WEBVTT physics/ap-physics-1-2/fullerton 00:00:00.000 --> 00:00:04.000 Hi, folks. I am Dan Fullerton and I would like to welcome you back to Educator.com. 00:00:04.000 --> 00:00:07.000 Today's lesson -- gravity and gravitation. 00:00:07.000 --> 00:00:16.000 Our objectives are going to be to utilize Newton's Law of Universal Gravitation to determine the gravitational force of attraction between two objects. 00:00:16.000 --> 00:00:27.000 We are going to determine the acceleration due to gravity near the surface of the earth, calculate gravitational field strength and explain apparent weightlessness for objects in orbit. 00:00:27.000 --> 00:00:29.000 So with that, why not dive right in? 00:00:29.000 --> 00:00:36.000 Universal gravitation -- All objects that have mass attract each other with a gravitational force. 00:00:36.000 --> 00:00:39.000 For example, right now you are attracted to me. 00:00:39.000 --> 00:00:47.000 Yes, I know, that is kind of creepy, but any two objects that have mass, no matter how far apart they are, all have some level of attraction. 00:00:47.000 --> 00:00:59.000 The bigger the masses, the more the attraction and the closer the masses are to each other, the closer the attraction, which is why we have a very, very, very, very tiny amount of attraction between us at the moment... 00:00:59.000 --> 00:01:05.000 ...or probably a long way away, our masses are relatively small and there is not much gravitational force there. 00:01:05.000 --> 00:01:16.000 Between you and the earth, for example, the earth has a very big mass and you are relatively close to it, so you have a very measurable gravitational force of attraction there. 00:01:16.000 --> 00:01:30.000 If we wanted to look at this in terms of our math, the force of gravity is gm1m2/r² in the direction of our hat and the negative just says that it is an attractive force. 00:01:30.000 --> 00:01:34.000 If we have one object over here -- let us call this mass 1 -- over here, we have some object, mass 2, and the distance between their centers of mass, we are going to call (r). 00:01:34.000 --> 00:01:49.000 In this case, (r) is not specifically a radius; it is a distance between the two centers of mass. 00:01:49.000 --> 00:02:04.000 Then you are going to have a force of Object 2 on Object 1 and you are going to have a force of Object 1 on Object 2 and they will be equal in magnitude and opposite in direction. 00:02:04.000 --> 00:02:08.000 We know that because of Newton's Third Law. 00:02:08.000 --> 00:02:19.000 If we wanted to get just the magnitude of the force, which is typically how this relationship is used, we say that the force of gravity is equal to mass 1 × mass 2/r². 00:02:19.000 --> 00:02:38.000 If you do that and your masses are in kilograms and your distance is in meters, the units do not work out to anything overly useful, so we put in this fudge factor, this universal gravitational constant (G). 0152 That is equal to 6.67 × 10^-11 Nm²/kg². 00:02:38.000 --> 00:02:42.000 It is there to make the units work out. 00:02:42.000 --> 00:02:48.000 So, how do we calculate g, the acceleration due to gravity? 00:02:48.000 --> 00:02:51.000 Let us see if we cannot use what we know to find out. 00:02:51.000 --> 00:02:59.000 The mass of the earth is approximately 6 × 10^24 kg and its radius is about 6.38 million meters. 00:02:59.000 --> 00:03:06.000 The force of gravity -- we typically write -- is mg in a constant gravitational field. 00:03:06.000 --> 00:03:14.000 Universal Gravitational Law says that G times the first mass times the second mass divided by the square of the distance between them. 00:03:14.000 --> 00:03:20.000 Over here, we are assuming the second mass is already the mass of the earth. 