https://youtu.be/B77rjaRLzTo
PHYS 1101: Lecture Fourteen, Part Ten
Okay. Back to our problem now. Once we’re away from the surface of the planet Earth, we’ve got to go back to that more general equation. We have to use that FG is actually this universal constant times the mass, in this problem, of Mars, times the mass of the satellite, I’ll call it Msat, divided by r squared. What’s r again? This has to be the separation or the distance between the centers of these two masses.
The distance between these two centers is going to be from the center of Mars, the radius of Mars plus the altitude. That’s r. That also happens to be the radius of our circular orbit.
Okay, so we’ve thought about a for this object, the satellite. It points to the center. We’ve thought now about the magnitude of the force that’s responsible for this a. We’re ready to do our list of knowns. Let’s start that with the radius, as I just said, has to be the radius of Mars plus the altitude. Right? That’s sum of these two pieces gets me the total radius.
As an aside, that’s common in a satellite problem that you’re told an altitude above the planet’s surface. And to get the proper r, you’ve got to add the radius of that planet. So I have a radius of 3397 times 10 to the 3rd meters, and I have to add the altitude, which is 488 times 10 to the 3rd meters. That combination turns out to be 3885 times 10 to the 3rd meters. Okay, there’s my r.
What other facts do I know? It looks like only the mass of Mars. That’s, like that planet Earth, is big. 6.4 times 10 to the 23rd kilograms, according to the problem.
What variable do I want? They asked for the orbital period of the satellite, okay? I’ve got to pick a variable. What variable or symbol were we using to represent that? That period, the time to go around once, I’m going to use a fancy T to represent. That’s the variable I’ve got to focus on. Now I’m ready to start with my second law equations and customize them. Work with them until I can arrive at a value for T.
Here’s my picture again. I’m going to add a coordinate system here. I only have vertical forces. I’m going to go ahead and call the y-axis positive downward, just so I can make all of my vectors positive. So I’m working with Ay is equal to the sum of the y forces divided by the mass of this object, the satellite. Okay, Ay I know is not 0. As usual I’m going to make the substitution that the magnitude is the speed squared divided by r. That’s positive. It’s downward. That’s equal to all of my y forces. I only have one.
I’m substituting that in, and then I still have the mass of the satellite in the denominator. Okay, well let’s keep peeling this away. Let me work on the right side for awhile. I don’t yet see T, the period, in my equation. I have v’s and things. V, for example, which I don’t have a value for. I’ve got to keep modifying this equation. I keep working it until I get to the thing I want. Well, I know I can substitute in here for the gravitational force.
I know that’s the universal constant times the mass of Mars, mass of the satellite divided by r squared. Let’s make that substitution just into the right side there. Notice I have to be careful, and I’m substituting that in for the numerator on the right side. In other words, this is all of this. I still have to divide by m of the satellite. But notice or recall that this denominator effectively is in this denominator. So this variable cancels with that variable.
Okay, now I end up with v squared over r is equal to G. Mass of Mars divided by r squared. And let’s see. I can simplify here a little bit. I see a lot of r’s that are the same variable. I multiply both sides by 1r just to simplify that a bit. It’s looking pretty good. I have a number for the mass of Mars. I know the value for G. I know the value for r. I don’t this, and I don’t know the period. But you know what? If I go back to those kinematic relationships for circular motion, I remember that the speed is related to the circumference of the circular path divided by the time to go around once, the period.
This is an excellent substitution to make because I know the radius and I now then have the variable I need to solve for. So let’s substitute this in. Okay, so I’m again going back to my main starting equation, Newton’s second law, and I’m massaging it some more. For v, I’m going to plug in 2 pi r over the period, but notice I have to square that to make that substitution correct. I’m substituting in for this, and on the right I have G, mass of Mars, divided by r.
Now I’m in great shape. I have the variable that I want, and I know the values of everything else. I just have to do some algebra. I’m going to multiply both sides by the period squared just to get the period out of the denominator. And now I’m going to move all of this over to the other side, which means multiplying by the inverse of that. And this then has been moved to the other side.
What do I have now? I’m going to switch the order here and write that I end up with T squared is equal to the product of all of this stuff. What do I have left? 2 pi r squared, but that’s another factor of r, so I’m going to write 2 pi squared times r cubed, then I have to divide that by G times the mass of Mars. All this stuff now will be equal to the period directly if I simply take the square root. So the period should be given by… It’s kind of a complicated equation, but it’s nonetheless, if we’ve not made any algebra mistakes, it should give us the right answer.
I have to take 4 times pi squared. I have to multiply by the radius of the circular orbit cubed. That’s 3,885 times 10 to the third meters cubed because I do have to work in SI units. I have to do that conversion of kilometers to meters. Then I have to divide by this universal constant G and divide by the mass of Mars, 6.4 times 10 to the 23rd kilograms.
Okay, here’s a word of caution. When you start doing satellite problems, you’re going to end up working with very large and very small numbers. To input these into your calculator, you need to use the ee button, or it might look slightly different to you. It might say times 10 to the x or something. Do not use the e to the x button. So e to the x, don’t use that. That’s not what you want when you enter these large and small numbers. You don’t want to type in all these zeroes.
Okay, so you just have to take your time, multiply it out carefully. Be sure, for example, that you cube this whole number, etc. When I plugged that into my calculator, I ended up with a period of 7,360 seconds and there’s 3,600 seconds in an hour. So when I convert that I get just a little over two hours. Takes a long time for the satellite to get around once. Okay, that brings us to the end of Lecture 15. I’m sorry. Okay, that brings us to the end of Lecture 14.