Energy of satellite systems

0:00- [Instructor] When scientists look at a comet, they can measure its position and speed and tell whether we will see it again or not. For example, this is the 3I Atlas discovered in 2025. It's an interstellar comet, meaning it came from deep intergalactic space into our solar system, and we know that we will never see it again.

0:21It's gonna go back into interstellar space. In contrast, this is the Halley's Comet last seen in 1986, and we know we'll see it again sometime in 2061. So the big question is, how can we just measure the position and the speed of a comet and tell whether we'll see it again or not?

0:41Well, it all has to do with the energy of orbiting systems. So let's dig into it. Let's consider a planet going around a star like Earth going around the Sun, for example. You can also consider a satellite going around a planet, or you can also consider a comet orbiting the Sun, whatever you want, okay?

0:59Take any orbiting system. The question we're gonna try and answer is, what is the energy of that system? The total energy of this system? Well, let's think about it. First of all, because the planet and the star are interacting gravitationally, there must be some gravitational potential energy.

1:16What's the expression for that? Well, the gravitational potential energy between any two masses is given by this expression where M and m are the masses, and r is the distance between them. But wait, why is the gravitational potential energy negative? Well, because when masses are infinitely far apart, the gravitational potential energy is maximum, right?

1:39Well, we choose that maximum value to be zero. That's our choice; that's our reference. And since the maximum gravitational potential energy itself is zero for any other configuration, when the distance between them is finite, the potential energy ends up becoming less than zero.

1:56So it becomes negative. Note that at this particular scale, we can pretty much treat both the celestial bodies as just point masses, so we don't have to worry about their radius and stuff, and secondly, remember that this is the energy of the entire system. We don't have to consider the energy of the star separately and the potential energy of the Earth separately.

2:16We always consider potential energy for a system. So this is the potential energy of this entire system. Okay, what else? Well, there's kinetic energy. Now, in this particular case, we are assuming that the star is at rest, and it's the planet that's moving. So all the kinetic energy is, you know, in the orbiting objects.

2:34So it'll be the kinetic energy of the planet or the kinetic energy of the satellite, whatever you wanna call that. So the total energy is going to be the sum of these two, right? Okay, so what will that be? Well, if the speed of that satellite or the planet is v, then we can say the total energy becomes the potential energy plus the kinetic energy of that satellite or the planet, which is just half mv squared.

2:59Okay, so let's see what we can do with this. Well, first of all, let's assume that there are no other planets or stars in the neighborhood, or let's assume they're very far away so we can ignore their gravitational influence. Then as the planet moves around, the total energy of the system must stay conserved because there is no external work being done on our system.

3:21So the total energy must stay the same, right? However, the potential and the kinetic energies can change. So we can use this to predict some stuff. So, for example, a few months later, when the Earth is somewhere over here, can we predict what's going to happen to its speed?

3:39Will it increase compared to this one, or decrease, or stay the same? Why don't you pause the video and think about this? Alright, so if you look at this expression, let's look at what happens to the potential energy. Well, the M and the small m and the G, that stays the same, but the r value has increased.

3:55So the denominator increases, which means this number goes down, but there is a negative sign, which means the potential energy actually increases. Does that make sense? I mean, a more intuitive way to think about it is you probably know when you throw a ball up, potential energy of the ball-Earth system increases, right?

4:13When masses go farther away, the gravitational potential energy increases, and the same thing is happening over here. Look, the Earth is going farther away from the Sun, so the gravitational potential energy increases, but the total energy must stay the same, which means the kinetic energy reduces.

4:29So you'd expect the speed over here to be lower than over here. So as the planet goes farther away, its speed reduces, and then as it comes closer, its speed increases. And so you should have the maximum speed when it's, you know, when it's closest to our star and so on and so forth.

4:44The same thing would apply to an orbiting satellite as well, for example. But remember, Earth's orbit around the Sun is pretty circular. So let's consider circular orbits now. Well, then the distance of the planet from the star stays exactly the same everywhere, right?

