Angular momentum of satellites

0:00- [Instructor] Let's talk about angular momentum of orbiting stuff, like planets and satellites. To do that, let's start with a simple-sounding question. I have a stone of mass m moving with some velocity v in a straight line. Does this stone have angular momentum?

0:16Well, we know angular momentum of any rigid body is given as the product of its rotational inertia times angular velocity. So our instincts might be, "Hey, this stone is not rotating. It's just moving in a straight line. So angular momentum is zero. It doesn't have any angular momentum."

0:30(laughing) But that would be wrong. Why? Well, think of it this way. Imagine there was a thin light rod, which was fixed at one end over here, so that, you know, it can freely rotate like this. Now, what do you think would happen if this stone were to go and hit this rod so that its velocity is perfectly perpendicular to the rod and it sticks to it?

0:52Well, you can imagine it's going to start spinning with some speed, right? And to figure out that speed, we would need to know the rotational inertia of this rod. But if you assume that the rod's mass is negligible compared to that of the stone, and if you also assume ideal conditions like say no friction and no air resistance and all of that, then once it's sticks to this rod, it will start spinning with the stone having that same linear speed forever, which means now the system has angular momentum.

1:24The system of rod and the stone together, it has some angular momentum, right? But think about where did that angular momentum come from? Remember, angular momentum of system cannot change unless there is an external torque acting on the system. Now, if we consider the rod and the stone as part of our system, then as the stone hits that rod and sticks to it, the forces and the torques are all internal to our system.

1:50There are no external torques, which means the angular momentum couldn't have changed. So if the system has angular momentum here, the system must also have angular momentum here as well. And all of that angular momentum must be part of this stone, because remember, this rod is not moving.

2:07I mean, we have assumed it's mass to be negligible, but even if it did have mass, if it's at rest, it wouldn't contribute to the angular momentum, which means the initial angular momentum of the system, all of it must be part of just that stone, which means even though this stone is moving in a straight line, it's not spinning at all, it still has angular momentum.

2:28Okay, but how much is that angular momentum? Well, the angular movement of the stone here should be the same as the angular momentum of the stone over here. Remember, we are assuming that the rod has negligible mass, so it has negligible rotational inertia. So all the angular momentum is due to the stone.

2:46Okay, so what's the angular momentum of the stone? Well, let's just stick to its magnitude. We can say the angular momentum is the momentum of inertia of the stone times its angular speed. I'm using the word speed because we are sticking to magnitude. We'll talk about its direction later.

2:59Okay, how much is the rotational inertia of the stone? Well, if the stone is pretty tiny, then we can say all of that mass is concentrated at say the distance r from the axis of rotation. So then the rotational inertia of the stone would be mr squared. But what's the angular speed of the stone?

3:16Well, angular speed equals just the linear speed v divided by r. So over here, omega is v divide by r. And so one r cancels and we get mvr. And mv represents the magnitude of the momentum of that stone, right? And so now we can write the angular momentum magnitude as just r times p.

3:37And since the angular momentum is conserved because there are no external torques acting on our system, the magnitude of the stone's angular momentum even here must be r times p. But here's a question. Remember this thin rod that we imagined? Well, that's just an imagination, that's not real, right?

3:54So then what exactly is this r? What does r truly represent in our angular momentum formula? Well, think about r as the radius of the would-be circle that the stone would be moving in if it was rotating. The whole point is that, even though it's not rotating right now, it has the ability to rotate sometime in the future.

4:16That's why we say it has angular momentum because it can rotate sometime in the future. That's the whole idea. So it can rotate about this point, and that's why we say it has angular momentum about this point. Okay, but saying that r is the radius of the would-be circle along which that stone would rotate in the future, (laughing) that's not very technical, right?

4:41So how can we put it in more technical terms? Well, we can say that this distance r represents the perpendicular distance from this point on to the direction in which that stone is moving, look. And so we can call this the r perpendicular times p. Makes sense, right?

