Video: Escape Velocity

In this video we learn how to calculate an object’s escape velocity from a spherically-symmetric object by balancing kinetic energy with gravitational potential energy.

10:19

Video Transcript

In this video, we’re going to learn about escape velocity. We’ll see what this term means. And we’ll also find out how to calculate it in different scenarios.

To start out, imagine that you’re in a model rocket competition, where in this competition, a prize is awarded to the person whose rocket travels highest. In order to guarantee victory, you want your rocket not only to travel higher than the others, but never to return to Earth. You want it to completely escape Earth’s gravitational pull.

In order to understand what will be needed for your rocket to do this, it will be helpful to learn about escape velocity. This term “escape velocity” comes out of a context of considering how to escape a planet’s gravitational field. Imagine that, in the entire universe, there’s just one planet in existence. Everything outside of that planet is just empty space.

We know that a planet having mass will create a gravitational field around itself, which will tend to draw other massive objects in towards itself. If we have some other massive object then on the surface of the planet, the tendency of that object is to stay there on the surface. Energetically, that’s what the object wants to do.

There’s a helpful way of showing this graphically. Imagine that we zoomed way out from our planet so that it looked like this. If we were to make a graph of the gravitational potential energy 𝑢 created by the planet, as a function of the distance 𝑟 from the planet’s surface, that curve would look something like this. Following the functional form GPE is equal to negative the universal gravitational constant times the mass of our planet times the mass of the object it’s attracting, all divided by the distance between their centers.

We can notice that this curve, as it moves farther and farther away from the planet, approaches zero but never reaches it. We can think of this curve almost as though it’s a hill, something that starts at the surface of the planet we’re talking of and climbs up and up and up, requiring energy to move further away from the planet. Let’s say that some object which started out on the surface of the planet was energetic enough to climb up this curve farther and farther until eventually, as it continues to move out, it effectively escapes the gravitational pull of the planet.

In order to climb out of this gravitational well or to climb up to the top of this gravitational hill, our object will need to start out on the surface of the planet with a certain minimum speed. That speed, which along with the mass of the object gives it a kinetic energy, is what is paid out over its ascent up this hill, until finally it escapes the gravity of the planet. This object by the way doesn’t need to be infinitely far away from the planet in order to have achieved an escape velocity. It’s more a question of the energy gap to be covered which is not infinite but finite.

Escape velocity then, which we can call 𝑣 sub 𝑒, is all about energy. It’s about an object’s kinetic energy being enough to meet or overcome a planet’s gravitational potential energy. We can say that an object’s escape velocity is found by balancing the object’s kinetic energy with a planet’s gravitational potential energy. Since escape velocity, 𝑣 sub 𝑒, is actually a speed rather than a velocity, we’ll equate the kinetic energy of the escaping object with the magnitude of the planet’s gravitational potential energy.

In this mathematical relationship, there’s something very interesting and perhaps counterintuitive. We see that the mass of our object, which we want to escape the gravitational pull of the planet, appears on both sides of the equation. Meaning that, mathematically, the mass cancels out so that, apparently, an object’s escape velocity has nothing to do with its own mass.

Canceling that mass out and dropping the absolute value bars, we can rearrange now to solve for 𝑣 sub 𝑒, the escape velocity. We find that it’s equal to the square root of two times the universal gravitational constant times the mass of the spherical body it’s trying to escape from divided by the radius of that body. Notice that the right side of this equation has nothing in it about the object that’s escaping but is only about the object being escaped from. And in particular, for a spherically symmetric body, it only depends on that body’s mass and its radius.

It turns out when we enter in the numbers for planet Earth that the escape velocity for an object to escape Earth’s gravitational field is about 11 kilometers per second. If an object can achieve that speed, it will be able to leave Earth’s gravity and escape the planet permanently. We can calculate escape velocity though for any spherically symmetric object.