00:03:20.000 --> 00:03:22.000 Let us rearrange this a little bit. 00:03:22.000 --> 00:03:27.000 What we can do is realize that we have the mass of the object here. 00:03:27.000 --> 00:03:43.000 That is a mass of the object, so we are left with the mass of the earth, therefore g equals G, that constant, times the mass of Object 2, the earth, divided by the square of the distance between the objects between their centers of mass. 00:03:43.000 --> 00:04:04.000 Therefore, g = 6.67 × 10^-11Nm²/kg² × the mass of the earth -- 6 × 10^24 kg divided by the square of the distance between them -- 6.38 × 10^6 m -- roughly the radius of the earth. 00:04:04.000 --> 00:04:07.000 Do not forget to square that. That is a big mistake that students make. 00:04:07.000 --> 00:04:19.000 Go through that and you should get an answer right around 9.8 m/s² or 9.8N/kg; the units are equivalent. 00:04:19.000 --> 00:04:24.000 Of course, that is what we expect. That is the acceleration due to gravity we have been using here on earth. 00:04:24.000 --> 00:04:33.000 For the AP test, we typically round that to 10 to make the math a little simpler but you can see that it works out. 00:04:33.000 --> 00:04:39.000 The force of gravity decreases with the square of the distance between the centers of the masses. 0277 This is called an inverse square law. 00:04:39.000 --> 00:04:41.000 The force of gravity is gm1m2/r². 00:04:41.000 --> 00:04:49.000 We are going to see lots of relationships in physics that have this inverse square relationship based on the distance between them. 00:04:49.000 --> 00:04:58.000 As the distance gets bigger, the force gets smaller and it gets smaller by the square of that distance between the objects. 00:04:58.000 --> 00:05:08.000 Graph of force versus distance is distance gets bigger -- the force tails off very, very quickly. 0304 That distance is an important factor because it is squared. 00:05:08.000 --> 00:05:13.000 So this graph would be proportional to 1/r². 00:05:13.000 --> 00:05:40.000 So then the question then, what happens to the force of gravity if you double the distance from the centers of mass? 00:05:40.000 --> 00:05:44.000 Let us take a look at how we could answer that. 00:05:44.000 --> 00:06:01.000 If the initial force of gravity Fgiinitial is gm1m2/r², the final is going to be gm1m2 over... 00:06:01.000 --> 00:06:23.000 We are going to double that distance, so this becomes 2 times whatever our initial was squared and that becomes gm1m2/initial r², but the 2 is squared there too over 4. 00:06:23.000 --> 00:06:42.000 So if you rewrite this a little bit, you could write this as 1/4 × gm1m2/r², but notice this is the force of gravity initial. 00:06:42.000 --> 00:06:52.000 So the final gravitational force is 1/4th the initial gravitational force. 00:06:52.000 --> 00:06:57.000 If you double the distance, you get 1/4th the gravitational force. 00:06:57.000 --> 00:07:00.000 If you halve the distance, you get 4 times the gravitational force. 00:07:00.000 --> 00:07:06.000 If you triple the distance between objects -- 1/9th the gravitational force. 00:07:06.000 --> 00:07:10.000 If you cut the distance between them into 1/3rd -- 9 times the gravitational force. 00:07:10.000 --> 00:07:22.000 Whatever that factor is that you change the distance by, you square it in order to find out what happens with that new force. 00:07:22.000 --> 00:07:26.000 Here are some problem-solving hints as we go through a lot of these gravity problems. 00:07:26.000 --> 00:07:30.000 Try and substitute values in for variables at the end of the problem only. 00:07:30.000 --> 00:07:38.000 Because you oftentimes have some pretty unwieldy numbers, the longer you can keep the formula in terms of variables, the fewer opportunities there are to make mistakes. 00:07:38.000 --> 00:07:50.000 Secondly, before using your calculator to find an answer, it is oftentimes valuable to try and estimate the order of magnitude of the answer. 