The table shows the mean (average) distances $ d $ of the planets from the sun (taking the unit of measurement to be the distance from the earth to the sun) and their periods $ T $ (time of revolution in years).
(a) Fit a power model to the data.
(b) Kepler's Third Law of Planetary Motion states that " The square of the period of revolution of a planet
is proportional to the cube of its mean distance from the sun."
Does your model corroborate Kepler's Third Law?
June 23, 2020
how is the graph of y = 2 sin x related to the graphof y =sin x
all right. Here's another great problem to do on a graphing calculator. And so we're going to go into the statistics menu and go into edit, and then we're going to type our numbers into list one endless, too, Where List one stands for the distance. The average distance of the planet from the Sun analyst to stands for the period. The time of revolution around the sun. Okay, once we have those numbers typed in, we want to find the power model. So we go into the stat menu, and then we go over to calculate, and we go down until we find power. It's a little ways down in the list. They're ago. Power regression, repress. Enter and we are Lee using List one and list, too. We don't need to worry about frequency list. We don't need to store the equation anywhere we could just calculate. So here's our power regression model. It's approximately y equals one times X to the 1.5 power. Okay, let's move on to Part B. So Part B talks about Kepler's third Law of planetary motion, saying that the square of the period of the revolution of a planet is proportional to the cube of its main distance from the sun. Well, that can be translated into this equation. The square of the period would be t squared is proportional to means we have some constant of proportionality K. And then the cube of the mean distance would be the Distance Cube. We want to know if that equation is similar to this equation that we got well in the equation we got from the calculator. RT is actually why, and it's not squared. It's just tea. Just why? So what we're going to do is square root both sides of the equation to get t equals instead of t squared equals. And when we do that, when we square root both sides of the equation, we end up with some other K. This K may not be the same as the other K because we've square rooted it, but it's still just some constant K times D to the three halves power. Now, the three halves power is equivalent to the 1.5 power. So I would say that the model we found is consistent with Kepler's third Law