WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I'm Dan Fullerton and in this lesson we are going to talk about gravity and orbits.
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Our objectives include determining the force that one, spherical asymmetric mass inserted on another.
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Determining the strength of the gravitational field at a specified point outside a spherically symmetric mass.
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Describing the gravitational force inside and outside the uniform sphere.
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Calculating how to feel that the surfaced depends on the radius and the density of the sphere.
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Stating Kepler’s laws of planetary motion and using them to describe the motion of an object in orbit.
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Finally, applying, conservation of the angular momentum to determine the velocity and radial distance at any point in that object’s orbit.
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Let us start by taking a look at Newton’s law of universal gravitation.
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The force of gravity, the force of attraction between any 2 objects with mass is given by this formula -G a constant,
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a fudge factor to make the units work out, × the mass of your first object × the mass of your second object ÷ the square of the distance between their centers.
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R hat this means it is in the direction of r hat.
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If we were to draw this, if we have object 1 and object 2 here, and we define from 1 to 2 as the R vector that means our unit vector in that direction, r hat is that direction.
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The force of 1 on 2 was going to be opposite the direction of r hat, so it is going to attract 2 force of 1 on 2.
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That is where the negative sign comes in.
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Usually, when we are dealing with this, we will worry about magnitudes and use common sense to realize they are attracting each other
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but the negative sign has to do with this defining the direction from 1 to 2 as in the r hat direction.
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Now that fudge factor G, the universal gravitational constant that is equal to 6.67 × 10⁻¹¹ N m² / kg².
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If they want to make a plot of the gravitational force between 2 objects as a function of the distance between their centers,
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we would have this in versus square law, another place where as the distance increases between objects,
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we see the force go down as the square of the distance.
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Let us do a quick sample here, find the magnitude of the gravitational force exerted on the earth by the sun given the mass of the earth,
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the mass of the sun, and the distance between them.
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Let us start the force of gravity, it is going to be - GM 1M2 /r² in the direction of r hat, but since I'm worried about the magnitude,
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the magnitude of the gravitational force is just going to be 6.67 × 10⁻¹¹ N m² / kg².
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G, the universal gravitational constant × our first math, the mass of the earth about 6 × 10²⁴ kg ×
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the mass of the sun about 2 × 10³⁸ kg ÷ the square of the distance between them 1.5 × 10¹¹ N m².
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Put all that in my calculator and I come up with the force of about 3.5 × 10²² N.
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Let us do another example, the diagram below represents 2 satellites of equal mass A and B and circular orbits around the planet.
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Compared to the magnitude of the gravitational force of attraction between A in the planet,
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the magnitude of the gravitational force of attraction between B and the planet is what?
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Notice that B is twice as far from the center of the planet as A is.
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Well, because it is twice as far and we have this inverse square law relationship, where the gravitational force is proportional to 1 /
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the square of the distance between the objects.
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If we double the distance, we get ¼ the force, answer is C because of the inverse square law relationship.
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As we are talking about gravity, let us take a look at the gravitational field.
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Anywhere around the earth or any other object with mass are going to have some force of gravity.
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To help us think about that, to visualize that as we talk about the gravitational field or the force that an object would feel if placed at that position.
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What we are going to define is if the force of gravity is GM 1M2 /r² and if we are uniform gravitational field, we can write that as MG.
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We can do a little simplification here and cancel out one of our masses to find that the gravitational field strength is GM 1 /r².
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That would be the force an object would feel per unit mass.
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As you look at these gravitational field lines, as they our more sparse you have less force.
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As they are more dense, you have a greater gravitational field strength you would have more force per unit mass.
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Now little G, we have been talking about is the acceleration due to gravity.
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On the surface of the earth that is 9.8 m / s².
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If we measure it this way and use these units we are going to come up with N / kg, m/ s² unit is equivalent to 1 N / kg.
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Gravitational field strength g is the same thing as the acceleration due to gravity.