00:07:50.000 --> 00:07:55.000 We will have to go through and calculate the whole thing but try and get a guess as to roughly where your answer is going to be and that way, if you make a goofy calculator error, it is pretty easy to pick up. 00:07:55.000 --> 00:08:03.000 Finally, once your calculations are complete, take a second to make sure your answer makes sense by comparing your answer to some sort of known or similar quantity where you can. 00:08:03.000 --> 00:08:20.000 If your answer does not make sense, stop, take just a second and see if you made a goofy calculator error or math mistake because lots of the problems I see are not with the physics here, it is with making goofy mistakes on calculators and calculations. 00:08:20.000 --> 00:08:33.000 Example 1 -- What is the gravitational force of attraction between two asteroids in space if each has a mass of 50,000 kg and they are separated by a distance of 3800 m? 00:08:33.000 --> 00:08:47.000 The force of gravity -- we are going to worry about the magnitude -- is equal to gm1m2/r², where g is 6.67 × 10^-11Nm²/kg²... 00:08:47.000 --> 00:09:04.000 ...that is given to you for the exam -- × the first mass, m1 (50,000 kg) × the second mass, also 50,000 kg divided by the square of the distance between their centers of mass, 3800 m². 00:09:04.000 --> 00:09:15.000 When I go through and do this, I get an answer of around 1.15 × 10^-8N. 00:09:15.000 --> 00:09:24.000 Why so small a force? You need a very, very, very big mass in order to have an appreciable gravitational force. 00:09:24.000 --> 00:09:36.000 If we wanted to take this problem and do a quick order of magnitude estimation -- just to show you how you have done that -- what I do, is I would look at this expression here and try and estimate it quickly. 00:09:36.000 --> 00:09:46.000 We have 10^-11. We have -- that is something times 10^4, so I would say times 10^4, that is times 10^4 divided by... 00:09:46.000 --> 00:09:57.000 ...Well, those are 10³, so 10³². Okay, 10^8. Then, 10^-11, 10^-3. 00:09:57.000 --> 00:10:07.000 And you have 10^6 down here, so I would say you are roughly talking in the order of magnitude of something in the 10^-9 and look you are only off by a factor of 10. 00:10:07.000 --> 00:10:11.000 You are in the ballpark. You probably did not make a really goofy calculator error. 00:10:11.000 --> 00:10:18.000 So that is how I would do an order of magnitude estimation here. 00:10:18.000 --> 00:10:30.000 Example 2 -- Meteor and earth -- As a meteor moves from a distance of 16 earth radii to a distance of 2 earth radii from the center of earth, the magnitude of the gravitational force between the meteor and the earth becomes... 00:10:30.000 --> 00:10:34.000 We have a couple of different solutions to choose from. 00:10:34.000 --> 00:10:42.000 The biggest problem I see students having with questions like this has to do with reading the question and understanding what it is talking about. 00:10:42.000 --> 00:10:43.000 Let us draw Earth here. 00:10:43.000 --> 00:10:50.000 The meteor starts at a distance of 16 earth radii away, so it is going to be way over there. 00:10:50.000 --> 00:10:56.000 There is its initial position -- 16 r's away. 00:10:56.000 --> 00:11:08.000 Now if this is one (r) right there, then when it is 2 earth radii away from the center of the earth, there is 1 (r), there is the second (r), so it is moving from 16r to 2r. 00:11:08.000 --> 00:11:17.000 The distance (r) is going from 16r to 2r -- the distance is 1/8th -- that's great. 00:11:17.000 --> 00:11:30.000 The first thing I do here, say, is the force going to get bigger or smaller? As it gets closer together, you expect a bigger force. Right away, we can make answer 1 go away. 00:11:30.000 --> 00:11:41.000 Because we have got that inverse square law with distance, our factor is not going to be 1/8th, it is going to be 1/8th squared, which is 164 and we said this is going to be bigger. 