WEBVTT physics/ap-physics-1-2/fullerton
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Hi, folks. I am Dan Fullerton and I would like to welcome you back to Educator.com.
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Today's lesson -- gravity and gravitation.
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Our objectives are going to be to utilize Newton's Law of Universal Gravitation to determine the gravitational force of attraction between two objects.
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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.
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So with that, why not dive right in?
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Universal gravitation -- All objects that have mass attract each other with a gravitational force.
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For example, right now you are attracted to me.
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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.
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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...
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...or probably a long way away, our masses are relatively small and there is not much gravitational force there.
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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.
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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.
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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).
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In this case, (r) is not specifically a radius; it is a distance between the two centers of mass.
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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.
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We know that because of Newton's Third Law.
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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.
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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.
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It is there to make the units work out.
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So, how do we calculate g, the acceleration due to gravity?
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Let us see if we cannot use what we know to find out.
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The mass of the earth is approximately 6 × 10^24 kg and its radius is about 6.38 million meters.
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The force of gravity -- we typically write -- is mg in a constant gravitational field.
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Universal Gravitational Law says that G times the first mass times the second mass divided by the square of the distance between them.
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Over here, we are assuming the second mass is already the mass of the earth.
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Let us rearrange this a little bit.
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What we can do is realize that we have the mass of the object here.
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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.
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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.
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Do not forget to square that. That is a big mistake that students make.
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Go through that and you should get an answer right around 9.8 m/s² or 9.8N/kg; the units are equivalent.
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Of course, that is what we expect. That is the acceleration due to gravity we have been using here on earth.
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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.
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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.
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The force of gravity is gm1m2/r².
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We are going to see lots of relationships in physics that have this inverse square relationship based on the distance between them.
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As the distance gets bigger, the force gets smaller and it gets smaller by the square of that distance between the objects.
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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.
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So this graph would be proportional to 1/r².
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So then the question then, what happens to the force of gravity if you double the distance from the centers of mass?
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Let us take a look at how we could answer that.
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If the initial force of gravity Fgiinitial is gm1m2/r², the final is going to be gm1m2 over...
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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.
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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.
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So the final gravitational force is 1/4th the initial gravitational force.
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If you double the distance, you get 1/4th the gravitational force.
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If you halve the distance, you get 4 times the gravitational force.
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If you triple the distance between objects -- 1/9th the gravitational force.
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If you cut the distance between them into 1/3rd -- 9 times the gravitational force.
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Whatever that factor is that you change the distance by, you square it in order to find out what happens with that new force.
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Here are some problem-solving hints as we go through a lot of these gravity problems.
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Try and substitute values in for variables at the end of the problem only.
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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.
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Secondly, before using your calculator to find an answer, it is oftentimes valuable to try and estimate the order of magnitude of the answer.
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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.
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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.
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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.
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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?
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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...
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...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.
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When I go through and do this, I get an answer of around 1.15 × 10^-8N.
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Why so small a force? You need a very, very, very big mass in order to have an appreciable gravitational force.
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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.
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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...
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...Well, those are 10³, so 10³². Okay, 10^8. Then, 10^-11, 10^-3.
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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.
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You are in the ballpark. You probably did not make a really goofy calculator error.
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So that is how I would do an order of magnitude estimation here.
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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...
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We have a couple of different solutions to choose from.
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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.
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Let us draw Earth here.
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The meteor starts at a distance of 16 earth radii away, so it is going to be way over there.
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There is its initial position -- 16 r's away.
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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.
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The distance (r) is going from 16r to 2r -- the distance is 1/8th -- that's great.
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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.
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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.
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Because the distance is in the denominator, it is going to be 64 × that's great.
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Another way you could do this is you could say the initial gravitational force is gm1m2/16 r², which will be gm1m2/256r².
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What I am going to do is I am just going to take this gm1m2/r² and I am going to call that x.
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So my initial force is going to be 1/256x.
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Now the final gravitational force is gm1m2/2r² which is gm1m2/4r².
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I am going to pull the same trick again and call that x. So that is 1/4th x.
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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.
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Which diagram best represents the gravitational forces (Fg) between a satellite (s) in the earth?
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First thing -- gravity only attracts, it never repels.
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So over here in number 1, the satellite is being attracted, but earth is being repelled.
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Nope, that does not work.
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Number 2 -- they are both being repelled.
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Number 3 -- they are both being attracted -- that is looking promising and they are both being attracted with the same force.
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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.
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Let us talk for a minute about gravitational fields.
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Gravity is what is known as a non-contact or a field force.
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We cannot see it. We cannot go touch it. We cannot detect it with a special scope.
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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.
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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.
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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.
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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.
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Universally, this one always works -- gm1m2/r², which is why it is called Newton's Law of Universal Gravitation.
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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...
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...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.
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This is what we call gravitational field strength.
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Wait -- you might say -- We have been calling g the acceleration due to gravity.
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Yes, they are the same thing.
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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.
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So those are equivalent -- the acceleration due to gravity and gravitational field strength.
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Let us take a look at an example.
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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?
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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.
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What is the mass of the planet if the astronaut is 2 × 10^6 m from its center?