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Let us do a quick example here, if we say that the radius of the earth is about 6,378 km, you can actually solve given the mass of the earth
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and the radius and find g is 9.8 m / s² or 9.8 N / kg.
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Give it a try to confirm that reading.
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Let us do another example, if you weigh 600 N on earth, what will you weigh on a planet with twice the mass of earth and half the radius of earth?
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Of this is a Newton’s law of universal gravitation problem.
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Then what happens if we double the mass?
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If we double the masses, we are going to get twice the force, so we would double the force and you have 2 × what you had originally
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and if you half the radius because of the inverse square law you are going to quadruple the force.
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At the radius and 2 and the denominator and it is squared, you will get 4 × what you have.
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Put those two together, you are going to have an 8 × 8x increase.
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8 × 600 N is going to give us a 4,800 N as your weight on this new planet.
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Inside a hollow sphere the gravitational field is 0 and it is important to remember, inside a hollow sphere the gravitational field is 0.
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Outside the hollow sphere you can treat it as if all of the mass were combined into one massive point at the very center,
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pretend its entire mass is concentrated at the center and then calculate the gravitational field. If we were to do that, we can go and graph it something like this.
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If this is our gravitational field strength and there is our R, inside the radius of the object we have 0
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and outside we have a gravitational field strength which is equal -GM 1 /R² in the direction of r hat.
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What if we look at a solid sphere?
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Outside a solid sphere, we said you can treat the spheres of all the masses at the center of the sphere.
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Inside the sphere, however, you treat the sphere as if the mass inside the radius is all at the center, the only part that counts is what is inside the radius.
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Outside the radius of where you are in that sphere does not count.
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Only the mass inside the radius of interest counts.
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As an example, let us say we want to know the gravitational field strength at some radius inside this solid sphere.
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Let us draw another sphere inside and pretend that we are standing right there, somehow inside the solid sphere.
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To start analyzing this problem, I’m first going to define a volume mass density ρ,
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which is the total mass ÷ the total volume or M / the volume of our entire sphere 4/3 π R³.
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Defining the distance from the center to there is R.
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Inside our sphere of interest here, what the volume enclosed in that sphere of interest is 4/3 π r³.
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Our r is our radius out to the edge of our sphere of interest so then the mass enclosed by our sphere of interest
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is going to be our volume mass density × the volume of our sphere of interest which is going to be, we have for ρ, we have capital M /4/3 π R³.
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For our volume enclosed we have 4/3 π r³.
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Our mass enclosed is going to be, we can factor out 4/3, π, we are going to just get M, r³ /R³.
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Looking for the gravitational field strength, G is - G × our mass enclosed/r², where our r is the distance to the edge to the radius of our sphere of interest
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is going to be - G /r² × our mass enclosed M r³/R³, which is just going to be - GM /R³ × r, where GM and R³ are constants.
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What we have is a linear function of our radius and this is good for r is less than or equal to R.
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If I wanted to make a graph of our gravitational field strength vs. our radius, there is our G and we are going to start at some 0 right at that center point.
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We are going to have a linear increase and then when we get right to R, the edge of our total sphere are going to follow our inverse square log N.
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Here in this region we are proportional to -r and in this region we are proportional to 1/r².
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Let us talk about circular orbits.
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Here we have an object M2 in a circular orbit about object M1.
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The radius of our orbit is R, the force of gravity that is causing the centripetal force allowing it to move in the circle fg,
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at any point it has some tangential velocity V along with a rotational or angular velocity ω.
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As we look at this to analyze that the net centripetal force which is always MAC, M2AC for our object 2 going around object 1,
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that has to equal the force of gravity because the force of gravity is what is causing that.
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We know that the force of gravity by Newton's law of universal gravitation is GM 1M2 /r² and our centripetal acceleration we know is V² /r.
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We could rewrite this as M2 V² /R must be equal to GM 1 M2 / R², which implies then that V²
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must be equal 2 GM1 /R or if we want just the velocity, the speed at any point in time, the velocity is the √ GM 1 /r.
