by tens of minus 11 on the answer I got was 5.69 by 10 to the 26 Kimmons family. Significant figures out about three. So I leave. It looks, question d calculate the time taken by oscillators to complete a full orbit around Saturn. We have the distance between us lettuce and Saturn, and we're working for tea and say, Let us Well, we could use this because I have our have the mass of Saturn from the people's question so I could calculate Omega and only guys to pyre over the period. So let's rearrange this equation to get coming. So I only got squid equals G m of All Cube. So I think it's equivalent. I'm playing in the numbers castle Saturn and divided by the distance between Saturn and also Kid and in meters. And I found Omega is 5.38 by 10 to the minus five. Redden's second. But I also know that only guys two pi over the period. So I can you arrange for the period t equals two pi omega a plug this number in and obtaining seconds 118,000 if I wanted to. Days I divide this number by the number of seconds in a day which is 83,400 and I find 1.37 days in this exercise we actually have been proving and then applying the third law of Kepler, we will be discussing the third law of Kepler in the next video. So I do that 6. The 3rd Law of Kepler - Lesson (Video 5.5): combining equations off circle emotion with his equations off orbital motion can present some interesting results. Let me show you imagine a planet of Mass begin around which you have amassed little M, which is in orbit. If I want to know the speed off, the object in orbit can lie down it speed would be equal to the perimeter, divided by the time it takes for the object to go around. So that would be two pi r I'll being the distance I radius was a sort of emotion divided by T period. This circular motion is due to gravity. So I can also like that the speed of the object in orbit is equal to the square it off the universal gravitational constant supplied by the mass around which it is in orbit divided by the distance between the two masses where these two expressions are the expression off the same thing so I can equip them to pay off. Divided by t equals square root off G m of up. I would square both sides in order to get rid of the square root giving me four pi squared r squared over a squared equals G end of off. Now the trick is to constantly all the variables and all the constant. I'm going to put all the variables on one side and all the constants on the other, meaning the R is going to go here and the foot by square is going to go here. So I end up with our tube of a T squared equals G m over four pi squared. This relation is called the Third Law of Kepler. What's so special about it? Imagine that given system with a center of mass began allowing which loss of objects in orbit and certain orbit. You see that do radius off the orbit cubes divided by the period off the orbit squared is a constant. This relation is very interesting for astronomers. If a national number managed to figure out, say, the period of revolution year off planet, he can calculate easily the distance from its star. And he said, Elsa, if a national Norman knows the distance planet from its star, he can create It's you. Let me show you how practical this is to determine the duration of the young mercury in terms off Earth years. We will use the third law of Kepler that states death. The cube of the orbital radius, divided by the square of the orbital period, is a constant. That means that for all planets which orbiting the same star, this ratio is constant. Let's apply it the ratio of Berkeley. We'll be the stay than that for the Earth that we know that the period of the other way just one year we've got the distances. That's fine, the yield monkey. So for that, I will flip this to make it easier to put the period for the orbital emotional mercury as subject. Yeah, and the view, injustice and I should be going to take this quote at the same time and rearrange this even more. I can take this after the square root because it's a square anyway, one. And, uh, this is Alicia coupes, which is also the Cuba the ratio inside the square woods. There's gonna be a power three house, and that's one So yet the period from Mercury enough years will be 58 by 1 50 Power three halfs. Deploy that one now. I don't care about the units here because situation so the units just get canceled and I found 0.24. He is. Do you want to train your new knowledge about the third Law? Kipler Okay, prepare yourself. Take pen. Paper. Calculator on. Let's get started. 7. Geostationary Orbits - Exercise (Video 5.6): geo stationary satellite. It's a satellite that is located always about the same point off the surface of the earth. It's very useful, for example, for TV satellites. We are requested to find out at what height the satellite must be in order to satisfy this condition. The word geo stationary has information in it. Yes, in order for the satellite to remain, have the same twins on the surface of the earth, it needs to turn at the same rate as the earth. So when it carries a full orbit, while the Earth will also have carried a form rotation. So we know actually the period off the orbit of the satellite 24 hours one day now using the third law, keep their we can figure out what would be the distance between satellite on the center of the earth, are given of a T squared equals G end of foot pie, squid. I can view injustice and put his quick hill. And if I want to know Oh, I just take the coup boot that's like in numbers. So the mass of the F and the period would be 24 hours, so we need to put 24 by 30. 600 which is number of second. We need to be in second s I unit, you know, and divide all this by four squared. I found 42 million meters so I'm going to put this in kilometres now. Be careful. This is not the answer to the question. The question is the height. So don't forget to remove from this value the radius of the earth. This is typical off high school exams, you know, and you lose the mark because of this. So that I find 35,900 kilometers In the next exercise, we will travel beyond brutal in the Cuba belt. So prepare your rockets and let's start exploring. 8. A Visit to the Kuiper Belt - Exercise (Video 5.7): I'm pretty sure that many of you think that the solar system contains nine clients or weight +11 being brutal. Okay. The solar system contains probably nine big planets. The Ninth planet is not discovered. It school Planet X and dozens off school of planets of true of planets like Peter. One of them is called home. Bulimia is located 43 astronomical units from the sun, meaning 43 times the distance between the sun and the earth Only I was discovered in 2000 and four by Spanish researchers. Question is assuming that Umea is in circular orbit. In what? Here? Where we find Romeo again at the position weight was discovered. To answer this, we need the orbital period of Julia third low key player to the rescue. Obey to appear of the earth. We're going to use the earth as reference divided by says that squared, divided by the orbital radius of the earth coop. Well, this ratio will be the same for home, and we're looking for This will be a total period of home so we can rearrange this and then take the square root and I plug in the numbers immediately. that that's 43 divided by 1 43 cooped, but then go with it but applied by one. So it's just 40 feet at the power with the house, which is 282 years. So we will see me at the same position that it was discovered in 2000 for plus 282 so in the year 2000 286. 9. The 3rd Law of Kepler used in Animations - Application Example (Video 5.8):