00:11:41.000 --> 00:11:49.000 Because the distance is in the denominator, it is going to be 64 × that's great. 00:11:49.000 --> 00:12:08.000 Another way you could do this is you could say the initial gravitational force is gm1m2/16 r², which will be gm1m2/256r². 00:12:08.000 --> 00:12:14.000 What I am going to do is I am just going to take this gm1m2/r² and I am going to call that x. 00:12:14.000 --> 00:12:21.000 So my initial force is going to be 1/256x. 00:12:21.000 --> 00:12:37.000 Now the final gravitational force is gm1m2/2r² which is gm1m2/4r². 00:12:37.000 --> 00:12:44.000 I am going to pull the same trick again and call that x. So that is 1/4th x. 00:12:44.000 --> 00:13:00.000 If we want to know the ratio then -- what happens -- we will take the final gravitational force over the initial gravitational force, which is 1/4th x/256x or 256/4 which is a factor of 64 times larger. 00:13:00.000 --> 00:13:21.000 Which diagram best represents the gravitational forces (Fg) between a satellite (s) in the earth? 00:13:21.000 --> 00:13:24.000 First thing -- gravity only attracts, it never repels. 00:13:24.000 --> 00:13:29.000 So over here in number 1, the satellite is being attracted, but earth is being repelled. 00:13:29.000 --> 00:13:31.000 Nope, that does not work. 00:13:31.000 --> 00:13:33.000 Number 2 -- they are both being repelled. 00:13:33.000 --> 00:13:36.000 Number 3 -- they are both being attracted -- that is looking promising and they are both being attracted with the same force. 00:13:36.000 --> 00:13:51.000 Even more promising, Newton's Third Law says that the force on one must be equal in magnitude to the force on the other just opposite in direction, so 3 must be our answer. 00:13:51.000 --> 00:13:54.000 Let us talk for a minute about gravitational fields. 00:13:54.000 --> 00:13:58.000 Gravity is what is known as a non-contact or a field force. 00:13:58.000 --> 00:14:04.000 We cannot see it. We cannot go touch it. We cannot detect it with a special scope. 00:14:04.000 --> 00:14:14.000 We just know it is there by putting an object there and then seeing what happens to it -- observing the force on some test particle that we would put out in space to see if there is a field there. 00:14:14.000 --> 00:14:27.000 The closer objects are to large masses, the more gravitational force they experience and the denser the force vectors, as shown here, the force that you would see on a test object, the stronger the gravitational force. 00:14:27.000 --> 00:14:39.000 So we could say that the gravitational field is weaker the further away you are if the lines are less dense and stronger as you get closer, where the lines are closer together. 00:14:39.000 --> 00:14:59.000 Now, you can use that the gravitational force or the weight of an object is mg when you are close to earth -- where the change in the radius is negligible or really what we are talking about is a constant gravitational field strength. 00:14:59.000 --> 00:15:09.000 Universally, this one always works -- gm1m2/r², which is why it is called Newton's Law of Universal Gravitation. 00:15:09.000 --> 00:15:27.000 Going a little bit further into this gravitational field strength concept, if the magnitude of the gravitational force is gm1m2/r² and that is equal to m1g, assuming that we do not have a big change in that distance -- that we are in a constant gravitational field... 00:15:27.000 --> 00:15:45.000 ...then in that instance, we could take a look and say that g therefore must equal gm2/r² and the units of that are going to be N/kg or m/s². 00:15:45.000 --> 00:15:52.000 This is what we call gravitational field strength. 00:15:52.000 --> 00:15:57.000 Wait -- you might say -- We have been calling g the acceleration due to gravity. 