# Video: Determining Escape Speed at a Point in a Gravitational Field

Find the escape speed of a satellite located at the Moon’s orbit about Earth, assuming the Moon is sufficiently far from the satellite not to influence its escape speed. Use a value of 385 × 10³ km for the Moon’s orbital radius and use a value of 5.97 × 10²⁴ kg for the Earth’s mass.

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### Video Transcript

Find the escape speed of a satellite located at the Moon’s orbit about Earth assuming the Moon is sufficiently far from the satellite not to influence its escape speed. Use a value of 385 times 10 to the third kilometers for the Moon’s orbital radius, and use a value of 5.97 times 10 to the 24th kilograms for the Earth’s mass.

Let’s call the given orbital radius of the Moon, 385 times 10 to the third kilometers, 𝑟. And we’ll call the mass of the Earth, 5.97 times 10 to the 24th kilograms, 𝑚. We want to find the escape speed of a satellite located a distance 𝑟 from the center of the Earth; we’ll call that escape speed 𝑣 sub 𝑒.

Let’s draw a sketch of the satellite orbiting around the Earth. As the satellite moves in a circular orbit around the Earth, it’s kept in its orbit by the attraction of the Earth’s gravity. This attraction can be expressed in terms of potential energy. We want to find the speed, 𝑣 sub 𝑒, that the satellite would need to achieve to escape the gravitational pull of the Earth. If the satellite achieved this speed or higher, its path would diverge from the circular orbital path it’s currently on.

To solve for that minimum speed the satellite would need to escape Earth’s gravity, let’s recall an equation for escape velocity. For an object to have enough energy to escape the gravitational attractional pull, its speed must be equal to or greater than the square root of two times the universal gravitational constant 𝐺 multiplied by the mass of the body attracting the satellite all divided by the distance between the two bodies.

We can apply this relationship to our scenario. In the problem statement, we’ve been given values for 𝑚 and 𝑟, and capital 𝐺 is the universal gravitational constant, which we assume to be exactly equal to 6.67 times 10 to the negative 11th cubic meters per kilogram-second squared.

When we plug these values into this equation, we’re careful to give the value of 𝑟 in units of meters; that is, 3.85 times 10 to the eighth meters. When we calculate this speed, it is, to three significant figures, 1440 meters per second. This is the speed the satellite would need to achieve in order to escape the Earth’s gravity permanently.