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To do that, let us go to the Universal Law of Gravitation -- Fg = gm1m2/r².
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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.
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Our force is 700N.
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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...
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...therefore, I come out with a mass of the planet of about 4.2 × 10^23 kg.
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Now, what happens if we talk about gravitational potential energy?
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Two masses separated by some distance exhibit an attractive force on each other.
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They want to move closer together because that gives them gravitational potential energy.
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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.
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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.
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That is -gm1m2/r. What does that minus mean?
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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.
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Practically, you cannot get there; theoretically, you can.
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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.
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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.
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It is almost like it is in energy debt before it is free, while it is trapped in the gravitational field here.
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That is where the negative sign comes from.
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Let us take a look at how orbits work.
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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."
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No, they are not weightless and to understand that, you really have to know how orbits work.
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We are going to go back to a thought experiment that Isaac Newton proposed many years ago.
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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."
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You do not have any friction because there is no air to slow anything down.
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At the top of this mountain, we are going to place a cannon.
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I know the cannon is not going to work without an atmosphere, but just hang with me for the purposes of the thought experiment.
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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.
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But if we shot it a little bit faster, it is going to travel a little bit further as it follows that parabolic trajectory.
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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.
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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.
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That is what happens in orbit.
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They are not weightless. They are falling.
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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.
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Let us take a look and see if we cannot prove that a little bit.
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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?
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At what speed does the shuttle have to travel to maintain that orbit?
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Let us start with the gravitational field strength.
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The force of gravity is mg, which equals gm1m2/r².
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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.
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The mass of the earth is 6 × 10^24 kg over the distance between their centers.
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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.
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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.
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Compare that to 9.8, what we have here on the surface of the earth.
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That is not a huge reduction. There is still an awful lot of gravitational field out there where they are orbiting.
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What speed does the shuttle travel to maintain that orbit?
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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.
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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.
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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.
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If that is v², then v itself must be g times the mass of the earth over (r) square root.
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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.
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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).
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To put that in perspective, that is more than 23 times the speed of sound at sea level.
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That is fast!
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Let us take a look at another example.
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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.
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Use that to determine the mass of the sun.
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Let us start out with the magnitude of the centripetal force.
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Centripetal force is mv²/r or 6 × 10^24 × our velocity, 3 × 10^4)²/1.5 × 10^11...
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... which gives me a value of about 3.6 × 10^22N.
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Let us use that to determine the mass of the sun.
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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.
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Therefore, we could say the mass of the sun is equal to 3.6 × 10^22N × r²/G × the mass of the earth.
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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.
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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.
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So you can see we are using the same equations and relationships over and over again.
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The tricky part is keeping all of your values well taken care of, being careful with the calculator -- very fastidious in your calculations.
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Example 7 -- The diagram shows two bowling balls, A and B.
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Each has a mass of 7 kg and they are 2 m apart.
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Find the magnitude of the gravitational force exerted by ball A on ball B.
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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.
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Example 8 -- A 2 kg object is falling freely near earth's surface.
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What is the magnitude of the gravitational force that earth exerts on the object?
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If it is near earth's surface, we can do this one a simple way.
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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.
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Nice, simple, straightforward because it is near the earth's surface.
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Let us do an example finding g.
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What is the acceleration due to gravity at a location where a 15 kg mass weighs 45N?
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Weight, mg = 45N, therefore, g must equal 45N/mass (15 kg) or 3 m/s².
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Just some very simple interpretation problems.
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Let us take a look at a space vehicle on Mars.
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A 1200 kg space vehicle travels at 4.8 m/s along the level surface of Mars.
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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.
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When I see normal force, right away I start thinking FBD.
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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.
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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.
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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.
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Let us take a look at a graphical analysis problem.
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This graph represents the relationship between gravitational force and mass for objects near the surface of the earth.
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What does the slope represent? The slope is rise/run.
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Rise is going to be change in gravitational force and our run is going to be change in mass.
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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.
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Then the slope is the acceleration due to gravity.
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All right. Let us go back to Mars. A 2 kg object weighs 19.6N on Earth.
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If the acceleration due to gravity on Mars is 3.71 m/s², what is the object's mass on Mars?
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I love these questions! They are so simple but meant to trick you and it is so easy to fall into the trap.
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It asks you what is the object's *mass* on Mars. The mass has not changed.
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The weight may have changed, but its mass is still 2 kg. Do not get suckered into those tricks!
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Your find is the same as your given.
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One more -- Here we have two satellites.
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The diagram shows the two satellites, both of equal mass, A and B, in circular orbits around a planet here.
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Compare the magnitude of the gravitational force of attraction between A and the planet.
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Find the magnitude of the gravitational force of attraction between B and the planet.
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First thing -- since B is further away, it should be pretty obvious that it is going to have a smaller force. Okay?
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Right away -- twice as great, four times as great we can eliminate.
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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.
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The answer is number 3: Inverse Square Law.
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There are lots of different ways you can go through and solve that.
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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.
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Hopefully, that gets you a great start on gravity and Newton's Law of Universal Gravitation.
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Thank you so much for your time and make it a great day!