00:15:57.000 --> 00:15:59.000 Yes, they are the same thing. 00:15:59.000 --> 00:16:11.000 N/kg, m/s², gravitational field strength, acceleration due to gravity -- they are the same thing, just different ways, different approaches of looking at the same phenomenon. 00:16:11.000 --> 00:16:19.000 So those are equivalent -- the acceleration due to gravity and gravitational field strength. 00:16:19.000 --> 00:16:21.000 Let us take a look at an example. 00:16:21.000 --> 00:16:36.000 Suppose we have 100 kg astronaut feeling a gravitational force of 700N when placed in the gravitational field of a planet. What is the gravitational field strength at the location of the astronaut? 00:16:36.000 --> 00:16:58.000 The force of gravity is mg, therefore, we could find gravitational field strength -- the force of gravity divided by the mass or 700N/100 kg, should be 7N/kg or 7 m/s². 00:16:58.000 --> 00:17:03.000 What is the mass of the planet if the astronaut is 2 × 10^6 m from its center? 00:17:03.000 --> 00:17:12.000 To do that, let us go to the Universal Law of Gravitation -- Fg = gm1m2/r². 00:17:12.000 --> 00:17:28.000 If we want the mass of the planet, that is going to be the force of gravity times the square of the distance between their centers of mass divided by G times the mass of our astronaut. 00:17:28.000 --> 00:17:31.000 Our force is 700N. 00:17:31.000 --> 00:17:47.000 Our distance is going to be 2 x 10^6 -- do not forget to square that-- divided by G, 6.67 × 10^-11Nm²/kg² × the mass of our astronaut, 100 kg... 00:17:47.000 --> 00:18:05.000 ...therefore, I come out with a mass of the planet of about 4.2 × 10^23 kg. 00:18:05.000 --> 00:18:11.000 Now, what happens if we talk about gravitational potential energy? 00:18:11.000 --> 00:18:15.000 Two masses separated by some distance exhibit an attractive force on each other. 00:18:15.000 --> 00:18:21.000 They want to move closer together because that gives them gravitational potential energy. 00:18:21.000 --> 00:18:32.000 In a uniform gravitational field, the gravitational potential energy can be found by mg -- the weight of the object times the height, and we will talk about that more when we get to energy and work and a couple of other topics. 00:18:32.000 --> 00:18:43.000 If the height is varying significantly to where we are not looking at a uniform gravitational field, we need something more general, a Universal Law for Gravitational Potential Energy. 00:18:43.000 --> 00:18:50.000 That is -gm1m2/r. What does that minus mean? 00:18:50.000 --> 00:18:59.000 Typically, we assume that potential energy equals 0 when you are infinitely far away from all other objects -- a long, long, ways away, you do not have any other influences. 00:18:59.000 --> 00:19:03.000 Practically, you cannot get there; theoretically, you can. 00:19:03.000 --> 00:19:13.000 If you were to take -- and we have a planet here and we have an object infinitely far away and we bring it closer and closer and closer and closer and closer, it wants to get sucked in -- gravity attracts. 00:19:13.000 --> 00:19:24.000 If it had 0 potential energy way out there -- well, to get it back to the point where it is completely free of this planet's influence, you would have to add energy to free it. 00:19:24.000 --> 00:19:28.000 It is almost like it is in energy debt before it is free, while it is trapped in the gravitational field here. 00:19:28.000 --> 00:19:36.000 That is where the negative sign comes from. 00:19:36.000 --> 00:19:40.000 Let us take a look at how orbits work. 00:19:40.000 --> 00:19:57.000 This is a very interesting discussion problem because lots of folks have seen videos of astronauts and the space shuttle and they are floating around and the question often comes up, "Why are they floating around? They must be weightless." 