5:00But remember, angular momentum is a vector quantity. This is just its magnitude. It also has direction. How do we think about the direction of the angular momentum of this stone right now? Well, we can use our right-hand thumb rule. You take your right hand and you curl your four fingers along the direction of the rotation.

5:18Then the thumb points in the direction of the angular momentum. So if you curl your fingers over here this way in the direction of rotation, look, the thumb points out of the screen, so the angular momentum here is out of the screen. But this means, and this is a very important thing about angular momentum, that this angular momentum depends on our reference point.

5:38Because remember, the value of our perpendicular depends on our reference point. If I choose a different reference point, the angular momentum of the stone would be different. For example, consider the angular momentum of the same stone with respect to this point.

5:55Notice now the r perpendicular is much bigger. So now the angular momentum magnitude is higher, even though nothing has physically changed, it's the same stone, it's the same situation, but if I consider it's angular momentum with respect to this point, it's much higher.

6:12Okay, what about angular momentum of that same stone with respect to this point? Why don't you pause the video and think about its magnitude? Would it be bigger, smaller? And think about the direction as well. Okay, so you can now see our perpendicular is much smaller.

6:27So the angular momentum magnitude would be smaller. But think about the direction of the rotation. Now, if it were to rotate about this point, it would rotate this way, right? Which means the angular momentum is not only smaller, but it changes its direction. So both its magnitude and direction depends on the point of reference.

6:44Okay, finally, what do you think is the angular momentum of this stone about this point? Well, now our perpendicular is zero because that point lies along this direction, which means the angular momentum is zero. Does that make sense? Well, yeah, it's kind of like stone going and hitting a hinge of a door, for example.

7:03You can't make it rotate, right? So now the angular momentum is zero, which means any point mass moving with some velocity will have angular momentum given by this expression. And that angular momentum depends on the choice of your reference point, or you can think of that as your origin.

7:20And now, since angular momentum depends on this reference point, which we can think of it as origin, it would make a lot of sense to define it in terms of a position vector that we draw from that reference point or from that origin to our point mass. That way, we don't have to think about angular momentum in terms of radius of some future imaginary circle.

7:41Instead, we can talk about it in terms of its current position. So our next question would be, can we find an expression for r perpendicular in terms of its current position? So we have a right-angle triangle. So if I call this angle as theta, then I can write this r perpendicular in terms of r and theta using some trigonometric ratio, right?

8:06So which trigonometric ratio would we use? Well, we can say sine theta is r perpendicular divided by the hypotenuse, which is r. And so from here, r perpendicular is just r times sine theta. So I can say angular momentum equals r times sine theta times the momentum p.

8:25And finally, we can now write this in its full glory, the angular momentum as the vector product or the cross product of r and p. Let's think about why do we write it this way. First of all, whenever you take cross product of two vectors, the magnitude of that cross product will be the magnitude of the first vector times the magnitude of the second vector times sine of the angle between them.

8:52And look, that's exactly what we have, magnitude of r times magnitude of p times sine of the angle between them. And so it makes sense to write this as a cross product. And one question I used to usually have when I think about this angle is, "Wait, is the angle this angle between these two vectors or should we consider this angle between the two vectors?"

9:09Well, technically it should be this angle because when we say, "Angle between vectors," we have to consider them tail to tail. However, here, it doesn't matter because this angle is just 180 degrees minus theta, and sine of 180 degrees minus theta is just sine theta.

9:27So when you're dealing with cross products, because you're dealing with sine theta, either of them gives you the same result. So for our purposes, we can just take this smaller angle. But writing it this way also encodes the direction. How do you think about the direction of the vector product?

9:42Well, again, you use the right-hand rule. Here, you start with your right hand, and the four fingers should be in the direction of the first vector. So your four fingers will be in the direction of the r vector, and then you cross it from there to the second vector.

9:58So we cross it from there to our p vector, which is the same direction as v vector. And when you do that, the thumb represents the direction of the cross product. And notice that is in the same direction as the angular momentum. And so look, this also gives us the right direction, and that's why we write the angular momentum as a cross product.

