Video: Escape Speed

In this video, we will learn how to calculate the vertical speed that an object must have in order to escape the gravitational pull of a planet.

17:05

Video Transcript

In this video, we’re going to be looking at escape speed. Escape speed is the speed at which a projectile must be launched in order to escape from the gravitational pull of a massive object such as a planet. For example, if this amateur astronaut were to leave the ramp at escape speed, they would completely escape the gravitational pull of the Earth. Although unfortunately for them, as we’ll find out later, they would have to be cycling at around 11,000 meters per second in order to achieve this. In this video, we’ll look at how we can calculate escape speeds like this. And we’ll also look at how the motion of projectiles changes as their launch speed approaches escape speed.

So, to start off with, let’s consider exactly what it means for a projectile to escape the gravitational pull of a planet. So, here’s a planet. And let’s say that we have some test mass which is initially at rest on the surface of the planet. Now, let’s say that we project our test mass directly away from the planet with some initial speed 𝑣. Well, as anyone who’s ever thrown an object in the air will attest to, would expect our test mass to travel away from the planet before slowing down coming momentarily to a stop and then falling back down to the surface of the planet. Now, if we were to increase the initial speed 𝑣 at which we projected our test mass, then we’d see that once again our object would travel away from the planet, this time, reaching a greater distance before it stopped but then, once more, falling back down to where it started.

In fact, for a planetary mass of big 𝑀 and a test mass of little 𝑚, we know that there will always be an attractive gravitational force between the two masses. And the size of this force is given by this expression, where big 𝐺 is the gravitational constant equal to 6.67 times 10 to the negative 11 meters cubed per kilogram per second squared and 𝑟 is the distance between the centers of mass of the two objects. So, we’ve seen that increasing the speed at which we project our test mass away from the planet will increase its maximum distance from the planet. But this equation makes it clear that no matter how far away our test mass gets from the planet, it will always experience some attractive force that pulls it back toward the planet.

Given this, we might start to wonder whether it really is possible for our test mass to escape the gravitational pull of the planet. However, looking at this equation, we can see that the attractive force between the two masses is inversely proportional to the distance between the masses squared. In other words, increasing the distance between them decreases the attractive force. This means that if we could put an infinite distance between our planet and our test mass, then the size of the attractive force between them would be reduced to zero, at which point we could say that it’s escaped from the planet. In other words, in order for our test mass to escape the gravitational pull of our planet, we would have to project it from the planet’s surface with sufficient speed that it could be carried to an infinite distance.

So, how is this possible? Well, it turns out that this question is easily answered if we consider how potential energy works in a radial gravitational field such as the one that exists around our planet. The potential energy of the test mass, 𝐸 p, when separated from the planetary mass by a distance of 𝑟 is given by this expression. Now, gravitational potential energy is defined in such a way that when the separation between the two masses is infinite, the gravitational potential energy is zero. And for any finite value of 𝑟, the potential energy is negative. And to be precise, 𝑟 is actually the distance between the centers of mass of the two masses.

Now, it might seem strange to say that an object can have a negative potential energy. However, throughout physics, we often defined potential energies as negative to indicate something being bound to something else. For example, in an atom, electrons are often described as having negative potential energy to signify that they’re bound to the nucleus. In this case, our test mess having a negative potential energy signifies that it is bound by gravitational attraction to the planet. The fact that its gravitational potential energy is negative signifies that we would have to give it energy in order to free it, at which point its potential energy would be zero. And the way that we can do this is by projecting our test mass at a certain speed, in other words, by giving it kinetic energy.

So, if our test mass is initially on the surface of the planet, we can work out how much energy we’d need to give it to free it from the planet’s gravitational pull by writing down a simple equation to show the energy changes that take place. The test mass initially has potential energy 𝐸 p, and we want to give it just enough kinetic energy so that it’s freed from the gravitational pull of the planet. In other words, we want to increase its overall energy to zero, signifying that it’s no longer bound to the planet. In place of 𝐸 p, we can substitute negative 𝐺 times big 𝑀 times little 𝑚 over 𝑟. And in place of the kinetic energy 𝐸 k, we can substitute a half 𝑚𝑣 squared, where 𝑣 is the speed at which we project the test mass.

And so, we obtain this expression, which we can rearrange by adding 𝐺 times big 𝑀 times little 𝑚 over 𝑟 to both sides. This shows us that the amount of kinetic energy we would need to give our test mass in order to free it from the gravitational pull of the planet is equal to the magnitude of its gravitational potential energy. And solving this equation for 𝑣 will tell us the escape speed. To do this, we can start by dividing both sides of the equation by little 𝑚, which leaves us with a half 𝑣 squared equals 𝐺 times big 𝑀 over 𝑟. Then, to get rid of this fraction, we can multiply both sides of the equation by two, giving us 𝑣 squared equals two 𝐺𝑀 over 𝑟. And finally, we can take the square root of both sides of this equation to give us 𝑣 equals the square root of two times 𝐺 times big 𝑀 over 𝑟.