00:19:57.000 --> 00:20:01.000 No, they are not weightless and to understand that, you really have to know how orbits work. 00:20:01.000 --> 00:20:07.000 We are going to go back to a thought experiment that Isaac Newton proposed many years ago. 00:20:07.000 --> 00:20:16.000 He said, "Let us imagine that we have this hypothetical mountain, huge mountain, so high that at the very top of it, you are above the atmosphere of the earth." 00:20:16.000 --> 00:20:19.000 You do not have any friction because there is no air to slow anything down. 00:20:19.000 --> 00:20:22.000 At the top of this mountain, we are going to place a cannon. 00:20:22.000 --> 00:20:28.000 I know the cannon is not going to work without an atmosphere, but just hang with me for the purposes of the thought experiment. 00:20:28.000 --> 00:20:37.000 While we are up there, if we were to shoot a cannon ball, it is going to follow some projectile path down to the earth. 00:20:37.000 --> 00:20:47.000 But if we shot it a little bit faster, it is going to travel a little bit further as it follows that parabolic trajectory. 00:20:47.000 --> 00:21:02.000 Give it a little bit more velocity, it is going to travel even further, but eventually you are going to come to a point where you shoot it fast enough that at the rate it is falling, it is also falling around the earth because the earth is a circular path. 00:21:02.000 --> 00:21:15.000 Yes, it is constantly falling. It is falling all the time, but it is moving so fast horizontally that by the time it falls, the earth has moved underneath it and it stays at the same altitude above the earth. 00:21:15.000 --> 00:21:18.000 That is what happens in orbit. 00:21:18.000 --> 00:21:20.000 They are not weightless. They are falling. 00:21:20.000 --> 00:21:35.000 They are just moving so fast horizontally that by the time they fall and the earth has moved around underneath them and because the earth is a sphere, they maintain the same altitude. 00:21:35.000 --> 00:21:38.000 Let us take a look and see if we cannot prove that a little bit. 00:21:38.000 --> 00:21:46.000 If the space shuttle orbits the earth at an altitude of 380 km above the surface of the earth, what is the gravitational field strength due to earth at that altitude? 00:21:46.000 --> 00:21:51.000 At what speed does the shuttle have to travel to maintain that orbit? 00:21:51.000 --> 00:21:55.000 Let us start with the gravitational field strength. 00:21:55.000 --> 00:22:03.000 The force of gravity is mg, which equals gm1m2/r². 00:22:03.000 --> 00:22:19.000 Therefore, the gravitational field strength (g) must be g times mass, which is going to be the mass of the earth divided by r², where g, we know is that constant 6.67 times10^-11Nm²/kg². 00:22:19.000 --> 00:22:27.000 The mass of the earth is 6 × 10^24 kg over the distance between their centers. 00:22:27.000 --> 00:22:36.000 To find the distance between their centers -- if this is 380 km above the surface of the earth, we also have to account for the radius of the earth. 00:22:36.000 --> 00:22:57.000 The radius of the earth is 6.37 × 10^6 m roughly + 380,000 m² or about 8.78 m/s² or 8.78N/kg. 00:22:57.000 --> 00:23:01.000 Compare that to 9.8, what we have here on the surface of the earth. 00:23:01.000 --> 00:23:09.000 That is not a huge reduction. There is still an awful lot of gravitational field out there where they are orbiting. 00:23:09.000 --> 00:23:13.000 What speed does the shuttle travel to maintain that orbit? 00:23:13.000 --> 00:23:30.000 To do that one, let us take a look at the force of gravity, which is gm1m2/r² = mv2, mv²/r because it is moving in a circular path -- centripetal force. 