### Video Transcript

Find the escape velocity from the surface of Mars. Use a value of 6.4 times 10 to the 23rd kilograms for the mass of Mars and a value of 3390 kilometers for its radius.

Let’s call the term that we’re searching for, escape velocity, 𝑣 sub 𝑒. We’ll assume in this problem that capital 𝐺, the universal gravitational constant, is exactly 6.67 times 10 to the negative 11th meters cubed per kilograms second squared. In this problem, we’re given the mass of the planet Mars, and we’re also given a value to use for its radius.

If we draw a sketch of the planet Mars, we can see that as with any mass, this mass creates gravitational attraction towards it. It’s this gravitational potential energy created by Mars that needs to be exceeded, or overcome, in order for an object on the surface of this planet to be able to escape without returning.

Let’s symbolize the mass of Mars as capital 𝑀, which we’re told is 6.4 times 10 to the 23rd kilograms. And the radius of the planet Mars, which we’ll call lower case 𝑟, is also given as 3390 kilometers.

Going back to this notion of gravitational potential energy, for an object to reach the escape velocity, or 𝑣 sub 𝑒, it’s essentially an energy balance problem. The gravitational field of Mars pulls objects in towards itself. That gravitational potential between the mass of Mars and a mass that it pulls in towards itself can be written in an equation. The gravitational potential created by 𝑚 one on 𝑚 two is equal to the product of those masses multiplied by 𝐺, the universal gravitational constant, divided by the distance between them.

If we apply this equation to our scenario, we will write it as capital 𝐺 times capital 𝑀, the mass of Mars, multiplied by 𝑚, our test mass, divided by 𝑟, the distance between the center of Mars and our test mass 𝑚. When this test mass is on the surface of the planet, that distance is 𝑟, the given radius of Mars. It’s this gravitational potential energy that needs to be overcome, and it will be overcome by a high enough escape velocity 𝑣 sub 𝑒. This potential term that we’ve written will be countered by the kinetic energy of our test mass 𝑚 as it leaves the planet. So we write that this potential is equal to one-half our test mass 𝑚 times the escape velocity, 𝑣 sub 𝑒 squared.

In setting up this equality, we are solving for the minimum velocity a test mass on the surface of Mars needs to have to escape its gravitational field. Let’s now rearrange this equation to solve for 𝑣 sub 𝑒, the escape velocity. The test mass 𝑚 appears on both sides, so we can cancel out that term. If we then multiply both sides by two, then the one-half and two on the right side of the equation cancel. Finally, we take the square root of both sides. And when we do that, the squared and square root term on the right side of the equation cancel.

We find ourselves with an equation for escape velocity, 𝑣 sub 𝑒 equals the square root of two times capital 𝐺 times capital 𝑀 divided by 𝑟. Since capital 𝐺 is a constant and 𝑀 and 𝑟 are given, we can now plug in to this equation to solve for the escape velocity from Mars. When we enter these values into our equation, we find that the escape velocity equals the square root of two times 6.67 times 10 to the negative 11th meters cubed per kilogram second squared times 6.4 times 10 to the 23rd kilograms divided by 3.390 times 10 to the sixth meters, the distance between the center of Mars and its surface in units of meters.

Entering these values in our calculator, we find that, to two significant digits, the escape velocity from the surface of Mars is 5000 meters per second. This is how fast an object on Mars’s surface would need to be moving in order to completely escape the gravitational field created by Mars.