So, there we have it. This equation tells us the escape speed 𝑣 for a projectile at distance 𝑟 from the center of mass of an object with mass capital 𝑀. And we can change the symbol 𝑣 to 𝑣 sub e to signify that this is the escape speed. Now, any projectile will always experience an attractive gravitational force pulling it back towards the object that it was launched away from. And this force causes it to accelerate towards that mass. However, a projectile traveling at the escape speed will never experience a large enough attractive force to pull it back towards the other mass, which enables the projectile to travel away indefinitely.

One interesting thing we can notice about this equation is that it’s independent of the mass of the projectile, and it’s also independent of the angle at which that projectile is launched. In fact, the escape speed of a projectile only depends on two things, the mass of the other object — in this case, a planet — and its distance from the center of mass of this object — in this case, the radius of a planet. So, a relatively small projectile traveling directly away from the planet and a relatively large object launched at a tangent to the planet’s surface would both have exactly the same escape speed. Note that, in this case, we’re ignoring any gravitational interactions between our two projectiles.

In this example, the object travelling directly away from the planet would follow a straight path and the projectile launched parallel to the planet’s surface would follow a curved path. However, both objects would be able to travel an infinite distance away from the planet. Now, let’s zoom in on the surface of this planet and consider what the distance–time graph would look like for a projectile launched vertically at escape speed. Specifically, we’ll have the projectile’s height on the 𝑦-axis, that is, its distance away from the surface of the planet. And we’ll put time on the 𝑥-axis.

For a projectile launched at escape speed, we get a graph that looks something like this. We can see that initially the gradient of the line is very steep and that this gradient decreases as time passes. This indicates that initially the height of the object is increasing at a very high rate, corresponding to a very high speed, and at a later time the speed has decreased somewhat. However, the height continues to increase indefinitely. Now, a distance–time graph like this might seem somewhat unusual for a projectile. Usually, projectile problems involving an object that travels away from the ground but then dropped back down to the ground. So, we’re used to seeing a certain type of symmetrical curve like this one. And the mathematical name for this type of curve is a parabola.

The distance–time graph for a projectile traveling at escape speed, however, is very much not a parabola. Even without looking at this curve closely, the fact that it’s not symmetrical means that it can’t be a parabola. This object just keeps getting higher and higher as time goes on. The reason for this discrepancy is that often when we’re looking at projectiles, we’re dealing with initial speeds that are much lower than the escape speed. So, if we call the initial speed of our projectile 𝑢, then we could say that 𝑢 is much less than the escape speed 𝑣 e.

If we launch a projectile at a relatively low speed like this, this means it only reaches a relatively low height before falling back down to Earth. If we just look at this very small length scale, we can see that the curvature of the planet in this region is actually not very obvious. And in fact, a lot of the time when we deal with projectiles, we can simplify our calculations by assuming that the Earth is flat. Effectively for small length scales, the gravitational field of the Earth is approximately uniform, which means that the gravitational force acting on objects is approximately constant. In fact, in these cases, we often calculate the gravitational force acting on an object simply by multiplying its mass by a constant, lowercase 𝑔.

In fact, this is Just an approximation that works for small length scales. And it’s this approximation that gives us our characteristic parabolic distance–time graph for projectiles moving at low speeds. In fact, we know that the gravitational force experienced by a projectile is not constant. Planets have radial magnetic fields, which means that the force decreases with increasing distance from the planet, as given by this equation, leading to distance–time graphs for projectiles which are not parabola shapes. So, projectiles that are launched at speeds much lower than the escape speed will have approximately parabola-shaped distance–time graphs. But as the launch speed of a projectile starts to approach the escape speed, its distance–time graph will deviate more and more from a true parabola.

One other useful graph that we can draw to show the behavior of our projectile is a graph of energy against time. And on this one set of axes, we can represent both the kinetic energy, which we’ll draw in red, and the gravitational potential energy, which we’ll draw in blue. Let’s start by considering a projectile whose initial speed is much less than the escape speed. And let’s say that this projectile is launched at time zero. In this case, this is what our graph would look like.

Looking first at the red line showing kinetic energy, we can see that at the instant the projectile is launched, the kinetic energy takes a maximum value. As the projectile travels away from the planet, its kinetic energy is converted into gravitational potential energy. So, we can see that over this time interval the gravitational potential energy, which was initially at some negative minimum value, increases to a maximum value. And the kinetic energy decreases to zero. It’s important to note here that because kinetic energy is converted directly into gravitational potential energy and there are no other forms of energy loss in this system, the size of this change in kinetic energy, which we could call Δ𝐸 k, is exactly the same as the size in the change of the potential energy, which we can call Δ𝐸 p.

At this moment in time, the gravitational potential energy takes a maximum value and the kinetic energy is zero. This means that here the projectile has reached the top of its trajectory. And after this, it starts to fall back down to the ground. This means that gravitational potential energy is converted back into kinetic energy. So, we see the gravitational potential energy decreasing back down to its minimum value and the kinetic energy increases back up to its initial value as it accelerates towards the ground. Finally, the projectile hits the ground, at which point the kinetic energy decreases immediately to zero and the gravitational potential energy has returned to its initial value.