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It is in a circular orbit, the radius and the mass of the planet or whether we are orbiting about determines the speed.
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Notice there is no dependence on the mass of the object that is orbiting.
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Taking this a bit further, we can look at the period and frequency for a circular orbit.
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Once around the entire orbit is the circumference or 2 π r and that has to equal the distance traveled which is going to be the velocity × the period, the speed × the period,
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which implies then that period is 2 π r / V.
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Due to our amazing physics skills, we just found out that the velocity for something in circular orbit is the √GM 1 /r.
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Therefore, we can write that the period is 2 π r / our velocity or speed √GM 1 /r.
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A little bit of rearrangement we can write that as 2 π √r / √GM 1 and all those square roots are making things a little messy.
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If we square both sides, we can write this as T² = 4 π² r² and √r² is going to be an r.
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In the numerator we will have r³ ÷ GM 1, that is known as Kepler’s third law for circular orbits.
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What is pretty cool about this, is if we take the ratio of T² / r³, that is equal to 4π² / GM 1, that is known as Kepler’s constant.
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For all the planets in our solar system, we have got the same M1 so we have roughly, assuming the roughly circular orbits and they are pretty close.
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We have the same ratio of the square of their periods to the radius so that is Kepler’s constant.
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It has a value of about 3 × 10⁻¹⁹ s² / m³.
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We did a pseudo derivation of Kepler’s third law.
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We could also look at the energy for objects moving in circular orbits.
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Our total energy is the kinetic + the potential energy and we know for an object that is moving in a circular orbit we will first offer kinetic energy is ½ M2 V².
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The gravitational potential energy is - GM 1M2 /r, this implies then that we could write the energy as equal to ½ M2 V² - GM 1M2 /r.
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We also know that V² = GM 1 /r so we can write this as E = ½ M2 GM1 /r - GM 1 M2 /r, which is just going to give us - GM 1M2 /2r.
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We have an expression for the total energy for an object in the circular orbit.
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As we are doing our analysis of things in circular orbits, we can also talk about escape velocity.
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Now escape velocity is a concept that describes the velocity an object needs while it is in caught by a gravitational field,
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the amount of speed it needs in order to completely escape the influence of that objects gravity.
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We know practically, you cannot do that because you cannot get infinitely away from an object.
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We can calculate how much energy that would require if you did everything perfectly.
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It is a conservation of energy approach we are going to take here.
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Let us start by looking at the total energy when we hit this point of being completely out of the influence of an object's gravity.
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We will do that at the point where its gravitational potential energy is 0 and right at the point where it is kinetic energy is 0, where this completely stopped.
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Anything past that, you would not need in order to escape that objects gravity that will be left over energy.
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If we set that as our beginning condition and that means that ½ M2 V² - GM 1M2 /r must equal 0 or ½ M2 V² must equal GM 1M2 /r
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or solving for V² that is going to be 2 GM1 /r, which implies then that the escape velocity is going to be √2GM1 /r.
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Since we have been talking about all these orbits, let us get into Kepler’s laws.
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Kepler’s first law of planetary motion states that the orbits of planetary bodies or ellipses with a sun at 1 of the foe side of the ellipse.
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We have 2 foe side, the sun is one of them, there is our first planetary law by Kepler.
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The second one gets a little bit more mathematically in depth.
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It says if you were to draw a line from the sun to the orbiting body, the body would sweep out equal areas along the ellipse in equal amounts of time.
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That means, let us assume that we have an object in orbit, right now it is at 0.1 and at some point it gets here to 0.2.
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As it does that, it sweeps out this amount of area, we will call that area 1.
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At a later point in time, it is at P3.
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If we let it go at the same amount of time that elapsed between P1 and P2, it will move from P3 to P4 and it will sweep out that this area of the ellipse.
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As long as the time between P1 and P2 is the same as the time between P3 and P4, the area swept out in each of these cases is the exact same.