00:23:30.000 --> 00:23:48.000 Therefore, the square of our velocity if we rearrange these is going to be -- we have rgm1m2/mr² and I can do a little bit of simplifying here. 00:23:48.000 --> 00:24:00.000 We have rn and r², we have a mass and a mass, so that will leave me with g times the mass of the earth divided by r. 00:24:00.000 --> 00:24:09.000 If that is v², then v itself must be g times the mass of the earth over (r) square root. 00:24:09.000 --> 00:24:33.000 When I substitute in my values, that is 6.67 × 10^-11Nm²/kg², -- mass of the earth is about 6 × 10^24 kg and the distance between their centers, 6.37 × 10^6 radius of the earth + 380,000 m above the surface of the earth. 00:24:33.000 --> 00:24:54.000 The square root of all that and I come up with a velocity of about 7700 m/s or that is greater than 17,000 miles per hour (mph). 00:24:54.000 --> 00:25:05.000 To put that in perspective, that is more than 23 times the speed of sound at sea level. 00:25:05.000 --> 00:25:13.000 That is fast! 00:25:13.000 --> 00:25:15.000 Let us take a look at another example. 00:25:15.000 --> 00:25:28.000 Calculate the magnitude of the centripetal force acting on earth as it orbits the sun, assuming a circular orbit of radius 1.5 × 10^11 m in an orbital speed of 3 × 10^4 m/s. 00:25:28.000 --> 00:25:32.000 Use that to determine the mass of the sun. 00:25:32.000 --> 00:25:36.000 Let us start out with the magnitude of the centripetal force. 00:25:36.000 --> 00:25:52.000 Centripetal force is mv²/r or 6 × 10^24 × our velocity, 3 × 10^4)²/1.5 × 10^11... 00:25:52.000 --> 00:26:02.000 ... which gives me a value of about 3.6 × 10^22N. 00:26:02.000 --> 00:26:05.000 Let us use that to determine the mass of the sun. 00:26:05.000 --> 00:26:20.000 If that is the force, we know gravitational force is gm1m2/r², where one of those is mass of the sun -- one is mass of the earth and that is equal to 3.6 × 10^22N. 00:26:20.000 --> 00:26:33.000 Therefore, we could say the mass of the sun is equal to 3.6 × 10^22N × r²/G × the mass of the earth. 00:26:33.000 --> 00:26:56.000 Or 3.6 × 10^22 or 1.5 × 10^11²/G, 6.67 × 10^-11Nm²/kg² × the mass of the earth, about 6 × 10^24 kg. 00:26:56.000 --> 00:27:08.000 If I plug that all into my calculator very carefully and I find that the mass of the sun is right around 2 × 10^30 kg. 00:27:08.000 --> 00:27:12.000 So you can see we are using the same equations and relationships over and over again. 00:27:12.000 --> 00:27:25.000 The tricky part is keeping all of your values well taken care of, being careful with the calculator -- very fastidious in your calculations. 00:27:25.000 --> 00:27:29.000 Example 7 -- The diagram shows two bowling balls, A and B. 00:27:29.000 --> 00:27:33.000 Each has a mass of 7 kg and they are 2 m apart. 00:27:33.000 --> 00:27:40.000 Find the magnitude of the gravitational force exerted by ball A on ball B. 00:27:40.000 --> 00:28:00.000 The gravitational force is gm1m2/r² where 6.67 × 10^-11 × mass 1 (7) mass 2 (7)/the square of the distance between them -- 2 m²) or about 8.2 × 10^-10N. 00:28:00.000 --> 00:28:13.000 Example 8 -- A 2 kg object is falling freely near earth's surface. 00:28:13.000 --> 00:28:18.000 What is the magnitude of the gravitational force that earth exerts on the object? 00:28:18.000 --> 00:28:22.000 If it is near earth's surface, we can do this one a simple way. 00:28:22.000 --> 00:28:34.000 Force of gravity or the object's weight is mg, which is going to be 2 kg; g is 9.8 or let us round that to 10 to make it easy -- about 20 N. 00:28:34.000 --> 00:28:39.000 Nice, simple, straightforward because it is near the earth's surface. 00:28:39.000 --> 00:28:42.000 Let us do an example finding g. 00:28:42.000 --> 00:28:48.000 What is the acceleration due to gravity at a location where a 15 kg mass weighs 45N? 00:28:48.000 --> 00:29:04.000 Weight, mg = 45N, therefore, g must equal 45N/mass (15 kg) or 3 m/s². 00:29:04.000 --> 00:29:09.000 Just some very simple interpretation problems. 