4:56Which means the potential energy is now a constant. Therefore, the kinetic energy of the orbiting planet or the satellite must also be a constant. In other words, the speed of the orbiting object should stay the same. So this tells us that in circular orbits, the object's speed must stay a constant.

5:16But wait, what should be the expression for that speed in circular orbits? I mean, we intuitively know if you're closer to a star, it should be orbiting faster. But if it's farther away from the star, the planet would be orbiting much slower. So can we find an expression for it?

5:33Yeah, we can use Newton's second law, which says net force acting on an object equals mass times acceleration. So over here, if we apply Newton's second law on the orbiting object like the Earth over here, then the net force over here is just the gravitational force that the Sun is, you know, exerting on the planet.

5:50So that's just Fg. That equals the mass of our Earth, the mass of the orbiting object times the acceleration. Now, which acceleration are we dealing with over here? If the force is in this direction, acceleration is also in this direction. That is the centripetal acceleration.

6:07This is a very special case. The speed is a constant, and therefore the acceleration is always perpendicular to the direction of the velocity. That is the centripetal acceleration. It's directed towards the center, right? So we can plug in what is the expression for the gravitational force?

6:21Well, that can be found by Newton's universal law of gravitation. That is GMm divided by r squared. That equals the mass m times... What's the expression for centripetal acceleration? Well, it's given by v squared over r. So the m cancels out. One of the r over here cancels out, and if we rearrange, we get v squared equals GM divided by r.

6:48So this shows that the orbital speed is independent of the mass of the orbiting satellite, but it does depend upon the mass of the parent star. Again, that makes intuitive sense. If this was more massive, then you would need a higher speed to maintain your circular orbit, and notice how it's inversely related to the distance, which means the farther you go, the lower is the orbital speed to maintain a circular orbit.

7:13Okay, so now that we know an expression for the orbital speed, we can plug this in over here and simplify the expression for our energy. Again, great idea to just pause the video and see if you can simplify and gain some insights. All right, so if we plug in for v squared GM over r, we end up with this expression.

7:35So let's clean this place up a little bit and think about what this tells us. Well, first of all, I can simplify this now. I can write this as GMm divided by R times minus one plus half, which is minus half. So my total energy now becomes minus half GMm divided by r.

7:56Okay. So the first thing I see is that my energy of this entire system is negative because remember, these numbers are all positive. Okay? So that's the first thing, but there's something else. Look at the expression for kinetic energy for circular orbits. What do you notice?

8:15Well, it's similar to potential energy, except we don't have a negative sign and we have a half, which means the kinetic energy is negative half of potential energy. And similarly, even the total energy can be written in terms of potential energy. We can say the total energy is half of potential energy.

8:38At this point, we might have even more questions. So again, let's clean this up a little more. Okay, the first question we could be having is how can total energy be half of potential energy? I mean, shouldn't it be more than potential energy? Because total energy is potential energy plus kinetic.

8:52So how can it be half of potential energy? Well, remember both these are negative. So, for example, if your potential energy was, you know, minus two units, for example, the total energy would be minus one unit; minus one is bigger than minus two. So yes, the total energy is always bigger than the potential energy.

9:10But the second thing that we see is that the kinetic energy will always be negative half of the potential energy. Again, that makes sense because for circular orbits, v is fixed; the planet needs to orbit with a very specific speed. So it makes sense that its kinetic energy should also be very specific, negative half of the potential energy.

9:31But the thing is, this only applies to circular orbits. For elliptical orbits... Remember, the values of the potential energy and the kinetic energy changes over the orbit. However, what does not change is this fact: the total energy, even for elliptical orbits, would be less than zero.

9:51So whether you consider circular orbits or elliptical orbits, the total energy should be less than zero. In other words, if the planet is bound to the star, then the total energy will always be negative, less than zero. And so this represents the bound state. Okay, so what do you think would happen if the total energy was more than zero?