Now, if we increase the initial speed of this projectile but keep it lower than the escape speed, then our graph instead looks like this. So, we can see that this isn’t drastically different. Because the projectile is launched to the highest speed, it has a higher initial kinetic energy. Because more kinetic energy is being converted into gravitational potential energy, this means the maximum gravitational potential energy is higher by a corresponding amount. And because this means the projectile was launched further, the entire process takes a longer amount of time.

Crucially, we can see that the initial amount of kinetic energy given to the projectile is still smaller than the magnitude of the gravitational potential energy. And we can recall that in order to escape from the gravitational pull of a planet, our projectile needs kinetic energy equal to the magnitude of the gravitational potential energy.

So, finally, let’s now see what this graph looks like when the initial launch speed is equal to the escape speed. This time, we can see that the initial kinetic energy of the projectile is equal in magnitude to the initial gravitational potential energy. This means that the conditions are met for the projectile to escape the gravitational pull of the planet. In other words, the projectile has enough energy to travel to an infinite distance away from the planet. Kinetic energy is still being converted into gravitational potential energy. But because the projectile is effectively traveling to an infinite distance away, this will take an infinite amount of time. So, on our graph, we simply see that kinetic energy tends to a minimum value and gravitational potential energy tends to a maximum value, but we never actually see them get there.

Okay, so far, in this lesson, we’ve seen how we can calculate the escape speed of a projectile and how we can represent the trajectories of projectiles with various speeds on both distance–time graphs and energy–time graphs. Now we’re armed with this information; let’s try answering a practice question.

Europa is a moon of Jupiter that has a mass of 4.80 times 10 to the power of 22 kilograms and a radius of 1,560 kilometers. At what speed would an object have to be launched vertically off the surface of Europa in order to escape its gravity? Give your answer to three significant figures.

So, in this question, we’re thinking about an object on the surface of one of Jupiter’s moons, known as Europa. And we’re being asked to find a speed, let’s call this 𝑣, at which we could launch this object vertically off the surface of Europa that would enable it to escape its gravity. Let’s start by thinking about exactly what it means for this object to escape Europa’s gravity.

Well, our object on the surface of Europa, or indeed any object on the surface of any planet, has a certain gravitational potential energy 𝐸 p, which is given by this equation, where 𝐺 is the gravitational constant, big 𝑀 is the mass of the planet or moon in question — in this case, it would be Europa’s mass — little 𝑚 is the mass of the object, and 𝑟 is the distance between the centers of mass of the two masses that we’re considering. So, in this case, it would be the distance between the center of mass of Europa and the center of mass of our object.

Now, we can see that this potential energy is negative. In physics, we use negative potential energies to represent when something is bound to something else. For example, in this question, our object is bound to Europa by a gravitational force. The fact that this is negative represents the fact that we would need to give the object energy in order to free it. Effectively, the magnitude of all this is the amount of kinetic energy we would need to give our object in order for it to be freed from the gravitational pull of Europa. So, we can say that our object will escape the gravitational pull of Europa when its kinetic energy, given by a half 𝑚𝑣 squared, is equal to 𝐺 times big 𝑀 times little 𝑚 divided by 𝑟.

Now, if we rearrange this expression to make 𝑣 the subject, we obtain this expression. This tells us the speed at which an object must be launched in order to escape the gravity of some other mass, represented by big 𝑀. The speed is known as the escape speed. And it’s often denoted as 𝑣 sub e. We can use this equation directly to work out the escape speed of our object on Europa. In this case, the value of big 𝑀 is the mass of Europa, which we’re told is 4.80 times 10 to the power of 22 kilograms. And 𝑟 in this equation is the distance between the center of mass of our object and the center of mass of Europa.

Now, we haven’t been given this quantity exactly. However, we have been told the radius of Europa is 1,560 kilometers. Because planets and moons are pretty much spherical, this means that center of mass is located in the middle. And as long as our object isn’t too big, its center of mass will be very close to the surface of Europa. This means that we can use the radius of Europa as the distance between our two centers of mass. And since this value has been given in kilometers, we’ll multiply it by 1,000 to give us the value in meters.

Finally, the other value that we need in our equation is 𝐺, the gravitational constant. This has a value of 6.67 times 10 to the power of negative 11 meters cubed per kilogram per second squared. Substituting these three values into our equation gives us this expression. And if we enter all of this into our calculator, we obtain a value of 2,025.99 and so on meters per second, which to three significant figures is 2,030 meters per second. This is the speed at which an object with any mass would have to be launched off the surface of Europa in order to escape its gravity. In other words, it’s the escape speed.

Let’s finish with a summary of the key points that we’ve looked at in this video. We’ve seen that an object travelling at or faster than a planet’s escape speed will be able to escape that planet’s gravity and travel away from the planet indefinitely. We’ve seen that we can calculate an object’s escape speed using this equation, where 𝐺 is the gravitational constant, 𝑀 is the mass of the planet that the object is traveling away from, and 𝑟 is the distance between the centers of mass of the object and the planet. Finally, we looked at position–time and energy–time graphs for projectiles traveling in a range of speeds. And we saw that the faster a projectile is launched, the more the shape of its position–time graph deviates from a parabola. This is a summary of escape speed.

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