Hey, guys. We saw that the orbital velocity of a satellite was how fast the satellite travels in its orbit. There's another variable you need to know to solve problems, and that's the orbital period, which is the time that it takes to complete one orbit. So we saw that the satellite has a tangential velocity as it's going around in its orbit, but it takes some amount of time to actually complete one full orbit. That's called t. We actually relate that velocity to the period just by using circular motion and basic kinematics. How do we get t? How do we get time from velocity? We know from kinematics that v=distancetime. What's the distance that this thing completes in one orbit? Well, it's traveling in a circle of radius r, so that's just the circumference of that circle, 2πr. And if you do that in one full circle, then this t just becomes a capital T. This equation is really useful for us in satellite motion, so I'm actually going to write it here. −2πrT. Now we can actually solve for this T, this capital T right here. Most of the time you'll see it in its T squared form. This is 4πr3∕gm. You can actually get to it pretty quickly by using this equation, so let's go ahead and do it real quickly. V_sat=2πrT. So what happens is if you want T, you can get T=2πr∕V_sat. So we have this V_sat, but we already have another equation that will tell us. We can actually stick that in here, and we're going to get 2πr∕gm∕r. So this nasty formula with square roots and fractions; what'll happen is if you square it, it'll become a lot cleaner: 2 becomes the 4, π becomes π2, r becomes r2. And what happens to the square root is that when you square it, the whole thing just goes away, or the square root goes away. So you get gm over the little r. So we've got this situation where you have a fraction on a fraction, and what happens is that this denominator of the bottom fraction will actually go up and merge with that top little r. So what'll happen is 4πr3∕gm. So you get that equation. This equation actually has a name. It is called Kepler's third law. Kepler was a guy who was studying the motions of planets in our solar system, and he noticed a relationship between the orbital period of all the planets and the distances from the sun. It's pretty cool. So we've got these three equations and these are all of our satellite motion equations. So what happens if the distance increases? Alright. So what happens if r increases? Well, we can use, how does v change? So we have these 2 V_sat equations, and if the little r is changing, how does this equation change? It's a little bit tricky to tell just by using this one because you have all 3 variables present in these, and we don't know how T changes yet. So instead, let's use the other V_sat equation. So as the r increases in the denominator, the V_sat has to decrease. Alright. And again, let's not use this equation for T because we have a relationship just between T and r. So it's going to be easier to relate those 2 by this equation. Now r is in the numerator here. So if r goes up, then T has to increase as well. So that means that as r increases, your velocity decreases, but your period increases. This should make some sense because as you're going farther away, the force of gravity gets weaker on you, which means you don't have to go as fast to stay in a circle. On the other hand, if you were to go really close to the earth, the earth would be pulling on you really, really hard, but you have to go really fast not to crash into the earth. Because you're traveling in a larger circle as r increases, the time that it takes also should be increasing. And that's basically it. So let's go ahead and take a look at an example and use all of these equations together. So we've got the orbital period and speed of the International Space Station. So first thing is we're going to be calculating the orbital period. So let's take a look at our equations. So T equals w...
8. Centripetal Forces & Gravitation
Satellite Motion: Speed & Period
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