### Video Transcript

IO is one of the four Galilean
moons of Jupiter. IO makes one complete orbit of
Jupiter every 1.77 days. Assuming that IO’s orbit is
circular with a radius of 422,000 kilometers, calculate the mass of Jupiter. Use a value of 6.67 times 10 to the
negative 11 meters cubed per kilogram second squared for the universal gravitational
constant. Give your answer in scientific
notation to two decimal places.

Now, calculating the mass of a
planet might seem like a wild thing to do. But we can do it using pretty
simple math that we already know. Let’s begin by recalling the
orbital speed formula for the special case of circular orbit. 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 at
the center of orbit. Here, that’s Jupiter. And 𝑟 is orbital radius, which
measures between Jupiter’s and IO’s centers of gravity.

Because we want to figure out the
mass of Jupiter, let’s solve this formula for 𝑀. First, we’ll square both sides to
undo the radical that 𝑀 appears under. And then we’ll multiply both sides
by 𝑟 divided by 𝐺. And now after simplifying, 𝑀
equals 𝑟𝑣 squared divided by 𝐺. Now, we’ve been given values for 𝑟
and 𝐺, but we don’t know what 𝑣 is yet, so we’ll need to calculate it.

To do so, recall that speed is
simply a distance traveled over time. We know the time it takes IO to
complete one full orbit. That’s its orbital period,
represented by capital 𝑇. Because we’re assuming the orbit to
be circular and we know that circle’s radius, we can calculate the circumference of
IO’s circular path. So the distance is the
circumference of the circle, two 𝜋𝑟.

But to calculate 𝑣, we first need
𝑟 and 𝑇 expressed in base SI units. Recall that one kilometer is equal
to 1,000 meters. So 𝑟 is equal to 422 million
meters, or 4.22 times 10 to the eight meters. And to convert 𝑇 from days to
seconds, recall that there are 60 seconds in a minute, 60 minutes in an hour, and 24
hours in a day. So canceling units and multiplying
through, we have found that the orbital period of IO is 1.53 times 10 to the five
seconds.

Now, substituting these values into
our formula for 𝑣, we have that 𝑣 equals two 𝜋 times 4.22 times 10 to the eight
meters divided by 1.53 times 10 to the five seconds. So we found that IO’s orbital speed
equals 1.73 times 10 to the four meters per second. And now we’re ready to solve for
the mass of Jupiter. Let’s copy the formula down below
and substitute in the values for 𝑟, 𝑣, and 𝐺. So 𝑀 equals 4.22 times 10 to the
eight meters times 1.73 times 10 to the four meters per second quantity squared
divided by 6.67 times 10 to the negative 11 meters cubed per kilogram second
squared.

Thus, in scientific notation to two
decimal places, we have found that the mass of Jupiter is 1.90 times 10 to the 27
kilograms.