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. Consider a 1000 kg communications satellite that needs to be boosted from an orbit 300 km above the earth to a geosynchronous orbit 35,900 km above the earth. Find the velocity v'₁ on the inner circular orbit and the velocity v'₁ at the low point on the elliptical orbit that spans the two circular orbits.

Accepted Answer

Everyone. Let's take a look at this practice problem dealing with orbital motion. So in this problem, we have a weather monitoring satellite with a mass of around 2000 kg. And it needs to move from an initial circular orbit which is approximately located at a distance of 500 kilometers above the earth surface to another circular orbit. That is approximately located at a distance of 37,000 kilometers above the earth's surface. We need to determine its speed in its first orbit. We also need to find its speed when it reaches the lowest point in an ellipse that's connecting both orbits. We're given a hint that says to use the viva equation which is V is equal to the square root of the quantity of GM multiplying the quantity of two divided by R minus one divided by A where A is the semi major axis of the ellipse which is equal to R one plus R two divided by two. We given four possible choices as our answers. A V circular is equal to 7.6 kilometers per second. The ellipse is equal to 10 kilometers per second choice B the circular is equal to 10 kilometers per second. The ellipse is equal to 9.2 kilometers per second. Choice C the circular is equal to 7.6 kilometers per second. The ellipse is equal to 9.2 kilometers per second. TDV. Circular is equal to 10 kilometers per second and the ellipse is equal to 10 kilometers per second. Now, in this problem, um I'm gonna be dealing with orbits and with orbits, we always deal with the radius of the orbit which is going to be the distance from the center of the planet. In this case, the center of the earth to the object that's or orbiting. However, in this problem, we were given the altitudes basically the distance above earth's surface. So what we need to do first is to calculate the different radii for the orbits. So we're gonna start with our lower orbit, which is the um one that's 500 kilometers above the earth's surface and we'll call that radius of the orbit R one. And so this is going to be equal to the distance above earth's surface, which is gonna be the 500 kilometers, but we'll need that meter. So we need to multiply that by 10 to the three to convert it to meters. And then we'll have to add to that the radius of the earth which is 6.37 multiplied by 10 to the 6 m. And that's a value that I can look up And so when I add those two together, I get R one is equal to 6.8 seven, multiplied by 10 of the 6 m. We'll do the same thing for the other orbital radius. We'll call that R two and this is gonna be equal to, we'll have the 37,000 kilometers. We need to convert that 2 m by multiplying by 10 to the three. And then we'll add to that the radius of the earth, which is the 6.37 multiplied by 10 to the 6 m. And so that gives me R two is gonna be equal to 43.37 multiplied by 10 to the 6 m. So now I have the two different radii, I can now use my equations to calculate the speeds. So the first thing I want to do is calculate the speed of the uh initial circular orbit. And so for that, we can recall our formula for the speed of the circular orbit. We'll call that VC. And this is gonna be equal to the square root of GM divided by R where G here is the gravitational constant M is the mass of the planet and R is the radio orbit. And in this case, um the radius of the orbit is just gonna be R one. So we can plug in values for everything on the right hand side. So we'll have VC gonna be equal to square root the gravitational constant. That is 6.67 multiplied by 10 to the negative 11 Newton meters squared per kilogram squared for the mass of the earth. That again is a concept that I can look up that is 5.97 multiplied by 10 to the 24 mg. And I'm going to divide this by R one which we just calculated which is the 6.87 multiplied by 10 to the 6 m. So I can put those values into my calculator and I get VC is going to be equal to 7.6 multiplied by 10 to the 3 m per second. And here I've just kept two significant figures. Now, if I look at my answer choices, all my answer choices have units of kilometers per second. So I'll need to convert this into kilometers per second by multiplying by the conversion factor of one kilometer divided by 1000 meters. And so that would give me VC or the V circular is gonna be equal to 7.6 kilometers per second. So that's the first half of the problem answered. Now, we're gonna do the same thing for the elliptical orbit. Now, for the elliptical orbit, we're gonna use the vis viva equation. So we'll have that as be for the um V ellipse and that's gonna be equal to the square root of GM multiplying the quantity of two divided by R. In this case, it's gonna be the point that's gonna be closest to the earth, which is gonna be its lower orbit, which is gonna be R one. Then we have minus the one divided by a. Now here A is gonna be my semi major axis. So I need to calculate it. So I have A is equal to R one plus R two divided by two. And we were given that information in the hint. So we know what R one and R two are for this case. So we'll have a equal two for R one, I'll have the 6.87 multiplied by 10 to the 6 m. They'll have plus or two will have the 43.37 multiplied by 10 to the six meters. And that sum gets divided by two. So that gives me a is equal to 25.12 multiplied by 10 to 6 meters. So now I have values for all the quantities that I have in my uh vis viva equation. So I can go ahead and plug them in. They'll have ve is equal to the square root. The gravitational constant will again have the 6.67 multiplied by 10 to the negative 11 Newton meter squared per kilogram squared mass of the earth again is the 5.97 multiplied by 10 to the 24 kg. Then this gets multiplied by the quantity of two divided by R one which is the 6.87 multiplied by 10 to the 6 m. Then we'll have minus one divided by a which is the 25.12 multiplied by 10 to the six meters. And so we plug all those values into my calculator. I get ve is equal to 1.0 multiplied by 10 to the 4 m per second. And here again, I've just kept two significant figures and just like with the other um speed, we need to convert this into kilometers per second. We multiply it again by the same conversion factor of one kilometer divided by 1000 meters. And that means be is gonna be equal to 10 kilometers per second. And that's the second part to my answer. And so if I look at my answer choices, this actually corresponds with answer choice. A so just a quick little recap of what we've done here. In order to find the speed of its initial orbit, we just use our formula for the speed of a circular orbit. And here we had to actually calculate the radius of this orbit or one by adding in the radius of the earth to the altitude, the distance above our surface that we were given in the problem. For the second half of this problem. We had to use the vis viva equation. And here we also had to calculate the semi major axis. And this required us to also calculate the radius of the second circular orbit again by adding in the radius of the earth to that of the distance above the earth surface that we were given. So once we had the semi major axis calculated, we could plug in the values to calculate our speed for the ellipse at this point. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Orbital Mechanics

Velocity in Circular Orbits

Elliptical Orbits and Hohmann Transfer

Watch next