In this video, our topic is orbital mechanics. Orbital mechanics is all about how the motion of objects that move in orbits, objects like planets or moons or satellites, can be described mathematically. We’re going to discover as well as use two equations that describe that motion. So, let’s get right into it. We can recall from what we already know, that, in general, there are two types of orbits. One is circular where the orbiting object is at constant distance from the object that orbits around. And there’re also elliptical orbits where that distance between the two objects is constantly changing.
In this lesson, our focus is going to be specifically on circular orbits. These are situations where the orbital radius, we can call it 𝑟, is a constant value all throughout the orbit. That’s what it means for the orbital path to be circular. Now, as we consider this case of a circular orbit more closely, we can see that there are a number of parameters to it. We’ve already seen that there’s an orbital radius and then, along with this, we know the object in orbit will have some orbital speed. We can call that speed 𝑠.
Now, the reason we say that this object is moving in an orbit is because it moves in a pathway that it repeats over and over and over again. Once the object has finished going all the way around the circle, it then starts in on that same pathway once more. We can see that the distance of that pathway is equal to the circumference of the circle with a radius value of 𝑟. And this brings us to our first mathematical relation for describing orbital motion.
If we want to know the circumference of a circular orbit, we’ll call it 𝑐, then what we do is we take two and multiply that by 𝜋 and multiply that by the radius of the orbit. Looking at this equation, we may recall that it’s the equation for the circumference of a circle, whether orbital motion is going on or not. The useful thing about this equation, for our purposes, is that it tells us how much distance our object travels in going through one revolution of its orbit.
Now, we’ve said that our object has some constant speed, we’ve called it 𝑠, as it moves through this orbit. And we now know that as it goes through one revolution, it travels a distance we can call 𝑐. So, we have a constant speed and a distance travelled by an object moving at that speed. And this can remind us of an equation that connects speed, distance, and a third variable, time. We can remember that in general, the average speed of an object is equal to the distance that object travels divided by the time it takes to travel that distance.
In our case, we have a speed 𝑠. That’s the speed of our orbiting object. And we can say that the distance, 𝑑, that our object travels is equal to the circumference of the circle it moves through. So, if we were to rewrite this equation for average speed to apply to the case of circular orbital motion, we could write that speed is equal to distance, which is two times 𝜋 times 𝑟 divided by time. And when it comes to time in this equation, we can see that this time represents how long it takes for the object to move around the circular path once. But there is a particular name for that amount of time. It’s called the orbital period. And it’s usually represented not with a lowercase 𝑡 but with an uppercase 𝑇.
To figure out what the orbital period of an orbiting object is, we could do something like this. Given a certain position of that object that we can call its starting position, we can begin a stopwatch. And then, as the object moves around its circular orbit, wait until it gets back to the place where it began. At that instant, we stop our stopwatch, and the total elapsed time is the orbital period of the object. This means we now have a mathematical relationship that connects speed, distance, and time but in terms of orbital motion. Here, 𝑇 is the orbital period, two times 𝜋 times 𝑟 is the path length of the orbit, the circumference of the circle representing it, and 𝑠, as we saw, is the orbital speed.
One reason we talk about orbital speed rather than orbital velocity is that speed can be constant as this object moves through its orbit. That’s because speed, we recall, is a scaler quantity. There is no direction associated with it but only size or magnitude. This means that even as our orbiting object moves around the circular path, it will have a constant speed. It will always be moving at the same number of, say, meters per second or kilometers per second. But velocity, on the other hand, will be constantly changing for this object. And that’s because velocity, as a vector, takes into account direction changes. And we can see that the direction of this orbiting object is always changing.
So, for a circular orbit like this one we have here, the speed of the orbiting object is constant, but the velocity is not. And that speed is equal to two times 𝜋 times the orbital radius divided by the orbital period, the time it takes the object to go around one complete revolution. Now that we have these mathematical descriptions for the way that objects that move in circular orbits travel, let’s get some practice working with these equations through an example.
Let’s say that we have this scenario. A planet moves around a star in a circular orbit. The radius of that orbit is 6.0 times 10 to the seventh kilometers, and the time it takes this planet to go completely around its orbit once is 210 days. That’s the orbital period. Knowing all this, let’s say that we want to solve for the speed of this orbiting object, we can call that orbital speed 𝑠.
To begin calculating 𝑠, we can recall a general equation for objects that move with a constant speed. This relationship says that the speed of an object is equal to the distance the object travels divided by the time taken. In our example, we want to calculate the orbital speed 𝑠. And to do it, we’ll need to know the distance the object moves in going through one complete orbit and the time taken to do that. Well, we’re given the time. That’s the orbital period, capital 𝑇. But we don’t yet know the orbital distance travelled. But looking again at our orbit, we can see that that distance will equal the circumference of the circle with a radius 𝑟.
