Video Transcript
A satellite follows a circular
orbit around Earth at a radial distance ๐
and with an orbital speed ๐ฃ. If the satellite were moved closer
to Earth so that it followed a circular orbit with the radius of ๐
divided by nine,
at what speed, in terms of ๐ฃ, would it have to move in order to maintain its
orbit?
In this question, weโre going to
consider two different setups for this system, the satellite orbiting Earth. In the first setup, the satellite
has an orbital radius of ๐
and an orbital speed ๐ฃ. Then, in the second setup, weโre
imagining that the satellite is moved much closer to Earth to a radial distance
one-ninth of ๐
. And itโs our job to figure out what
speed it needs to stay in circular orbit.
Letโs begin by recalling a formula
involving orbital radius and speed in a special case of circular orbit like we have
here. Itโs the orbital speed formula ๐ฃ
equals the square root of ๐บ๐ divided by ๐, where ๐ฃ is orbital speed. ๐บ is the universal gravitational
constant. ๐ is the mass of the large body
being orbited, which is Earth in this case. And ๐ is orbital radius.
Notice that we havenโt been given
any numeric values for orbital speed or orbital radius. However, we can still use this
formula to learn how ๐ฃ responds to a change in ๐. Another thing to note is that while
๐บ and ๐ represent real constant quantities, we wonโt need to know or use their
exact values either. Since we wonโt be calculating a
numeric answer, itโs not necessary to write out their entire values. And itโll be simpler to just leave
them written as ๐บ and ๐.
Now, letโs 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
๐ one equals ๐
and the initial orbital speed ๐ฃ one equals ๐ฃ. 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 ๐ฃ equals the square root of ๐บ๐ divided by the initial orbital
radius ๐
.
Now letโs move on to the proposed
new orbit, which weโll indicate using the subscript two. We know that the new orbital radius
๐ two equals ๐
divided by nine. And we wanna find the new orbital
speed ๐ฃ two that meets the conditions of circular orbit and therefore satisfies
this formula. So letโs substitute these values in
and see what happens. We still donโt know the satelliteโs
final speed, so weโll leave it written as ๐ฃ two, which equals the square root of
๐บ๐ divided by ๐
over nine.
We can simplify the math, first by
writing it as the square root of nine ๐บ๐ divided by ๐
. 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 ๐ฃ
two equals three times the square root of ๐บ๐ divided by ๐
, the initial
radius. And remember, weโve already
established that the square root of ๐บ๐ divided by ๐
equals ๐ฃ, the satelliteโs
initial orbital speed.
So letโs make this substitution and
write the result up here. The satelliteโs final speed ๐ฃ two
equals three times ๐ฃ, the satelliteโs initial speed. And so we have our answer. In terms of ๐ฃ, we found that for
the satellite to maintain circular orbit at radius ๐
over nine, it must move at
orbital speed three ๐ฃ.