10:17It's basically writing the same way in a much more compact form. It encodes both the magnitude and the direction in it. And one final thing about the cross product is r cross p is not the same as p cross r. I mean, the magnitudes would be the same, but if you were to do p cross r, then you would start with your forefingers along this vector, the p vector, and then you would cross it towards the r vector.

10:39So now your thumb would point inwards. So r cross p is actually equal to the negative of p cross r, so it's not commutative, okay? So we need to be careful. So angular momentum is not p cross r, it has to be r cross p. Okay, now let's apply this to orbiting satellites and planets.

10:56Let's consider the simple case of Earth going around the Sun, and let's assume the orbit to be perfectly circular. So the Earth has some mass m, it has some velocity v. The question is, what happens to the angular momentum of the Earth as it moves around the orbit?

11:11The moment you hear the question, the first thing you should say is, "About which point?" Because remember, angular momentum depends on the reference point. So let's consider the Sun's center to be the reference point. What happens to the angular momentum of the Earth about this point as it goes around the orbit?

11:26Does it change? Does it stay the same? That's the question. Well, one way to answer that question is, we know the angular momentum of a system can only change if there's an external torque acting on the system. And how do we calculate torque? In a similar way, we calculate torque as the cross product of the r vector, the position vector, and the force.

11:45So over here, our position vector from the origin to the planet would look like this. And is there a force acting on our planet? Yes, force of gravity, right? But does that force produce a torque? That's the question. Because if it does, then the angular momentum would change.

11:59Well, what is the angle between r and f? It's 180 degrees. Sine of 180 degrees is zero, so the torque is zero about this point, the force will be in the opposite direction of the position vector, which means throughout the orbit, the torque produced by the force of gravity about this point is always zero.

12:19So its angular momentum stays conserved. But at this point, you might say, "Well, isn't it obvious? Like at every point the value of r, m, v, and theta, everywhere is the same, so obviously the angular momentum must stay the same." Well, in circular orbits, it is quite obvious.

12:37But what about elliptical orbits? Would the angular momentum stay the same? Well, now look, the value of r changes, and even the magnitude of velocity changes at every point. The angle between r and v, the theta, that also changes. So all of these are changing.

12:54Now, it's not so straightforward. Now, we can't tell just by looking at this whether the angular momentum, you know, of this planet changes. But if you think in terms of torque, we can see it because, remember, the torque is still zero, right, because the force of gravity is always at the opposite direction of the position vector.

13:13And therefore, we can immediately say, if the torque is zero everywhere, then the angular momentum of the planet should stay conserved. and this is a beautiful result. We can use this to predict properties. For example, if I knew the velocity of Earth at this point, I can use that to predict what the velocity would be at this point.

13:31It's a powerful principle. And what's more important is that this is not just true for Earth and Sun, this is true for any orbiting object, whether you're a planet orbiting a star or a satellite orbiting a planet, the force of gravity, because it is acting towards the center, we call that as a central force, it cannot produce torque about the center.

13:53And so the angular momentum of the orbiting planet or satellite about the center of the parent star will always, always stay conserved. That's a powerful principle. Now, having said that, remember, in our solar system, there are other planets that are pulling on Earth as well, and therefore they will produce a toque on Earth with respect to this point.

14:14And so the Earth's angular momentum does change because of the other planets. But if you consider all the planets as part of our system, then again, the angular momentum of that new system would stay conserved because gravity is a central force. But of course, now if an interstellar asteroid came in, again, that would change the angular momentum of our system.

14:36And another thing is that remember that our planets are not just revolving, they're also spinning about their own axis. So that also adds to the angular momentum. But of course, at this scale, the size of the planet is so minuscule, we can just assume it to be a point object, and we can consider its spin around its axis negligible at this scale.

14:54So long story short, any mass having some velocity v will have angular momentum given by this expression. And that angular momentum depends on your point of reference. And the cool thing is that the angular momentum of any orbiting planet or a satellite about the center of the parent body, that's important, will always be conserved, because gravity is a central force, it cannot produce a torque about this point.