10:15What would that represent? For that, let's do a thought experiment. Suppose we give a kick to our Earth, increasing its speed such that it escapes the Sun's gravitational influence and reaches infinity and stops. What I mean is, let's say Earth has just enough speed to reach infinity; then what would be the total energy of the system?

10:39Well, when you're at infinity, we know that the potential energy of the system is zero because that's our definition of zero potential. When two masses are infinitely spaced out, that's what we consider as zero potential energy. That's our reference. So the potential energy of the system is zero at infinity, and we are also saying that the Earth just reaches infinity with zero speed, meaning the kinetic energy of the Earth is also zero.

11:01So in this particular situation, the energy of the system at infinity is zero. So what happened is that after giving the hypothetical kick, we did work on the system, and so the total energy of the system increased. From negative, it became zero. So when the planet or satellite has just enough speed to escape to infinity, the total energy of the system is zero.

11:26And this minimum speed needed to just escape the gravitational clutches of a star or a planet or whatever, we call that the escape speed. Mathematically, you can think of it as the speed at which the total energy of the system goes to zero. So how do we derive an expression for the escape speed?

11:48Well, we know the total energy of the system is zero. At any point, the total energy must be zero. So when the planet is over here, the total energy, which is zero, that should equal the potential energy minus GMm by r plus the kinetic energy half mv squared. And so if we rearrange this, we get this expression, and from here again, we can cancel m.

12:11Rearranging, we get this expression. Taking square root on both sides, we get the expression for the escape speed. Again, notice the escape speed is independent of the mass that is escaping, right? But it does depend upon the mass of the parent body. So in this particular case, the mass of the star; that makes sense.

12:29If the star was massive, it's harder to escape that because it has more gravity. But notice that it is inversely related to the distance from the star. That also makes sense. If you're farther away from the star to begin with, it takes less energy to escape it, so less speed to escape it.

12:44And normally when we're talking about the escape speed of a planet or a star for that matter, what we usually mean is its surface escape speed. So in that case, the small r just becomes the radius of the planet or the star from which you're trying to escape. So if you plug the values in for Earth, you will find the surface escape speed to be 11.2 kilometers per second.

13:07That's the minimum speed with which you need to kick an object from the Earth's surface so that it never returns back to Earth. Of course, all of that is assuming no atmosphere and drag and all of that stuff. The surface escape speed from the Sun is about 618 kilometers per second.

13:25Anyways, we now understand if an orbiting satellite or an orbiting planet has this particular speed at a given position, then the total energy is zero. And the object would just escape the parent body, right? Which means, what if its speed was more than the escape speed?

13:46Well, then it'll have more than enough speed to escape, which means it'll reach infinity, and it won't stop. It'll have some residual speed left as well. So now the total energy would be more than zero, right? And this represents an unbound state. So the planet is no longer bound to the star, or the satellite is no longer bound to the planet.

14:08And so we now have the full picture. If the speed of the orbiting object is less than the escape speed, then the total energy will be less than zero. We are in the bound state. In a special case, the orbit will be circular, but in general, the orbit will be elliptical; it'll be closed, bound state.

14:26In contrast, if the speed of the orbiting satellite is more than the escape speed at that particular given position, then the total energy of the system will be more than zero. Now it'll be unbound. Now the path will be a hyperbola. It'll take a hyperbolic path, and it'll never come back.

14:44And so look, if you can measure the speed of an object and you know its position, you can tell it's in the bound state or the unbound state. And from that, we can tell whether we'll see that object again or it will escape the gravitational influence of the parent body.

15:01So if you go back to the 3I Atlas, it turns out that the total energy is more than zero. It's in the unbound state, and therefore we are sure that it'll never turn back. It takes a hyperbolic trajectory and escapes our solar system and goes back into deep interstellar space.

15:21In contrast, if you calculate the energy of the Halley's Comet, you'll find that it's less than zero. And that's how we know it's a bound system. It is a very elliptical orbit, and it takes about 76 years to complete that. And that's why you will see Halley's Comet about every 76 years.