Realizing that, we remember that the circumference, 𝑐, of a circle is equal to two times 𝜋 times the circle’s radius. And this circumference here is the distance 𝑑 that we want to use in our equation for orbital speed. So, here is what we can write. The orbital speed of our object, that’s what we want to solve for, is equal to the distance it travels, that’s one time around the circular orbit, two times 𝜋 times 𝑟, all divided by the orbital period, capital 𝑇. And as we look at the known information, we can see that we’re given 𝑟 and we’re given capital 𝑇. So, we can now substitute those two values in to our equation for orbital speed 𝑠.
Having done that, we can see that if we calculate this fraction right now, we’ll get units of kilometers per day. Those will be the units in terms of which we calculate our orbital speed. Now, when we talk about astronomical bodies such as this planet orbiting a star, it’s not unusual to leave the distances in terms of kilometers because distances on an astronomical scale are so large. But it is uncommon to leave time in units of days. So, instead of expressing time that way, let’s do it in terms of the SI base unit of time, the second.
To do that, we’ll need to convert this unit of time. And here is what we can recall to help us with that. We can recall that one day is equal to 24 hours of time and one hour is equal to 3600 seconds. And the reason we know that one hour is 3600 seconds is one hour is 60 minutes and there are 60 seconds in every minute. So, if we multiply 60 minutes by 60 seconds per minute, we get 3600 seconds. Now, we want to reexpress 210 days as some number of seconds. We can do that this way. We can take our original time value, 210 days, and multiply it by the number of hours in a day and then multiply that by the number of seconds in one hour.
Notice that these two fractions, 24 hours in a day and 3600 seconds in an hour, came from a recollection of how to convert different units of time. And when we multiply our original time value of 210 days by these two fractions, look at what happens to the units. First, consider the units of days. We have that unit in the numerator and now also in the denominator, which means that multiplying through causes that to cancel. And as well, we have units of hours in the numerator and denominator. So, this unit of time cancels too. The unit we’re left with, when all the dust settles, is simply units of seconds, the time unit we wanted.
So, to get this orbital period of 210 days expressed in seconds, we’ll need to multiply 210 by 24 by 3600. Rewriting our expression this way, we can now see that when we calculate this fraction, we’ll indeed get a speed 𝑠 in units of kilometers per second. And when we enter all these values on our calculator, we find that to two significant figures, the orbital speed of this orbiting planet is 21 kilometers per second. That’s how fast this planet travels as it moves around its orbit.
Let’s now consider a second example involving orbital mechanics.
In this case, a satellite is in circular orbit around the earth. As it moves through one complete revolution, the satellite travels a distance of 4.0 times 10 to the 10th meters. What is the radius of its orbit?
So, in this scenario, if this here is planet Earth, then we’re told a satellite is in circular orbit around the earth. And that as it goes through one complete revolution of its orbit, that is one complete turnaround this circle, it travels a total distance given as 4.0 times 10 to the 10th meters. Based on this information, we want to solve for the orbital radius, what we’ve called 𝑟. To solve for this radius, a key realization is that there is a mathematical relationship between the circumference of a circle and its radius. That relationship says that circle’s circumference, 𝑐, is equal to two times 𝜋 multiplied by circle radius, 𝑟.
Now, since our satellite is moving in a circular orbit, that means this distance travelled is equal to the circumference of the circular path it moves along. And that means we can use this relationship to solve for the radius of the circle, 𝑟. Here is how we can do that. We’ll begin by rearranging this equation so that 𝑟 is the subject of the equation. That is, it’s on one side of the equation all by itself. To do that, we can divide both sides of the equation by two times 𝜋. And when we do this, that means that two times 𝜋 cancels out on the right-hand side. We’re left with an equation that says that 𝑐 over two 𝜋, the circumference of the circle divided by two 𝜋, is equal to the circle radius. And that’s exactly what we want to calculate.
Now, as we think about substituting in to the left-hand side of this equation so we can calculate 𝑟, we can remember that 𝜋 in the denominator is a constant. 𝜋 is equal to 3.1415..., and it goes on and on. So, we’ll use an approximate value for 𝜋. We’ll treat it as exactly 3.14. Then, see the circumference of the circle is given to us in the problem statement. It’s 4.0 times 10 to the 10th meters. Before we go ahead and calculate 𝑟, notice that the units in this expression are meters, units of distance. So then, the radius we calculate will be in units of meters. When we compute this fraction to two significant figures, we find a result of 6.4 times 10 to the ninth meters. That’s the radius of this satellite’s orbit.
Now, let’s summarize what we’ve learned about orbital mechanics. Starting off, we saw that objects in orbit can move either in circles like here or in ellipses like here. For the case of circular orbits, we saw that if 𝑐 is the circumference of that circular orbit, the distance the orbiting object travels in one complete revolution of its orbit, and 𝑟 is the orbital radius, then 𝑐 is equal to two times 𝜋 times 𝑟.
And then, as we thought about the orbiting object moving with a speed that we can call 𝑠, we saw that that orbital speed is equal to the circumference of the orbit, two times 𝜋 times its radius, all divided by capital 𝑇, the orbital period, which is the time it takes the orbiting object to move through one complete revolution. These two relationships specifically apply to circular orbits and we can use them to solve for orbital period, orbital speed, orbital radius, and the distance an object travels in moving through its orbit.