Let's look in more detail at how a satellite is moved from one...

Question

Let's look in more detail at how a satellite is moved from one circular orbit to another. FIGURE CP13.70 shows two circular orbits, of radii r₁ and r₂, and an elliptical orbit that connects them. Points 1 and 2 are at the ends of the semimajor axis of the ellipse. How much work must the rocket motor do to transfer the satellite from the circular orbit to the elliptical orbit?

Accepted Answer

Hello, let's go through this practice problem. Consider a probe of mass 750 kg in a low Venus orbit circular orbit A that needs to move to a higher orbit circular orbit. B as shown in the figure to move from orbit A to orbit B. The satellite might must travel through an intermediate elliptical orbit known as a transfer orbit. Calculate the work produced by the probe's engine to move the probe from orbit A to the elliptical orbit option A 1.44 multiplied by 10 to the power of seven jewels B 2.88 multiplied by 10 to the power of seven C, 1.08 multiplied by 10 to the power of 10 and D 2.16 multiplied by 10 to the power of 10. So this problem is asking us to find the amount of work necessary to move the probe from orbit A the inner orbit to the elliptical orbit, which as we can see from the diagram will involve the probe of being able to reach the radius for orbit B. So one way we can think of this problem is that we're trying to find the amount of energy necessary or the amount of work necessary to move the probe from an inner point point one to the outer radius orbit B at 0.2. So we'll label the inner radius as 0.1 and the outer radius is 0.2 both of which are points along the elliptical orbit. Now recall that according to the work kinetic energy theorem, the amount of work done on a moving object is equal to the change in kinetic energy. And recall the kinetic energy is represented by one half multiplied by the mass of the object in question multiplied by the square of its speed. So the change in kinetic energy for this problem for the probe can be thought of as one half multiplied by the mass of the probe multiplied by the speed of the probe at its final position. So I'll say the speed at 0.2 squared and then we're subtracting the same kinetic energy term but for the inner radius at 0.1 so one half M the sub one squared. So to find the change in kinetic energy and ultimately, the work we need to find the speeds V sub one and V sub two. And we can do this by recalling our other conservation laws specifically the law of conservation of angular momentum and the law of conservation of energy. Let's start by talking about angular momentum. Recall that for an orbital motion, angular momentum L is equal to the mass of the object multiplied by the rotational radius, the orbital radius multiplied by the speed of the object. The law of conservation of angular momentum tells us that this quantity should be conserved. So L sub one should be equal to L sub two or more precisely the mass of the, of the probe at position one multiplied by the smaller radius. R sub a multiplied by the speed at 0.1 is equal to the mass multiplied by R sub B for the outer radius multiplied by the second speed because the mass of the probe isn't changing M is constant. And we can cancel the MS out. And we can solve for either of the speeds with some pretty basic algebra. So solving for V sub one, for example, we just divide both sides of this equation by our sub A. This becomes our sub B divided by our sub A multiplied by V sub two. This gives us one relationship between the two speeds. But we still need to know at least one of the speeds in order to find the other. So let's now talk about the law of conservation of mechanical energy. According to the law of conservation of energy energy should also be conserved. So E sub one is equal to E sub two. And recall that mechanical energy is potential energy plus kinetic energy. So you sub one for potential energy plus case of one hm is equal to the final potential energy U sub two plus the outer kinetic energy K sub two. So now let's expand this out a little bit further using the actual formulas for potential and kinetic energy that are relevant to this case recall that the potential energy of a case like this where we're looking at the gravitational potential energy between two masses. In this case, between Venus and the probe and the potential energy is negative of the gravitational constant. Big G multiplied by the product of the two masses. So I will use capital M to represent the mass of Venus and still using a lowercase M to represent the mass of the probe divided by the radius the distance. So that's our sub A and that's the potential energy between the, between Venus and the probe. When it's at the inner radius, then we'll add the kinetic energy. Once again, it's just one half multiplied by the mass of the probe multiplied by the sub one squared. This is equal to the potential energy at the outer radius plus the kinetic energy at the outer radius. So that's once again, negative big G multiplied by capital M small M divided by R sub B plus the kinetic energy of one half MV sub two squared doing some simplification on this. Now, all the lower case MS will cancel out because it's one is in every single term, these one halves in front of the kinetic energy terms are a little annoying to deal with. So they'll also multiply every single term by two that becomes negative two G capital A A divided by R sub A. And I will also bring the other big G term to the left hand side of the equation so that we can keep all the like variables together. So that becomes plus two big G M divided by R sub B and this is equal to and I'll bring the V squared terms to the left hand side of the equation to the right hand side of the equation by subtracting. So that is V sub two squared minus V sub one squared. And then to simplify the left hand side of the equation a bit more, I'll factor out the two GM and put the one divided by R terms in a parenthesis. So this is equal to two G big M multiplied by one divided by R sub B minus one divided by R sub A. So now that we've got two different equations that give us a relationship between speed one and speed two, now we can solve them together as a system of equations to find those speeds. So