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That is Kepler’s second law of planetary motion.
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If we take a look at his third law and we also did this for circular orbits, it is a bit more complex.
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It says the ratio of the squares of the periods, squares of the periods of 2 planets is equal to the ratio of the cubes of their semi major axis.
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Instead of r³, we are doing this for a circular path, now we are going to use the semi major axis to find here for the ellipse.
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This would be for an elliptical path T² /a³ = 4 π² / GM 1 or you might see this written as T² = 4π² a³ /GM 1.
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The ratio of the squares of the period for the cubes of their semi major axis that is referred to as Kepler’s constant again.
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There is Kepler’s third law of planetary motion, this time for elliptical orbits.
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Let us take a look at total mechanical energy for elliptical orbit.
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We are not going to go through the entire derivation, we did it for a circular orbit.
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This would get a little more complex but the total energy is the kinetic + the potential which would is ½ M2 V² - GM 1M2 /r, when you are talking about circular orbits.
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We get to elliptical orbits, when we do all the math and simplify it, we come out with the total energy as - GM 1M2 /2a, that is semi major axis instead of the radius of the circle again.
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How about the velocity and radius from elliptical orbit?
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As we look at this one, we have an object M2 and elliptical orbit about M1 and we are going to look into the couple points here.
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As we analyze it, we can take a look starting with the derivative of the angular momentum about point P with respect to time has to equal the net torque about point B.
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Assuming we have no external forces causing any net external torques, that has to be 0.
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Therefore, our angular momentum about point P, I should say equal must be concert.
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That means that the magnitude of the angular momentum about point P which will be at different points, if we look at that point A for example,
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that will be M2 velocity of A, radius at A, which would be the semi major axis × the sin θ A has to equal the same thing when it is at point B, M2 VB RB sin θ B.
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And because the mass is the same, the masses cancel out, VA RA sin θ = VB RB sin θ B and that is true for any of the points on the ellipse.
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Specifically, when you are at points A and B, the perigee point and what is known as the apogee point, at those points θ A = θ B those are 90°,
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which implies then that VA RA must equal VB RB, my simplification at those specific points.
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Let us finish up with a couple problems here.
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A rocket is launched vertically from the surface of the earth with an initial velocity of 10 km /s, what maximum height does it reached neglecting air resistance?
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Some things that might be helpful, the mass of the earth, the radius of the earth, and we may not assume that the acceleration due to gravity is constant,
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it is going to be moving so far away from the surface of the earth.
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The g is going to be changing in this problem.
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We can solve this with the conservation of energy approach.
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The initial kinetic energy + the initial gravitational potential energy must equal the final kinetic + final gravitational potential energies.
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This implies then that ½ × M2, the mass of our object × the square of its velocity - GM 1M2 / the radius of the earth.
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Its gravitational potential energy when it is on the surface of the earth because now our sending arbitrary 0 point is infinity has to equal to,
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its highest point kinetic is 0 and its gravitational potential energy will be - GM 1M2 /r.
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So then this implies, if we solve for the square of the velocity V² - 2 GM 1 /E = -2 GM1 /R.
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Notice that M2 has completely have been eliminated from the equation but we kind of expected it.
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A little bit more rearrangement, 1 /r = -V² / 2 GM1 + 2 GM 1/2 GM 1 RE, radius of the earth which implies that
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1 /r = - V² / 2 G M1 + 1 / the radius of the earth,
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which implies that 1 /r = we have our negative square the initial velocity 10,000 m / s² ÷ 2 × G 6.67 × 10⁻¹¹ N m² / kg² × the mass of the earth
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6 × 10²⁴ kg + 1 /the radius of the earth 6.37 × 10⁶ m, which implies then that r is approximately equal to 3.12 × 10 ⁺7m.
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Now it is important here to realize that, that is the total distance from the center of the earth.
