 Let's try another example. How does the period of Mars compare to the period of Earth? The mass of the Sun is equal to 1.988 times 10 to the 30 kilograms. The distance from Mars to the Sun is equal to 2.111 times 10 to the 8 kilometers. We have previously derived an equation for the period of a satellite undergoing uniform circular motion, and this equation is T the period squared divided by r cubed is equal to 4 pi squared divided by Big G Big M. For this problem, we would like to find T, the period. So rearranging the equation above, we get that T is equal to 4 pi squared r cubed divided by Big G Big M, square rooted. We know that Big G is equal to the universal constant of gravitation. We know that r is equal to d is equal to 2.111 times 10 to the 8 kilometers, or 2.111 times 10 to the 11 meters. And we know that the mass of the Sun, which is the mass that is being orbited, is 1.988 times 10 to the 30 kilograms. Let's plug in these numbers. And this is equal to 5.29 times 10 to the 7 multiplied by the square root of meters over Newtons, which are kilogram meters per second squared, divided by kilograms. We cancel the kilograms, the meters, and the square of the second. We see that this is 1 over 1 over 1 second, which is equal to seconds, and therefore T is equal to 5.29 times 10 to the 7 seconds. What is this in days? If we multiply our answer in seconds by 1 minute over 60 seconds, which is the same as multiplying it by 1, and we multiply that by 1 hour over 60 minutes, and we multiply that by 1 day over 24 hours, we can convert it into days. Plugging this in, we get an answer of T equal to 612.53 days. Thank you.