because the first equation we found as V sub one on the left hand side, let's substitute that in for V sub one in the second equation. So the left hand side of the equation is V sub two squared minus the square of V sub one. But we found V sub one to be equal to R sub B divided by R sub A multiplied by V sub two. Don't forget to square that. And for the sake of simplification, since we then want to solve for V sub two, let's distribute out the squares and then factor out the V sub two so that we can solve for V sub two. So this is equal to B sub two squared minus R sub B squared divided by R sub A squared multiplied by V sub two squared. And then we can factor out the V sub two squared as V sub two squared multiplied by one minus R sub V squared divided by R sub A squared. And then we can rewrite the term in the parentheses as a more simple fraction by changing the one to be an R sub A squared divided by R sub A squared. So that term becomes you subdue the sub two squared multiplied by rub A squared minus R sub B squared divided by R sub A squared. And this is all equal to the right hand side of the equation which is still two G big M multiplied by. And then we can combine the fractions in a similar way. So instead of writing one divided by R sub D minus one divided by R sub A, we can instead multiply the denominators together to find the least common denominator. So the term in the parenthesis for the right hand side of the equation becomes R sub A minus R sub B divided by R sub A multiplied by R sub B. So now we can see that one of the R sub A's in the denominators will cancel out. And the difference between the two squares on the left hand side with the R A and the RB cancels out with the one on the right hand side. And so bringing the R values over to the right hand side of the equation. We can see that V sub two squared is equal to two GM multiplied by are sub A divided by R sub B multiplied by R sub A plus our sub D using the difference of two squares method. And then from here, solving for V sub two is just a matter of taking the square root. So V sub two is equal to the square root of two G big M multiplied by our sub A divided by R sub B multiplied by R sub A plus R sub B. We can also turn this into an equation for V sub one by plugging it into the earlier equation. We found the earlier relationship we found between the two speeds. So the sub one is equal to R sub B divided by R sub A multiplied by the square root of two GM multiplied by R sub A divided by R sub B multiplied by R sub A plus R sub B. And we can simplify this down a bit further by imagining that the R sub B divided by R sub A outside is squared and inside the square root which causes the mi the R is inside the square root to do a flip. So this whole thing is equal to the square root of two GM multiplied by our sub B divided by R sub A multiplied by, in parentheses, R sub A plus R sub B. And that is our formula for V sub one. So now to find out how much work is done on the probe to move it from the circular orbit to the elliptical orbit. Let's find the speed that the probe would have in the circular orbit and then find the speed that it would have on the elliptical orbit. Using the equation we just found, let's start by finding the speed for the probe while it's in a circular orbit at position one. So V sub one C is what I'll call it. Now recall that for a circular orbit, the speed is equal to the gravitational uh the universal gravitational constant big G multiplied by the mass of Venus divided by the distance the orbital radius. So let's put in our values into a calculator. This is the square root of 6.67 multiplied by 10 to the power of a negative 11 Newton meters squared per kilogram squared multiplied by the mass of Venus or 4.87 multiplied by 10 to the power of 24 kg divided by the orbital radius as our sub A in this case, which is 6500 kilometers, 6500 multiplied by 10 to the power of 3 m. Put that into a calculator and we find a speed of about 7069 m per second. Now, let's do the same thing. But now we're gonna find the elliptical speed at 0.1. So V sub one e using the equation we just found a moment ago. So the square root of two G two multiplied by the constant 6.67 multiplied by 10 to the power of negative 11 Newton meters squared per kilogram squared multiplied by the mass of Venus. 4.87 multiplied by 10 to the power of 24 kg multiplied by the larger orbital radius. So that's 24,000 multiplied by 10 to the power of 3 m. And then we divide by our sub A. So that's 6500 multiplied by 10 to the power of three meters and then multiplied by R sub A plus R sub B. So that's 6500 plus 24,000. Then outside the parentheses multiply by 10 to the power of 3 m. And putting this into a calculator gives us a speed of about 8868 m per second. So now that we have the two speeds, now we're gonna go back all the way back to our first formula for the kinetic energy theorem to find out the change in energy. So that's one half multiplied by the mass of the probe multiplied by the square of the elliptical speed at 0.1 one E squared minus the kinetic energy for this, for the circular speed at 0.1. So minus one half M sub P multiplied by V sub one C squared. And if you want, you can simplify this equation a bit by factoring out the one half and the sub P. So one half M sub P multiplied by V sub one E squared minus V sub one C squared. So now we'll just plug in our values one half multiplied by the mass of the probe which is 750 kg multiplied by the difference of the squares of the speeds. So that's 8868 m per second squared minus the square of 7069 m per second, putting all that into a calculator. And we find a change in energy of 1.8 or 1.08 multiplied by 10 to the power of 10 joules. And so that is how much work is produced by the probe's engine to move the probe from orbit A to the elliptical orbit, which corresponds to option C and that is it for this problem. I hope this video helped you out and please consider checking out our other tutoring videos which will give you more experience with these types of problems, that's all for now and I hope you have a lovely day. Bye bye.

Key Concepts

Work-Energy Principle

Gravitational Potential Energy

Elliptical Orbits and Hohmann Transfer

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