### 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.