### 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 two 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.