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If we want to know the maximum height it reaches, how far above the surface of the earth, what we have calculated here,
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if there is earth, there is radius of the earth, we have calculated how high it goes there.
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We are really want to know this distance so that means we have to go a little bit further, the height above the surface of the earth,
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let us call that H = the radius - the radius of the earth which is going to be our 3.12 × 10⁷ m - the radius of the earth
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6.37 × 10⁶ m or 2.48 × 10 ⁺7m above the surface of the earth.
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Let us finish out by doing an old AP free response problem.
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We are going to start with a 2007 exam free response question number 2.
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We are looking at the Mars global surveyor, entered its final orbit about Mars and gives us some data about that,
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we are asked to calculate the radius of the global surveyor's orbit.
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For part A, it is going in some orbit where we have a radius of Mars of 3.43 × 10⁶ m /s, I should probably should put that in red for Mars.
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The period it tells us is 118 min or 7080 s, our mass is 930 kg and our velocity is 3400 m/s, calculate the radius then.
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Period is the distance traveled ÷ the velocity so that is going to be 2 π × the radius ÷ the speed of which implies then
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that r is going to be equal the VT / 2 π which is 3400 m /s × our period 7080 s / 2 π or I come up with a radius of our orbit of about 3.83 × 10 ⁺6m.
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Looking at part B, calculate the mass of Mars.
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We can go back to our Newton’s second law to write that the net centripetal force which is going to be MV² /r must equal the gravitational force,
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Newton’s law of universal gravitation because the gravitational force is providing that centripetal force GM 1M2 /r²,
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which implies then that V² is going to be equal 2 G × the mass of Mars ÷ R,
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which implies then that the mass of Mars must equal the radius × to square root of velocity ÷
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the gravitational constant which is 3.83 × 10 ⁺6m, we just found that out up there.
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The square of our velocity is going to be 3400 m /s² ÷ the universal gravitational constant 6.67 × 10⁻¹¹ N m² / kg².
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It comes out to be about 6.64 × 10²³ kg.
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For C, it is asking us to find the total mechanical energy of the global surveyor in that orbit.
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Our total energy is the kinetic energy + the potential energy which is going to be ½ MV² – G mass of the satellite,
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Mass of Mars over the radius which will be ½ × our mass 930 kg × our speed 3400² – G 6.67 × 10⁻¹¹ ×
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the mass of our satellite 930 × the mass of Mars 6.64 × 10²³ ÷ 3.83 × 10⁶.
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I come up with a total energy then, when I put that in my calculator of about -5.38 × 10⁹ joules.
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Let us take a look at D, if the global surveyor was replaced in the lower circular orbit, will the new orbital period be greater than or less than a given period?
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That is definitely going to be less than.
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The velocity must go up, the shorter circumference to travel so the distance it is traveling is going down.
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And if velocity is displacement /time, distance traveled / time, and time is displacement/ velocity.
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We have a distance going down and the velocity going up, both those factors lead to a shorter time period /orbit.
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I might explain that out in a little more detail and sentences but that is the idea,
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And for part E, we have a question about the global surveyor entered an elliptical orbit, if the speed of its closest approach was 3400 m/s,
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find the speed at the furthest point of its orbit.
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We can do that by conservation of spin angular momentum and state that the mass of our satellite × the velocity position 1 × the radius of position 1 =
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the mass of the satellite × the velocity of position 2 × the radius of position 2 or V1 r1 = V2 r2.
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If we want the velocity at that new point, V2 is just V1 r1/r2 which would be our 3400 m/s × our radius 3.71 + 34.3 × 10⁵ m.
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I'm just going to use a short hand, we do not need to put all of the thousands in there since they are all going to factor out here, over 4.36 + 34.3 × 10⁵ m.
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We will have this ratio × 3400, gives us a velocity of about 3343 m /s.
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Hopefully, that gets you a great start on gravity and orbits.
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Thank you for joining us today at www.educator.com.
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I look forward to seeing you again real soon and make it a great day everybody.