00:29:09.000 --> 00:29:13.000 Let us take a look at a space vehicle on Mars. 00:29:13.000 --> 00:29:19.000 A 1200 kg space vehicle travels at 4.8 m/s along the level surface of Mars. 00:29:19.000 --> 00:29:31.000 If the magnitude of the gravitational field strength on the surface of Mars is 3.7 N/kg -- that is g -- find the magnitude of the normal force acting on the vehicle. 00:29:31.000 --> 00:29:34.000 When I see normal force, right away I start thinking FBD. 00:29:34.000 --> 00:29:44.000 We have the weight down (mg) -- normal force which we will call Fn -- pointing up -- and they must be balanced -- we call that +y direction. 00:29:44.000 --> 00:30:03.000 It is not accelerating spontaneously up off the surface of the planet or going down through it. 1790 Therefore, the net force in the y direction must be 0 and the normal force and mg must be matched, therefore net force in the y direction is the normal force minus mg must equal 0. 00:30:03.000 --> 00:30:25.000 Therefore, the normal force equals the object's weight (mg) or its mass (1200 kg) × g (3.7 N/kg) for a force of around 4,440N. 00:30:25.000 --> 00:30:28.000 Let us take a look at a graphical analysis problem. 00:30:28.000 --> 00:30:34.000 This graph represents the relationship between gravitational force and mass for objects near the surface of the earth. 00:30:34.000 --> 00:30:44.000 What does the slope represent? The slope is rise/run. 00:30:44.000 --> 00:30:51.000 Rise is going to be change in gravitational force and our run is going to be change in mass. 00:30:51.000 --> 00:31:01.000 Change in gravitational force, as long as we are near the surface of the earth is Δmg/Δm and that is just going to give us g. 00:31:01.000 --> 00:31:14.000 Then the slope is the acceleration due to gravity. 00:31:14.000 --> 00:31:18.000 All right. Let us go back to Mars. A 2 kg object weighs 19.6N on Earth. 00:31:18.000 --> 00:31:26.000 If the acceleration due to gravity on Mars is 3.71 m/s², what is the object's mass on Mars? 00:31:26.000 --> 00:31:31.000 I love these questions! They are so simple but meant to trick you and it is so easy to fall into the trap. 00:31:31.000 --> 00:31:37.000 It asks you what is the object's mass on Mars. The mass has not changed. 00:31:37.000 --> 00:31:45.000 The weight may have changed, but its mass is still 2 kg. Do not get suckered into those tricks! 00:31:45.000 --> 00:31:50.000 Your find is the same as your given. 00:31:50.000 --> 00:31:53.000 One more -- Here we have two satellites. 00:31:53.000 --> 00:31:59.000 The diagram shows the two satellites, both of equal mass, A and B, in circular orbits around a planet here. 00:31:59.000 --> 00:32:05.000 Compare the magnitude of the gravitational force of attraction between A and the planet. 00:32:05.000 --> 00:32:09.000 Find the magnitude of the gravitational force of attraction between B and the planet. 00:32:09.000 --> 00:32:18.000 First thing -- since B is further away, it should be pretty obvious that it is going to have a smaller force. Okay? 00:32:18.000 --> 00:32:22.000 Right away -- twice as great, four times as great we can eliminate. 00:32:22.000 --> 00:32:30.000 Because of that Inverse Square Law, we are going from radius (r) to 2r as we are doubling the distance and we must have 1/4th the force. 00:32:30.000 --> 00:32:33.000 The answer is number 3: Inverse Square Law. 00:32:33.000 --> 00:32:36.000 There are lots of different ways you can go through and solve that. 00:32:36.000 --> 00:32:48.000 You can go through and do it analytically or you can make up numbers for them, but the easiest way is if you understand the Inverse Square Law, you can realize right away if the distance doubles, the force becomes 1/4th. 00:32:48.000 --> 00:32:53.000 Hopefully, that gets you a great start on gravity and Newton's Law of Universal Gravitation. 00:32:53.000 --> 00:32:56.000 Thank you so much for your time and make it a great day!