Say that we have a tennis ball. The mass of the tennis ball is about 0.058 kilograms. And it has a radius of a little over three centimeters. If we enter these values in SI base units into our equation for escape velocity along with the universal gravitational constant, we find that, for an object to escape from the surface of a tennis ball, it needs to have a minimum speed of about 1.5 times 10 to the negative fifth meters per second, which is roughly an object moving at one twentieth of a meter every hour.

This just shows that escape velocity can be calculated for any spherically symmetric object. But of course, it was developed in a context of planetary scales. Let’s get some more practice calculating escape velocity using another example.

What is the speed needed to escape from the Earth–Moon system from a point on the surface of Earth? Assume there are no other bodies involved. And do not account for the fact that Earth and the Moon are moving in their orbits. Use a value of 385.0 times 10 to the third kilometers for the distance between the centers of the Moon and Earth, 5.97 times 10 to the 24th kilograms for Earth’s mass, and 7.35 times 10 to the 22 kilograms for the Moon’s mass.

We’re asked to solve for the speed needed to escape the Earth–Moon system. That’s an escape velocity which we’ll call 𝑣 sub 𝑒. And we can start on our solution by drawing a sketch of this situation. Considering this system of the Earth and the Moon which orbits around it, when we consider an object somewhere on Earth’s surface escaping the gravitational pull of this system, we’d want to position it on Earth so that we minimize the escape speed or escape velocity needed to do this.

The position on Earth’s surface then that we pick for our object would be directly opposite the position of the Moon. That way, our mass is already as far as possible from the center of mass of the Earth–Moon system. So if our mass was to blast off perpendicularly from Earth’s surface, in order to escape this system, what escape velocity would it need?

We recognize this as an energy balance situation. The kinetic energy of our escaping object must be at least equal to the gravitational potential energy created by the Earth and the Moon on that object. Recalling that kinetic energy is equal to an object’s mass times its speed squared divided by two. And that the magnitude of the gravitational potential energy between two masses, capital 𝑀 and lowercase 𝑚, is equal to their product times the universal gravitational constant divided by the distance between their centers of mass. From an energy balance perspective, we can write that the kinetic energy of our escaping mass is equal to the gravitational potential energy between that mass and the Earth plus the gravitational potential energy between that mass and the Moon.

Looking over this expression, we see that our escaping object’s mass appears in every term. And therefore, it cancels out. If we then multiply both sides of the equation by two and take the square root of both sides, we find that the escape velocity of our object is equal to the square root of two times big 𝐺, all multiplied by the mass of the Earth over its radius plus the mass of the moon over the radius of the Earth plus the distance between the center of the Moon and the Earth. Big 𝐺, the universal gravitational constant, we set to be exactly equal to 6.67 times 10 to the negative 11th cubic meters per kilogram second squared.

Looking over this equation, we see that we know all of the values in it except for 𝑟 sub 𝐸, the radius of the Earth. That’s a value we can look up. And when we do, we find it’s equal to approximately 6.37 times 10 to the sixth meters. We’re now ready to plug in and solve for 𝑣 sub 𝑒, the escape velocity of our object.

When we do plug in, we’re careful to convert our distance between the centers of the Moon and the Earth into units of meters so it agrees with the units of the rest of this expression. To three significant figures, 𝑣 sub 𝑒 is 11.2 times 10 to the third meters per second. That’s how fast an object would need to be moving off of Earth’s surface in order to escape the Earth–Moon system.

Let’s summarize what we’ve learned so far about escape velocity. We’ve seen that escape velocity, often symbolized 𝑣 sub 𝑒, is the speed needed to leave an object’s gravitational field. We’ve also seen that escape velocity is found using energy balancing. We say that when the initial kinetic energy of our object that wants to escape is equal to negative the gravitational potential energy of the object that draws it, then that escaping object’s speed will be equal to its escape velocity.

For a spherically symmetric object, this relationship implies that the escape velocity is equal to the square root of two times the universal gravitational constant times the mass of the spherical body being escaped from, all divided by its radius. And finally, we’ve seen that an object’s escape velocity is independent of its mass. And it only varies with the planet it’s escaping from’s mass and radius.

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