# Question Video: Understanding Variable Dependence in the Orbital Speed Equation Physics • 9th Grade

A satellite follows a circular orbit around Earth at a radial distance 𝑅 and with an orbital speed 𝑣. If the satellite were moved closer to Earth, so that it followed a circular orbit with the radius of 𝑅/9, at what speed, in terms of 𝑣, would it have to move in order to maintain its orbit?

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### Video Transcript

A satellite follows a circular orbit around Earth at a radial distance 𝑅 and with an orbital speed 𝑣. If the satellite were moved closer to Earth so that it followed a circular orbit with the radius of 𝑅 divided by nine, at what speed, in terms of 𝑣, would it have to move in order to maintain its orbit?

In this question, we’re going to consider two different setups for this system, the satellite orbiting Earth. In the first setup, the satellite has an orbital radius of 𝑅 and an orbital speed 𝑣. Then, in the second setup, we’re imagining that the satellite is moved much closer to Earth to a radial distance one-ninth of 𝑅. And it’s our job to figure out what speed it needs to stay in circular orbit.

Let’s begin by recalling a formula involving orbital radius and speed in a special case of circular orbit like we have here. It’s the orbital speed formula 𝑣 equals the square root of 𝐺𝑀 divided by 𝑟, where 𝑣 is orbital speed. 𝐺 is the universal gravitational constant. 𝑀 is the mass of the large body being orbited, which is Earth in this case. And 𝑟 is orbital radius.

Notice that we haven’t been given any numeric values for orbital speed or orbital radius. However, we can still use this formula to learn how 𝑣 responds to a change in 𝑟. Another thing to note is that while 𝐺 and 𝑀 represent real constant quantities, we won’t need to know or use their exact values either. Since we won’t be calculating a numeric answer, it’s not necessary to write out their entire values. And it’ll be simpler to just leave them written as 𝐺 and 𝑀.

Now, let’s get organized and write out what we know, beginning with the first orbit and using the subscript one to refer to this initial setup. We know the initial orbital radius 𝑟 one equals 𝑅 and the initial orbital speed 𝑣 one equals 𝑣. And since we also know that the satellite is in circular orbit here, we know that these values must satisfy this formula. In other words, if we substitute these values into the orbital speed formula, we know that the resulting relationship must be true. Therefore, we know that the initial orbital speed 𝑣 equals the square root of 𝐺𝑀 divided by the initial orbital radius 𝑅.

Now let’s move on to the proposed new orbit, which we’ll indicate using the subscript two. We know that the new orbital radius 𝑟 two equals 𝑅 divided by nine. And we wanna find the new orbital speed 𝑣 two that meets the conditions of circular orbit and therefore satisfies this formula. So let’s substitute these values in and see what happens. We still don’t know the satellite’s final speed, so we’ll leave it written as 𝑣 two, which equals the square root of 𝐺𝑀 divided by 𝑅 over nine.

We can simplify the math, first by writing it as the square root of nine 𝐺𝑀 divided by 𝑅. And then we can take the square root of nine to move it out from under the radical. So now we know the final speed 𝑣 two equals three times the square root of 𝐺𝑀 divided by 𝑅, the initial radius. And remember, we’ve already established that the square root of 𝐺𝑀 divided by 𝑅 equals 𝑣, the satellite’s initial orbital speed.

So let’s make this substitution and write the result up here. The satellite’s final speed 𝑣 two equals three times 𝑣, the satellite’s initial speed. And so we have our answer. In terms of 𝑣, we found that for the satellite to maintain circular orbit at radius 𝑅 over nine, it must move at orbital speed three 𝑣.